Temperature and energy of photons [Archive] - Bad Astronomy and Universe Today Forum

tusenfem

11-April-2007, 03:11 PM

1. Balmer

But 11000K is 1eV in kinetic theory. I don't know if you could say a photon has kinetic energy.

And the point I was making before I was misunderstood, was that a kinetic heat source is harder to come by than electricity. As a matter of fact you can't ionize hydrogen with a kinetic heat source(outside a star).

1. the 650 nm photon from the Balmer series has absolutely nothing to do with ionization. The Balmer series are produced by electrons hopping from one shell to another, in this case from n=3 to n=2. These are bound states of the electron with the proton. If any, you should have chosen the n=infinity "line" at 364 nm, because that is the wavelength of a photon that, when absorbed by an electron in orbit n=2, will give it enough energy to escape from the proton, or the ionization energy.

It has already been pointed out that a photon has energy (h nu) and it has momentum (h nu / c). But that has nothing to do with temperature, because temperature is defined for a body with mass through kT = 0.5 m v2, and a photon is massless, so the definition is useless.

A black body spectrum, like of the sun, at 5700 K does not just emit photons with an energy h nu = 5700 k. It emits photons of all frequencies, and the sun is a pretty good black body radiator. That means that there are photons around that can interact with a neutral hydrogen and kick the electron away, h nu = 13.5 eV => nu = 13.5 / 4.14e-15 = 3.3e15 Hz => lambda = c/nu = 92 nm, which I think falls in the far UV range.

No, if you compress a H gas, the temperature will rise. This is a very nice homework assignment. Higher temperature, higher velocity. Higher velocity and higher density, stronger collisions. How much will the temperarture have to rise in order that H-H collisions knock out an electron and ionize and H?

upriver

12-April-2007, 12:58 AM

1. the 650 nm photon from the Balmer series has absolutely nothing to do with ionization. The Balmer series are produced by electrons hopping from one shell to another, in this case from n=3 to n=2. These are bound states of the electron with the proton. If any, you should have chosen the n=infinity "line" at 364 nm, because that is the wavelength of a photon that, when absorbed by an electron in orbit n=2, will give it enough energy to escape from the proton, or the ionization energy.

Yes, you are correct.

It has already been pointed out that a photon has energy (h nu) and it has momentum (h nu / c). But that has nothing to do with temperature, because temperature is defined for a body with mass through kT = 0.5 m v2, and a photon is massless, so the definition is useless.

Ok, I'll go along with that.

A black body spectrum, like of the sun, at 5700 K does not just emit photons with an energy h nu = 5700 k. It emits photons of all frequencies, and the sun is a pretty good black body radiator. That means that there are photons around that can interact with a neutral hydrogen and kick the electron away, h nu = 13.5 eV => nu = 13.5 / 4.14e-15 = 3.3e15 Hz => lambda = c/nu = 92 nm, which I think falls in the far UV range.

Are you saying that all the ionized H on the surface of the sun and above the photosphere comes from UV photon ionization? What about the FeXV?

No, if you compress a H gas, the temperature will rise. This is a very nice homework assignment. Higher temperature, higher velocity. Higher velocity and higher density, stronger collisions. How much will the temperarture have to rise in order that H-H collisions knock out an electron and ionize and H?

That has nothing to do with my original point.

My original point is this.

1eV is equal to 11,000oK in kinetic theory, yes or no?

That means one electron volt in of acceleration by an electric field is equal to 11,000oK of combustion or compression (kinetic)heat, yes or no?

So 13.6eV is equal to 11,000oK X 13.6eV = 149490oK.

Is there a combustion or compression(besides shock waves) process outside a star that burns that hot? NO.

What else can ionize a plasma that is as common as all the ionized plasma?

Electricity.

tusenfem

12-April-2007, 07:19 AM

Are you saying that all the ionized H on the surface of the sun and above the photosphere comes from UV photon ionization? What about the FeXV?

No, I am not saying that, I was just giving you an example that there are (ample) photons that will ionize a neutral H when in the solar atmosphere.

That has nothing to do with my original point.
My original point is this.
1eV is equal to 11,000oK in kinetic theory, yes or no?
That means one electron volt in of acceleration by an electric field is equal to 11,000oK of combustion or compression (kinetic)heat, yes or no?

So 13.6eV is equal to 11,000oK X 13.6eV = 149490oK.

Is there a combustion or compression(besides shock waves) process outside a star that burns that hot? NO.

What else can ionize a plasma that is as common as all the ionized plasma?

Electricity.

To be completely correct, 13,6eV is not equal to 11000K (by the way Kelvin does not have degrees, it is just K) it is equivalent (one being an energy, the other being a temperature, different units, so no equal sign).

Now the center of the Sun (according to a plot by Alfven, so you will have to believe it) the plasma has energies of 104 to 105 eV, so, yes there is compression where the "temperature goes above 11000 K", otherwise we would not have fusion in the sun.

UV radiation can ionize plasma, that is the most common cause of ionization in the universe.

So, why don't you read up on stellar formation, see how the cloud starts to collapse on itself and the pressure inside keeps on growing as well as the temperature, etc. etc. I would recomment Bowers & Deeming "Astrophysics I" as a good introduction.

korjik

12-April-2007, 03:33 PM

Yes, you are correct.

Ok, I'll go along with that.

Are you saying that all the ionized H on the surface of the sun and above the photosphere comes from UV photon ionization? What about the FeXV?

That has nothing to do with my original point.

My original point is this.

1eV is equal to 11,000oK in kinetic theory, yes or no?

technically? No. It is the temperature where the average kinetic energy equals 1eV. Maxwellian distributions have alot of spread, so the actual KE of any one particle could (and most likely will) be quite a bit different

That means one electron volt in of acceleration by an electric field is equal to 11,000oK of combustion or compression (kinetic)heat, yes or no?

Definitely not. A monoenergenic beam will have a temp far different than the KE. You could have a beam with 1eV of temp (maxwellian distribution spread in velocity caused by collision between particles), and 100 eV of bulk flow velocity. They dont directly translate

So 13.6eV is equal to 11,000oK X 13.6eV = 149490oK.

No. KE dosent directly translate.

Is there a combustion or compression(besides shock waves) process outside a star that burns that hot? NO.

What else can ionize a plasma that is as common as all the ionized plasma?

Electricity.

Except that you dont need to get that hot to ionize. Like Tusenfem said, you should get and read some basic astrophysics books. They explain how the ionization occurs.

The simple explanation is that in any gas at thermodynamic equilibrium, there will be a large variation of individual particle kinetic energies. For H, when you get up around 10000K, the high energy tail will have more than enough energy to ionize the gas (by the way, the gas will start emmitting lines when the tail gets energetic enough to cause excitations). If the temp of the gas is high enough, the tail will be big enough that ionization will beat recombination, and the gas will completely ionize.

korjik

12-April-2007, 03:36 PM

1. the 650 nm photon from the Balmer series has absolutely nothing to do with ionization. The Balmer series are produced by electrons hopping from one shell to another, in this case from n=3 to n=2. These are bound states of the electron with the proton. If any, you should have chosen the n=infinity "line" at 364 nm, because that is the wavelength of a photon that, when absorbed by an electron in orbit n=2, will give it enough energy to escape from the proton, or the ionization energy.

It has already been pointed out that a photon has energy (h nu) and it has momentum (h nu / c). But that has nothing to do with temperature, because temperature is defined for a body with mass through kT = 0.5 m v2, and a photon is massless, so the definition is useless.

A black body spectrum, like of the sun, at 5700 K does not just emit photons with an energy h nu = 5700 k. It emits photons of all frequencies, and the sun is a pretty good black body radiator. That means that there are photons around that can interact with a neutral hydrogen and kick the electron away, h nu = 13.5 eV => nu = 13.5 / 4.14e-15 = 3.3e15 Hz => lambda = c/nu = 92 nm, which I think falls in the far UV range.

No, if you compress a H gas, the temperature will rise. This is a very nice homework assignment. Higher temperature, higher velocity. Higher velocity and higher density, stronger collisions. How much will the temperarture have to rise in order that H-H collisions knock out an electron and ionize and H?

Dude you completely stole my thunder :D

Then again, my last post answered your homework. Sorry

tusenfem

12-April-2007, 08:54 PM

Dude you completely stole my thunder :D

Then again, my last post answered your homework. Sorry

Great minds think alike *grin*

Nereid

13-April-2007, 04:14 PM

[snip]

What else can ionize a plasma that is as common as all the ionized plasma?

Electricity.I already asked you (http://www.bautforum.com/showpost.php?p=965954&postcount=10) about this; I will ask again (in a slightly different way): how does "electricity" ionize a plasma?

In your answer, please include at least some details of the mechanism, at the molecule/atom/ion/particle level.

upriver

13-April-2007, 11:54 PM

Now the center of the Sun (according to a plot by Alfven, so you will have to believe it) the plasma has energies of 104 to 105 eV, so, yes there is compression where the "temperature goes above 11000 K", otherwise we would not have fusion in the sun.

I have said before, how does the inside of the sun ionize all the plasma outside the sun especially the plasma outside the heliosphere?

UV radiation can ionize plasma, that is the most common cause of ionization in the universe.

And you are saying that all plasma not ionized by shockwaves is ionized by UV?

So, why don't you read up on stellar formation, see how the cloud starts to collapse on itself and the pressure inside keeps on growing as well as the temperature, etc. etc. I would recomment Bowers & Deeming "Astrophysics I" as a good introduction.

Conjecture.

Davidlpf

14-April-2007, 12:34 AM

Conjecture.
Your one to talk about conjecture.

upriver

14-April-2007, 02:16 AM

I already asked you (http://www.bautforum.com/showpost.php?p=965954&postcount=10) about this; I will ask again (in a slightly different way): how does "electricity" ionize a plasma?

In your answer, please include at least some details of the mechanism, at the molecule/atom/ion/particle level.

Ionization is when an atom or molecule has an imbalance of charges. Positive ionization is when an there are an unequal number of electrons and protons, there being more protons, negative ionization is when an electron gets caught and radiates some of its energy away.

The most interesting form of ionization that I studied(?) is K-shell hole ionization.

In my experiments I made a plasma tube, a simple 12 inch sealed glass tube with copper electrodes at each end.
It was driven by a 15Kv power supply that could supply 10ma. As I turned the voltage up, it would ionize at about 5Kv and then the voltage would stabilize. As you turned the voltage control at this point, only the current would increase as there was now ionization.
As you varied the pressure and voltage of the D2 you would get things like multiple double layers(striations), flashes, electrode melting, etc.

The idea with electrical ionization is that the cathode begins to emit electrons and as these electrons "bump" into atoms they ionize the gas, producing a electric field wave that travels the length of the gas forming a plasma as it moves.

Now that only talks about the kinetic energy that the electrons have coming off of the electrode, I cant seem to find any mention of the role charges in this.

Originally Posted by Nereid
Then what ionised the hydrogen in this object? and what for this one (other than the EUV from the star)? And what ionised the hydrogen in the Gum Nebula?

Here is a caption from one of the examples that you asked about.

"M1: The Crab Nebula from NOT"
"This is the mess that is left when a star explodes. The Crab Nebula, the result of a supernova seen in 1054 AD, is filled with mysterious filaments. The filaments are not only tremendously complex, and a higher speed than expected from a free explosion."
http://antwrp.gsfc.nasa.gov/apod/ap050920.html

"When a star explodes."
Conjecture.
"Mysterious filaments."
Plasma filaments.
"but appear to have less mass than expelled in the original supernova"
Because they didnt come from a supernova as expected.
"Higher speed than expected from a free explosion."
Electrical acceleration.

Need I say more?

Oh, the Gum Nebula contains a supernova/pulsar/neutron star, an indication of elevated levels of electrical activity.
I think its also part of Goulds Belt which I've talked about.

Originally Posted by korjik
Except that you dont need to get that hot to ionize. Like Tusenfem said, you should get and read some basic astrophysics books. They explain how the ionization occurs.

The plasma's average temperature is different than the atoms ionization energy or the driving voltage.

The atom has an electron that needs another electron with the same or more energy, or a photon with the same or more energy to knock it loose.

You cant get around that(except for Penning Mixtures?).

For example, you have a driving voltage connected to a plasma tube.
And because at a certain value of drive there is not enough current flow to ionize the rest of the gas as said, recombination is the dominant process.

You still have electrons in the tube that are above 13.6eV,(because your drive is 5Kv).
At a certain value of drive just enough of them ionize some of the volume of gas, and your average temperature reads 10,000K.
That 10000K has nothing to do with the ionization energy of an atom. It is the average temp of the plasma. Each atom still requires a 13.6 eV electron or photon to ionize.

Now you mean to tell me that that the 1eV to 11000K conversion depends on the ambient properties of the gas and electrode distance?

If you look at the energy a single atom. Does the 1eV to 11000K directly hold?

The simple explanation is that in any gas at thermodynamic equilibrium, there will be a large variation of individual particle kinetic energies.

Yes.

For H, when you get up around 10000K, (by the way, the gas will start emmitting lines when the tail gets energetic enough to cause excitations).

I will assume that when you say (10,000)1eV "the high energy tail will have more than enough energy to ionize the gas" you are talking about the temperature of the plasma and not the driving voltage or energy of individual particles.

If the temp of the gas is high enough, the tail will be big enough that ionization will beat recombination, and the gas will completely ionize.

You mean, if the voltage is turned up and there are enough electrons flowing out of the electrode to dominate recombination, and ionize the volume of gas, then the volume of gas ionizes.

Which has nothing to do with the ionization energy of an atom. You are talking about the energy required to ionize a volume of gas.

upriver

14-April-2007, 02:48 AM

"The emphasis of this early theoretical work was on producing data for modeling collisional ionization equilibrium (sometimes also called coronal equilibrium). Under these conditions an ion forms at a temperature about an order of magnitude higher than the temperature where it forms in photoionized plasmas
(Kallman & Bautista 2001)."
http://arxiv.org/PS_cache/arxiv/pdf/0704/0704.0905v1.pdf

Because there is no such thing as a mechanical means of producing temperatures that high, or I should say, there are no mechanical temperatures that high, and theory is only predicting and modeling imagination.

Electrons and photons are the only way to produce a high temperature plasma.

Nereid

14-April-2007, 03:58 PM

[Moderator Note]

Thread (temporarily) locked, to give upriver a chance to respond to the open questions here (http://www.bautforum.com/showthread.php?p=968246#post968246).

[/Moderator Note]

Nereid

14-April-2007, 10:11 PM

Thread re-opened, as the other ATM one is now closed (http://www.bautforum.com/showthread.php?p=968387#post968387), per upriver's request.

Nereid

15-April-2007, 12:05 AM

I already asked you about this; I will ask again (in a slightly different way): how does "electricity" ionize a plasma?

In your answer, please include at least some details of the mechanism, at the molecule/atom/ion/particle level.Ionization is when an atom or molecule has an imbalance of charges. Positive ionization is when an there are an unequal number of electrons and protons, there being more protons, negative ionization is when an electron gets caught and radiates some of its energy away.

The most interesting form of ionization that I studied(?) is K-shell hole ionization.

In my experiments I made a plasma tube, a simple 12 inch sealed glass tube with copper electrodes at each end.
It was driven by a 15Kv power supply that could supply 10ma. As I turned the voltage up, it would ionize at about 5Kv and then the voltage would stabilize. As you turned the voltage control at this point, only the current would increase as there was now ionization.
As you varied the pressure and voltage of the D2 you would get things like multiple double layers(striations), flashes, electrode melting, etc.

The idea with electrical ionization is that the cathode begins to emit electrons and as these electrons "bump" into atoms they ionize the gas, producing a electric field wave that travels the length of the gas forming a plasma as it moves.So, if I may summarise the first part: according to upriver (or is it standard physics? please clarify), '"electricity" ionises neutral hydrogen' at the atom/particle level, is due to a collision between an electron and a hydrogen atom, in which there is an exchange of energy such that the hydrogen atom's electron acquires sufficient energy to 'escape from' the atom (and the electron which collided with the atom loses the same energy).

Is that an accurate summary?

Please clarify the second part ("producing a electric field wave that travels the length of the gas forming a plasma as it moves"); specifically:
a) what is "a electric field wave"?
b) how does "a electric field wave" form a plasma?Now that only talks about the kinetic energy that the electrons have coming off of the electrode, I cant seem to find any mention of the role charges in this. Then what ionised the hydrogen in this object? and what for this one (other than the EUV from the star)? And what ionised the hydrogen in the Gum Nebula?
Here is a caption from one of the examples that you asked about.

"M1: The Crab Nebula from NOT"
"This is the mess that is left when a star explodes. The Crab Nebula, the result of a supernova seen in 1054 AD, is filled with mysterious filaments. The filaments are not only tremendously complex, and a higher speed than expected from a free explosion."
http://antwrp.gsfc.nasa.gov/apod/ap050920.html

"When a star explodes."
Conjecture.Please clarify.

Specifically:
a) What phenomenon did the Chinese, Korean, etc writers record (in 1054)?
b) What is the Crab Pulsar (http://en.wikipedia.org/wiki/Crab_Pulsar)?
c) What, in the upriver view, are supernovae?
"Mysterious filaments."
Plasma filaments.How did you determine this?"but appear to have less mass than expelled in the original supernova"
Because they didnt come from a supernova as expected.What alternative(s) has (have) been proposed?"Higher speed than expected from a free explosion."
Electrical acceleration.How did you arrive at this conclusion?Need I say more?Yes.

Please provide a quantitative account - at least to an OOM level - of how "[e]lectrical acceleration" produced the speeds observed.Oh, the Gum Nebula contains a supernova/pulsar/neutron star, an indication of elevated levels of electrical activity.
I think its also part of Goulds Belt which I've talked about.

[snip]Please provide a reference, so any anyone reading this may check for themselves.

korjik

15-April-2007, 01:06 AM

The plasma's average temperature is different than the atoms ionization energy or the driving voltage.

The atom has an electron that needs another electron with the same or more energy, or a photon with the same or more energy to knock it loose.

You cant get around that(except for Penning Mixtures?).

Actually, you dont need to have 13 eV. For hydrogen, you only need 10.6 eV and have the average time between collisions be less than the time it takes for the atom to return to the ground state. It is even easier for heavier atoms.

For example, you have a driving voltage connected to a plasma tube.
And because at a certain value of drive there is not enough current flow to ionize the rest of the gas as said, recombination is the dominant process.

You still have electrons in the tube that are above 13.6eV,(because your drive is 5Kv).
At a certain value of drive just enough of them ionize some of the volume of gas, and your average temperature reads 10,000K.
That 10000K has nothing to do with the ionization energy of an atom. It is the average temp of the plasma. Each atom still requires a 13.6 eV electron or photon to ionize.

Now you mean to tell me that that the 1eV to 11000K conversion depends on the ambient properties of the gas and electrode distance?

What I am talking about is basic properties of gases. There are no electrodes, or voltages or any electrical considerations.

I want to make this abundantly clear: THERE ARE NO VOLTAGES OR CURRENTS INVLOLVED

If you look at the energy a single atom. Does the 1eV to 11000K directly hold?

No. Not in the slightest. A single atom with a KE of 1 eV does not have a temperature.

Yes.

I will assume that when you say (10,000)1eV "the high energy tail will have more than enough energy to ionize the gas" you are talking about the temperature of the plasma and not the driving voltage or energy of individual particles.

You would be assuming wrong. I am talking about the temperature of the gas. The neutral gas. NOT the plasma.

When you have a gas at a temperature T, you will have an average kinetic energy of 3/2kT. This is the AVERAGE kinetic energy. Some of the individual particles will have less kinetic energy, and some will have more. Some will have a much higher energy, and this is called a high energy tail.

One thing I want to note here: IF YOU CANNOT GET AN AVERAGE YOU CANNOT CORRELATE TEMPERATURE AND KINETIC ENERGY.

So, when you have a hot enough gas, the tail will have enough kinetic energy to ionize the gas.

You mean, if the voltage is turned up and there are enough electrons flowing out of the electrode to dominate recombination, and ionize the volume of gas, then the volume of gas ionizes.

Which has nothing to do with the ionization energy of an atom. You are talking about the energy required to ionize a volume of gas.

No. That is exactly what I am not talking about. There is no electrodes or electrics or currents or anything. I am talking about a blob of gas at a temperature. Nothing else. Once it is warm enough, the tail will have enough energy to start knocking electrons loose. If the rate of ionization due to the temp of the gas is greater than the rate of recombination, the gas will ionize. There is no outside influence, no electrodes, no extra electrons.

As for the ionization energy of an atom, it doesnt take a temperature to ionize it takes energy. You could have 2K (temperature) gas accelerated somehow to 15eV of kinetic energy slam into another 2K gas, and there will be all sorts of ionization. The temp is nowhere near the 150000K you seem to think is necesary, but there is plenty of energy.

upriver

15-April-2007, 03:30 AM

So, if I may summarise the first part: according to upriver (or is it standard physics? please clarify), '"electricity" ionises neutral hydrogen' at the atom/particle level, is due to a collision between an electron and a hydrogen atom, in which there is an exchange of energy such that the hydrogen atom's electron acquires sufficient energy to 'escape from' the atom (and the electron which collided with the atom loses the same energy).

Is that an accurate summary?

Yes. Its pretty much a standard observation.

Please clarify the second part ("producing a electric field wave that travels the length of the gas forming a plasma as it moves"); specifically:
a) what is "a electric field wave"?
b) how does "a electric field wave" form a plasma?Please clarify.

"Xenon on the Verge of an Electric Breakdown"
"In carefully designed lab experiments the region between a pair of electrodes fills with plasma smoothly, starting at the positive end, with a wave front that sweeps quickly across to the negative end."

"At the leading edge of this so-called ionization front is a narrow band of enhanced electric field, according to theory and indirect experiments, but theorists only vaguely understand what determines the field's profile."
http://focus.aps.org/story/v19/st4

Specifically:
a) What phenomenon did the Chinese, Korean, etc writers record (in 1054)?

An ejection as opposed to an explosion.

b) What is the Crab Pulsar (http://en.wikipedia.org/wiki/Crab_Pulsar)?

From that wiki article.

"The same set of observations suggested that since the primary radio pulse coming from the pulsar lasts only 0.4 nanoseconds it is being emitted from a cloud of plasma on the surface of the neutron star only 12 centimeters across - the smallest feature ever to be observed in astronomy.[2]"

That would be a 99% confirmation of a plasma pinch(or focus). I cant imagine any thing else that could produce those energies of particles in a small space like that. But as compared to earthly pinches, that one is huge.

This paper talks about the energies that could generated in a plasma pinch.

"Cosmic ray spectrum above 1015eV (a new approach)"
http://hepg.sdu.edu.cn/Chinese_2003/literatures/29icrc/PAPERS/OG12/rus-petrukhin-AA-abs1-og12-oral.pdf

There seems to be an excess of electrons in the pulsar winds models;

"Although no model has been developed so far to explain the details of this component, one may verify that the total number of the electrons responsible for it is in agreement with what predicted by the classical pulsar-wind models, which otherwise are known to fail in accounting for the number of radio emitting electrons"
http://www.aanda.org/index.php?option=com_base_ora&url=articles/aa/full/2002/18/aah3375/aah3375.right.html&access=standard&Itemid=81

Of course there may be other types of oscillatory phenomena relating to plasma.

c) What, in the upriver view, are supernovae?

Bigger pinches. Not spherical, barrel shaped remnants, the controversy about light curves, GRB's, double exploding supernovas. You could almost say that each one was different.....

Originally Posted by upriver
"Mysterious filaments."
Plasma filaments.

How did you determine this?

If you have a large volume of plasma like in a TOKAMAK, which is the one of the largest continuous volumes of plasma on earth, when you excite it, it always seems to form filaments called instabilities, that give off x-rays(pinch), and travel to the wall, in it. Apparently when you reach a certain energy density in the plasma it spontaneously forms filaments to discharge. Maybe someday they will solve that problem but right now it looks insurmountable even at 6 seconds burn time.
The experiments talked about in some of the reconnection threads have filaments in them. Birkeland currents exist. Plasma filaments exist. The DNA Nebula, space slinkies, etc. In the lab they exist.

Not one of them is gravity created.

Just because I cant explain how the central generator works does not mean that it does not exist. We definitely have observed high energy particles leaving the center of our galaxy.

Quote:
"but appear to have less mass than expelled in the original supernova"
Because they didnt come from a supernova as expected

What alternative(s) has (have) been proposed?

If a supernova is really a pinch that is bigger than say a pulsar, then it will definitely have smaller mass expulsion but extremely high energies.
This is one of those time where I wish was better with math.

