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  #61 (permalink)  
Old 12-May-2005, 01:07 PM
lyndonashmore lyndonashmore is offline
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Default Re: Ashmore's "paradox" Reloaded.

Sylas wrote
Quote:
It is precisely because I have heard of and actually understand and apply the conservation of momentum and energy that can I recognize immediately why the Lyndon effect is a violation of basic school level physics.

The Lyndon effect cannot conserve momentum, or it cannot conserve energy. To prove this, assume a Lyndon interaction between a photon and electron with conservation of energy and momentum, and derive a contradiction.

You have an initial photon, with frequency f, and hence with energy hf and momentum hf/c.

There is some kind of interaction with an electron, from which the photon emerges redshifted to frequency F. This means that the electron must have a boost in energy of h(f-F), to conserve energy.

In the Lyndon effect, there is no scattering of the photon, so the direction for the photon is as before. The recoil in the electron must be along the same line, or else conservation of momentum is violated.

Consider a case where the electron is initially at rest. You can always reduce to this case by a simple change of reference frame. By conservation of energy, the electron ends up with a small amount of kinetic energy E equal to h(f-F). Since there is no scatter and everything is straight line motion, we can also infer that the momentum difference in the photon is the momentum p of the electron, being h(f-F)/c.

Hence E = pc.

For those used to Newtonian physics, which would actually apply here for small velocities, we have E = mv^2/2 and p = mv, where v is the electron velocity.

E = pc becomes mv^2 = 2mvc, or v*(v-2c) = 0. That is, either v is zero (no recoil, no redshift) or v is twice the speed of light (not possible).

If we use relativity, we use the total energy equation to get
(mc^2)^2 + (pc)^2 = (mc^2 + E)^2 = (mc^2 + pc)^2
and hence
0 = 2mpc^3
Hence p is zero. The final electron must have zero momentum; no recoil, no redshift.

This isn't hard. It's very elementary. But I don't think Lyndon even understands the equations, and years of attempts to explain it to him have had no effect.

Cheers -- Sylas
[/quote]

Are you feeling OK Sylas?

I will try to make it simple for you and other board members reading your posts, so lets think of a frog jumping across a pond from one lily leaf to the next.
Frog leaps forward, it has both energy and momentum, and lands on a stationary lily leaf. The frog plus leaf move forward as the momentum and some energy is transferred to the leaf/frog combination. Its a type of recoil.
Now the frog jumps off in the forward direction. To do this the frog pushes back on the leaf - recoil once again, and the leaf gains momentum in the opposite direction to before. Since the leaf was already moving forwards the effect of the work done on the leaf by the
Frog is to bring the leaf back to rest. The frog has lost energy because it has done two lots of work on the leaf, one lot to get it moving and a second lot to bring it back to rest.
Now think of a person stood on the bank watching the whole thing from start to finish. The person says "OOH! look, a contradiction. The frog has lost energy but the leaf is still at rest. Overall, the leaf has gained no momentum. "
Now Sylas, could you give me some advice as to how I would explain the Physics of the situation to this person?
I would be grateful of your advice.
Cheers,
Lyndon
PS if the leaf is fixed firmly to the pond then it will not recoil and so no energy will be transferred to it. The Frog loses no energy and so is not redshifted. ie Kermit the frog's mossbauer effect.
Edited at the request of Papageno though it was perfectly polite to Sylas in any case!
  #62 (permalink)  
Old 12-May-2005, 01:15 PM
lyndonashmore lyndonashmore is offline
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Default Re: Ashmore's "paradox" Reloaded.

Quote:
Originally Posted by Lurker
Quote:
Originally Posted by papageno
Quote:
Originally Posted by lyndonashmore
No its fine Eta C, you forget that there are restraining forces acting on our electrons in the plasma from other charges in the plasma which restrict sideways motion. That is the restoring forces that we are talking about that enable it to perform SHm stop it moving sideways.
Even your beloved Compton scater allows the photon to come back on itself - conservation of linear momentum
Provide quantitative estimates for these "restraining forces" and show that they make an electron oscillate about a position.

And don't come up with out-of-context quotes from sources that deal with high-powered laser in high-density plasma, as you have done before.
OK... papageno I have quietly followed this discussion through many incarnations and have to ask; does he listen to anything you say?? I only ask because the longer this goes on the more it seems that you are having to repeat the same ideas over and over in different words.
As long as papageno continues to treat particles in a plasma as neutral atoms in a gas, then I will not listen.
Cheers,
Lyndon.
  #63 (permalink)  
Old 12-May-2005, 01:20 PM
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Default Re: Ashmore's "paradox" Reloaded.