Originally Posted by upriver
"Higher speed than expected from a free explosion."
Electrical acceleration.

How did you arrive at this conclusion?

What else is there?

Originally Posted by upriver
Need I say more?

Yes.

Please provide a quantitative account - at least to an OOM level - of how "[e]lectrical acceleration" produced the speeds observed.

The filaments have an average temperature of 1eV, while the pulsar has average particles energies of 35MeV.
The problem is trying to figure out which of the effects they are measuring when they say the nebula is expanding at 1500m/s. They say proper motion of the filaments. Is that the speed of the ionization wave or the particle motion. I will try to get you a better answer.

Wiki on Birkeland currents.

"Professor Emeritus of the Alfvén Laboratory in Sweden, Carl-Gunne Fälthammar wrote (1986): "A reason why Birkeland currents are particularly interesting is that, in the plasma forced to carry them, they cause a number of plasma physical processes to occur (waves, instabilities, fine structure formation). These in turn lead to consequences such as acceleration of charged particles, both positive and negative, and element separation (such as preferential ejection of oxygen ions). "

wiki on acceleration of charged particles.

"For example, one experimental device at the Lawrence Berkeley National Laboratory accelerates electrons to 1 GeV over about 3.3 cm, whereas the SLAC conventional accelerator requires 64 m to reach the same energy. A recent experiment performed by a team at SLAC achieved an energy gain to 42 GeV over 85 cm using a plasma wakefield accelerator.[1]"
http://en.wikipedia.org/wiki/Plasma_acceleration

"1 GeV over about 3.3 cm," Is smaller than 12cm.

Only with the modern miracle of electricity.

Quote:
Oh, the Gum Nebula contains a supernova/pulsar/neutron star, an indication of elevated levels of electrical activity.
I think its also part of Goulds Belt which I've talked about.

[snip]

Please provide a reference, so any anyone reading this may check for themselves.

I'm am making the relation between electrical activity, star forming regions and brightness.

From this post here (http://www.bautforum.com/showpost.php?p=926927&postcount=403).

Originally posted by upriver

A little bit on our local neighborhood.

A survey placing Orion in Goulds belt.
http://www.jach.hawaii.edu/JCMT/surveys/gb/

A map of our local neighborhood.

"Mythological enemies Orion (left) and Scorpius (right) owe their brilliance to Gould's belt."

"Gould's belt is the most prominent starry feature in the Sun's neighborhood, contributing most of the bright young stars nearby. Nearly two thirds of the massive stars within 2,000 light-years of the Sun belong to Gould's belt. If I were kidnapped by an alien spaceship and taken to some remote corner of the Galaxy, Gould's belt is what I'd look for to find my way back home."
Goulds Belt.
http://kencroswell.com/GouldBelt.html

"The Gum Catalog"
"This vast 2.6 million year old supernova remnant is usually simply called "the Gum nebula". Located at the collision between the Gould Belt and the Vela region, the Gum nebula is a major part of the local galaxy."
http://galaxymap.org/cgi-bin/gum.py?s=11

Of course the Vela region contains the Vela pulsar.

And in the UEU brightness is related to electrical activity.

upriver

16-April-2007, 02:39 AM

Here is a paper on the Crab Nebula jet.
They use the movement of the knot in the wall of the jets to determine the "speed" of the jet. That is definitely a different speed than the matter inside of the "tubular structure".
Page 6. Different than the filaments(I think), but still one of the methods.
http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1993MNRAS.263...69F&data_type=PDF_H IGH&whole_paper=YES&type=PRINTER&filetype=.pdf

Here they talk about its unusual collimated filamentary structure.
http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1982ApJ...260L..75G&data_type=PDF_H IGH&whole_paper=YES&type=PRINTER&filetype=.pdf

It is a Birkeland current. It is the power source for the pulsar.

Now here they actual use the words "electric voltages" but they say the are generated by the action of the star. I'm saying that the external electric voltages from the center of the galaxy(?)in the form of Birkeland currents drive it.

The Crab Nebula Pulsar Shrugs
http://antwrp.gsfc.nasa.gov/apod/ap020920.html

Here they say
"The Circinus pulsar itself, thought to generate more than 7 quadrillion volts"
http://antwrp.gsfc.nasa.gov/apod/ap010913.html

Now I have been entertaining a thought that says "For some reason the density of local space is such that it starts emitting electricity." Either that or everything has to have a invisible filament connected to it. The Crab has a filament that is only partially visible.
We shall see.

upriver

17-April-2007, 03:40 AM

"Determination of the Distance to the Crab Nebula"
"The sources of these nebular emission lines are the
intricate knots of the nebula. An interesting fact about these
knots is revealed by looking closely at the necklace shape
of the emission lines. No lines are detected between the
two strands, indicating that all of the visible knots lie on the
outer edge of the nebula either moving toward or away from
Earth. No knots lie in between; only the synchrotron radiation
continally emitted from the pulsar exists there."

http://www.hcs.harvard.edu/~jus/0302/bester.pdf

tusenfem

17-April-2007, 08:21 AM

I find it hard to find the relevance of all this talking about pulsars and whatevers on the simple discussion of energy of photons and temperature of plasmas, which was what all this was about.

tusenfem

17-April-2007, 01:38 PM

Well, in between sessions (I got a bit tired of listening to talks about the clouds of Venus, however interesting they might be), I'd thought I'd write something about the production of plasmas and their temperatures and photons.

Definition
A plasma is a gas that has a sufficiently high ionizations level for collective behaviour to occur. Basically this means that the number of charged particles in the deBye sphere needs to be much greater than 1.

Ionization
How does one achieve ionization? Knock off an electron from the atom! This can be done in several ways:
1. By (multiple) photon absorption. There is an ionization energy (the binding energy of the electron to the nucleus). When the atom absorbs a photon, the atom gets excited and the electron moves to a higher shell, or the atom becomes ionized when the photon has enough energy. If the excited atom absorbs another photon before the electron returned to its ground state its get more excited or ionized, depending on the sum of the energies of the photons. This is the method mostly preferred by interstellar galactic nebula.
2. By (multiple) collisions. This can be collisions between neutrals or colisions of atoms with electrons or ions. There is a slight differnce, especially with the ion collisions. Again, basically it is the same as above, but instead that a photon gets absorbed, now a collision takes place, with either enough or too little energy and the story just goes on like in 1. This is what usually takes place in laboratory plasmas. I remember standing at the glass tube with gas and voltage and using a magical tool called a "Tesla" to start the plasma glowing in a nice pink color. The Tesla functioned as the "starter" you have in the frame of your fluorescent light.
The special case is when a neutral collides with an ion, then something called charge-exchange can happen, where the ion steals an electron from the atom, and then usually you end up with a energetic neutral atom (ENA) and a thermal ion.

So various ways of making plasmas, but what about the temperature of the plasma? The temperature of the plasma is determined by the distribution of the kinetic energies of all plasma particles. Now electrons are much lighter than ions, so basically you have a "light gas" and a "heavy gas". Because of collisions and stuff you will find that the two temperatures of these gasses are not equal, the electron temperature is much larger than the ion temperature. If the plasma is unmagnetized, you just have a normal Maxwell/Boltzmann distribution. If the plasma is magnetized, however, there is an asymmetry because the particles cling to the field. Then you will find you will have to slit the temperature into parallel to the field and perpendicular to the field.

Well, gotta go again, gonna listen to a talk about the Saturnian current sheet.

More later

upriver

17-April-2007, 04:48 PM

I find it hard to find the relevance of all this talking about pulsars and whatevers on the simple discussion of energy of photons and temperature of plasmas, which was what all this was about.

Sorry, but Nereid asked me some direct questions.

upriver

18-April-2007, 04:08 AM

So various ways of making plasmas, but what about the temperature of the plasma? The temperature of the plasma is determined by the distribution of the kinetic energies of all plasma particles. Now electrons are much lighter than ions, so basically you have a "light gas" and a "heavy gas". Because of collisions and stuff you will find that the two temperatures of these gasses are not equal, the electron temperature is much larger than the ion temperature.

If the electrons are at a higher temperature than their ionization temperature and the ion are at their ionization temperature, that would mean the plasma is no longer neutral, even though there are the same number of particles, the distribution of energy of the plasma favors the electrons.

Is that true?

If the plasma is unmagnetized, you just have a normal Maxwell/Boltzmann distribution. If the plasma is magnetized, however, there is an asymmetry because the particles cling to the field. Then you will find you will have to slit the temperature into parallel to the field and perpendicular to the field.

Interesting. So you have 2 different physical distributions of plasma at the same time. Is it broken up into ions and electrons? Or low temp electrons and high temp electrons.

Well, gotta go again, gonna listen to a talk about the Saturnian current sheet.

More later

Sounds interesting.

So now if you have a plasma that is fully 100% ionized, its temperature is now characteristic of its ionization energy. At least the electrons are.
We do have 1 million degree plasmas coming from the sun, 100eV.

From wiki.
"If, instead of talking of room temperature as 300 K (27 °C or 80 °F), it were conventional to speak of the corresponding energy kT of 4.14×10−21 J, or 0.0259 eV, then Boltzmann's constant would simply be the dimensionless number 1."

So if you used a direct conversion then 1eV = 11000K

And now to the main point.
Since no elements have an ionization energy under 3eV(33,000K) that means that the photosphere has some amount of neutrals and some amount of ions/electrons, which by definition are hotter than 30,000K(at least the electrons).

So the average temperature is correct but the actual energy at that level of the sun 3eV, It would have to be to keep that plasma ionized.

According to the MS model all particles are cooled down to a 6000C temperature and then scattered at the change in plasma density called the photosphere.
How are there 30,000K particles that make it through the photosphere? All the ones that are ionized are at least that.

There has to be 30,000K to keep those particles ionized, yet it is claimed that the photosphere is 6000C.

tusenfem

18-April-2007, 09:02 AM

If the electrons are at a higher temperature than their ionization temperature and the ion are at their ionization temperature, that would mean the plasma is no longer neutral, even though there are the same number of particles, the distribution of energy of the plasma favors the electrons.

Is that true?

Totally wrong, the temperature of the various species in a plasma has nothing to do with the plasma being neutral or not. Neutrality is only dependend on the number of positive and negative charge being equal: charge = number electrons - number ions (assuming only 1 charge)
Distribution functions do not do anything, favouring or not. It is the number of electrons which is the integral over all velocity space of the distribution function, similarly for ions. Please please please, try al least to read up on the basic definitions in plasma physics. This makes the discussion very nerving!

Interesting. So you have 2 different physical distributions of plasma at the same time. Is it broken up into ions and electrons? Or low temp electrons and high temp electrons.

Just like I said, read up on the basics!!!!!!!!!!!
A distribution function describes both the low and the high temp electrons, for each species (except for very special cases) there is only ONE distribution function that describes the whole electron (or ion) population.
So, two distribution functions, one for the electrons and one for the ions (and there could be another for the neutrals, but we will keep those out of the picture).

So now if you have a plasma that is fully 100% ionized, its temperature is now characteristic of its ionization energy. At least the electrons are.
We do have 1 million degree plasmas coming from the sun, 100eV.

No nononononononononono. The temperature of the plasma has exactly NOTHING to do with the ionization energy of the atoms. Why would it? A gas has a temperature. Now a highly energetic atom collides with another atom and knocks out an electron. Of the whole energy in that collision, the knocking out of the electron (suppose it is H) takes out 13.5 eV, the remaining energy gets distributed over the three particles the H+ the e- and the Hneutral.

What you could say is that if you have a gas and put in a sufficiently large amount of energy E and the whole gas (H) of N particles is ionized, then of the energy that you have put in E-13.5N remains to heat up the particles.

And before we can discuss any further, you first have to show that you understand the above.

upriver

19-April-2007, 03:00 AM

Totally wrong, the temperature of the various species in a plasma has nothing to do with the plasma being neutral or not. Neutrality is only dependend on the number of positive and negative charge being equal: charge = number electrons - number ions (assuming only 1 charge)
Distribution functions do not do anything, favouring or not. It is the number of electrons which is the integral over all velocity space of the distribution function, similarly for ions. Please please please, try al least to read up on the basic definitions in plasma physics. This makes the discussion very nerving!

So even if the electrons have more energy(than their ionization energy), you would still term it a neutral plasma because of the equality of species.

Just like I said, read up on the basics!!!!!!!!!!!
A distribution function describes both the low and the high temp electrons, for each species (except for very special cases) there is only ONE distribution function that describes the whole electron (or ion) population.
So, two distribution functions, one for the electrons and one for the ions (and there could be another for the neutrals, but we will keep those out of the picture).

And the term "distribution function" is taken to apply to each species.

Quote:
So now if you have a plasma that is fully 100% ionized, its temperature is now characteristic of its ionization energy. At least the electrons are.

No nononononononononono. The temperature of the plasma has exactly NOTHING to do with the ionization energy of the atoms. Why would it? A gas has a temperature. Now a highly energetic atom collides with another atom and knocks out an electron. Of the whole energy in that collision, the knocking out of the electron (suppose it is H) takes out 13.5 eV, the remaining energy gets distributed over the three particles the H+ the e- and the Hneutral.

Ya, your right. I was thinking about if you are using electricity to ionize a gas, the the hottest electrons are going to be representative of the drive voltage.
Supposing that atom has just enough energy to knock the electron loose, I guess since its bigger, it can lose alot of energy compared to an electron, then the electron has an energy of 13.5eV while the atom now 13.5 eV less or cooler.

So now if you have a plasma composed of only 13.5 eV electrons, would the mono energetic plasma be 150,000K?

What you could say is that if you have a gas and put in a sufficiently large amount of energy E and the whole gas (H) of N particles is ionized, then of the energy that you have put in E-13.5N remains to heat up the particles.

If you put in more than enough energy to ionize a gas, what ever is leftover from kicking electrons off atoms, will go into the nuclei.

I kind of don't understand this part "then of the energy that you have put in E-13.5N remains to heat up the particles."

You have a 30eV electron that impacts a H atom, and knocks the electron loose, is the first electron captured and then imparts 17eV of energy to the ion?
Or does the first electron continue onward with 17eV of energy, so now you have 2 free electrons?

tusenfem

19-April-2007, 08:51 AM

1. So even if the electrons have more energy(than their ionization energy), you would still term it a neutral plasma because of the equality of species.

2. And the term "distribution function" is taken to apply to each species.

3. Ya, your right. I was thinking about if you are using electricity to ionize a gas, the the hottest electrons are going to be representative of the drive voltage.
Supposing that atom has just enough energy to knock the electron loose, I guess since its bigger, it can lose alot of energy compared to an electron, then the electron has an energy of 13.5eV while the atom now 13.5 eV less or cooler.

4. So now if you have a plasma composed of only 13.5 eV electrons, would the mono energetic plasma be 150,000K?

5. If you put in more than enough energy to ionize a gas, what ever is leftover from kicking electrons off atoms, will go into the nuclei.

6. I kind of don't understand this part "then of the energy that you have put in E-13.5N remains to heat up the particles."

7. You have a 30eV electron that impacts a H atom, and knocks the electron loose, is the first electron captured and then imparts 17eV of energy to the ion? Or does the first electron continue onward with 17eV of energy, so now you have 2 free electrons?

1. That is the DEFINITION!!!!!!!!!!!!!!!!!!!!!!!!!! OF NEUTRALITY. Sheesh

2. Get thee to a library, and take any undergraduate book on statistical
physics.

3. Then you are still not using "electricity", you use an electric field to get fast electrons which collide with the atoms and ionize them.

4. That totally depends, if all particles have the same energy, than the distribution function turns into a delta function and you cannot put a traditional temperature to it, but it could be considered as you say, depending on what you want to do.

5. Wrong again, if you e.g. use high energy photons to ionize, the electron absorbs the photon and all energy goes into the electron. so for H, have a 17eV photon, the electron when knocked off, will have a kinetic energy of 17-13.5=3.5eV. In collisions too, most energy always goes to the lightest particle. Simple basic physics, read up on it.

6. Explained in 5

7. also 5, but the first electron will have lost some energy because of the collision, it is all a question of energy balance in non-elastic collisions. And yes you have 2 free electrons.

upriver

19-April-2007, 07:03 PM

Originally Posted by tusenfem
Now a highly energetic atom collides with another atom and knocks out an electron. Of the whole energy in that collision, the knocking out of the electron (suppose it is H) takes out 13.5 eV, the remaining energy gets distributed over the [b]three particles the H+ the e- and the Hneutral.[b]

What you could say is that if you have a gas and put in a sufficiently large amount of energy E and the whole gas (H) of N particles is ionized, then of the energy that you have put in E-13.5N remains to heat up the particles.

Originally Posted by upriver
5. If you put in more than enough energy to ionize a gas, what ever is leftover from kicking electrons off atoms, will go into the nuclei.

5. Wrong again, if you e.g. use high energy photons to ionize, the electron absorbs the photon and all energy goes into the electron. so for H, have a 17eV photon, the electron when knocked off, will have a kinetic energy of 17-13.5=3.5eV. In collisions too, most energy always goes to the lightest particle. Simple basic physics, read up on it.

You mean like the bowling ball and cue ball on the pool table experiment.

You were talking about atoms, you said nothing about photons.

I have always assumed that photons were like electrons in their action.

Only an atom to atom or ion to atom collision may impart energy to the nucleus.

And compressing atoms to achieve ionization is a hypothetical situation which "may" exist only in the center of a star and not a lab.

So I base my ideas(partially) of ionization on the fact that electrons are generally the initiators which creates the moving double layer that sweeps across the gas turning it into plasma. Or in the case of photons that individually ionize atoms.

Actually, there may be an exception. In the work I was doing with sonoluminescence, from the spectrum it was estimated that the temperature of the plasma in the bubble at a size of 5 microns was about 1500000C because we were seeing ionized Argon(18eV)(I have reservations about that). Personally, I think there was another cause for the ionization. If they get fusion out of collapsing bubbles then there might be something to it. But in my estimation, I don't think they will ever find fusion in a bubble from compression.
That is my opinion from running hours and hours of experiments.

I will deny I ever said that, too, because I could be wrong.:)

Here is a link to the some of the work that I did(I taped all these movies).
And I ran all these devices. The glass one is a Rusi Cell. It is an exact replication of the Perdue bubble fusion device. The steel sphere was of a proprietary design.
http://www.impulsedevices.com/media.html

korjik

19-April-2007, 07:40 PM

You mean like the bowling ball and cue ball on the pool table experiment.

You were talking about atoms, you said nothing about photons.

I have always assumed that photons were like electrons in their action.

Only an atom to atom or ion to atom collision may impart energy to the nucleus.

And compressing atoms to achieve ionization is a hypothetical situation which "may" exist only in the center of a star and not a lab.

So I base my ideas(partially) of ionization on the fact that electrons are generally the initiators which creates the moving double layer that sweeps across the gas turning it into plasma. Or in the case of photons that individually ionize atoms.

Actually, there may be an exception. In the work I was doing with sonoluminescence, from the spectrum it was estimated that the temperature of the plasma in the bubble at a size of 5 microns was about 1500000C because we were seeing ionized Argon(18eV)(I have reservations about that). Personally, I think there was another cause for the ionization. If they get fusion out of collapsing bubbles then there might be something to it. But in my estimation, I don't think they will ever find fusion in a bubble from compression.
That is my opinion from running hours and hours of experiments.

I will deny I ever said that, too, because I could be wrong.:)

Here is a link to the some of the work that I did(I taped all these movies).
And I ran all these devices. The glass one is a Rusi Cell. It is an exact replication of the Perdue bubble fusion device. The steel sphere was of a proprietary design.
http://www.impulsedevices.com/media.html

If you have an atom with 20 eV of kinetic energy impact an H atom, 13eV goes to ionization and is lost, the rest is split between the ion, the free electron, and the neutral atom. Most goes to the electron.

Since when is PV=nkT hypothetical? Heck you gave an example of compression ionization in your experiment.

Your thoughts about ionization are wrong. Ionization will occur anytime there is an energy source with with quanta above the ionization energy. Wether it is photon energy, bulk flow kinetic energy, thermal maxwellian tail energy, or just bulk thermal isnt important. The only factor is wether the energy impulse exceeds the ionization energy.

upriver

20-April-2007, 02:48 AM

Since when is PV=nkT hypothetical? Heck you gave an example of compression ionization in your experiment.

It is the only example of compression ionization that I know of and there is still debate as to whether it really is compression and not some EM cavity effect.
And its not like you can stick a probe in a 5 micron bubble(at microsecond time scales), you can get shadow graphs with a streak camera, and analyze the light(spectrograph), but thats about it.
There's been a lot of work done of those bubbles and I'm sure I've read 90% of the papers in that field and for some reason there is still doubt in my mind as to the real cause of ionization.

I hope they prove me wrong, get fusion and clean energy, and everybody live happily ever after.

Your thoughts about ionization are wrong. Ionization will occur anytime there is an energy source with with quanta above the ionization energy. Wether it is photon energy, bulk flow kinetic energy, thermal maxwellian tail energy, or just bulk thermal isnt important. The only factor is wether the energy impulse exceeds the ionization energy.

I'm not arguing about the amount of energy required. I'm talking about the energy available in a particular process.

tusenfem

20-April-2007, 08:48 AM

upriver

20-April-2007, 09:33 PM

The energy available in a process can easily be calculated. With PV=nkT. This relationship has been tested and tested and tested, and the nicest example of it is using your fire extinguisher, shooting out gas that turns into CO2 snow because of the sudden change in PV.

Yes.

I am not familiar with the bubbles (except for hearing about them a little and that while collapsing the pressur goes up and it could maybe once be used for fusion). Is the temperature of the bubble only determined by the fact that you have ionized Argon?

Here are some papers.
"Plasma formation and temperature measurement during single-bubble cavitation"
http://www.nature.com/nature/journal/v434/n7029/full/nature03361.html;jsessionid=7D64338F2CC87BBB82E870 62C1F1203C

SONOFUSION – FACT OR FICTION?
http://www.rpi.edu/~laheyr/Sonofusion%20Paper-pdf_Lahey_NURETH-11.pdf

See what you think.

I guess I have confused you on the ionization by either photons or collisions. The photon gives all energy to the electron, the collision gives most energy to the electron and a little to the ion. It is not totally like the bowling ball and the billard ball, because there is a loss of energy so it is an inelastic collision. The loss of energy being the ionization energy.

The interesting thing about the bubbles is they were talking about convergence ratios of 1000 to 1 to achieve a million degrees. These bubbles do about 2/3rds that. They go from 2mm to about 4 microns. If you scale that up you have 1000ft to 1ft in 50 microseconds. That seems physically impossible on the macroscopic scale. Even if you said 250ft to 1 ft. to get 250,000 degrees. Diesels are about 25 to 1.
So the idea that there is a mechanical macroscopic system that can ionize a quantity of a neutral gas(atoms) seems untestable.

And I am not sure that the ionization in a lab plasma goes through a moving double layer, it is just an ionization front moving through the gas when energetic enough electrons come from the cathode. There might be a sheet around the cathode which accelerates the electrons. I never created a double layer in the plasma before it was completely iononized already.

"Xenon on the Verge of an Electric Breakdown"
"In carefully designed lab experiments the region between a pair of electrodes fills with plasma smoothly, starting at the positive end, with a wave front that sweeps quickly across to the negative end."

"At the leading edge of this so-called ionization front is a narrow band of enhanced electric field, according to theory and indirect experiments, but theorists only vaguely understand what determines the field's profile."
http://focus.aps.org/story/v19/st4

Wouldn't you call that a moving double layer?

korjik

20-April-2007, 10:46 PM

Yes.

Here are some papers.
"Plasma formation and temperature measurement during single-bubble cavitation"
http://www.nature.com/nature/journal/v434/n7029/full/nature03361.html;jsessionid=7D64338F2CC87BBB82E870 62C1F1203C

SONOFUSION – FACT OR FICTION?
http://www.rpi.edu/~laheyr/Sonofusion%20Paper-pdf_Lahey_NURETH-11.pdf

See what you think.

The interesting thing about the bubbles is they were talking about convergence ratios of 1000 to 1 to achieve a million degrees. These bubbles do about 2/3rds that. They go from 2mm to about 4 microns. If you scale that up you have 1000ft to 1ft in 50 microseconds. That seems physically impossible on the macroscopic scale. Even if you said 250ft to 1 ft. to get 250,000 degrees. Diesels are about 25 to 1.
So the idea that there is a mechanical macroscopic system that can ionize a quantity of a neutral gas(atoms) seems untestable.

Try 1 solar mass of 75% H 25% He compressed from the better part of a cubic light-year down to a ball a million kilometers wide.

"Xenon on the Verge of an Electric Breakdown"
"In carefully designed lab experiments the region between a pair of electrodes fills with plasma smoothly, starting at the positive end, with a wave front that sweeps quickly across to the negative end."

"At the leading edge of this so-called ionization front is a narrow band of enhanced electric field, according to theory and indirect experiments, but theorists only vaguely understand what determines the field's profile."
http://focus.aps.org/story/v19/st4

Wouldn't you call that a moving double layer?

That isnt every way to ionize

tusenfem

21-April-2007, 06:57 PM

Here are some papers.
"Plasma formation and temperature measurement during single-bubble cavitation"
http://www.nature.com/nature/journal/v434/n7029/full/nature03361.html;jsessionid=7D64338F2CC87BBB82E870 62C1F1203C

SONOFUSION – FACT OR FICTION?
http://www.rpi.edu/~laheyr/Sonofusion%20Paper-pdf_Lahey_NURETH-11.pdf

See what you think.

The interesting thing about the bubbles is they were talking about convergence ratios of 1000 to 1 to achieve a million degrees. These bubbles do about 2/3rds that. They go from 2mm to about 4 microns. If you scale that up you have 1000ft to 1ft in 50 microseconds. That seems physically impossible on the macroscopic scale. Even if you said 250ft to 1 ft. to get 250,000 degrees. Diesels are about 25 to 1.
So the idea that there is a mechanical macroscopic system that can ionize a quantity of a neutral gas(atoms) seems untestable.

"Xenon on the Verge of an Electric Breakdown"
"In carefully designed lab experiments the region between a pair of electrodes fills with plasma smoothly, starting at the positive end, with a wave front that sweeps quickly across to the negative end."

"At the leading edge of this so-called ionization front is a narrow band of enhanced electric field, according to theory and indirect experiments, but theorists only vaguely understand what determines the field's profile."
http://focus.aps.org/story/v19/st4

Wouldn't you call that a moving double layer?

Well, the SBSL almost goes above my hat. I see that there was a mix up, there is no ionized but excited Ar. For the rest, it is pretty difficult stuff. With the gas and the sulphuric acid in water and then the sonic pressure pulses of several bar. I don't think the nature paper is the place to start, after a quick read. The other paper seems to start introductory, but then does not do that and quickly goes into the fusion stuff. So, I cannot judge here at the moment. I did see that the emissions of the collapsing bubbles were indeed the indicators of the temperature of the Ar. Guess it is time to edumacate myself end get mee to a library :-)

The other thing about the collapse, and Korjik already commented on this is at least similar to the collapse of a cloud into a star.

The last thing, reading about the measurements of the electric field. Interesting, how they were able to measure it, but I would hardly say it is a double layer. Not every electric field in a plasma or a gas is a double layer. It has to answer to specific characteristics, e.g. there are a negative and a positive layer, but in the experiment it appears there are only highly energetic electrons that perform the ionization. It is, however, an electric front that moves, maybe numerical simulations can give new insight into the process, now that they have discovered some of the details of an ionization front.

upriver

22-April-2007, 12:48 AM

Try 1 solar mass of 75% H 25% He compressed from the better part of a cubic light-year down to a ball a million kilometers wide.

The issue is not the compression ratio but how fast you compress the gas and how much radiation there is. With the bubbles how fast they compress is a major issue. Shortening the compression time by raising the drive frequency or other tricks(complex wave forms), increases the brightness of the flash.
Too slow and you get no flash..

This is with a volume of gas bounded by water!

How could gravity(very weak) cause gas or plasma! to clump together, when their natural inclination is to repel, with the electrical force being 1039 stronger. Especially a plasma.... The only thing I have ever seen that causes a plasma to clump is magnetic field. The gas would have to be compressed and really cold, just like making liquid air to clump. Wait a minute it has to be hot. And star forming regions are areas of great activity, hot plasma...
I'm sure they accounted for the rate of collapse in the models but I have a hard time seeing how in an unbounded model(density decrease as you go outward).

Now do you see my skepticism?

That isnt every way to ionize

Your right. The other way to ionize is with a high energy light source, like a UV lamp, or arc lamp. The photons need to be at least 3eV depending on the element. That would ionize a whole volume pretty much evenly at once, where as the electron ionization proceeds in a linear movement across a volume. (Actually, you probably could measure the light wave front as it moves across a volume.)

Now there are some shock wave experiments, but everyone that I've seen uses some form of pre-ionization like an explosion or wire pinch.

So that leaves compression ionization which cannot be done on the human macro scale, and I believe on the astronomical macro scale.

If you know of a good solar collapse model paper, I would like to read it.

upriver

22-April-2007, 02:10 AM

Well, the SBSL almost goes above my hat. I see that there was a mix up, there is no ionized but excited Ar.

here is a quote from the Nature article.

"The strong neutral Ar atom emission lines in the red and near-infrared region of the spectra arise from the 4s–4p manifold. The 4s state is situated 11.5–11.8 eV above the ground state (3p), whereas the 4p state lies 13.1–13.5 eV above the ground state. b, Ar atom line emission (2.8 bar; solid line) compared to a calculated Ar atom emission spectrum at 15,200 K (dashed line)."

The energies are comparable to the first ionization level of H which is why I was using H as an example.

The other important point from that articles is this;

"The emissive excited states observed from both Ar and O2 + are inconsistent with any thermal process. The Ar excited states involved are extremely high in energy (>13 eV) and cannot be thermally populated at the measured Ar emission temperatures (4,000–15,000 K); the ionization energy of O2 is more than twice its bond dissociation energy, so O2 + likewise cannot be thermally produced. We therefore conclude that these emitting species must originate from collisions with high-energy electrons, ions or particles from a hot plasma core."

What they are saying is that at the measured blackbody temperatures, you cannot excite(ionize) Argon.

I'm saying the same thing about the photosphere and hydrogen.

For the rest, it is pretty difficult stuff. With the gas and the sulphuric acid in water and then the sonic pressure pulses of several bar. I don't think the nature paper is the place to start, after a quick read. The other paper seems to start introductory, but then does not do that and quickly goes into the fusion stuff. So, I cannot judge here at the moment. I did see that the emissions of the collapsing bubbles were indeed the indicators of the temperature of the Ar. Guess it is time to edumacate myself end get mee to a library :-)

Here is a good basic educational by Dr. Brenner. The Dr Gaitan in the first paragraph, is the Chief Scientist at where I worked. He and I were helping Dr. Hiller move today....
http://www.seas.harvard.edu/brenner/Single_bubble.pdf

SBSL single bubble sonoluminiscence
MBSL multi bubble sonoluminiscence

The same apparatus works with water. The reason they use SO was for increased light output and they could identify the known reactions by their spectrum. From this they can determine the temperatures and some of processes. Since the light from these flashes can be on the order of 50,000 photons for a small flash, its crucial to have as much light as possible for taking a spectrum.
You have to use a black enclosed light tight optical bench and integrate for a long time(seconds).

I would be glad to answer questions about this. Of course don't take me as the last word.:)

The other thing about the collapse, and Korjik already commented on this is at least similar to the collapse of a cloud into a star.

Unbounded vs bounded.

The last thing, reading about the measurements of the electric field. Interesting, how they were able to measure it, but I would hardly say it is a double layer. Not every electric field in a plasma or a gas is a double layer. It has to answer to specific characteristics, e.g. there are a negative and a positive layer, but in the experiment it appears there are only highly energetic electrons that perform the ionization. It is, however, an electric front that moves, maybe numerical simulations can give new insight into the process, now that they have discovered some of the details of an ionization front.

"Current-free double layers"
"Current-free double layers occur at the boundary between plasma regions with different plasma properties. We explain how they form (neglecting the ions which are considered solely as a neutralizing background). Consider a plasma divided into two regions by a plane, which has a higher electron temperature on one side than on the other (the same analysis can also be done for different densities)."
http://en.wikipedia.org/wiki/Double_layer#Current-free_double_layers

I think this is correct.

Van Rijn

22-April-2007, 02:49 AM

How could gravity(very weak) cause gas or plasma! to clump together, when their natural inclination is to repel, with the electrical force being 1039 stronger.

Why would there be a natural inclination to repel in a neutral gas or plasma?

Now do you see my skepticism?

No.

tusenfem

22-April-2007, 08:40 AM

here is a quote from the Nature article.

"The strong neutral Ar atom emission lines in the red and near-infrared region of the spectra arise from the 4s–4p manifold. The 4s state is situated 11.5–11.8 eV above the ground state (3p), whereas the 4p state lies 13.1–13.5 eV above the ground state. b, Ar atom line emission (2.8 bar; solid line) compared to a calculated Ar atom emission spectrum at 15,200 K (dashed line)."

The energies are comparable to the first ionization level of H which is why I was using H as an example.

The other important point from that articles is this;

"The emissive excited states observed from both Ar and O2 + are inconsistent with any thermal process. The Ar excited states involved are extremely high in energy (>13 eV) and cannot be thermally populated at the measured Ar emission temperatures (4,000–15,000 K); the ionization energy of O2 is more than twice its bond dissociation energy, so O2 + likewise cannot be thermally produced. We therefore conclude that these emitting species must originate from collisions with high-energy electrons, ions or particles from a hot plasma core."

What they are saying is that at the measured blackbody temperatures, you cannot excite(ionize) Argon.

I'm saying the same thing about the photosphere and hydrogen.

Here is a good basic educational by Dr. Brenner. The Dr Gaitan in the first paragraph, is the Chief Scientist at where I worked. He and I were helping Dr. Hiller move today....
http://www.seas.harvard.edu/brenner/Single_bubble.pdf

SBSL single bubble sonoluminiscence
MBSL multi bubble sonoluminiscence

The same apparatus works with water. The reason they use SO was for increased light output and they could identify the known reactions by their spectrum. From this they can determine the temperatures and some of processes. Since the light from these flashes can be on the order of 50,000 photons for a small flash, its crucial to have as much light as possible for taking a spectrum.
You have to use a black enclosed light tight optical bench and integrate for a long time(seconds).

I would be glad to answer questions about this. Of course don't take me as the last word.:)

Unbounded vs bounded.

"Current-free double layers"
"Current-free double layers occur at the boundary between plasma regions with different plasma properties. We explain how they form (neglecting the ions which are considered solely as a neutralizing background). Consider a plasma divided into two regions by a plane, which has a higher electron temperature on one side than on the other (the same analysis can also be done for different densities)."
http://en.wikipedia.org/wiki/Double_layer#Current-free_double_layers

I think this is correct.

I will see if I find the time to look into this stuff. Got another meeting the next few days so not much time. I do not think I understand the thing about the O2+ because I assume it comes from the desolved sulphuric acid and the SO42+ might well be the source of O2+, unless they also desolved molecular oxygen in the water.

The thing that the bubbles are in water is just a technique to keep things bound. In the universe gravity (albeit a "weak" force) acts at this force. There is no problem for gravity to keep a cloud together and to let it collapse on itself. The comment that electrostatic force is a bezillion times stronger has been shown repeatedly to be a false argument. First of all the cloud is neutral gas, secondly there is a pressure gradient in the gas cloud towards the center, just like when you are on the bottom of the ocean you better have a pressure suit or submarine.

You still mix up excitation and ionization in your comments. And as excited lines are seen something is happening, some high energy electrons was talked about in the paper. They say this comes from a plasma, well I gotta look into it. But this has nothing to do with the photosphere at the sun, this is something completely different, as we have already explained to you in previous mails. Maybe in a bubble there is not time enough for multiple excitation/ionization but in the photosphere there is plenty time for multiple event ionizations.

Glad you like my DL page on Wiki and read it and ofcourse the description that you copied is correct. But this is no DL going through, because at one side there is no ionized gas and at the other side there is. It is more like an EM soliton wave that slides through the gas, causing the ionization with the electrons that are accelerated by it (or something like that). Not every electric field in a plasma is a double layer. And I doubt that you will find that this is a current free situation, when over the column of gas is a strong potential drop and electrons from the cathode speed towards the anode, definitely a current!

upriver

23-April-2007, 04:23 AM

I do not think I understand the thing about the O2+ because I assume it comes from the desolved sulphuric acid and the SO42+ might well be the source of O2+, unless they also desolved molecular oxygen in the water.

Well the first thing is that there is no or not much chemistry that happens at those energies. Its mostly ionization(or excitation).
SO42+ is the source of the O2+. They did not add O2. Those 2 reactions have know energies and therefore are known indications of temperatures inside the bubble.

originally posted by Plasma formation and temperature measurement during single-bubble cavitation

"the ionization energy of O2 is more than twice its bond dissociation energy, so O2 + likewise cannot be thermally produced."

That is the whole thing, trying to figure out how hot it is inside and the reactions. And what are the conditions required to achieve higher temperatures. As I said I've taken lots of pictures of those bubbles and we have not figured out how to directly probe the inside of the bubbles.
One of the biggest issues with measuring the light from the bubbles is the UV extinction rate in water. Shorter than 180nm, you get nothing after the first half millimeter. And that is the only way to tell if you are heading towards soft x-rays in temperature(energy).

The other part of the reason they use sulfuric acid is for the known reactions. The type of gas used in these experiments is usually a noble gas except helium. Xenon, krypton, argon are the most common and produce the most light. N seems to quench the light.

You still mix up excitation and ionization in your comments.

There must be an issue with language here. You are still not understanding what I'm saying.

H ionizes at 13.6 eV. The Ar excitation energies are from 11 to 13 eV. Those are the same energies. They both require a 11 to 13.6 eV electron(or photon). They are saying that 15,000 degree plasma is not hot enough to ionize(18eV) or excite the Ar at 11 to 13.6 eV so it must be fast electrons.

I'm saying the same thing about the photosphere. Now if you are saying that the Maxwellian tail is responsible, then what heats the tail?

But this has nothing to do with the photosphere at the sun, this is something completely different, as we have already explained to you in previous mails.

I must have missed it.

If the photosphere is the surface of last scattering and its 6000 degrees, all the photons and particles from the inside have cooled down to 6000 degrees by the time they reach the surface. Where does the energy come from to ionize the photosphere? At least 3 eV or 33,000 degrees. Or for H 13.6 eV or 150000 degrees.

Maybe in a bubble there is not time enough for multiple excitation/ionization but in the photosphere there is plenty time for multiple event ionizations.

Don't multiple event ionizations require fast electrons? And if its multiple then the electron has to be really fast!!

And as excited lines are seen something is happening, some high energy electrons was talked about in the paper. They say this comes from a plasma, well I gotta look into it.

This where I have some what of an issue with the current interpretation.
I think the when the cavity collapses there is a formation of a electric field that accelerates a free electron ionizing the gas into a plasma?.

As opposed to it collapsing, getting hot enough(colliding atoms) to make a plasma of the required temperature, then shooting out a fast electron to excite the Argon.

This all has to happen in a pulse width of less than 50 pico seconds, in a space of 5 microns.

Glad you like my DL page on Wiki and read it and ofcourse the description that you copied is correct. But this is no DL going through, because at one side there is no ionized gas and at the other side there is. It is more like an EM soliton wave that slides through the gas, causing the ionization with the electrons that are accelerated by it (or something like that).

If there was a very thin layer of ions on the leading edge of the breakdown would you consider it a double layer?

Not every electric field in a plasma is a double layer. And I doubt that you will find that this is a current free situation, when over the column of gas is a strong potential drop and electrons from the cathode speed towards the anode, definitely a current!

Thats what I thought, but I wasn't sure.

korjik

23-April-2007, 03:20 PM

The issue is not the compression ratio but how fast you compress the gas and how much radiation there is. With the bubbles how fast they compress is a major issue. Shortening the compression time by raising the drive frequency or other tricks(complex wave forms), increases the brightness of the flash.
Too slow and you get no flash..

This is with a volume of gas bounded by water!

How could gravity(very weak) cause gas or plasma! to clump together, when their natural inclination is to repel, with the electrical force being 1039 stronger. Especially a plasma.... The only thing I have ever seen that causes a plasma to clump is magnetic field. The gas would have to be compressed and really cold, just like making liquid air to clump. Wait a minute it has to be hot. And star forming regions are areas of great activity, hot plasma...
I'm sure they accounted for the rate of collapse in the models but I have a hard time seeing how in an unbounded model(density decrease as you go outward).

Now do you see my skepticism?

Your skepticism is due to ignorance.

This is a question to be anwered in the ATM forum:

What is the force on an electron .5m away from 1 kg, 1m diameter ball of NEUTRAL plasma?

I will make it even easier: You can treat the ball as a point source (gauss' law) for both gravity and EM.

I want the total force, the gravitational force and the electromagnetic force.

Your right. The other way to ionize is with a high energy light source, like a UV lamp, or arc lamp. The photons need to be at least 3eV depending on the element. That would ionize a whole volume pretty much evenly at once, where as the electron ionization proceeds in a linear movement across a volume. (Actually, you probably could measure the light wave front as it moves across a volume.)

Now there are some shock wave experiments, but everyone that I've seen uses some form of pre-ionization like an explosion or wire pinch.

So that leaves compression ionization which cannot be done on the human macro scale, and I believe on the astronomical macro scale.

If you know of a good solar collapse model paper, I would like to read it.

Try almost any introductory astrophysics text, like both tusenfem and I have told you repeatedly.

upriver

24-April-2007, 07:55 PM

Your skepticism is due to ignorance.

No. There is nothing to support your view except calculations.
It's not like you have seen a whole life cycle of a star.

This is a question to be anwered in the ATM forum:

What is the force on an electron .5m away from 1 kg, 1m diameter ball of NEUTRAL plasma?

I give up. What is it?

I will make it even easier: You can treat the ball as a point source (gauss' law) for both gravity and EM.

I want the total force, the gravitational force and the electromagnetic force.

"Plasma physicists Hannes Alfvén and Carl-Gunne Fälthammar, noted in their book, Cosmic Electrodynamics"
"The basic reason why electromagnetic phenomena are so important in cosmical physics is that there exist celestial magnetic fields which affect the motion of charged particles in space. Under certain conditions electromagnetic forces are much stronger than gravitation. In order to illustrate this, let us suppose that a particle moves at the earth's solar distance RE ((the position vector being RE) with the earth's orbital velocity v. If the particle is a neutral hydrogen atom, it is acted upon only by the solar gravitation (the effect of a magnetic field upon a possible atomic magnetic moment being negligible). If M is the solar and m, the atomic mass, and γ is the constant of gravitation, this force is f = -γMm RE/RE3. If the atom becomes singly ionized, the ion as well as the electron (charge e = ± 4.8 x 10-10 e.s.u.) is subject to the force fm = e(v/c) x B from an interplanetary magnetic field which near the earth's orbit is B. The strength of the interplanetary magnetic field is of the order of 10-4 gauss, which gives fm/f ≈ 107. This illustrates the enormous importance of interplanetary and interstellar magnetic fields, compared to gravitation, as long as the matter is ionized."

Try almost any introductory astrophysics text, like both tusenfem and I have told you repeatedly.

What happens if you have a tank of neutral gas the size of a galaxy under some pressure.
You open this tank of gas into an area of the universe where there is no matter, only space.
The tank is emptied into space.

What would be the end result?

I would say that the gas dissipates.

Same with a plasma.

korjik

24-April-2007, 08:30 PM

No. There is nothing to support your view except calculations.
It's not like you have seen a whole life cycle of a star.

I give up. What is it?

Your thread. ATM rules say you must answer pertinent questions. If figuring the force on a bit of matter isnt pertinent to how a plasma can gravitationally collapse, I dont know what is.

"Plasma physicists Hannes Alfvén and Carl-Gunne Fälthammar, noted in their book, Cosmic Electrodynamics"
"The basic reason why electromagnetic phenomena are so important in cosmical physics is that there exist celestial magnetic fields which affect the motion of charged particles in space. Under certain conditions electromagnetic forces are much stronger than gravitation. In order to illustrate this, let us suppose that a particle moves at the earth's solar distance RE ((the position vector being RE) with the earth's orbital velocity v. If the particle is a neutral hydrogen atom, it is acted upon only by the solar gravitation (the effect of a magnetic field upon a possible atomic magnetic moment being negligible). If M is the solar and m, the atomic mass, and γ is the constant of gravitation, this force is f = -γMm RE/RE3. If the atom becomes singly ionized, the ion as well as the electron (charge e = ± 4.8 x 10-10 e.s.u.) is subject to the force fm = e(v/c) x B from an interplanetary magnetic field which near the earth's orbit is B. The strength of the interplanetary magnetic field is of the order of 10-4 gauss, which gives fm/f ≈ 107. This illustrates the enormous importance of interplanetary and interstellar magnetic fields, compared to gravitation, as long as the matter is ionized."

Wow, you just explained one of the forces on solar wind. Unfortunately, my question was about a charged particle near a neutral non-magnetized plasma. So basically this whole paragraph is misunderstood word salad that has no bearing whatsoever to the discussion at hand.

What happens if you have a tank of neutral gas the size of a galaxy under some pressure.
You open this tank of gas into an area of the universe where there is no matter, only space.
The tank is emptied into space.

What would be the end result?

I would say that the gas dissipates.

Same with a plasma.

Really? Why dosent the earth's atmosphere dissipate then?

tusenfem

24-April-2007, 09:10 PM

I'm saying the same thing about the photosphere. Now if you are saying that the Maxwellian tail is responsible, then what heats the tail?

I must have missed it.

If the photosphere is the surface of last scattering and its 6000 degrees, all the photons and particles from the inside have cooled down to 6000 degrees by the time they reach the surface. Where does the energy come from to ionize the photosphere? At least 3 eV or 33,000 degrees. Or for H 13.6 eV or 150000 degrees.

Don't multiple event ionizations require fast electrons? And if its multiple then the electron has to be really fast!!

I will let the bubbles be for right now, because I cannot find the time to read some introductory papers on it, being too busy professionally. So, I skip to the part about which I do know something.

"What heats the tail?" This shows that you have not had any statistical physics taught. A Maxwellian distribution deals with particles with a speed of 0 to (theoretically) a speed of infinity. Any gas (gas not plasma, gas) that you have and study you will find that there is a large spread of speeds of the particles (note that I use speed, which is the length of the 3D velocity vector). You will find that there are few particles with low speed, and few with very high speed, and lot of particles in between. This spread of speeds is natural and is created by inter-particle processes, collisions. It may seem unlikely to you, but it occurs that a particle just gets the right kick all the time an goes way out there in speed. Now as a plasma is created from a gas ... I guess you can fill in the blanks here (I hope).

"I must have missed that." Down below I talked about the photosphere, and in other places too. I discussed ionization etc. I have explained to you (as have others) that the photosphere with attributed temperature or 5700K does not mean that all photons have an energy of 5700k (note how it is important here to make the difference between K and k). I gave an explanation about how in the photosphere a neutral H can get ionized.

Multiple electron events has nothing to do with the fact whether the electron is very fast or not. It collides with the neutral H (to keep that example) and then if it is energetic enough it will bump the electron of the H into a higher shell. Now, if the collision time is smaller than the de-excitation time, another collision will give another bump to an even higher shell, and as soon as the total transferred energie of all those bumping electrons is greater than 13.5eV the H will be ionized and you have a p and an e. Now in the photospere the collision time may already be too long, however, there is a very intense black body radiation coming from the photosphere (which you don't believe in, but nevertheless). The whole story, as explained above, can also be done with absorbing multiple photons.

This is all very very basic atomic and plasma physics. And if you really do those experiments, I think it would be time to get some knowledge of the basics.

upriver

25-April-2007, 05:05 AM

"What heats the tail?" This shows that you have not had any statistical physics taught. A Maxwellian distribution deals with particles with a speed of 0 to (theoretically) a speed of infinity. Any gas (gas not plasma, gas) that you have and study you will find that there is a large spread of speeds of the particles (note that I use speed, which is the length of the 3D velocity vector). You will find that there are few particles with low speed, and few with very high speed, and lot of particles in between. This spread of speeds is natural and is created by inter-particle processes, collisions. It may seem unlikely to you, but it occurs that a particle just gets the right kick all the time an goes way out there in speed.

Thats interesting. I always thought that the faster particle speeds up the slower particle.

The reason there is Maxwellian distribution in any gas or plasma has to do with drive conditions.
You can change the distribution so the comment that the velocity of the electrons/ion goes to infinity is nonsense. If you have a cold bottle of pure H, the distribution is tighter, than inside a room at 72F where there are free ions from combustion, sparks(static, lighters, unplugging stuff), and lightning.
A highly energetic plasma has a tighter distribution.
If you have a small number of molecules/atom(BOSE-Einstein), that definitely does not apply.