Quote:
Originally Posted by lyndonashmore
Quote:
Originally Posted by Lurker
Quote:
Originally Posted by papageno
Quote:
Originally Posted by lyndonashmore
No its fine Eta C, you forget that there are restraining forces acting on our electrons in the plasma from other charges in the plasma which restrict sideways motion. That is the restoring forces that we are talking about that enable it to perform SHm stop it moving sideways.
Even your beloved Compton scater allows the photon to come back on itself - conservation of linear momentum
Provide quantitative estimates for these "restraining forces" and show that they make an electron oscillate about a position.

And don't come up with out-of-context quotes from sources that deal with high-powered laser in high-density plasma, as you have done before.
OK... papageno I have quietly followed this discussion through many incarnations and have to ask; does he listen to anything you say?? I only ask because the longer this goes on the more it seems that you are having to repeat the same ideas over and over in different words.
As long as papageno continues to treat particles in a plasma as neutral atoms in a gas, then I will not listen.
Wrong.
You keep equating the interaction of light with the charged particles in a plasma with the interaction of light with the neutral atoms in glass.
Obviously you cannot keep your story straight, and end up contradicting yourself.
So, where the quantitative estimates?
(I guess you have none, since you have to resort to diversionary tactics to avoid the issue.)


By the way, I urge you -- again -- to edit your last post addressed to Sylas.
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  #64 (permalink)  
Old 12-May-2005, 01:38 PM
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Yes. That was definately in bad tatse Lyndon.
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  #65 (permalink)  
Old 12-May-2005, 02:12 PM
akirabakabaka akirabakabaka is offline
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Quote:
Originally Posted by papageno
Quote:
Originally Posted by akirabakabaka
Quote:
Originally Posted by papageno
Quote:
Originally Posted by akirabakabaka
In his ill-fated analogy, he is equating the number of collisions of photons with particles. He has calculated that the lens of a camera is about equal to 5mly worth of low density IG plasma in this respect.
Except that he assumed a uniform density of particles, and that the red-shift is the same for each collision.
Both are over-simplified assumptions.
This is a good point, and is specifically taken into account on Ashmore's website.
How?
Take a look at the website already.

Quote:
Originally Posted by Ashmore's Tired Light Theory
Full determination of the Hubble Constant For all Galaxies Near and Far.

The determination of H given above included an approximation. I assumed there that the collision cross-section of the photon was constant throughout its entire journey. This is clearly not the case. The reason being that the collision cross-section is proportional to the wavelength and each time a photon interacts with an electron it is redshifted and undergoes an increase in wavelength. That is, as the photon travels along and is absorbed and re-emitted by the electrons in the plasma of intergalactic space, the wavelength will get bigger and bigger and so the chances of it making a collision will also become bigger and bigger. The distance travelled by the photon between collisions becomes less and less. The rate at which the photon is redshifted becomes greater and greater as the photon travels further and further.
As for why we don't see redshift in glass...

Quote:
Originally Posted by papageno
Quote:
Originally Posted by akirabakabaka
It has been explained to death why the redshift in Ashmore's model does not appear in either glass or dense plasma. The mass/density is too great for the photon to have much effect on it, so little energy is lost in the collisions.
Why would the mass/density be too great?
Ashmore's point is that the electron loses energy while it keep the photon in his pocket.
The higher the density, the larger the energy loss to the other particles.
Now to jog your memory again, let's take a look at the first page of the first Ashmore's Paradox thread:

Quote:
Originally Posted by lyndonashmore
Quote:
Originally Posted by papageno
Mössbauer effect is the recoil-less emission of gamma-photons by excited nuclei of atoms in a lattice (there is no recoil because there is not enough energy to excite a quantum of lattice vibration).
What has this to do with electrons in a plasma?
What you described looks more like the Compton effect (scattering of a photon by an electron
Glad you mentioned this, as this point answers Gritmongers objection to tired light in a post on another topic i.e. why don’t we get a redshift when light passes through glass? As I show, the Mossbauer effect produces a shift in wavelength of h/mc each time a photon is absorbed and reemitted. However, if the electron is bound in a single atom then it is the whole atom that recoils and m is the mass of the atom (redshift is already much smaller). For an electron in glass it is the whole glass block that recoils and so m is the mass of the block of glass and we can forget the effect, as the redshift is too small. In the same way, you only get my redshift mechanism when the plasma is very rarefied. In dense plasmas the effective mass of the electron increases and so one gets less redshift not more. Mossbauer effect is only recoilless for atoms bound in a solid. Under no way of the imagination is the effect described in my paper remotely the same as the Compton effect as there is no shift in wavelength in the forward direction with this effect.
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  #66 (permalink)  
Old 12-May-2005, 02:27 PM
akirabakabaka akirabakabaka is offline
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Quote:
Originally Posted by Fram
Quote:
Originally Posted by akirabakabaka
He is making an analogy here (which he explained several posts ago).
No. The blooming of the camera lens just appears here out of nowhere (the 5 million ly was mentioned before when he needed to clarify what he meant by nearby galaxies, as he again was very unclear), and comparing a visual effect with a distance is wrong. Lyndon clearly has a huge problem when it comes to explaining things in a comprehensible and logically consistent way, which is rather sad for someone trying to have a paradox and unit named after him.
If you read any of the discussions or any of the material on Ashmore's website, you will see this analogy has been explained several times over. It is not our fault that you are unfamiliar with the material, please do not pass the blame onto others..