"I must have missed that." Down below I talked about the photosphere, and in other places too. I discussed ionization etc. I have explained to you (as have others) that the photosphere with attributed temperature or 5700K does not mean that all photons have an energy of 5700k (note how it is important here to make the difference between K and k). I gave an explanation about how in the photosphere a neutral H can get ionized.

Blackbody 6000k means the average temperature or distribution of the particles is such that the photosphere has a spectrum of light that indicates the temperature is 6000k. The excitation spectrum indicates its this hot.

Interestingly enough, all these spectrum's are taken with electrically excited plasma. We dont even know what a 6000k non-electrical plasma looks like in a lab!!!

Multiple electron events has nothing to do with the fact whether the electron is very fast or not.

I thought you were talking about multiple glancing collisions.

It collides with the neutral H (to keep that example) and then if it is energetic enough it will bump the electron of the H into a higher shell. Now, if the collision time is smaller than the de-excitation time, another collision will give another bump to an even higher shell, and as soon as the total transferred energie of all those bumping electrons is greater than 13.5eV the H will be ionized and you have a p and an e.

The average temperature of the photosphere is .6eV, less in some places.

Now in the photospere the collision time may already be too long, however, there is a very intense black body radiation coming from the photosphere (which you don't believe in, but nevertheless).

Actually, the MS view is that that the blackbody spectrum comes from inside the photosphere.

I just don't think that the MS interpretation of its cause is correct.
It has that little UV hump at the end that is characteristic of an arc discharge. Why would it have that?

And I think that a true blackbody only comes from a solid surface.
It does not come from a decreasing density plasma ball. That would generate line emission.

The whole story, as explained above, can also be done with absorbing multiple photons. This is all very very basic atomic and plasma physics. And if you really do those experiments, I think it would be time to get some knowledge of the basics.

Hmm. What do I look for to explain what you are talking about. You say multiple electron events. Do you have a reference(book, article), that talks about how H gets ionized that way in the photosphere?

upriver

25-April-2007, 05:11 AM

Originally Posted by upriver

What happens if you have a tank of neutral gas the size of a galaxy under some pressure.
You open this tank of gas into an area of the universe where there is no matter, only space.
The tank is emptied into space.

What would be the end result?

I would say that the gas dissipates.

Same with a plasma.

Really? Why dosent the earth's atmosphere dissipate then?

????? Because the earth is a pre-existing solid body with gravity.

Not a gas cloud!!!!

Forskern

25-April-2007, 05:21 AM

????? Because the earth is a pre-existing solid body with gravity.

So only solids have gravity?

tusenfem

25-April-2007, 10:21 AM

Thats interesting. I always thought that the faster particle speeds up the slower particle.

The reason there is Maxwellian distribution in any gas or plasma has to do with drive conditions. You can change the distribution so the comment that the velocity of the electrons/ion goes to infinity is nonsense. If you have a cold bottle of pure H, the distribution is tighter, than inside a room at 72F where there are free ions from combustion, sparks(static, lighters, unplugging stuff), and lightning. A highly energetic plasma has a tighter distribution.
If you have a small number of molecules/atom(BOSE-Einstein), that definitely does not apply.

Blackbody 6000k means the average temperature or distribution of the particles is such that the photosphere has a spectrum of light that indicates the temperature is 6000k. The excitation spectrum indicates its this hot.

Interestingly enough, all these spectrum's are taken with electrically excited plasma. We dont even know what a 6000k non-electrical plasma looks like in a lab!!!

I thought you were talking about multiple glancing collisions.

The average temperature of the photosphere is .6eV, less in some places.

Actually, the MS view is that that the blackbody spectrum comes from inside the photosphere.

I just don't think that the MS interpretation of its cause is correct.
It has that little UV hump at the end that is characteristic of an arc discharge. Why would it have that?

And I think that a true blackbody only comes from a solid surface.
It does not come from a decreasing density plasma ball. That would generate line emission.

Hmm. What do I look for to explain what you are talking about. You say multiple electron events. Do you have a reference(book, article), that talks about how H gets ionized that way in the photosphere?

Upriver, you are blatandly showing how little you know about physics, I am sorry to be so harsh, but if you do not even know the basics about statistical physics, about distribution functions, about what it means that the Sun has a black body temperature of 5700K, then what are we discussing here.

You keep on misinterpreting everything I explain in great detail, like "I thought you were talking about multiple glancing collisions." which I was! And it still does not have anything to do with whether it is a fast or a slow electron. Well, it has to have some energy to deliver to the electron in the neutral H.

How would I be able exactly to "change the distribution function"? Whatever do you mean by that? Ever seen a plot of a Maxwellian function? Naturally I exaggerated with infinity, I thought that at least would be clear (as no particle will go faster than c). And what the frag is "tighter distribution"? You still seem to think that a plasma at temperature T only has particles with a kinetic energy given by kT. This is nonsense! A plasma also has a spread in kinetic energy, the temperature alotted to the species in the plasma is an ensemble average over the kinetic energies of all particles, basically it is where the Maxwellian distribution peaks, when plotted in the correct units.

What excitation spectrum. The temperature of the sun is (and this is the very very very very basics of astronomy and thermodynamics) determined by measuring the electromagnetic spectrum from IR to EUV and beyond. This again shows a sort of Maxwellian curve which is then fitted with a Maxwellian to obtain the temperature T of a black body which emits the same radiation as the sun. For the sun, this temperature of a BB is 5700K, but look at a picture of the solar spectrum and the fit for a BB, you will see that there are all kinds of photons emitted with either a much smaller or much higher energy than 5700k. Be assured that the greatest amount of the photons from the sun does not have an energy of 5700k.

If you go into details of the solar spectrum then indeed you will find that there are deviations from the BB, indeed there is a hump in the UV, which is created by some processes and is ubiquitous through the universe. Indeed the BB is produced from the photosphere down, the photosphere is called such because that is the location in the sun that the photons have a finite chance to escape the sun, whereas below it, that chance is zero and they will be absorbed and reemmited.

If you want to know everything about radiation and such, you should get the book by Rybicky and Lightmann Radiative processes in astrophysics. And as has been told to you again and again and again, you do not need a solid surface for a BB spectrum, you can also have a plasma that is in thermal equilibrium (or more complicated local thermal equilibrium), which means that the particles and the photons are interacting so well with eachother that they both can be described with a Maxwell-Boltzmann distribution.

korjik

25-April-2007, 05:55 PM

????? Because the earth is a pre-existing solid body with gravity.

Not a gas cloud!!!!

Then how did it get the atmosphere in the first place? If you say the Earth outgassed the atmosphere, then why is the earth different from your giant air tank? If the atmo came from elsewhere, then why goes gas stick to the earth because of gravity but it dosent feel gravity from other gas?

You still havent answered my question about forces.

upriver

26-April-2007, 05:23 AM

Quote:
Originally Posted by upriver View Post
Thats interesting. I always thought that the faster particle speeds up the slower particle.

The reason there is Maxwellian distribution in any gas or plasma has to do with drive conditions. You can change the distribution so the comment that the velocity of the electrons/ion goes to infinity is nonsense. If you have a cold bottle of pure H, the distribution is tighter, than inside a room at 72F where there are free ions from combustion, sparks(static, lighters, unplugging stuff), and lightning. A highly energetic plasma has a tighter distribution.
If you have a small number of molecules/atom(BOSE-Einstein), that definitely does not apply.

Blackbody 6000k means the average temperature or distribution of the particles is such that the photosphere has a spectrum of light that indicates the temperature is 6000k. The excitation spectrum indicates its this hot.

Interestingly enough, all these spectrum's are taken with electrically excited plasma. We dont even know what a 6000k non-electrical plasma looks like in a lab!!!

I thought you were talking about multiple glancing collisions.

The average temperature of the photosphere is .6eV, less in some places.

Actually, the MS view is that that the blackbody spectrum comes from inside the photosphere.

I just don't think that the MS interpretation of its cause is correct.
It has that little UV hump at the end that is characteristic of an arc discharge. Why would it have that?

And I think that a true blackbody only comes from a solid surface.
It does not come from a decreasing density plasma ball. That would generate line emission.

Hmm. What do I look for to explain what you are talking about. You say multiple electron events. Do you have a reference(book, article), that talks about how H gets ionized that way in the photosphere?

Upriver, you are blatandly showing how little you know about physics, I am sorry to be so harsh, but if you do not even know the basics about statistical physics, about distribution functions, about what it means that the Sun has a black body temperature of 5700K, then what are we discussing here.

See below.

You keep on misinterpreting everything I explain in great detail, like "I thought you were talking about multiple glancing collisions." which I was! And it still does not have anything to do with whether it is a fast or a slow electron. Well, it has to have some energy to deliver to the electron in the neutral H.

A fast electron is an electron with more than approximately 15eV of energy(or a beta particle). But I've seen them classified from 10eV to 50eV.

So what your telling me is that if you have 25 1/2 eV collisions it will ionize an atom that requires 13.6 eV. But this all has to happen before de-excitation.

And this is the way H is ionized in the photosphere.

How would I be able exactly to "change the distribution function"? Whatever do you mean by that? Ever seen a plot of a Maxwellian function?

Wiki
"Nevertheless, for a large number of particles, the fraction of particles within a particular velocity range is practically constant. The Maxwell distribution of velocities specifies what this fraction is for any velocity range as a function of the temperature of the system."

The more you constrain the temperatures of the particles(electrons, ions not photons) "the closer your function will be to a line". (Or, you will have to use a function that more closely represents a narrower(tighter) distribution of velocities.)

Naturally I exaggerated with infinity, I thought that at least would be clear (as no particle will go faster than c). And what the frag is "tighter distribution"?

See above.

No swearing.

You still seem to think that a plasma at temperature T only has particles with a kinetic energy given by kT. This is nonsense! A plasma also has a spread in kinetic energy, the temperature alotted to the species in the plasma is an ensemble average over the kinetic energies of all particles, basically it is where the Maxwellian distribution peaks, when plotted in the correct units.

I think I used that word average before. Maybe I should have said a peak where there is the greatest number of particles of any given energy.

What excitation spectrum. The temperature of the sun is (and this is the very very very very basics of astronomy and thermodynamics) determined by measuring the electromagnetic spectrum from IR to EUV and beyond.

Ya. That is the blackbody and it has the greatest number of photons in a specific range, usually shorter wavelengths.

This again shows a sort of Maxwellian curve which is then fitted with a Maxwellian to obtain the temperature T of a black body which emits the same radiation as the sun.

Yes, sort of Maxwellian.

For the sun, this temperature of a BB is 5700K, but look at a picture of the solar spectrum and the fit for a BB, you will see that there are all kinds of photons emitted with either a much smaller or much higher energy than 5700k. Be assured that the greatest amount of the photons from the sun does not have an energy of 5700k.

Blackbody 6000k means the average temperature or distribution of the particles is such that the photosphere has a spectrum of light that indicates the temperature is 6000k.

As you can see I said particles. Particles are electrons and ions. Photons are photons, and I don't consider them particles.

If you go into details of the solar spectrum then indeed you will find that there are deviations from the BB, indeed there is a hump in the UV, which is created by some processes and is ubiquitous through the universe.

Yes. The only example of that feature in any spectrum that you can find anywhere in a lab, is an arc discharge. Nothing else.
And you cant explain that feature.

Here is a webpage having to do with film and movie lighting, that has a bunch of arc source spectrums.
http://www.mole.com/aboutus/history/smpte/1943-06p333.html

Hot in the UV for a high voltage and current.
Notice for the lower current (http://www.mole.com/aboutus/history/images/smpte/1943-06_fig-13.jpg) its pretty close to the solar spectrum. It's got a little UV hump.
Solar spectrum showing the UV hump (http://climate.gsfc.nasa.gov/~cahalan/Radiation/Images/SolarIrrVblackbody.gif) as compared to a true blackbody.

Solar spectroscopic data.
http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1995ASPC...81...74G&data_type=PDF_H IGH&whole_paper=YES&type=PRINTER&filetype=.pdf

And as has been told to you again and again and again, you do not need a solid surface for a BB spectrum, you can also have a plasma that is in thermal equilibrium (or more complicated local thermal equilibrium), which means that the particles and the photons are interacting so well with eachother that they both can be described with a Maxwell-Boltzmann distribution.

I would love to see a BB thin plasma on earth.

Every plasma that I have seen makes lines. From my work with bubbles the only time that the spectrum became more blackbody like is under pressure.
If you look at this plot here you can see as the pressure goes up, it becomes more blackbody like.
http://www.nature.com/nature/journal/v434/n7029/fig_tab/nature03361_F2.html

Would a photon traveling a short distance in a dense plasma be the same as a photon traveling a long distance in a thin plasma? Would dexcitation time enter the picture?

Its clear from the Nature article that I posted, that as you increase the drive pressure in the bubble, the spectrum changes from lines to a more blackbody like shape known as a quasi continuum.

Also from what I remember, at the indicated 2.5bar drive pressure there was a 8000bar actual pressure at the center of the bubble at stagnation.

upriver

26-April-2007, 05:36 AM

Quote:
Originally Posted by upriver View Post
????? Because the earth is a pre-existing solid body with gravity.

Not a gas cloud!

Then how did it get the atmosphere in the first place? If you say the Earth outgassed the atmosphere, then why is the earth different from your giant air tank? If the atmo came from elsewhere, then why goes gas stick to the earth because of gravity but it dosent feel gravity from other gas?

We are not even to that point.

I was just trying to see how a cloud of gas would collapse and form a sun in the middle of the vacuum of space.
The equivalent of that would be if I released a galaxy sized cloud of gas or plasma into an area of space devoid of matter.
It would dissipate with no solid body for it to gravitate to.

If there was a body in the middle of the gas or maybe a plasma with a magnetic field to get it started. Then gravity takes over. That might be acceptable.

I'm not arguing with gravity's existence.

There are other ways of coming up with an atmosphere on a barren planet.

You still havent answered my question about forces.

For now, I dont know.

tusenfem

26-April-2007, 08:50 AM

I would love to see a BB thin plasma on earth.

Ofcourse you will not see a BB from a thin plasma on Earth, because one of the requirements is that the plasma has to be optically thick to create a BB.

Would a photon traveling a short distance in a dense plasma be the same as a photon traveling a long distance in a thin plasma? Would dexcitation time enter the picture?

Well, that depends on what happens to the photon, now doesn't it? De-excitation has nothing to do with this. A photon has a greater chance to be absorbed by a dense optically thick plasma then by a dilute optically thin plasma.

A fast electron is an electron with more than approximately 15eV of energy(or a beta particle). But I've seen them classified from 10eV to 50eV.
So what your telling me is that if you have 25 1/2 eV collisions it will ionize an atom that requires 13.6 eV. But this all has to happen before de-excitation.

Fast is whatever you define it to be, usually that it has more energy than thermal. But that are details.
And no, you cannot have 25 0.5eV collisions, because the excitation is quantized, so the collision by the electron or the absorbed photon needs to have at least an energy compatible to one of the shell-differences.

Wiki
"Nevertheless, for a large number of particles, the fraction of particles within a particular velocity range is practically constant. The Maxwell distribution of velocities specifies what this fraction is for any velocity range as a function of the temperature of the system."

The more you constrain the temperatures of the particles(electrons, ions not photons) "the closer your function will be to a line". (Or, you will have to use a function that more closely represents a narrower(tighter) distribution of velocities.)

I don't understand the wiki quote, do they mean that f(v)dv is constant?
If you "constrain" the energy of the particles you get a "tighter" distribution (in your words) but if you constrain them too much you can no longer define a temperature and the plasma is cold. A delta function does no longer give a temperature as we know it.

And particles are particles and photons are photons. However above the photosphere, where I was giving the example, there are very little collisions, but therefore a very strong photon field, and therefore I was talking about photons which are exciting and ionizing the H near the sun.

There are all kinds of excitation and ionization processes, collisions and photons play the important roles.

And from your UV lamps that mimic daylight, you should notice that the MIMIC daylight, and are basically concearned with the visible, which is needed for making movies.

Van Rijn

26-April-2007, 09:25 AM

korjik

26-April-2007, 03:11 PM

We are not even to that point.

I was just trying to see how a cloud of gas would collapse and form a sun in the middle of the vacuum of space.
The equivalent of that would be if I released a galaxy sized cloud of gas or plasma into an area of space devoid of matter.
It would dissipate with no solid body for it to gravitate to.

If there was a body in the middle of the gas or maybe a plasma with a magnetic field to get it started. Then gravity takes over. That might be acceptable.

I'm not arguing with gravity's existence.

There are other ways of coming up with an atmosphere on a barren planet.

For now, I dont know.

The pressure in a gas is caused by collisions in the gas. Very basic kinetic theory of gases. The rate that collisions occur is dependent on the density of the gas. Most space gases, wether plasma or not are extremely thin and dont collide very often. This means that the primary driver of motions in the gas will be gravity and electromagnetic. For any neutral atom, EM is weak and the driver of the motion will be gravity.

All of this is freshman physics.

Now you take a roughly spherical blob of neutral gas with a roughly constant pressure throughout. For any atom of this gas to leave the blob, it has to have a kinetic energy greater than the total gravitational potential of the rest of the blob. For the blob to be stable, the average kinetic energy should be roughly equal to the gravitational potential.

This is upper level undergraduate physics. Lagrangian/Hamiltonian formulation of physics to be exact.

If the gas is dense enough to be collisional, the same rules still apply. For any atom to escape, its KE > GPE must be true. Either the gas is warm enough and it expands, or it is dense enough and contracts. The derivation of the Jeans Criteria is a good explanation of the conditions needed for collapse. This should be in any introductory astrophysics book. Which is one of the reasons that you keep getting asked to read one.

upriver

29-April-2007, 10:19 PM

Ofcourse you will not see a BB from a thin plasma on Earth, because one of the requirements is that the plasma has to be optically thick to create a BB.

Or said another way. You would have to have a tube 500 miles long with a plasma of the same density as the photosphere to obtain a blackbody spectrum for "thin" plasma.

Originally Posted by upriver View Post
Would a photon traveling a short distance in a dense plasma be the same as a photon traveling a long distance in a thin plasma? Would dexcitation time enter the picture?

Well, that depends on what happens to the photon, now doesn't it? De-excitation has nothing to do with this. A photon has a greater chance to be absorbed by a dense optically thick plasma then by a dilute optically thin plasma.

If you had a ion that was in a thin plasma, it would have a greater chance of deexciting before the next photon impact, as opposed to a thick plasma.

So for a give energy of photon entering a plasma I would expect more ions
in a dense plasma for a given density vs absorption rate.

Fast is whatever you define it to be, usually that it has more energy than thermal. But that are details.
And no, you cannot have 25 0.5eV collisions, because the excitation is quantized, so the collision by the electron or the absorbed photon needs to have at least an energy compatible to one of the shell-differences.

Yes. Fast electrons are any electrons with more energy than thermal.

For H that is a minimum of 3.4eV.

Originally Posted by upriver View Post
Wiki
"Nevertheless, for a large number of particles, the fraction of particles within a particular velocity range is practically constant. The Maxwell distribution of velocities specifies what this fraction is for any velocity range as a function of the temperature of the system."

The more you constrain the temperatures of the particles(electrons, ions not photons) "the closer your function will be to a line". (Or, you will have to use a function that more closely represents a narrower(tighter) distribution of velocities.)

I don't understand the wiki quote, do they mean that f(v)dv is constant?

Thats what I thought they meant at first, but that doesn't make sense because you can change the distribution so that Maxwellian no longer represents the distribution. I think thats right.

If you "constrain" the energy of the particles you get a "tighter" distribution (in your words) but if you constrain them too much you can no longer define a temperature and the plasma is cold. A delta function does no longer give a temperature as we know it.

If you ultimately confine the plasma you would have a Bose-Einstein condensate.

And particles are particles and photons are photons. However above the photosphere, where I was giving the example, there are very little collisions, but therefore a very strong photon field, and therefore I was talking about photons which are exciting and ionizing the H near the sun.

There are all kinds of excitation and ionization processes, collisions and photons play the important roles.

The mainstream model says that blackbody comes from an optically thick plasma.
So where does the UV come from? In the MS model all UV(192,171,1600, etc.) pictures are above the photosphere. Those UV photons come from inside the photosphere?
Obviously you have 2 different areas generating the blackbody spectrum of the sun.
Visible photons from the photosphere are not strong enough to excite H to emit anything so it has to be UV from the transition region that ionizes everything underneath(MS model).
http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html#c3

originally posted by NASA
Heat flows down from the corona into the chromosphere and in the process produces this thin region where the temperature changes rapidly from 1,000,000ºC (1,800,000ºF) down to about 20,000ºC (40,000ºF). Hydrogen is ionized (stripped of its electron) at these temperatures and is therefore difficult to see. Instead of hydrogen, the light emitted by the transition region is dominated by such ions as C IV, O IV, and Si IV (carbon, oxygen, and silicon each with three electrons stripped off). http://solarscience.msfc.nasa.gov/t_region.shtml

There are a few things wrong here.
"Heat flows down from the corona into the chromosphere"
Kinetic or photons?
It would have to be photons since the electrons and ions are moving outward(solar wind).
Million(12.3nm EUV) degree photons are heating the transition region?
Heat also flows up from the photosphere.

Its like the oven example give by Van Rijn.

H is stripped of it electrons. Not possible at these temperatures.
I understand that the temperature is a measurement of the average of the plasma.

But there has to be a process that is equivalent of at least 3.4 eV(33,000K).
They only way that plasma can be driven is electrically because that is the only way to produce a plasma with a lower temperature than the ionization energy. Something has to drive the electron tail otherwise it would cool off.

Of course the there is photo ionization, but that cant account for all of the ionization above the photosphere including the corona(>100eV).

And from your UV lamps that mimic daylight, you should notice that the MIMIC daylight, and are basically concearned with the visible, which is needed for making movies.

Carbon arc lamps at different voltages do not MIMIC the sun.
It is a natural process unto itself, unlike blackbody calculations above 3000K.

"Canada's Ultimate Light Ruler"
"The core of the high-temperature blackbody is a hollow tube of a special form of graphite that can withstand intense heating. In order to produce the required UV radiation, the graphite core is heated to approximately 3230 ºC, a temperature at which almost all metals melt – hence the name an ultra high-temperature blackbody. The graphite is gradually heated over the course of several hours in the same way a stove element is heated – by running an electrical current through it. The graphite core is insulated with many concentric layers of carbon cloth which are water-cooled.

"At 3230 ºC any oxygen would react instantly with the graphite, causing a fire. So, during operation the entire core is flushed with argon, a non-reactive gas."
http://www.nrc-cnrc.gc.ca/highlights/2006/0602blackbody_e.html

I called them up and talked to them about this device.
The reason they use Argon also is that it produces no measurable spectrum at that temperature to distort the solid blackbody spectrum.

If you can get a spectrum from an arc that looks like the sun spectrum with that little UV hump, then you have to say "Why would the sun have a feature that appears in an electrical process?"

An arc is an irreducible process. It is a basic feature of the universe!
They appear everywhere on our planet.

You can use it as an explanation for other spectra.

tusenfem

30-April-2007, 07:09 AM

1. Or said another way. You would have to have a tube 500 miles long with a plasma of the same density as the photosphere to obtain a blackbody spectrum for "thin" plasma.

2. If you had a ion that was in a thin plasma, it would have a greater chance of deexciting before the next photon impact, as opposed to a thick plasma.

3. So for a give energy of photon entering a plasma I would expect more ions
in a dense plasma for a given density vs absorption rate.

4. Yes. Fast electrons are any electrons with more energy than thermal. For H that is a minimum of 3.4eV.

5. Thats what I thought they meant at first, but that doesn't make sense because you can change the distribution so that Maxwellian no longer represents the distribution. I think thats right.

6. If you ultimately confine the plasma you would have a Bose-Einstein condensate.

7. The mainstream model says that blackbody comes from an optically thick plasma. So where does the UV come from? In the MS model all UV(192,171,1600, etc.) pictures are above the photosphere. Those UV photons come from inside the photosphere? Obviously you have 2 different areas generating the blackbody spectrum of the sun. Visible photons from the photosphere are not strong enough to excite H to emit anything so it has to be UV from the transition region that ionizes everything underneath(MS model).
http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html#c3

8. There are a few things wrong here. "Heat flows down from the corona into the chromosphere" Kinetic or photons?

9. Carbon arc lamps at different voltages do not MIMIC the sun. It is a natural process unto itself, unlike blackbody calculations above 3000K.

1. Whatever! Could you please try to understand that the photosphere is defined by the fact that photos can escape. The BB is created below the photosphere. So the density of the photosphere is PER DEFINITION not optically thick.

2. Well, that would depend on the density of the photon field, now wouldn't it. A excited ion in a dilute plasma has:
a: a bigger chance to de-excite than to collisionally excite to a higher state
b: in a dilute photon field a bigger chance to de-excite than to absorb a photon to go to a higher state
c: in a dense photon field a bigger chance to absorb a photon before de-excitation.

3. HUH? If you mean that a photon has a bigger chance of interacting with a ion in a dense plasma, then yes, but it has nothing to do with the energy of the photon, unless you want it to be absorbed, and thus has to have an energy greater or equal to one of the excitation levels.

4. For example, but these are details, first get the basics.

5. To go into detail, you can have a plasma with a maxwellian and then you can e.g. use a laser to cool the plasma. This is nothing else than that the spread in velocities of the ions (electrons) is being reduced. This will "narrow" your maxwell distribution. Now you can have this "cold" plasma travel extremely fast in one direction, but it will still be cold.
What the wiki quote means I think is N(energy interval)/N(total) is approximately constant. I think it is a strange sentence.

6. Sure BEC, let's make this discussion even more complicated. I would advise to first get the basics of plasma physics right and then you can have a look into BEC.

7. UV is created by the BB underneath the photosphere. But if you would look at a spectrum of a BB you would find that the number of photons falls off. Naturally there are lots of other processes in the sun, all having to do with particle acceleration that create also high energy photons. Now, these processes are more efficient (because the mainly radiate in these frequencies) and therefore, these are much brighter than the "background" coming from the photosphere. Take a picture of the sun in a clear blue sky and you will see that the sky will be black. That is how photography works, if you want to look at the details of something bright.

8. Why not read a book, instead of looking at a 10 line website. These are only for quick info, if you want the details you need to go to a library or a astrophysics course.

9. Keep dreaming. They use it to produce daylight in the movies, so, the lamp MIMICS daylight, it cannot be daylight, because it comes from a lamp, so it is lamplight, a lamp mimicking the sun.

upriver

01-May-2007, 01:30 AM

1. Whatever! Could you please try to understand that the photosphere is defined by the fact that photos can escape. The BB is created below the photosphere. So the density of the photosphere is PER DEFINITION not optically thick.

7. UV is created by the BB underneath the photosphere. But if you would look at a spectrum of a BB you would find that the number of photons falls off. Naturally there are lots of other processes in the sun, all having to do with particle acceleration that create also high energy photons. Now, these processes are more efficient (because the mainly radiate in these frequencies) and therefore, these are much brighter than the "background" coming from the photosphere.

You are claiming that the pictures taken at 171, 192, transition layer and Lyman alpha are from under the photosphere?

Are you claiming that all of the energy IR to EUV from the sun is from under the photosphere? Because we know that x-ray and higher energies, come from above the photosphere.

9. Keep dreaming. They use it to produce daylight in the movies, so, the lamp MIMICS daylight, it cannot be daylight, because it comes from a lamp, so it is lamplight, a lamp mimicking the sun.

Again. That lamp is not MIMICKING sunlight. That is what they say. It is its own process. It's not like they added gas or mercury or something. It is an arc with a spectrum that is close to sunlight. I never said it was daylight. It is just a carbon arc. No gas, nothing else. Look at the chart, the only thing they change is the voltage.
I said it has a spectral feature that is from an electrical process.
The sun has the same feature.
It proves that the sun is an electrical process. You cannot explain that feature any other way even though you may try.
It is a process unto itself. Just because its a lamp doesn't mean it is not its own process.
If you can get a spectrum from an arc that looks like the sun spectrum with that little UV hump, then you have to say "Why would the sun have a feature that appears in an electrical process?"

An arc is an irreducible process. It is a basic feature of the universe!
They appear everywhere on our planet.

You can use it as an explanation for other spectra.

Take a picture of the sun in a clear blue sky and you will see that the sky will be black. That is how photography works, if you want to look at the details of something bright.

Are you saying that cameras see something different than our eyes?
I heard of UFO's being captured on camera that our eyes didn't see.
Every time I video the sun, I get the sky. Photography is a family tradition.

Karen Callahan(mom) Wildflowers.
http://www.amazon.com/500-Sierra-Wildflowers-Larry-Norris/dp/1890266345
Rose Callahan photography.(sister)
http://www.rosecallahan.com/
Brant Callahan high speed video(me)
http://www.impulsedevices.com/media.html

If I was out in space photographing the sun then the background around the sun would be black. Or maybe using a telescope.....

tusenfem

01-May-2007, 09:51 AM

You are claiming that the pictures taken at 171, 192, transition layer and Lyman alpha are from under the photosphere?

Are you claiming that all of the energy IR to EUV from the sun is from under the photosphere? Because we know that x-ray and higher energies, come from above the photosphere.

Did you read at all, everything I wrote? All energy of the sun is created in its center by fusion (which you don't believe, but that's an aside). Therefore, all photons emitted from the sun are eventually "created in the center of the sun". The photosphere is called such, because that is the layer that photons can escape from the sun, that is where the sun gets optically thin. This has all to do with the BB spectrum from the sun.

Now you might want to look at specific lines (which you like because then you can claim if you see black it means there is nothing there and an iron shell is located there, but that's an aside).

I claimed that the BB of the sun is created in the optically thick part of the sun under the photosphere. Apart from that I wrote that there are numerous processes above the photosphere that also create UV and X-rays, most of them to do with particle acceleration, and bremsstrahlung. However, the BB of the sun ALSO contains UV and Xray wavelengths.

Again. That lamp is not MIMICKING sunlight. That is what they say. It is its own process. It's not like they added gas or mercury or something. It is an arc with a spectrum that is close to sunlight. I never said it was daylight. It is just a carbon arc. No gas, nothing else. Look at the chart, the only thing they change is the voltage.
I said it has a spectral feature that is from an electrical process.
The sun has the same feature.
It proves that the sun is an electrical process. You cannot explain that feature any other way even though you may try.
It is a process unto itself. Just because its a lamp doesn't mean it is not its own process.
If you can get a spectrum from an arc that looks like the sun spectrum with that little UV hump, then you have to say "Why would the sun have a feature that appears in an electrical process?"

An arc is an irreducible process. It is a basic feature of the universe!
They appear everywhere on our planet.

You can use it as an explanation for other spectra.

My bolding in the 3rd line. So, it produces a spectrum that looks like that of the sun and thus MIMICs the spectrum of the sun. From Webster-Mirriam dictionary the definition of "to mimic":

Main Entry: 3mimic
Function: transitive verb
Inflected Form(s): mim·icked /-mikt/; mim·ick·ing
1 : to imitate closely : APE
2 : to ridicule by imitation
3 : SIMULATE
4 : to resemble by biological

I think that meaning 1 and 3 fit closely to what I bolded.

From the link you gave about the arc and spectrum etc.:

At 3230 ºC any oxygen would react instantly with the graphite, causing a fire. So, during operation the entire core is flushed with argon, a non-reactive gas. At operating temperature, the high-temperature blackbody produces an intense beam of light that's emitted from a tiny eight millimetre hole – a beam of light that will soon be Canada's ultimate light ruler.

Well well, the arc heats up the carbon which gives off a BB radiation because it is so hot. Surprise? not really. And with the sun there is no arc but hot fusion inside and the optically thick plasma takes over the role of the carbon and whoopie there is a BB spectrum.

And I have no idea what you mean with that an arc is an "irreducible process".

Are you saying that cameras see something different than our eyes?
I heard of UFO's being captured on camera that our eyes didn't see.
Every time I video the sun, I get the sky. Photography is a family tradition.

If I was out in space photographing the sun then the background around the sun would be black. Or maybe using a telescope.....

I mean that if you photograph something with great intensity contrasts, and you want to concentrate on the stuff that gives the greatest intensity (because on the sun that is the most interesting) then you will not see the low intensity stuff. It is like the photographs of the astronauts on the moon, where you do not see the background stars because they are too weak, the intensity contrast is too great and the shutter time is too short to resolve this contrast. If you want to look at coronal loops e.g. you do not want them to be washed out by over exposing them.

korjik

01-May-2007, 02:56 PM

upriver

01-May-2007, 04:58 PM

Originally Posted by upriver View Post
You are claiming that the pictures taken at 171, 192, transition layer and Lyman alpha are from under the photosphere?

Are you claiming that all of the energy IR to EUV from the sun is from under the photosphere? Because we know that x-ray and higher energies, come from above the photosphere.

emission lines arent blackbody radiation. You should learn the difference.

Yes, that is true. However those emission lines are EUV at 17.1nm. EUV is from the shorter end of the blackbody curve. And tusenfem is saying that EUV comes from under/in the photosphere. Yet NASA says that it is from above the photosphere.
I have maintained that the images taken at 171A are from under the photosphere, and are of the surface of the sun.
I was told that was incorrect.
I was then told that the photosphere would absorb UV. So I went to the TOPS data base and did a little research. It turns out that for certain conditions the photosphere is transparent to UV at certain wavelengths.

"However, the BB of the sun ALSO contains UV and Xray wavelengths."

6000K is enough to make x-rays?

From under the photosphere?

I for one, am curious how you determine what UV comes from where.

Let's see: There is no such thing as a perfect vaccum and when you run a carbon arc lamp some of the carbon is vaporized and the vapor is able to make emission lines.

Wow, I did it in one runon sentence. I guess that the sun isnt a huge carbon arc lamp.

I agree. The spectrum from the arc is a combination of local conditions(gas, carbon, voltage) but it still has certain features for a default setting that make it recognizably an arc.

Show me the full spectrum emission of both please.

Carbon arc.
http://www.mole.com/aboutus/history/images/smpte/1943-06_fig-13.jpg

Sun.
http://climate.gsfc.nasa.gov/~cahalan/Radiation/SolarIrrVblackbody.html

Funny, I dont remember Maxwells Law of Arcing. Seems reducible to me.

If you cranked up a plasma tube voltage high enough you would have an arc in it. An arc is a basic process. I dont mean on a microscopic level even though microscopic sparks/arcs exist. I mean as a macroscopic phenomena.

Are you honestly suggesting that cameras see the same as eyes? Try again. Even the best cameras have limits different than the human eye.

Come on. Lets not go there. Do you honestly think I would not have that level of understanding?

Actually, if you reduce the apeture of the camera to the point where you can get details of the sun, the sky is going to be too dark to see. Try getting a solar filter and look up.

Actually, it all depends on the dynamic range of the detector. Infinite dynamic range means you can resolve all light and all dark.

You still havent answered my questions about forces.

Refresh me.

tusenfem

01-May-2007, 06:58 PM

Yes, that is true. However those emission lines are EUV at 17.1nm. EUV is from the shorter end of the blackbody curve. And tusenfem is saying that EUV comes from under/in the photosphere. Yet NASA says that it is from above the photosphere.
I have maintained that the images taken at 171A are from under the photosphere, and are of the surface of the sun.
I was told that was incorrect.
I was then told that the photosphere would absorb UV. So I went to the TOPS data base and did a little research. It turns out that for certain conditions the photosphere is transparent to UV at certain wavelengths.

"However, the BB of the sun ALSO contains UV and Xray wavelengths."

6000K is enough to make x-rays?

From under the photosphere?

I for one, am curious how you determine what UV comes from where.

Please try to understand the difference between the BB spectrum created by the sun and the emission lines and other non-thermal emission processes on the sun.

I told you that the UV, EUV and Xray parts of the BB spectrum, OF THE BB SPECTRUM, comes from the sun. And yes, a 6000 K body will also have Xray emission, take a look at a BB spectrum of a 6000 K body.

there are other emission lines in the EUV that indeed are created above the photosphere. I told you that often these emissions have to do with particle acceleration, and the Xrays that are created are usually associated to bremsstrahlung, the quick braking of highly accelerated electrons by the denser plasma at the footpoints of magnetic loops.

One cannot see under the photosphere, because that is the level at which photons can escape. Any photons created slightly deeper will be absorbed again and reemitted, forgetting where they came from. Definition by optical terms (taking optical in a very broad meaning) the photosphere IS the "surface" of the sun.

I cannot remember that you were told that the photosphere would absorb the UV. I do remember pointing to the opacity project so you could look at absorption coefficients and such. You tried to calculate whether a photospheric plasma would be optically thick or thin. But you did not take into account that the density of the plasma increases when you "look further into the sun."

Please try first to get the definitions of terms correct, like what exactly the photosphere is, what a BB spectrum is, what emission lines and non-thermal emissions are above the photosphere, etc. etc.

upriver

02-May-2007, 06:40 AM

Please try to understand the difference between the BB spectrum created by the sun and the emission lines and other non-thermal emission processes on the sun.

Let me see if I can clear this up.

171, 192A are emission lines that are Extreme Ultra Violet(EUV) in nature. In other words 171A is 17.1nm in wavelength.

A plasma blackbody spectrum extends from depending on the temperature, x-rays to IR.

17.1nm falls in the upper end of the black body spectrum.
It is included in the spectrum.

We "know" that 17.1nm comes from the transition layer above the photosphere.
Is there 17.1nm that comes from inside the photosphere that is also part of the BB spectrum?

I told you that the UV, EUV and Xray parts of the BB spectrum, OF THE BB SPECTRUM, comes from the sun. And yes, a 6000 K body will also have Xray emission, take a look at a BB spectrum of a 6000 K body.

x-rays begin at 10nm.
http://eosweb.larc.nasa.gov/EDDOCS/images/Erb/wavelength_figure.jpg

20,000K before x-rays here.
http://www.inchem.org/documents/ehc/ehc/e23_3.gif

30,000K before x-rays here.
http://www.britastro.org/vss/gifs/gs1.gif
http://www.britastro.org/vss/gs-cofp1.html

They even say things like gas of millions of degrees is required for x-rays.:)

What they really mean in that they need electrons of a certain energy to make x-rays.

Interestingly enough, x-rays are a k-shell phenomena available only with high energy electrons. In a solid you electrically accelerate electron into a target.
In plasma you electrically excite the plasma.

there are other emission lines in the EUV that indeed are created above the photosphere. I told you that often these emissions have to do with particle acceleration, and the Xrays that are created are usually associated to bremsstrahlung, the quick braking of highly accelerated electrons by the denser plasma at the footpoints of magnetic loops.

Where do the highly accelerated electron come from? And isnt that just another name for electricity.

One cannot see under the photosphere, because that is the level at which photons can escape. Any photons created slightly deeper will be absorbed again and reemitted, forgetting where they came from. Definition by optical terms (taking optical in a very broad meaning) the photosphere IS the "surface" of the sun.

Yes, I understand that. But if you have a gamma ray and it takes its 1 million or so years to get the surface, losing energy all the way so that by the time it gets to the surface its a photon that falls anywhere in the range from x-ray to IR, are there x-ray images of the photosphere?

Here is an x-ray image of the sun. It speaks for itself.
http://www.solarviews.com/raw/sun/xsun.jpg

wiki
"X-rays in the 0.5 to 5 keV (80 to 800 aJ) range, where most celestial sources give off the bulk of their energy, can be stopped by a few sheets of paper; ninety percent of the photons in a beam of 3 keV (480 aJ) X-rays are absorbed by traveling through just 10 cm of air."

I cannot remember that you were told that the photosphere would absorb the UV. I do remember pointing to the opacity project so you could look at absorption coefficients and such. You tried to calculate whether a photospheric plasma would be optically thick or thin. But you did not take into account that the density of the plasma increases when you "look further into the sun."

I claimed that the BB of the sun is created in the optically thick part of the sun under the photosphere.

"Optical thickness is the depth of a material or medium in which the intensity of light (or other radiation) of a given frequency is reduced by a factor of 1 / e. In rough terms ⅔ of the light is absorbed within one optical thickness depth."

If you look at it they way the photons are coming, the plasma is decreasing in thickness. So from a certain depth, it becomes easier for the photons to travel, meaning less absorption.

The photosphere is defined as being 500km in depth because I would assume that is the calculated optical thickness depth.

Would photons of specific energies escape at different depths?

What about the fact that the photosphere has 300km high features on it.
http://www.lmsal.com/Press/SPD2003.html

Please try first to get the definitions of terms correct, like what exactly the photosphere is, what a BB spectrum is, what emission lines and non-thermal emissions are above the photosphere, etc. etc.

Si, above.

tusenfem

02-May-2007, 07:39 AM

Let me see if I can clear this up.

171, 192A are emission lines that are Extreme Ultra Violet(EUV) in nature. In other words 171A is 17.1nm in wavelength.

A plasma blackbody spectrum extends from depending on the temperature, x-rays to IR.

17.1nm falls in the upper end of the black body spectrum.
It is included in the spectrum.

We "know" that 17.1nm comes from the transition layer above the photosphere.
Is there 17.1nm that comes from inside the photosphere that is also part of the BB spectrum?

171 A (A for Angstrom with a little circle on the A and an umlaut on the o) is 17.1 nm (nanometer) which shows you that Angstroms are 10-10 m, both are lengths so both are wavelengths.

A BB extends from IR to X true and will peak somewhere in between. The very high and the very low wavelength regions will have very little intensity.

We know that the bulk of the 17.1 nm EUV emission is coming from above the photosphere, but there is a small portion of it that comes from the BB of the sun.

x-rays begin at 10nm.
http://eosweb.larc.nasa.gov/EDDOCS/images/Erb/wavelength_figure.jpg

20,000K before x-rays here.
http://www.inchem.org/documents/ehc/ehc/e23_3.gif

30,000K before x-rays here.
http://www.britastro.org/vss/gifs/gs1.gif
http://www.britastro.org/vss/gs-cofp1.html

They even say things like gas of millions of degrees is required for x-rays.:)

I told you the far regions of the BB spectrum were weak in intensity, as show the plots you have linked to. Indeed, if you would have million degrees plasma the peak might even shift into the Xray.

What they really mean in that they need electrons of a certain energy to make x-rays.

Interestingly enough, x-rays are a k-shell phenomena available only with high energy electrons. In a solid you electrically accelerate electron into a target.
In plasma you electrically excite the plasma.

Why would they be a k-shell phenomenon? High energy electrons along a magnetic field, that suddenly enter at the footpoints of the field into a denser plasma are braked, a sudden decrease of velocity. They emit bremsstrahlung (a german word for braking-radiation), there is nothing k-shell about it.

I guess you are thinking about heavy ions, where one might knock out an electron from the k-shell and then the electrons falling back emitting X-rays (???).

Do you mean excite or accelerate in your last sentence?

Where do the highly accelerated electron come from? And isnt that just another name for electricity.

The electrons are accelerated by: 1. electric fields; 2. shocks.
The emission is created by taking the energy of the electrons.
Being an engineer, I guess you want to call moving electrons electricity, because they might (or might not) constitute a current.

Yes, I understand that. But if you have a gamma ray and it takes its 1 million or so years to get the surface, losing energy all the way so that by the time it gets to the surface its a photon that falls anywhere in the range from x-ray to IR, are there x-ray images of the photosphere?

Here is an x-ray image of the sun. It speaks for itself.
http://www.solarviews.com/raw/sun/xsun.jpg

wiki
"X-rays in the 0.5 to 5 keV (80 to 800 aJ) range, where most celestial sources give off the bulk of their energy, can be stopped by a few sheets of paper; ninety percent of the photons in a beam of 3 keV (480 aJ) X-rays are absorbed by traveling through just 10 cm of air."

"Optical thickness is the depth of a material or medium in which the intensity of light (or other radiation) of a given frequency is reduced by a factor of 1 / e. In rough terms ⅔ of the light is absorbed within one optical thickness depth."

If you look at it they way the photons are coming, the plasma is decreasing in thickness. So from a certain depth, it becomes easier for the photons to travel, meaning less absorption.

The photosphere is defined as being 500km in depth because I would assume that is the calculated optical thickness depth.

Would photons of specific energies escape at different depths?

What about the fact that the photosphere has 300km high features on it.
http://www.lmsal.com/Press/SPD2003.html

The optical depth of tau=2/3 naturally depends on wavelength, because the absorption coefficients depend on wavelength. So you will not see exactly the same layer in different wavelengths.

I will not claim that an Xray from the BB will be created in the center of the sun. There is just a finite chance for an Xray to be emitted by a BB.

Indeed, the Xray pic shows that the "surface" of the sun is dark, meaning very little intensity.

What about the structure on the photosphere? As the surface of the sun resembles a boiling pot of water, I am not really surprised.

korjik

02-May-2007, 08:04 PM

upriver

05-May-2007, 01:51 AM

We know that the bulk of the 17.1 nm EUV emission is coming from above the photosphere, but there is a small portion of it that comes from the BB of the sun.

The point I was trying to make is that even though statistically you could have x-rays in a 6000K blackbody, is that reality? In a 6000K blackbody you dont even get the top of UV. 10000K gets you about 60nm.

And as I asked before, how do you separate BB from other emissions(assuming the same process).

I told you the far regions of the BB spectrum were weak in intensity, as show the plots you have linked to. Indeed, if you would have million degrees plasma the peak might even shift into the Xray.

In a million degrees plasma on earth I would expect lines, unless under pressure, or in a 500km long tube, as you would say 2/3 tau.

Why would they be a k-shell phenomenon? High energy electrons along a magnetic field, that suddenly enter at the footpoints of the field into a denser plasma are braked, a sudden decrease of velocity. They emit bremsstrahlung (a german word for braking-radiation), there is nothing k-shell about it.

I guess you are thinking about heavy ions, where one might knock out an electron from the k-shell and then the electrons falling back emitting X-rays (???).

Yes. I was thinking ionization processes. You are correct that Bremsstrahlung is the dominant x-ray process.

From some where;
"The interpretation of observations of the hard x-ray emission from flares has in fact suggested that all of the energy released in flares initially goes into accelerated electrons! It is difficult to see how this could happen."

I had to look it up. Basically, with a change in vector direction, the electron emits an x-ray with an energy depending on the speed of the electron.
It is interesting that the electron emits radiation, when in gravitational theory there is no work done in this same situation.
What is the difference?

Charged particles accelerated in electric or magnetic fields radiate.

"Mechanisms of particle acceleration in solar flares."
"Bremsstrahlung (plasma electrons accelerated by the electric field of positive ions)
Bremsstrahlung can be emitted at all wavelengths, from long radio waves to short X-rays. During solar flares it becomes very intense at X–ray wavelengths.
Two kinds of bremsstrahlung at X-ray wavelengths: thermal and nonthermal.
Thermal bremsstrahlung is produced by electrons described by a distribution (e.g. Maxwellian) characterized by a single temperature.
In solar flares plasma is heated to high temperatures of about 10 million degrees, producing thermal bremsstrahlung in soft X–rays (1-10 KeV).
Nonthermal bremsstrahlung radiation is emitted by electrons which are not part of a thermal gas. They cannot be described by a conventional temperature.
Hard X–ray (10-1000 KeV) bremsstrahlung is produced when highly energetic nonthermal electrons (at relativistic speeds) interact with ambient protons.
Gamma ray continuum emission is produced as the result of bremsstrahlung from electrons with energies 1 Mev.
Cyclotron or synchrotron radiation (charged particles gyrating in a magnetic field with gyrofrequency).
Cyclotron radiation: moderately relativistic electrons. Spectrum consists of discrete lines at frequencies given by the resonance condition.
Synchrotron radiation: highly relativistic electrons.
Overlap of higher harmonics spectrum is continuous."
http://www-solar.mcs.st-and.ac.uk/~paolo/talk1.pdf

Do you mean excite or accelerate in your last sentence?

I meant excite which means "to excite a gas" but that also means "to accelerate" in the case of the electrons in the plasma.

The electrons are accelerated by: 1. electric fields; 2. shocks.
The emission is created by taking the energy of the electrons.
Being an engineer, I guess you want to call moving electrons electricity, because they might (or might not) constitute a current.

Yes. It depends on the endpoints as to whether it is a current or not. If you have 100,000 electrons flowing from one piece of matter to another, it is a current for that moment. You are equalizing the charge differential between the 2 pieces of matter.

I would say, To digress for a moment, that the sun and the heliosphere are endpoints in a circuit.

The optical depth of tau=2/3 naturally depends on wavelength, because the absorption coefficients depend on wavelength. So you will not see exactly the same layer in different wavelengths.

As the TOPS database showed us.

I will not claim that an Xray from the BB will be created in the center of the sun. There is just a finite chance for an Xray to be emitted by a BB.

Ya, well I will say that there are no x-rays emitted by a blackbody of the suns temperature. According to the curves at that temperature it doesnt emit below 90nm(significantly).
The only way they could possibly be emitted is if there is electricity involved.

Indeed, the Xray pic shows that the "surface" of the sun is dark, meaning very little intensity.

No intensity except from flare activity.

What about the structure on the photosphere? As the surface of the sun resembles a boiling pot of water, I am not really surprised.

How could you have boiling in a gas?
Boiling is cavitation in a liquid. Sonoluminiscence is cavitation. I am intimately familiar with this.
It is when the steam/air dissolved in the liquid vaporizes/expands, it forms a bubble and rises to the top.

Think about turbulent flows in water. The only way you could tell the difference is if you have different properties like one glowed and one did not. Or one was dyed and one was not.
But to say that the change is density is responsible for the determination of the surface of last scattering, is not addressing the real question, what causes the difference in density as evidenced by helioseismology?

upriver

05-May-2007, 02:26 AM

The pressure in a gas is caused by collisions in the gas. Very basic kinetic theory of gases. The rate that collisions occur is dependent on the density of the gas. Most space gases, wether plasma or not are extremely thin and dont collide very often. This means that the primary driver of motions in the gas will be gravity and electromagnetic. For any neutral atom, EM is weak and the driver of the motion will be gravity.

All of this is freshman physics.

Now you take a roughly spherical blob of neutral gas with a roughly constant pressure throughout. For any atom of this gas to leave the blob, it has to have a kinetic energy greater than the total gravitational potential of the rest of the blob. For the blob to be stable, the average kinetic energy should be roughly equal to the gravitational potential.

This is upper level undergraduate physics. Lagrangian/Hamiltonian formulation of physics to be exact.

If the gas is dense enough to be collisional, the same rules still apply. For any atom to escape, its KE > GPE must be true. Either the gas is warm enough and it expands, or it is dense enough and contracts. The derivation of the Jeans Criteria is a good explanation of the conditions needed for collapse. This should be in any introductory astrophysics book. Which is one of the reasons that you keep getting asked to read one.

Well according to the sun, you would not have a constant pressure throughout "a roughly spherical blob of neutral gas" even though you could mathematically simulate that condition. I understand what you are saying but is it reality?

It is denser in the middle.
And a very rare condition.

Most of the literature on star formation is about triggered formation, not purely gravitational collapse formation.
They say 10,000F is sufficient to escape. And since temperature is an average, there will always be escapees....

tusenfem

05-May-2007, 05:53 PM

The point I was trying to make is that even though statistically you could have x-rays in a 6000K blackbody, is that reality? In a 6000K blackbody you dont even get the top of UV. 10000K gets you about 60nm.

And as I asked before, how do you separate BB from other emissions(assuming the same process).

yes it is reality, take a look at the total BB spectrum of the sun and you will see that indeed the Xray part is non-zero, low, but nonetheless.

Separating BB from other emissions is simple. First of all the BB is a continuum over all frequency space, whereas all other emissions mostly are line emissions or other non-thermal emissions. And finding out the difference between the two is in most cases not too difficult. I am not here to teach you all this, because it would take way too much time and space here on BAUT. Read up on it in basically ANY introductory astrophysics book.

In a million degrees plasma on earth I would expect lines, unless under pressure, or in a 500km long tube, as you would say 2/3 tau.

Yes. I was thinking ionization processes. You are correct that Bremsstrahlung is the dominant x-ray process.

From some where;
"The interpretation of observations of the hard x-ray emission from flares has in fact suggested that all of the energy released in flares initially goes into accelerated electrons! It is difficult to see how this could happen."

I had to look it up. Basically, with a change in vector direction, the electron emits an x-ray with an energy depending on the speed of the electron.
It is interesting that the electron emits radiation, when in gravitational theory there is no work done in this same situation.
What is the difference?

Oh brother, read the first two chapters of Rybicky and Lightman, they can explain it much better than I can here in this limited forum. You can build a 500 km long plasma tube that along its length may be optically thick, but not across its width. A plasma has to be totally optically thick and will have to be in Local Thermal Equilibrium (LTE), which will also not be the case in your tube. So, you will NOT get a BB spectrum out of your 500 km tube. It is al such basic plasma and radiative transport physics, that can be read in any introductory book.

Then the electrons rotating around the magnetic field and radiating cyclotron or synchrotron radiation. These particles lose energy,that is where the energy of the photons comes from.
Apart from that you cannot compare this situation with a body in a gravitational field, because the emission of the photon is specifically related to the fact that there is a charged particle gyrating along the field. And it needs not radiate an X-ray, that totally depends on the energy of the electron. Again, read up on it in e.g. Jackson "Classical Electrodynamics". It all has to do with your favourite topic that a moving charge constitutes a current and has a EM field related to it because of its motion. No such field is present in gravity (although I guess that gravitoelectrodynamics has somewhere a counterpart in it, but let's not go there)

"Mechanisms of particle acceleration in solar flares."
"Bremsstrahlung (plasma electrons accelerated by the electric field of positive ions)
Bremsstrahlung can be emitted at all wavelengths, from long radio waves to short X-rays. During solar flares it becomes very intense at X–ray wavelengths.
Two kinds of bremsstrahlung at X-ray wavelengths: thermal and nonthermal.
Thermal bremsstrahlung is produced by electrons described by a distribution (e.g. Maxwellian) characterized by a single temperature.
In solar flares plasma is heated to high temperatures of about 10 million degrees, producing thermal bremsstrahlung in soft X–rays (1-10 KeV).
Nonthermal bremsstrahlung radiation is emitted by electrons which are not part of a thermal gas. They cannot be described by a conventional temperature.
Hard X–ray (10-1000 KeV) bremsstrahlung is produced when highly energetic nonthermal electrons (at relativistic speeds) interact with ambient protons.
Gamma ray continuum emission is produced as the result of bremsstrahlung from electrons with energies 1 Mev.
Cyclotron or synchrotron radiation (charged particles gyrating in a magnetic field with gyrofrequency).
Cyclotron radiation: moderately relativistic electrons. Spectrum consists of discrete lines at frequencies given by the resonance condition.
Synchrotron radiation: highly relativistic electrons.
Overlap of higher harmonics spectrum is continuous."
http://www-solar.mcs.st-and.ac.uk/~paolo/talk1.pdf

thanks for the summation of all these effects

I meant excite which means "to excite a gas" but that also means "to accelerate" in the case of the electrons in the plasma.

Yes. It depends on the endpoints as to whether it is a current or not. If you have 100,000 electrons flowing from one piece of matter to another, it is a current for that moment. You are equalizing the charge differential between the 2 pieces of matter.

I would say, To digress for a moment, that the sun and the heliosphere are endpoints in a circuit.

As the TOPS database showed us.

Ya, well I will say that there are no x-rays emitted by a blackbody of the suns temperature. According to the curves at that temperature it doesnt emit below 90nm(significantly).
The only way they could possibly be emitted is if there is electricity involved.

No intensity except from flare activity.

Please keep separated excitation and acceleration. You may accelerate electrons, and with them excite atoms, but you can also accelerate ions and plasma.

All those electrons, a current, unless you have equal amounts of positive ions flowing in the same direction at the same speed.

I will not get into a discussion on the electric sun here, because this is suposed to be about the temperature and energy of photons.

(significantly) Yes, significantly, there is the crux!

How could you have boiling in a gas?
Boiling is cavitation in a liquid. Sonoluminiscence is cavitation. I am intimately familiar with this.
It is when the steam/air dissolved in the liquid vaporizes/expands, it forms a bubble and rises to the top.

Think about turbulent flows in water. The only way you could tell the difference is if you have different properties like one glowed and one did not. Or one was dyed and one was not.
But to say that the change is density is responsible for the determination of the surface of last scattering, is not addressing the real question, what causes the difference in density as evidenced by helioseismology?

Don't be silly, that is not what I said, I said that looking at the sun it RESEMBLES A BOILING POT OF WATER. I did not say that the "gas was boiling", but now that you get to it, the upper layers of the sun are totally convective, which means that hot cells rise to the surface of the sun and cold cells flow down again. This is so much like a boiling pot of water, you might almost say (as a metaphor) that the plasma is really boiling!

Could you make stuff a little more complicated, please? The increase in density in the sun from the surface to the center (that is past the iron shell that you visualize there) changes the properties of the optical depth. Absorption is a function of ion parameter AND DENSITY, the denser the plasma the more absorption will occur. Even common sense reasoning (of which I am not a fan in dealing with plasma physics) would come to this conclusion.
The difference in density is caused by gravity, more "gas" will be pulled to the center of the sun, and thus on top of every shell in the sun there is the weight of the other shells. And because we are dealing with a gas (plasma) it can be compressed (and no the electrostatic force vs gravity force being tenzilliontrillion has nothing to do with it), and therefore .... the density increases if you go radially into the sun.

Man, I will be glad when the 30 days are over. Maybe Korjik can add some more to my answers.

upriver

05-May-2007, 06:46 PM

Man, I will be glad when the 30 days are over.

Thats the problem with all you mainstreamers, your all so rude.

I have been nice the whole time.

upriver

05-May-2007, 09:01 PM

yes it is reality, take a look at the total BB spectrum of the sun and you will see that indeed the Xray part is non-zero, low, but nonetheless.

And you say that comes from under the photosphere.

Just because the x-ray part is non zero does not prove it is from the same blackbody process under the photosphere.

Oh brother, read the first two chapters of Rybicky and Lightman, they can explain it much better than I can here in this limited forum. You can build a 500 km long plasma tube that along its length may be optically thick, but not across its width. A plasma has to be totally optically thick and will have to be in Local Thermal Equilibrium (LTE), which will also not be the case in your tube. So, you will NOT get a BB spectrum out of your 500 km tube. It is al such basic plasma and radiative transport physics, that can be read in any introductory book.

The only laboratory plasma I have ever seen, that produced a blackbody spectrum, is a high pressure gas arc.

The rest is conjecture and calculations.

Then the electrons rotating around the magnetic field and radiating cyclotron or synchrotron radiation. These particles lose energy,that is where the energy of the photons comes from.
Apart from that you cannot compare this situation with a body in a gravitational field, because the emission of the photon is specifically related to the fact that there is a charged particle gyrating along the field. And it needs not radiate an X-ray, that totally depends on the energy of the electron. Again, read up on it in e.g. Jackson "Classical Electrodynamics". It all has to do with your favourite topic that a moving charge constitutes a current and has a EM field related to it because of its motion. No such field is present in gravity (although I guess that gravitoelectrodynamics has somewhere a counterpart in it, but let's not go there)

thanks for the summation of all these effects

If we work together, we will advance science that much quicker.

All those electrons, a current, unless you have equal amounts of positive ions flowing in the same direction at the same speed.

A neutral plasma is different than a plasma that has a imbalance of energy.

Even if there is the same amount of particles, the electrons can still carry energy. A million degree neutral plasma is carrying a current.

Originally Posted by upriver View Post
How could you have boiling in a gas?
Boiling is cavitation in a liquid. Sonoluminiscence is cavitation. I am intimately familiar with this.
It is when the steam/air dissolved in the liquid vaporizes/expands, it forms a bubble and rises to the top.

Think about turbulent flows in water. The only way you could tell the difference is if you have different properties like one glowed and one did not. Or one was dyed and one was not.
But to say that the change is density is responsible for the determination of the surface of last scattering, is not addressing the real question, what causes the difference in density as evidenced by helioseismology?

Don't be silly, that is not what I said, I said that looking at the sun it RESEMBLES A BOILING POT OF WATER. I did not say that the "gas was boiling", but now that you get to it, the upper layers of the sun are totally convective, which means that hot cells rise to the surface of the sun and cold cells flow down again. This is so much like a boiling pot of water, you might almost say (as a metaphor) that the plasma is really boiling!

Again, convection is different than boiling.

How do you get differentiation between the different elements of the filaments under the sunspots. Some parts are glowing and some are not. Just their shape alone brings forth questions in the context of surface of last scattering.

And I dont know how you get convection, blackbody and "surface of last scattering" out of these structures.
http://www.solarphysics.kva.se/gallery/images/2003/08Aug2003_4364.1805-1807_color.jpg

http://www.solarphysics.kva.se/gallery/images/2003

Could you make stuff a little more complicated, please? The increase in density in the sun from the surface to the center (that is past the iron shell that you visualize there) changes the properties of the optical depth. Absorption is a function of ion parameter AND DENSITY, the denser the plasma the more absorption will occur. Even common sense reasoning (of which I am not a fan in dealing with plasma physics) would come to this conclusion.

The only laboratory plasma I have ever seen, that produced a blackbody spectrum, is a high pressure gas(plasma) arc.

The difference in density is caused by gravity, more "gas" will be pulled to the center of the sun, and thus on top of every shell in the sun there is the weight of the other shells. And because we are dealing with a gas (plasma) it can be compressed (and no the electrostatic force vs gravity force being tenzilliontrillion has nothing to do with it), and therefore .... the density increases if you go radially into the sun.

I'm not arguing with the concept of increasing density in the sun(actually it only increases to a certain limit) but that doesn't explain stratification on the level that it is observed. Especially acting like a bounded sphere.

The solar O crisis? I'm really curious to see how that plays out.
The first clues are Deuterium synthesis and the CNO cycle on the surface.

I just found this. Had a really good laugh. This should be debunked!

"Helioseismology as a new constraint on supersymmetric dark matter"
http://www.blackwell-synergy.com/doi/abs/10.1046/j.1365-8711.2002.05238.x?journalCode=mnr

Yet when electricity is mentioned we are called unscientific.

papageno

05-May-2007, 09:51 PM

Thats the problem with all you mainstreamers, your all so rude.

I have been nice the whole time.

It is not about being nice. It is your apparent lack of interest in learning and understanding what mainstream science actually says.

(A clear example is the " How many watts does an electron use?" (http://www.bautforum.com/showthread.php?t=56968) thread in the Q&A forum, where you were fishing for loopholes in the mainstream physics large enough to fit through the ideas you presented in the " ATM gravity and EU Sun" (http://www.bautforum.com/showthread.php?t=56138) thread in the ATM forum.)

tusenfem

06-May-2007, 09:51 AM

The "rudeness" of my comment is because for 3 long pages we try to explain to you the concept of black body radiation in mainstream physics and you just ignore whatever is explained to you, in this respect. I am just tired of trying to come up with even more clear and simple explanations, because I know as soon as I will have explained it again, in a (I hope) more simple and comprehensive way, you come up with some incredably difficult thing, like Bose Einstein Condensation.

And you say that comes from under the photosphere.

Just because the x-ray part is non zero does not prove it is from the same blackbody process under the photosphere.

Because the photosphere is per definition optically thin, it itself cannot create a BB, so any BB that is emitted by the sun has to come from below the photosphere. Simple comme bonjour.

The only laboratory plasma I have ever seen, that produced a blackbody spectrum, is a high pressure gas arc.

The rest is conjecture and calculations.

And if you would have read what I wrote you would have seen that I claimed that your 500 km tube of plasma my be optically thick in the length but not in the width and for other obvious reasons that column will not emit a BB.

So, in your words you now claim that a high pressure gas arc can emit a BB. First you kept claiming that only solids could emit BBs. But the gas arc is used to heat a carbon cathode, which is the part that emits the BB, not the gas.

If we work together, we will advance science that much quicker.

A neutral plasma is different than a plasma that has a imbalance of energy.

Even if there is the same amount of particles, the electrons can still carry energy. A million degree neutral plasma is carrying a current.

If you were willing to accept some of mainstream physics then we could make progress.

What is that second point supposed to mean? A neutral plasma is a plasma that has equal positive and negative charge carriers.
What is a plasma that has an imbalance of energy? Do you mean that the electrons have more energy than the ions. Do you mean a plasma where the dirstribution function has a "bump in tail"?
You keep on making stuff mutually exclusive when they are not as in the next claim.

A million degree neutral plasma is carrying a current. WHY FOR GOODNESS SAKE????? The Sun's million degree corona is just a neutral plasma at a million degrees. There are currents in there because of surface processes on the sun, e.g. when a coronal loop is there. But that has nothing whatsoever to do with the temperature of the plasma.

Again, convection is different than boiling.

How do you get differentiation between the different elements of the filaments under the sunspots. Some parts are glowing and some are not. Just their shape alone brings forth questions in the context of surface of last scattering.

Convection in the upper layers of the sun, makes the sun look like a boiling kettle of water. Take a look at the movies, Upriver. For once you can do science by just looking at pictures and movies. It very well resembles what I saw this morning while cooking eggs for breakfast.

Because the background will be an even glow, whereas all other non-thermal emission will give localized emissions. The emissions from loops are not hampered with the surface of last scattering because the are created well above it.

And I dont know how you get convection, blackbody and "surface of last scattering" out of these structures.
http://www.solarphysics.kva.se/gallery/images/2003/08Aug2003_4364.1805-1807_color.jpg

http://www.solarphysics.kva.se/gallery/images/2003

Well, I guess you have to study like 5 years of physics and astrophysics and then train another couple of years as a PhD student and postdoc, and then you can interprete these images. I am not an expert in these images. Then again convection you can only see in images that make the flow patters visible, black body radiation can only be interpreted if you scan the whole frequency range. you cannot just look at one pic in one wavelength and distinguish what is BB and what is non-thermal (although the very bright spots at the footpoints make it a little easier). Neither can you find differences in "last scatter surfaces" from looking at just one wavelength.

I'm not arguing with the concept of increasing density in the sun(actually it only increases to a certain limit) but that doesn't explain stratification on the level that it is observed. Especially acting like a bounded sphere.

The solar O crisis? I'm really curious to see how that plays out.
The first clues are Deuterium synthesis and the CNO cycle on the surface.

Yet when electricity is mentioned we are called unscientific.

It most definitely explains stratification, if you would only be willing to read some introductory tests. There are different layers in the sun and ... read a book!

Solar O crisis, well let's keep on topic shall we? That is another thread.

Helioseismo. Well now you are complaining about an experiment that could be done to see if WIMPs exist and gather in the center of the sun? Now that is very scientific of you. If they don't find anything with what they propose, they will have debunked themselves.

And no electricity is very mainstream and loads and loads of papers are published dealing with electric currents in plasmas at the earth and at the sun. Specifically, I am looking at the substorm current wedge right now in the earth's magnetotail, where electric currents are flowing and are diverted and and and. Also on the sun, magnetic loops are changing and very strong currents are in those loops (mm how do they flow these currents). The problem is that ATM EU proponents do not come up with any models that can be tested. They keep on saying mainstream is wrong and for the EU the equations do not yet exist to describe what is going on in the electrically driven sun. Well I am diverting too much from the OP, so I better stop here.

upriver

07-May-2007, 01:20 AM

It is not about being nice. It is your apparent lack of interest in learning and understanding what mainstream science actually says.

(A clear example is the " How many watts does an electron use?" (http://www.bautforum.com/showthread.php?t=56968) thread in the Q&A forum, where you were fishing for loopholes in the mainstream physics large enough to fit through the ideas you presented in the " ATM gravity and EU Sun" (http://www.bautforum.com/showthread.php?t=56138) thread in the ATM forum.)

Papageno, I have heard what you said and went and looked up the equations even though I may not understand them. I understand what you said about mechanical equilibrium.

To me the thought of the motion(of an electron) with no energy input goes against every thing I have been taught about conservation of energy. I dont believe in perpetual motion.

Maybe someday I will believe it, but not today.

upriver

07-May-2007, 04:59 AM

The "rudeness" of my comment is because for 3 long pages we try to explain to you the concept of black body radiation in mainstream physics and you just ignore whatever is explained to you, in this respect. I am just tired of trying to come up with even more clear and simple explanations, because I know as soon as I will have explained it again, in a (I hope) more simple and comprehensive way, you come up with some incredably difficult thing, like Bose Einstein Condensation.

No, you mean the theory of blackbody as applied to the sun. And no I have not ignored you.

I understand blackbody as applied to solids. I understand how its being applied to the sun.

I notice they dont use plasma for a blackbody devices even though you can get much higher temperatures

Because the photosphere is per definition optically thin, it itself cannot create a BB, so any BB that is emitted by the sun has to come from below the photosphere. Simple comme bonjour.

Its coming from the pressurized plasma below. As I said pressurized plasma and solids.

And if you would have read what I wrote you would have seen that I claimed that your 500 km tube of plasma my be optically thick in the length but not in the width and for other obvious reasons that column will not emit a BB.

So, in your words you now claim that a high pressure gas arc can emit a BB. First you kept claiming that only solids could emit BBs. But the gas arc is used to heat a carbon cathode, which is the part that emits the BB, not the gas.

I have always said that a solid or high pressure plasma can emit blackbody spectrum. And I have provided examples even if they were not understood.

The carbon arc example was used for the UV hump. It produces a blackbody at certain current and voltages.

If you were willing to accept some of mainstream physics then we could make progress.

Which part do you think I should accept?

What is that second point supposed to mean? A neutral plasma is a plasma that has equal positive and negative charge carriers.
What is a plasma that has an imbalance of energy? Do you mean that the electrons have more energy than the ions. Do you mean a plasma where the dirstribution function has a "bump in tail"?
You keep on making stuff mutually exclusive when they are not as in the next claim.

Yes.

A million degree neutral plasma is carrying a current. WHY FOR GOODNESS SAKE????? The Sun's million degree corona is just a neutral plasma at a million degrees. There are currents in there because of surface processes on the sun, e.g. when a coronal loop is there. But that has nothing whatsoever to do with the temperature of the plasma.

Process of elimination. If we go back to our methods of making plasma, we find that we can rule 2 of them out immediately.

Compression. There is no source of compression on the surface of the sun except the z-pinch which you dont think happens on the surface of the sun.

Photoionization. You need photons of at least 114eV for a million degree Hydrogen electron plasma. Even higher for iron. Thats like 1nm soft x-rays.

Whats left? Electricity, because reconnection is not going to do it.
Ah, magnetic fields are driven by electrical currents.

So an ionized plasma coming from the sun is driven by electricity unless you know some new laws of physics.

The only way to make steady streams of million degree plasma is with electric currents.

Convection in the upper layers of the sun, makes the sun look like a boiling kettle of water. Take a look at the movies, Upriver. For once you can do science by just looking at pictures and movies. It very well resembles what I saw this morning while cooking eggs for breakfast.

It looks more like cell division than bottom up convection.

Hi-Node movie.
http://solar-b.nao.ac.jp/news/061127PressConference/movie/cagb_20061102.mpg

Close up. Look at the little glowing "sprites" between the granulations. They last for a long time and merge and break up.
http://solar-b.nao.ac.jp/news/061127PressConference/movie/cagb_20061102.mpg

Well, I guess you have to study like 5 years of physics and astrophysics and then train another couple of years as a PhD student and postdoc, and then you can interprete these images.

That would not guarantee that my interpretation is any more correct, just that I would be versed in the mainstream way of doing things.

Its pretty clear that there are structures. The are definitely contrasted with other parts. They have curves that give no indication of convection. Just because you see motion on the top of a plasma its called convection

See this APOD. They say that the H-alpha (ending)picture is from above the photosphere, but if you look at the sunspot it has to be under it.
http://antwrp.gsfc.nasa.gov/apod/ap050216.html

Then again convection you can only see in images that make the flow patters visible, black body radiation can only be interpreted if you scan the whole frequency range. you cannot just look at one pic in one wavelength and distinguish what is BB and what is non-thermal (although the very bright spots at the footpoints make it a little easier). Neither can you find differences in "last scatter surfaces" from looking at just one wavelength.

Yes. That what I've been try to say. The transition layer and the chromosphere are pretty continuous. So it would blend pretty well with BB. There may be a spike in the spectrum.

And no electricity is very mainstream and loads and loads of papers are published dealing with electric currents in plasmas at the earth and at the sun. Specifically, I am looking at the substorm current wedge right now in the earth's magnetotail, where electric currents are flowing and are diverted and and and. Also on the sun, magnetic loops are changing and very strong currents are in those loops (mm how do they flow these currents). The problem is that ATM EU proponents do not come up with any models that can be tested. They keep on saying mainstream is wrong and for the EU the equations do not yet exist to describe what is going on in the electrically driven sun. Well I am diverting too much from the OP, so I better stop here.

I think the difference of opinion is how the electricity is used in stellar processes, and where it comes from.

Again the only way to have a 20,000 degree plasma that can ionize H is with electricity.

tusenfem

07-May-2007, 08:00 AM

No, you mean the theory of blackbody as applied to the sun. And no I have not ignored you.
I understand blackbody as applied to solids. I understand how its being applied to the sun.
I notice they dont use plasma for a blackbody devices even though you can get much higher temperatures

And I explained you why plasmas in labs are not used to create a BB, because the plasma needs to be thick and in LTE

Its coming from the pressurized plasma below. As I said pressurized plasma and solids.

I have always said that a solid or high pressure plasma can emit blackbody spectrum. And I have provided examples even if they were not understood.

The carbon arc example was used for the UV hump. It produces a blackbody at certain current and voltages.

I guess I will let that go, but I do remember another thread where you were vehement about BB only being emitted by a solid.

Process of elimination. If we go back to our methods of making plasma, we find that we can rule 2 of them out immediately.

Compression. There is no source of compression on the surface of the sun except the z-pinch which you dont think happens on the surface of the sun.

Photoionization. You need photons of at least 114eV for a million degree Hydrogen electron plasma. Even higher for iron. Thats like 1nm soft x-rays.

Whats left? Electricity, because reconnection is not going to do it.
Ah, magnetic fields are driven by electrical currents.

So an ionized plasma coming from the sun is driven by electricity unless you know some new laws of physics.

The only way to make steady streams of million degree plasma is with electric currents.

Whether a plasma has a current or not has absolutely nothing to do with how the plasma was created. I will acknoledge that when a plasma is created in the lab, by cranking up the voltage, there is indeed a current.

About the photoionization, we have gone through that already 3 times now, I think. Photoionization needs not be done by one photon. Apart from that you seem to forget that the plasma is then still in a intense photon field, which will lead to photon-particle interactions that can heat up the plasma even more.

I don't need any new laws of physcis. And your statement is wrong.

It looks more like cell division than bottom up convection.

Hi-Node movie.
http://solar-b.nao.ac.jp/news/061127PressConference/movie/cagb_20061102.mpg

Close up. Look at the little glowing "sprites" between the granulations. They last for a long time and merge and break up.
http://solar-b.nao.ac.jp/news/061127PressConference/movie/cagb_20061102.mpg

That would not guarantee that my interpretation is any more correct, just that I would be versed in the mainstream way of doing things.

Its pretty clear that there are structures. The are definitely contrasted with other parts. They have curves that give no indication of convection. Just because you see motion on the top of a plasma its called convection

See this APOD. They say that the H-alpha (ending)picture is from above the photosphere, but if you look at the sunspot it has to be under it.
http://antwrp.gsfc.nasa.gov/apod/ap050216.html

I seem to have problems loading the movies here. But if one would look at the dopplergram of the movies one would see that there are regions of upflow and regions of downflow, i.e. convection.

Yes. That what I've been try to say. The transition layer and the chromosphere are pretty continuous. So it would blend pretty well with BB. There may be a spike in the spectrum.

I think the difference of opinion is how the electricity is used in stellar processes, and where it comes from.

Again the only way to have a 20,000 degree plasma that can ionize H is with electricity.

Continuous but do they emit BB? Is perhaps the chromosphere emission created by the BB below the photosphere? What is the chromosphere?

The electric sun does not work in my opinion (and with a little calculation I once did for the Juergens model), but you are free to build a model and present it here.

Again, you are wrong, your ideas on how ionization works are too simplistic. As are your ideas on the "motion" of the electron as you told Papageno:

To me the thought of the motion(of an electron) with no energy input goes against every thing I have been taught about conservation of energy. I dont believe in perpetual motion.

In fact the electron is NOT circling around the nucleus. The Bohr planetary orbit model is just a way of trying to visualize various energy levels that the electrons can have around the nucleus. There are "pictures" of the probability function of the electron around the nucleus. Don't know how they were made, but they were all but planetary orbits. I cannot find the experimental pics (which I think should be there, maybe our birdcatcher knows where) but [url=http://panda.unm.edu/courses/finley/P262/Hydrogen/WaveFcns.html]here[/b] is a mathematical calculation of the wave function of the electron.

papageno

07-May-2007, 08:44 PM

Papageno, I have heard what you said and went and looked up the equations even though I may not understand them. I understand what you said about mechanical equilibrium.

You withdrew your claims because you couldn't support them, not because you were convinced otherwise:

So I will withdraw my claims to whoever wants me to withdraw them.
Of course that does not mean I have given up.
As I continue on I'm beginning to see a big rip, big enough to drive a truck through. As soon as I figure out how to show that rip without question, I will.

There is no sign of understanding in your words.

To me the thought of the motion(of an electron) with no energy input goes against every thing I have been taught about conservation of energy. I dont believe in perpetual motion.

Pity that you cannot understand that the conservation laws are the reason why the planets keep orbiting the Sun, and that electrons do not bend to your wishes.

Maybe someday I will believe it, but not today.

It is not a matter of belief.

korjik

07-May-2007, 09:12 PM

Thats the problem with all you mainstreamers, your all so rude.

I have been nice the whole time.

Some of us consider being called a mainstreamer an insult. Not that I have ever seen anything here that was remotely accurate, but I am a scientist and I keep an open mind.

The reason I have been more lurking than participating, any why I doubt I will post to this thread again is that the original poster shows an extreme aversion to learning about what he dosent know. Several problems with his ideas have been pointed out repeatedly, and he still does not admit that the problem could possibly be that he dosent know enough to understand what is going on. It is always that the 'mainstreamers' just dont understand. There are many here, myself included, who are more than willing to teach, if on the other side there is someone willing to learn.

That is what is lacking here

upriver

08-May-2007, 05:20 AM

In fact the electron is NOT circling around the nucleus. The Bohr planetary orbit model is just a way of trying to visualize various energy levels that the electrons can have around the nucleus. There are "pictures" of the probability function of the electron around the nucleus. Don't know how they were made, but they were all but planetary orbits. I cannot find the experimental pics (which I think should be there, maybe our birdcatcher knows where) but here[/b] is a mathematical calculation of the wave function of the electron.

The model of the electron that I subscribe to is a wave model in that the electron is made of the underlying structure of the universe. I'm not sure I can get on board the quark model since that is more particles(actually I think they are pieces of enregy but thats a different thread)..
I dont believe in the "particle" circling. Thats just a convenient tool for motion of any sort.

And yes, I have seen just about every picture of an electron that you can find on the internet, along with all the theories.

Quantum corral. Atoms in a corral. My favorite.
http://images.google.com/images?q=quantum+corral&gbv=2&hl=en&safe=off&start=20&sa=N&ndsp=20 (http://panda.unm.edu/courses/finley/P262/Hydrogen/WaveFcns.html)
[url]http://images.google.com/images?q=atoms+in+corral&gbv=2&hl=en&safe=off&start=20&sa=N&ndsp=20
Electron wave function.
=http://images.google.com/images?q=electron+wave+function&btnG=Search+Images&hl=en&gbv=2&safe=off
Electron orbits
http://images.google.com/images?hl=en&gbv=2&safe=off&q=electron+orbits&btnG=Search+Images

The point is that any change in the wave structure, magnetic field etc interactions, the ability to maintain its location in space with a nucleus as reference, requires energy. If there is motion of a wave shape, an interaction of a field, or whatever.

I think everything is tied together at a most basic level the we dont(or barely can) see yet.

It all requires energy.

upriver

08-May-2007, 05:22 AM

Some of us consider being called a mainstreamer an insult. Not that I have ever seen anything here that was remotely accurate, but I am a scientist and I keep an open mind.

I apologize and will refrain from using that term. I was not aware.

upriver

08-May-2007, 06:33 AM

And I explained you why plasmas in labs are not used to create a BB, because the plasma needs to be thick and in LTE

I should have been a little more clear and said " Why dont they use a pressurized plasma in a blackbody device. That way you can get higher temperatures."

I guess I will let that go, but I do remember another thread where you were vehement about BB only being emitted by a solid.

Yes. That was the very first claim, that was quickly modified after talking to Dr. William Ott at NIST about high pressure Argon plasma. Then later, after certain sonoluminescence results I knew for sure, which I presented from the Nature article.

So to restate. A true(not with a UV hump, or sort of a black body) blackbody emission curve comes from a solid.

"A quasi continuum comes from a pressurized plasma." (These are the words of Dr. Ott at NIST.) The more you apply pressure the closer to a blackbody the emission curve becomes(the wings lift up).

Sonoluminescence bubbles.
As the drive pressure changes from 2.3 to 5.5 bar, the curve changes. That is the pressure measured near(as close as practical with a 3mm needle hydrophone) the bubble with a hydrophone. Then the voltage and current settings are calibrated from the hydrophone measurements. If you leave the hydrophone in the solution at large drive amplitudes, it disturbs the sound field, and the bubble is ejected from the antinode. So you must interpolate drive pressures from voltage settings.
The emission lines change to a continuum.
The actual pressure in the center of the bubble is closer to 8000 Atm.
http://www.nature.com/nature/journal/v434/n7029/fig_tab/nature03361_F2.html
http://www.nature.com/nature/journal/v434/n7029/full/nature03361.html;jsessionid=7D64338F2CC87BBB82E870 62C1F1203C

Now there is a parallel that you can draw between density and distance in the number of collisions.
But the only supporting observations are the sun itself, and you cant see into the area of interest.

Originally Posted by upriver
Process of elimination. If we go back to our methods of making plasma, we find that we can rule 2 of them out immediately.

Compression. There is no source of compression on the surface of the sun except the z-pinch which you dont think happens on the surface of the sun.

Photoionization. You need photons of at least 114eV for a million degree Hydrogen electron plasma. Even higher for iron. Thats like 1nm soft x-rays.

Whats left? Electricity, because reconnection is not going to do it.
Ah, magnetic fields are driven by electrical currents.

So an ionized plasma coming from the sun is driven by electricity unless you know some new laws of physics.

The only way to make steady streams of million degree plasma is with electric currents.

Whether a plasma has a current or not has absolutely nothing to do with how the plasma was created. I will acknoledge that when a plasma is created in the lab, by cranking up the voltage, there is indeed a current.

About the photoionization, we have gone through that already 3 times now, I think. Photoionization needs not be done by one photon. Apart from that you seem to forget that the plasma is then still in a intense photon field, which will lead to photon-particle interactions that can heat up the plasma even more.

I don't need any new laws of physcis. And your statement is wrong.

I have never seen anybody advance photoionization as the reason for the corona. When I say 114eV or 1 million degrees, I am talking about the corona, unless I say iron. Then I am talking about loops. I have never seen anybody advance photoionization as the reason for the loops.

Right there seem to be 2 competing models of coronal heating. Alfven waves and reconnection. We can rule out reconnection from our discussions.

I seem to have problems loading the movies here. But if one would look at the dopplergram of the movies one would see that there are regions of upflow and regions of downflow, i.e. convection.

Yes. Actually you are right. They talk them in terms of million degree plasma right next to cooler plasma flowing in the opposite direction. And that was in the filaments(I think) that connect the surface "granules". The same filaments you see in a sunspot hole.

Originally Posted by upriver
Yes. That what I've been try to say. The transition layer and the chromosphere are pretty continuous. So it would blend pretty well with BB. There may be a spike in the spectrum.

I think the difference of opinion is how the electricity is used in stellar processes, and where it comes from.

Again the only way to have a 20,000 degree plasma that can ionize H is with electricity.

Continuous but do they emit BB? Is perhaps the chromosphere emission created by the BB below the photosphere? What is the chromosphere?

"The chromosphere is an irregular layer above the photosphere where the temperature rises from 6000° C to about 20,000° C. At these higher temperatures hydrogen emits light that gives off a reddish color (H-alpha emission)."
http://solarscience.msfc.nasa.gov/chromos.shtml

Even at 20,000C you canot ionize H. The tail can ionize but where does that tail come from. And its not enough for the continuous ionization happening at the chromosphere.
I mean continuous in coverage of the solar surface as opposed to point source emission. The chromosphere is visible through its H Alpha emission at 656.281nm. With continuous coverage there should be a indication of it in the blackbody spectrum of the sun.

I tried to find detailed solar emission below 200nm. Pretty much nothing exists in terms of a 1nm resolution composite spectrum down to 1nm soft x-rays.

The electric sun does not work in my opinion (and with a little calculation I once did for the Juergens model), but you are free to build a model and present it here.

I appreciate your confidence. But we both know I have a ways to go.
The other issue is it gets unfocused rather quickly due to EUer newbies(sorry).

upriver

08-May-2007, 06:50 AM

You withdrew your claims because you couldn't support them, not because you were convinced otherwise:

Exactly.

There is no sign of understanding in your words.

I'm sorry you see it that way.

May my words someday reflect understanding in your eyes.

If I had believed we were somewhere near ultimate understanding, or nirvana.
Or that our race had the knowledge of millions or billions of years of existence, then you might have me, but for now we are still exploring.

tusenfem

08-May-2007, 07:43 AM

1. I should have been a little more clear and said " Why dont they use a pressurized plasma in a blackbody device. That way you can get higher temperatures."

2. Now there is a parallel that you can draw between density and distance in the number of collisions. But the only supporting observations are the sun itself, and you cant see into the area of interest.

3. I have never seen anybody advance photoionization as the reason for the corona. When I say 114eV or 1 million degrees, I am talking about the corona, unless I say iron. Then I am talking about loops. I have never seen anybody advance photoionization as the reason for the loops.

4. Right there seem to be 2 competing models of coronal heating. Alfven waves and reconnection. We can rule out reconnection from our discussions.

5. Even at 20,000C you canot ionize H. The tail can ionize but where does that tail come from. And its not enough for the continuous ionization happening at the chromosphere.
I mean continuous in coverage of the solar surface as opposed to point source emission. The chromosphere is visible through its H Alpha emission at 656.281nm. With continuous coverage there should be a indication of it in the blackbody spectrum of the sun.

6. I tried to find detailed solar emission below 200nm. Pretty much nothing exists in terms of a 1nm resolution composite spectrum down to 1nm soft x-rays.

1. They do not use pressurized plasma for a BB because there are better, easier and cheaper ways of creating a BB, e.g. by using carbon heated by an electric arc, as you described. I think that would be obvious, especially since you seem to work in sonolunimiscence. Wanna carry that aparatus around just to get a BB spectrum or would you rather carry a relatively small carbon cathode?

2. Well, I guess then that the sun is just a rather opague gass at 1 atmosphere throughout, glowing a little.

3/4. Nope, photoionization is not the way of heating up the corona. First of all everything is already ionized. Please remember why I wrote that stuff about an H above the photosphere. To show that in a dense photon field you can get multiple absorption of photons, leading to ionization.
The heating of the corona can also be done with soundwaves, there are lots of processes that were being looked at at e.g. the Astrophysical Institute of Utrecht University. Another one was nanoflares that occur at the surface of the sun. All of these processes are working at the same time, I am convinced that there is no one process that is heating the corona just by itself.
And I was definitely not talking about loops, where all kinds of processes lead to emission.

5. Even at 20000 K we CAN ionize H I have explained that to you three times already. It seems you do not want to believe me, okay, then don't.
What do you mean "with continuous coverage there should be an indication of it in the BB spectrum". This either shows you have no idea what a BB exactly is (continuous emission over the whole frequency range, described by the BB equation). Or do you mean that measuring the spectrum of the sun (not that I do not say BB) there may be a peak at Ly alpha in the spectrum?

6. Maybe that is because it is not interesting and the interesting stuff in soft Xrays does not come from the sun proper as BB but from loops and acceleration regions etc.

korjik

08-May-2007, 04:24 PM

snip..

Yes. That was the very first claim, that was quickly modified after talking to Dr. William Ott at NIST about high pressure Argon plasma. Then later, after certain sonoluminescence results I knew for sure, which I presented from the Nature article.

So to restate. A true(not with a UV hump, or sort of a black body) blackbody emission curve comes from a solid.

"A quasi continuum comes from a pressurized plasma." (These are the words of Dr. Ott at NIST.) The more you apply pressure the closer to a blackbody the emission curve becomes(the wings lift up).

Odd that apparently talking to someone is gospel to you but we are just plain wrong all the time.

Sonoluminescence bubbles.
As the drive pressure changes from 2.3 to 5.5 bar, the curve changes. That is the pressure measured near(as close as practical with a 3mm needle hydrophone) the bubble with a hydrophone. Then the voltage and current settings are calibrated from the hydrophone measurements. If you leave the hydrophone in the solution at large drive amplitudes, it disturbs the sound field, and the bubble is ejected from the antinode. So you must interpolate drive pressures from voltage settings.
The emission lines change to a continuum.
The actual pressure in the center of the bubble is closer to 8000 Atm.
http://www.nature.com/nature/journal/v434/n7029/fig_tab/nature03361_F2.html
http://www.nature.com/nature/journal/v434/n7029/full/nature03361.html;jsessionid=7D64338F2CC87BBB82E870 62C1F1203C

Ah. This graph again. Specifically the first graph of the first link for anyone who is reading. This graph has shown up before and it still shows a blackbody radiation curve on the left side of the graph and Argon emission on the right side. The second graph in the first link looks like the spectral decomposition of the emission lines.

Notice that the graph behaves as you would expect as the pressure and temp rise. The blackbody portion gets brighter and the peak of the emission heads towards shorter wavelengths, and the Ar emission has the lines smeared due to the effects of the temp and pressure on the argon.

snip...

papageno

08-May-2007, 07:39 PM

You withdrew your claims because you couldn't support them, not because you were convinced otherwise:

Exactly.

As if your complete inability to support your claim with evidence and reasoning were a good thing...

There is no sign of understanding in your words.

I'm sorry you see it that way.

And yet you are convinced that you are right and generations of thousands of active scientists are wrong.

May my words someday reflect understanding in your eyes.

If I had believed we were somewhere near ultimate understanding, or nirvana.
Or that our race had the knowledge of millions or billions of years of existence, then you might have me, but for now we are still exploring.

Ah yes, the typical ATM, CT and creationist excuse.

No, upriver, the fact that modern science does not know everything does not mean that what we know is wrong.
And with your attitude you are certainly not making a positive contribution to this exploration.

Forskern

08-May-2007, 07:40 PM

<delurk />

I have never seen anybody advance photoionization as the reason for the corona. When I say 114eV or 1 million degrees, I am talking about the corona, unless I say iron. Then I am talking about loops. I have never seen anybody advance photoionization as the reason for the loops.

You are correct in that the primary ionization process in the corona is not photoionization, as the radiation field has far too low energy. Actually, the ionization is mainly due to electron impacts (collisions), which is an effective process despite the extremely low coronal density because the low density also keeps the recombination rate low.

A third process, excitation of multiple electrons followed by autoionization, is effective in some cases, particularly near the temperature of the ionic species' maximum formation.

But to distinguish between the corona and coronal loops based on wavelength (remember that 171 angstrom does not necessarily imply iron) is most likely a fallacy. The entire corona can pretty much be regarded as one tangled mess of magnetic loops, and all of them contain the same abundance ratios.

Right there seem to be 2 competing models of coronal heating. Alfven waves and reconnection. We can rule out reconnection from our discussions.

Do you have any physical reason for ruling out reconnections? Acoustic heating is too weak by a factor at least five (iirc), while more and more evidence is pointing towards reconnections as the cause of the high coronal temperatures.

Check your local library for a copy of Aschwanden's "Physics of the Solar Corona" (Springer), he dedicates an entire chapter to coronal heating, in not too dense language.

Fortis

08-May-2007, 08:19 PM

So to restate. A true(not with a UV hump, or sort of a black body) blackbody emission curve comes from a solid.)I know of no solids that are "true" blackbodies. For a solid to be a true blackbody its emissivity would have to be 1 across the entire electromagnetic spectrum. That is why good blackbody sources (i.e. not "true" but good approximations in the wavelength region of interest) tend to be very expensive, and usually exploit geometric as well as material features.

upriver

08-May-2007, 08:45 PM

I know of no solids that are "true" blackbodies. For a solid to be a true blackbody its emissivity would have to be 1 across the entire electromagnetic spectrum. That is why good blackbody sources (i.e. not "true" but good approximations in the wavelength region of interest) tend to be very expensive, and usually exploit geometric as well as material features.

Here is the best approximation that I know of.
Canada's Ultimate Light Ruler
http://www.nrc-cnrc.gc.ca/highlights/2006/0602blackbody_e.html

Internal layout.
http://www.npl.co.uk/publications/news/opticalrm/issue16/

upriver

08-May-2007, 09:08 PM

As if your complete inability to support your claim with evidence and reasoning were a good thing...

When have I ever claimed it was a good thing? Its terrible and embarrassing.

And yet you are convinced that you are right and generations of thousands of active scientists are wrong.

You act like every scientist has fallen lock step into the same description of the universe.
In reality everybody's view is a little different, even if you went to the same school. I dont mean to say that a basic level of understanding is not required.

Ah yes, the typical ATM, CT and creationist excuse.

No, upriver, the fact that modern science does not know everything does not mean that what we know is wrong.

"CT and creationist" ?????

Don't count me into whatever that is.....

How old is current science 50 yrs, 20 yrs?
Just imagine the surety of science in 1000yrs.

And with your attitude you are certainly not making a positive contribution to this exploration.

Why? Because I question everything?
I know you just got your achievement award, but does that mean there is nothing left to question.

Like the gravity pendulum machine. Nobody has offered an published explanation that was nice and neat.
I brought up the one that I thought logically fit, even though I could not "show you" because of my lack of mathematical training(I think its more like a slight dyslexia).

I'm stubborn(Taurus:)) I admit, but I am always open to change.

korjik

08-May-2007, 09:21 PM

snip....

How old is current science 50 yrs, 20 yrs?
Just imagine the surety of science in 1000yrs.

In the year 1900 all of physics had been solved except for a couple little annoyingly persistent problems. Within 10 years, that view was proven to be wrong and whole new worlds of physics were opening up. Surety in physics now involves how good your proof is.

Why? Because I question everything?
I know you just got your achievement award, but does that mean there is nothing left to question.

Like the gravity pendulum machine. Nobody has offered an published explanation that was nice and neat.
I brought up the one that I thought logically fit, even though I could not "show you" because of my lack of mathematical training(I think its more like a slight dyslexia).

I'm stubborn(Taurus:)) I admit, but I am always open to change.

Because you feel the need to prove that the sun rises in the west, no matter how much proof is shown that it rises in the east

upriver

08-May-2007, 09:34 PM

<delurk />

But to distinguish between the corona and coronal loops based on wavelength (remember that 171 angstrom does not necessarily imply iron) is most likely a fallacy. The entire corona can pretty much be regarded as one tangled mess of magnetic loops, and all of them contain the same abundance ratios.

171 is an identification of particular wavelength that is emitted by a plasma of a certain temperature.
The reason they use 171 is because it is a known emission that can be correlated with a temperature.

Here is a summary of the TRACE passbands
http://trace.lmsal.com/Project/Instrument/inspass.htm

In this particular example it is correlated with FE IX.

Originally Posted by upriver
Right there seem to be 2 competing models of coronal heating. Alfven waves and reconnection. We can rule out reconnection from our discussions.

Do you have any physical reason for ruling out reconnections? Acoustic heating is too weak by a factor at least five (iirc), while more and more evidence is pointing towards reconnections as the cause of the high coronal temperatures.

Check your local library for a copy of Aschwanden's "Physics of the Solar Corona" (Springer), he dedicates an entire chapter to coronal heating, in not too dense language.

After doing the research into reconnection, I kind of understand the conditions required and frequency of reconnection.

Neither one of these is sufficient to continuously heat the corona.

"Magnetic reconnection throughout the solar atmosphere."
http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?2002solm.conf..285H&data_type=PDF_H IGH&whole_paper=YES&type=PRINTER&filetype=.pdf

The issue here is that reconnection is not the cause of anything(except jets of particles).
Reconnection is the result of 2 filaments or loops interacting(pinch).
Reconnection is driven by electric currents.

upriver

08-May-2007, 09:45 PM

Odd that apparently talking to someone is gospel to you but we are just plain wrong all the time.

He described the experiment that he did.
"William R. Ott, Deputy Director, Physics Laboratory, National Institute of Standards and Technology."
"His personal research has been in the fields of electron-atom collisions, plasma spectroscopy, and ultraviolet radiation physics and metrology."
http://physics.nist.gov/StaffOrg/div840/OttBio.html

Now if you think that your credentials are as good as his(isnt that how it works in science?) then show me and I will give your words equal weight.

Ah. This graph again. Specifically the first graph of the first link for anyone who is reading. This graph has shown up before and it still shows a blackbody radiation curve on the left side of the graph and Argon emission on the right side. The second graph in the first link looks like the spectral decomposition of the emission lines.

Yes, notice how the lines change into BB as the pressure goes up.

Notice that the graph behaves as you would expect as the pressure and temp rise. The blackbody portion gets brighter and the peak of the emission heads towards shorter wavelengths, and the Ar emission has the lines smeared due to the effects of the temp and pressure on the argon.

snip...

Its not temp. Its only pressure that causes the smearing.

upriver

08-May-2007, 09:46 PM

Because you feel the need to prove that the sun rises in the west, no matter how much proof is shown that it rises in the east

No, the sun rises in the East. Anyone can see that.:)

korjik

08-May-2007, 10:09 PM

He described the experiment that he did.
"William R. Ott, Deputy Director, Physics Laboratory, National Institute of Standards and Technology."
"His personal research has been in the fields of electron-atom collisions, plasma spectroscopy, and ultraviolet radiation physics and metrology."
http://physics.nist.gov/StaffOrg/div840/OttBio.html

Now if you think that your credentials are as good as his(isnt that how it works in science?) then show me and I will give your words equal weight.

Yes, notice how the lines change into BB as the pressure goes up.

No, the very specifically DO NOT turn into a blackbody radiation curve. The figure shows the figured blackbody curve as the dotted line, and if you look closely, it does not turn upward on the right side of the graph. The Argon lines are smeared due to collisional and doppler (I think) effects and start to look like a continuum, but they do not look at all like a blackbody. Your understanding of your own graph is highly flawed.

Its not temp. Its only pressure that causes the smearing.

Hardly. There are a whole bucket load of effects that smear spectral lines.

upriver

08-May-2007, 10:34 PM

Originally Posted by upriver View Post
1. I should have been a little more clear and said " Why dont they use a pressurized plasma in a blackbody device. That way you can get higher temperatures."

2. Now there is a parallel that you can draw between density and distance in the number of collisions. But the only supporting observations are the sun itself, and you cant see into the area of interest.

3. I have never seen anybody advance photoionization as the reason for the corona. When I say 114eV or 1 million degrees, I am talking about the corona, unless I say iron. Then I am talking about loops. I have never seen anybody advance photoionization as the reason for the loops.

4. Right there seem to be 2 competing models of coronal heating. Alfven waves and reconnection. We can rule out reconnection from our discussions.

5. Even at 20,000C you canot ionize H. The tail can ionize but where does that tail come from. And its not enough for the continuous ionization happening at the chromosphere.
I mean continuous in coverage of the solar surface as opposed to point source emission. The chromosphere is visible through its H Alpha emission at 656.281nm. With continuous coverage there should be a indication of it in the blackbody spectrum of the sun.

6. I tried to find detailed solar emission below 200nm. Pretty much nothing exists in terms of a 1nm resolution composite spectrum down to 1nm soft x-rays.

1. They do not use pressurized plasma for a BB because there are better, easier and cheaper ways of creating a BB, e.g. by using carbon heated by an electric arc, as you described. I think that would be obvious, especially since you seem to work in sonolunimiscence. Wanna carry that aparatus around just to get a BB spectrum or would you rather carry a relatively small carbon cathode?

You have seen Canada's Ultimate Light Ruler....

2. Well, I guess then that the sun is just a rather opague gass at 1 atmosphere throughout, glowing a little.

My newest thoughts are, if the sun is a dense ball of blackbody gamma rays at the center, and by the time photons get to the surface, the whole spectrum is down shifted billions of eV's down by the time it gets to optical depth, which is another filter over the spectrum of the sun.
How could you possibility get so lucky to have a "nice" blackbody spectrum?

3/4. Nope, photoionization is not the way of heating up the corona. First of all everything is already ionized. Please remember why I wrote that stuff about an H above the photosphere. To show that in a dense photon field you can get multiple absorption of photons, leading to ionization.

"Already ionized" is not an explanation.

The heating of the corona can also be done with soundwaves, there are lots of processes that were being looked at at e.g. the Astrophysical Institute of Utrecht University. Another one was nanoflares that occur at the surface of the sun. All of these processes are working at the same time, I am convinced that there is no one process that is heating the corona just by itself.
And I was definitely not talking about loops, where all kinds of processes lead to emission.

I have read all about nanoflares, blinkers, impulsive events, all that stuff.
I used to think they were heating the corona, too. They are all impulsive. You would see some kind of spatial variation of rates. I have read everything I could find on that stuff.

Here I am back at the same process that creates a glow like this.
http://fel.web.psi.ch/images/GlowDischarge.JPG

I have had a hard time finding exact theoretical descriptions of this.

Here is one, but then I digress.
http://www.aetherometry.com/PAGD/PwrfromAEemissions.html

5. Even at 20000 K we CAN ionize H I have explained that to you three times already. It seems you do not want to believe me, okay, then don't.

I believe you.
But is it the 20,000K average plasma temperature that does the ionizing, the tail, or the electric current driving the tail?
Because if you translate 13.6eV into plasma temperature that is 149000K. To depend strictly on plasma temperature alone to ionize the H collisionally that is the temperature you would need.

If you use electricity, the plasma temperature can be a lot lower and still have ionization due to the energy of the drive electrons, and the fact that if you have very low current, you have very few ionization events per unit of time, leading to a lower apparent temperature.

What do you mean "with continuous coverage there should be an indication of it in the BB spectrum". This either shows you have no idea what a BB exactly is (continuous emission over the whole frequency range, described by the BB equation). Or do you mean that measuring the spectrum of the sun (not that I do not say BB) there may be a peak at Ly alpha in the spectrum?

Yes.

6. Maybe that is because it is not interesting and the interesting stuff in soft Xrays does not come from the sun proper as BB but from loops and acceleration regions etc.

So what about that finite chance of x-rays coming from a 6000K blackbody?

Thats not interesting?

upriver

08-May-2007, 10:47 PM

No, the very specifically DO NOT turn into a blackbody radiation curve. The figure shows the figured blackbody curve as the dotted line, and if you look closely, it does not turn upward on the right side of the graph. The Argon lines are smeared due to collisional and doppler (I think) effects and start to look like a continuum, but they do not look at all like a blackbody. Your understanding of your own graph is highly flawed.

Actually, its called a quasi continuum(Dr Ott). And your right, its not a true blackbody.
I'm glad somebody finally looked at that thing for real.
Therein lies the difference between a solid matter spectrum and a pressurized plasma spectrum.
But they fit it to a blackbody to try to determine temperatures inside the bubbles.
I have looked at many SBSL spectrum's with people with real credentials(vs mine) so I have a rough idea of what is and what is not.

Hardly. There are a whole bucket load of effects that smear spectral lines.

Yes, but grossly speaking the dominant effect here is pressure.

tusenfem

09-May-2007, 09:17 AM

You have seen Canada's Ultimate Light Ruler

seen your link, whatever, did not see a spectrum of the lamp.
and in the discussoin with Korjik you have shown you lack of knowledge about BB

My newest thoughts are, if the sun is a dense ball of blackbody gamma rays at the center, and by the time photons get to the surface, the whole spectrum is down shifted billions of eV's down by the time it gets to optical depth, which is another filter over the spectrum of the sun.
How could you possibility get so lucky to have a "nice" blackbody spectrum?

My newest idea is that the moon is made out of green cheese.
Your lack of understanding of BB shows through again and again and again. A dense ball of BB gammas rays? What is that supposed to mean?
We are so lucky to have a nice BB spectrum from the sun, because the sun is in Local Thermal Equilibrium and for the most part optically thick, THE necessary conditions for a plasma to emit an black body curve.

"Already ionized" is not an explanation.

You keep on forgettng here that that neutral H in the photosphere was an example!!!! to explain to you about how you can ionize an atom in a field with photons (and particles) that have not the necessary 13.6 eV to ionize it immediately.
The plasma in the corona comes from the sun, it escapes from the sun, the sun is a big ball of ionized gas, the velocities of the electrons and the ions are too big to significantly recombine while escaping, hence "already ionized" is a perfectly sound conclusion.

I have read all about nanoflares, blinkers, impulsive events, all that stuff.
I used to think they were heating the corona, too. They are all impulsive. You would see some kind of spatial variation of rates. I have read everything I could find on that stuff.

Here I am back at the same process that creates a glow like this.
http://fel.web.psi.ch/images/GlowDischarge.JPG

I have had a hard time finding exact theoretical descriptions of this.

Here is one, but then I digress.
http://www.aetherometry.com/PAGD/PwrfromAEemissions.html

As far as I recall from Utrecht University, nanoflares are all over the sun, going off all the times (Schadee came up with these things, maybe I should do a search). Reconnection converts magnetic energy into kinetic energy (heating), if everywhere on a small scale then there will be no spatial variation. Apart from that, I think the energy in the nanoflares is very quickly redistributed, it is not like a normal solar flare where you most definitely have localized energization.

I guess it would not be too hard to find an explanation for your glow discharge. I think in any book on plasma devices there will be an explanation. But please do not come up with the Correa's as a reference (for obvious reasons to be found in other threads).

I believe you.
But is it the 20,000K average plasma temperature that does the ionizing, the tail, or the electric current driving the tail?
Because if you translate 13.6eV into plasma temperature that is 149000K. To depend strictly on plasma temperature alone to ionize the H collisionally that is the temperature you would need.

If you use electricity, the plasma temperature can be a lot lower and still have ionization due to the energy of the drive electrons, and the fact that if you have very low current, you have very few ionization events per unit of time, leading to a lower apparent temperature.

Thanks for believing me and then returning to your own beliefs again about ionization.
WHO THE BLUBLU CARES WHAT DOES THE IONIZATION?? I have most specifically explained the ionization to you in the previous pages, using photons (but also having stated about collisions). So everything is doing the ionization, photon absorption, inelastic collisions between electrons-ions and ions-ions. Stop translating eV into T, you will get lost in translation! One needs not have a T of bezillion to ionize H, just a couple of bumps in the right time frame with fitting energies will do it.

Mebbie I should give my credentials, so you will believe me like you believe Dr. Ott.
Martin Volwerk
Studied Plasma Astrophyscis at Utrecht University, MSc in 1989 on the topic of "Magnetic flaring near black holes"/
PhD in Plasma Astrophysics at Utrech Universtiy, in 1993 on the topic of "Strong double layers in astrophysical plasmas". Research included a 1 year stay at the Alfven laboratory in Stockholm, Sweden, to do experiments on double layers in a double plasma device.
Postdoc in Paris, working on magnetospheric physics, Freja data, solitary kinetic Alfven waves
Postdoc in Tucson, working on UV emissions by the so called Io plasma torus.
Postdoc at UCLA, working on magnetic field data from Galileo, investigating the magnetospheric interaction of the Galilean moons with the Jovian magnetic field.
Researcher at the Space Research Institute of the Austrian Academy of Sciences, working on magnetic field and plasma data from the Cluster and DoubleStar missions in the Earth's magnetotail. Working on magnetic field data from the Venus Express mission
My papers can be found through ADS or ISI.

Yes.

And???

So what about that finite chance of x-rays coming from a 6000K blackbody?

Thats not interesting?

No, not that much, because it will tell us exactly nothing (except that the BB model for emission is correct, and we already know that), and apart from that, X-ray images of the sun will only show us the activity in the loops and at the footpoints, all the non-thermal emissions, and the WAY outshine any emission coming from the sun proper.

papageno

09-May-2007, 09:48 PM

When have I ever claimed it was a good thing? Its terrible and embarrassing.

So it is terrible and embarassing to persist in a belief for which you cannot provide evidence?
Then why do you keep believing you are right? Doesn't it occur to you that you might be wrong?

You act like every scientist has fallen lock step into the same description of the universe.

This is just the typical excuse of ATM proponents, creteationists and CTs: "The people that don't agree with are all mindless sheeple!".

In reality everybody's view is a little different, even if you went to the same school. I dont mean to say that a basic level of understanding is not required.

A basic understanding you failed to show. And yet it is the others that are wrong...

"CT and creationist" ?????

Don't count me into whatever that is....

You are the one using the same flawed arguments.

How old is current science 50 yrs, 20 yrs?
Just imagine the surety of science in 1000yrs.

You see, you don't even realize that modern science started with Galileo Galilei, four hundred years ago.

Why? Because I question everything?

No, because you persist in your belief in the face of evidence contradicting it.

I know you just got your achievement award, but does that mean there is nothing left to question.

As if the researchers worldwide didn't know that.

Who said that there is nothing to question?

Like the gravity pendulum machine. Nobody has offered an published explanation that was nice and neat.

I can understand, since you don't seem to go into a library...

I brought up the one that I thought logically fit, even though I could not "show you" because of my lack of mathematical training(I think its more like a slight dyslexia).

The Aetherometry website?

Your lack of mathematical training should be a reason to study and learn before making bold claims of knowing better.

I'm stubborn(Taurus:)) I admit, but I am always open to change.
Your actions sadly disprove your words.

upriver

10-May-2007, 12:31 AM

Thanks for believing me and then returning to your own beliefs again about ionization.
WHO THE BLUBLU CARES WHAT DOES THE IONIZATION?? I have most specifically explained the ionization to you in the previous pages, using photons (but also having stated about collisions). So everything is doing the ionization, photon absorption, inelastic collisions between electrons-ions and ions-ions.

WHO THE BLUBLU CARES WHAT DOES THE IONIZATION??

But that is the whole point.

There are 3 ways of initial ionization. Compression, electricity, and photoionization. Correct?

Stop translating eV into T, you will get lost in translation! One needs not have a T of bezillion to ionize H, just a couple of bumps in the right time frame with fitting energies will do it.

Lets see if I can do this.

"The degree of ionization of a plasma is the proportion of atoms which have lost (or gained) electrons, and is controlled mostly by the temperature."

"Plasma temperature is commonly measured in kelvins or electronvolts, and is (roughly speaking) a measure of the thermal kinetic energy per particle."

The Maxwellian Distribution of a plasma is an average of all the particles.
If you have a 20,000C plasma it has a bunch of neutral particles(unionized atoms) and some ionized atoms with their attendant electrons floating around.
This plasma has a tail that tapers off, in the upper part of the distribution that has electrons that have an energy level higher than the average.
There is a corresponding tail of low energy particles on the lower side of the distribution.

Now the only electrons that can ionize anything are the ones with enough energy. The ones in the tail(pretty much).
And by having a low energy electrons, the plasma temperature is low.

But if you increase the energy of the electrons, the plasma heats up.

That means add more electrons of a higher energy to the tail etc.

How do you do that? Electricity.

When you say that a 20,000C plasma ionizes H thats not exactly correct.

It is not the 20,000C(2eV electron) plasma that is doing it, it is the 15.6eV electron component of the tail that is ionizing the neutrals.

13.6eV to knock the electron off and 2Ev for the 20,000C.

So how do you keep this process happening?
You have to fed it more electrons otherwise everything goes away.

just a couple of bumps in the right time frame with fitting energies will do it.

The first "fitting energy" for a H atom is 3.4eV. 20,000C is 1.9eV.

And this all has to happen before deexcitation(right time frame)?

I think thats what your talking about.

Mebbie I should give my credentials, so you will believe me like you believe Dr. Ott.
Martin Volwerk
Studied Plasma Astrophyscis at Utrecht University, MSc in 1989 on the topic of "Magnetic flaring near black holes"/
PhD in Plasma Astrophysics at Utrech Universtiy, in 1993 on the topic of "Strong double layers in astrophysical plasmas". Research included a 1 year stay at the Alfven laboratory in Stockholm, Sweden, to do experiments on double layers in a double plasma device.
Postdoc in Paris, working on magnetospheric physics, Freja data, solitary kinetic Alfven waves
Postdoc in Tucson, working on UV emissions by the so called Io plasma torus.
Postdoc at UCLA, working on magnetic field data from Galileo, investigating the magnetospheric interaction of the Galilean moons with the Jovian magnetic field.
Researcher at the Space Research Institute of the Austrian Academy of Sciences, working on magnetic field and plasma data from the Cluster and DoubleStar missions in the Earth's magnetotail. Working on magnetic field data from the Venus Express mission
My papers can be found through ADS or ISI.

Martin, I already know your credentials from your web page(at least part of them).
You already have my respect even though it may not seem like it.
(You too, papageno.)

So now, given your credentials and Dr Ott's, who would you use as a reference, IF you had to choose?
I'm not trying to dis you but lets be a little real on this for a moment.

Have you ever done a plasma lab experiment without electricity?
How do you separate the effects of the electricity from that of the plasma?

upriver

10-May-2007, 12:48 AM

So it is terrible and embarassing to persist in a belief for which you cannot provide evidence?

It builds character. I can speak in front of anybody.
And I know the evidence exists. I just dont have the tools to correctly provide the evidence in a world accepted format.

Then why do you keep believing you are right?

I have no clue.

Doesn't it occur to you that you might be wrong?

Everyday.

This is just the typical excuse of ATM proponents, creteationists and CTs: "The people that don't agree with are all mindless sheeple!".

Well it is a system that strives to churn out plentiful copies of a doctrine.:)

A basic understanding you failed to show. And yet it is the others that are wrong...

You assume that I am calling everybody else wrong by not automatically accepting what I am told.
Am I wrong for deeply questioning certain aspects of science?

You see, you don't even realize that modern science started with Galileo Galilei, four hundred years ago.

I guess it all depends on how you view modern science.
I still think science is not very modern(sorry), especially medicine.

No, because you persist in your belief in the face of evidence contradicting it.

Is it possible, however unlikely that I saw or read something that gave me a legitimate question to ask.
The problem is that I may not be able to frame it correctly.

upriver

10-May-2007, 01:14 AM

A dense ball of BB gammas rays? What is that supposed to mean?

Well, the ball of fusion in the center of the sun emits gamma rays right?
And its very dense, denser than lead.
So I just made the distribution BB just for fun(sorry).

So now looking at the pressurized spectrum of the bubbles again.
See how it is a quasi continuum?

Do you think the sun would exhibit the same spectrum because its pressurized plasma?

Now add your optically thick plasma on top of that.

Is that how they determine the spectrum of the sun?

The plasma in the corona comes from the sun, it escapes from the sun, the sun is a big ball of ionized gas, the velocities of the electrons and the ions are too big to significantly recombine while escaping, hence "already ionized" is a perfectly sound conclusion.

So there are ions with greater energies than 6000C that come through the photosphere.
Those ions would be considered the tail of the distribution?
And how do they get to >100eV in the corona?

Forskern

10-May-2007, 07:27 AM

There are 3 ways of initial ionization. Compression, electricity, and photoionization. Correct?

There are three ways, but those are photo-, direct impact (collisional) and excitation-autoionization. In the sun's atmosphere they are important in very different regimes. The chromosphere and the corona represent two relatively simple regimes, with photoionization being important in the former and collisional in the latter. Just what do you mean by "electricity" and "compression" --- if you think of these as processes that accelerates the individual electrons and thus accelerates collisional ionization, then you should be careful about applying them to the chromosphere.

"The degree of ionization of a plasma is the proportion of atoms which have lost (or gained) electrons, and is controlled mostly by the temperature."

"Plasma temperature is commonly measured in kelvins or electronvolts, and is (roughly speaking) a measure of the thermal kinetic energy per particle."

There are many ways to defined the temperature in a plasma, and one is to choose a mean value of the kinetic energies of the constituent particles. Under non-dynamic conditions the different definitions will match, and the kinetic energies will be normally distributed about this temperature. Then you can make claims like you just did.

But you have to be careful about what you measure the temperature of. Photoionization depends on the temperature of the photosphere (6000K), while collisional depends on the kinetic temperature of the plasma itself.

The Maxwellian Distribution of a plasma is an average of all the particles.
If you have a 20,000C plasma it has a bunch of neutral particles(unionized atoms) and some ionized atoms with their attendant electrons floating around.
This plasma has a tail that tapers off, in the upper part of the distribution that has electrons that have an energy level higher than the average.
There is a corresponding tail of low energy particles on the lower side of the distribution.
Now the only electrons that can ionize anything are the ones with enough energy. The ones in the tail(pretty much).
And by having a low energy electrons, the plasma temperature is low.

If you are at 20000K, and in the solar atmosphere, you have already a fully ionized hydrogen plasma, with the rest divided between HeII and HeIII. Saha's tells us that most hydrogen is (photo)ionized above 10000K.

But if you increase the energy of the electrons, the plasma heats up.
That means add more electrons of a higher energy to the tail etc.
How do you do that? Electricity.

Or strong radiation, or a shock...

When you say that a 20,000C plasma ionizes H thats not exactly correct.
It is not the 20,000C(2eV electron) plasma that is doing it, it is the 15.6eV electron component of the tail that is ionizing the neutrals.

13.6eV to knock the electron off and 2Ev for the 20,000C.

This is wrong, try re-reading some of of tusenfem and korjiks posts.

tusenfem

10-May-2007, 08:06 AM

There are 3 ways of initial ionization. Compression, electricity, and photoionization. Correct?

as I said in the quote you gave, although electricity and compression all have to do with collisions, so they are one and the same, i.e. inelastic collision, but I doubt you will see it this way.

Lets see if I can do this.

... snip
When you say that a 20,000C plasma ionizes H thats not exactly correct.

It is not the 20,000C(2eV electron) plasma that is doing it, it is the 15.6eV electron component of the tail that is ionizing the neutrals.

13.6eV to knock the electron off and 2Ev for the 20,000C.

So how do you keep this process happening?
You have to fed it more electrons otherwise everything goes away.

The first "fitting energy" for a H atom is 3.4eV. 20,000C is 1.9eV.

And this all has to happen before deexcitation(right time frame)?

I think thats what your talking about.

YES THAT IS WHAT I WAS TALKING ABOUT, but you are still misinterpreting somehow. And it is the 20000 K plasma that is doing it, because a temperature is an ensemble average of the whole plasma to be specific, it is the second moment of the parcticle distribution function, i.e. the integral of (m v2 F(x,v,t))

Why would you need the extra 2 eV, ah I guess you don't want the plasma to loose energy. Did you consider that in the sun there is also the intense radiation? And that photon-particle interactions occur.

Have you ever done a plasma lab experiment without electricity?
How do you separate the effects of the electricity from that of the plasma?

Cute question, no I always use electricity, because that is the way we can produce a plasma most easily in the laboratory. Naturally I could use a nuclear explosion, but I do not think that Hannes Alfven would have allowed that in his laboratory.
And we can separate them, because your "electrical effects" will most often produce non-thermal emissions, at least at the sun. In the lab I have looked at actually the radiation emitted because of your electricity. And if you know what you are doing, and understand how plasmas behave you can measure everything and find out what is related to what.

tusenfem

10-May-2007, 08:21 AM

Well, the ball of fusion in the center of the sun emits gamma rays right?
And its very dense, denser than lead.
So I just made the distribution BB just for fun(sorry).

So now looking at the pressurized spectrum of the bubbles again.
See how it is a quasi continuum?

Do you think the sun would exhibit the same spectrum because its pressurized plasma?

Now add your optically thick plasma on top of that.

Is that how they determine the spectrum of the sun?

So there are ions with greater energies than 6000C that come through the photosphere.
Those ions would be considered the tail of the distribution?
And how do they get to >100eV in the corona?

There is fusion in the center of the sun where H gets converted to He. Please look up in a book on astrophysics how much energy gets released and how it gets released.

Dunno if it is a quasi continuum, if i look at the one plot of the nature paper I see a dashed curve which would be BB and a solid curve that shows the emission. quasi-continuum is something that is not really a continuum, for the plot shown, it looks pretty continuous to me, I guess you want to say that is is quasi BB.

I would expect the sun to show a BB, and then the very outer layers will create extra emission and absorption lines.

The BB comes from the optically thick region of the sun, anything that is put "on top of that" will have to be an optically thin region. Get your models straightened out!

They determine the spectrum of the sun by measuring it. The observatory of Utrecht Univesity was the first one to measure the greatest part of the solar spectrum and published a comprehensive "atlas" of the spectrum.

Well, they are obviously accelerated, but we will not get into acceleration, because it does not belong in this thread, which is about energy and temperature of photons.

papageno

10-May-2007, 07:50 PM

So it is terrible and embarassing to persist in a belief for which you cannot provide evidence?

It builds character. I can speak in front of anybody.
And I know the evidence exists. I just dont have the tools to correctly provide the evidence in a world accepted format.

And yet you are doing nothing to acquire those tools.

Then why do you keep believing you are right?

I have no clue.

So, you are not actually interested in finding out what is correct.

Doesn't it occur to you that you might be wrong?

Everyday.

And isn't that enough to push to learn?

This is just the typical excuse of ATM proponents, creteationists and CTs: "The people that don't agree with are all mindless sheeple!".

Well it is a system that strives to churn out plentiful copies of a doctrine.:)

:rolleyes:

Of course you have no evidence whatsoever to support that accusation...

What is is with the ATM proponents and their grudge against the scientists?

A basic understanding you failed to show. And yet it is the others that are wrong...

You assume that I am calling everybody else wrong by not automatically accepting what I am told.

Your lack of understanding is in plain view: no assumptions necessary; just simple observation.

Am I wrong for deeply questioning certain aspects of science?

You don't understand those aspects, and you questioning is based on that ignorance. Therefore it is unfounded.

No, because you persist in your belief in the face of evidence contradicting it.

Is it possible, however unlikely that I saw or read something that gave me a legitimate question to ask.
The problem is that I may not be able to frame it correctly.
And yet you are doing nothing to become able to frame it.