Quote:
Originally Posted by Fram
Take this quote from his press release:
Quote:
If we convert the Hubble constant into SI units, 64 km/s per Mpc becomes 2.1exp(-18 ) per sec.
Where have all the units gone? Only the seconds are left!
Oh, I see, because he pulls the same trick in the next line,
Quote:
But what Ashmore found was that the quantity ‘hr/m per cubic metre’ (h is the planck constant, m is the mass of the electron and r is the classical radius of the electron) is also equal to 2.1exp(-18 ) per sec! Or, to put it another way, ‘hr/m per cubic metre’ in astronomical units is 64 km/s per Mpc – the Hubble constant.
This is plain gibberish and pure numerology, and thus hardly fit as the basis for a scientific theory. Add to this that the mechanism he describes is equally ridiculous (papageno explains this way better than I ever can), and you get an idea of why I have a problem with Ashmore's Hat of Tricks and with the way he expresses his ideas.
Again, it is not Ashmore's fault that you can not follow the math. This issue is also touched on in the first page of the first Ashmore's Paradox thread. If you read his paper or his website, you can see how all of the units are derived and converted. If you think he has done something wrong, then please explain. Don't rely on others to provide your opinion for you.
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  #67 (permalink)  
Old 12-May-2005, 02:30 PM
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Quote:
Originally Posted by akirabakabaka
Quote:
Originally Posted by papageno
Quote:
Originally Posted by akirabakabaka
Quote:
Originally Posted by papageno
Quote:
Originally Posted by akirabakabaka
In his ill-fated analogy, he is equating the number of collisions of photons with particles. He has calculated that the lens of a camera is about equal to 5mly worth of low density IG plasma in this respect.
Except that he assumed a uniform density of particles, and that the red-shift is the same for each collision.
Both are over-simplified assumptions.
This is a good point, and is specifically taken into account on Ashmore's website.
How?
Take a look at the website already.
Do you think that I have not?
Since you seem convinced that Ashmore is right, you should not have a problem explaining his "theory" without resorting to quotes.

Quote:
Originally Posted by Ashmore's Tired Light Theory
Full determination of the Hubble Constant For all Galaxies Near and Far.

The determination of H given above included an approximation. I assumed there that the collision cross-section of the photon was constant throughout its entire journey.
That's the formula of collision cross-section for "off-resonance" absorption of photons, ifI remember correctly.
I already explained this absorption process works for atoms, so it cannot be applied to electrons in a plasma, because these are not bound to positive nuclei as in a n atom.


Quote:
Originally Posted by Ashmore's Tired Light Theory
This is clearly not the case. The reason being that the collision cross-section is proportional to the wavelength and each time a photon interacts with an electron it is redshifted and undergoes an increase in wavelength. That is, as the photon travels along and is absorbed and re-emitted by the electrons in the plasma of intergalactic space, the wavelength will get bigger and bigger and so the chances of it making a collision will also become bigger and bigger. The distance travelled by the photon between collisions becomes less and less. The rate at which the photon is redshifted becomes greater and greater as the photon travels further and further.
Still the same problem: Ashmore still wrongly assumes that he can apply that formula to plasma.

Quote:
Originally Posted by akirabakabaka
As for why we don't see redshift in glass...

Quote:
Originally Posted by papageno
Quote:
Originally Posted by akirabakabaka
It has been explained to death why the redshift in Ashmore's model does not appear in either glass or dense plasma. The mass/density is too great for the photon to have much effect on it, so little energy is lost in the collisions.
Why would the mass/density be too great?
Ashmore's point is that the electron loses energy while it keep the photon in his pocket.
The higher the density, the larger the energy loss to the other particles.
Now to jog your memory again, let's take a look at the first page of the first Ashmore's Paradox thread:

Quote:
Originally Posted by lyndonashmore
Quote:
Originally Posted by papageno
Mössbauer effect is the recoil-less emission of gamma-photons by excited nuclei of atoms in a lattice (there is no recoil because there is not enough energy to excite a quantum of lattice vibration).
What has this to do with electrons in a plasma?
What you described looks more like the Compton effect (scattering of a photon by an electron
Glad you mentioned this, as this point answers Gritmongers objection to tired light in a post on another topic i.e. why don’t we get a redshift when light passes through glass? As I show, the Mossbauer effect produces a shift in wavelength of h/mc each time a photon is absorbed and reemitted. However, if the electron is bound in a single atom then it is the whole atom that recoils and m is the mass of the atom (redshift is already much smaller). For an electron in glass it is the whole glass block that recoils and so m is the mass of the block of glass and we can forget the effect, as the redshift is too small. In the same way, you only get my redshift mechanism when the plasma is very rarefied. In dense plasmas the effective mass of the electron increases and so one gets less redshift not more. Mossbauer effect is only recoilless for atoms bound in a solid. Under no way of the imagination is the effect described in my paper remotely the same as the Compton effect as there is no shift in wavelength in the forward direction with this effect.
Did you actually read the following posts, or even just this thread?
Do you actually understand my criticism to Ashmore's ideas?
How many times do we need to explain that the "mechanism" describes does not happen?
Why don't you address this post?
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  #68 (permalink)  
Old 12-May-2005, 02:42 PM
lyndonashmore lyndonashmore is offline
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Quote:
TheAtomium

Like I said I don't know much about this side of physics so please understand if I'm asking questions that have obvious answers.

By what process does the photon loose energy to the electron? Does the electron just say I'll have some of that, strip the electron of some energy, and send it on its way? If this is right, then when does the electron blip out some microwave (CMB) energy?
Or is it lost in the 'rebounding' part where some momentum is transfered into the matrix of plasma electrons*? In that case how is the energy then re-emitted as microwave energy?
Basically, in the same way as the frog in my post to Sylas above. Electrons in plasma can oscillate. Th incoming photon sets an electron into oscillation. It is absorbed. The oscillating electron then re-emits a new photon. This is what happens in all transparent medium and the way light travels through them. But in plasma the electrons recoil on absorption and re-emission so some energy is transferred from the photon to the electron. The frequency of the photon reduces, the wavelength increases - it is redshifted. This does not happen in glass because the elctrons are 'fixed' and so cannot recoil.
The energy transferred to the recoiling electron is emitted by a process known as bremstrahhlung (German for 'braking radiation'). The recoiling electron nears other charges, decelerates and radiates the energy. I calculate the energy of this radiation and show it to be in the microwave region.



Quote:
Where does the SHM motion of plasma electrons come from? Is it because a displaced electron will want to slide back to a point of zero net electric force (terminology :oops:)?
Plasma likes to be neutral, when a recoiling electron moves backwards it moves towards other electrons behind of it and away from electrons in front of it. This leads to electrostatic restoring forces that try to bring it back. Its like a mass between two springs - displace it and it oscillates. Electrons in dense plasma are known to perform SHM when displaced. I assume it still happens in the Sparse plasma of IG space. I work out an expression for the Hubble constant, insert known data and get the correct value. As a result I believe that the assumption is correct.
  #69 (permalink)  
Old 12-May-2005, 02:56 PM
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Quote:
Originally Posted by akirabakabaka
Quote:
Originally Posted by Fram
Quote:
Originally Posted by akirabakabaka
He is making an analogy here (which he explained several posts ago).
No. The blooming of the camera lens just appears here out of nowhere (the 5 million ly was mentioned before when he needed to clarify what he meant by nearby galaxies, as he again was very unclear), and comparing a visual effect with a distance is wrong. Lyndon clearly has a huge problem when it comes to explaining things in a comprehensible and logically consistent way, which is rather sad for someone trying to have a paradox and unit named after him.
If you read any of the discussions or any of the material on Ashmore's website, you will see this analogy has been explained several times over. It is not our fault that you are unfamiliar with the material, please do not pass the blame onto others..
You said he explained it several posts ago. I said he didn't. You don't admit you were wrong but just make another claim. I haven't read everything on his website (thank goodness!), but I haven't come across this analogy in what I have read (including his main pages, his paper (can you tell me where it is published, by the way? The website only states that it is accepted, but doesn't say for what magazine or when it is published), and the press release).
I did not pass the blame onto others, you gave a wrong statement while trying to defend Lyndon after his wrong statement. Two wrongs don't make a right, you know...

Quote:
Originally Posted by akirabakabaka
Quote:
Originally Posted by Fram
Take this quote from his press release:
Quote:
If we convert the Hubble constant into SI units, 64 km/s per Mpc becomes 2.1exp(-18 ) per sec.
Where have all the units gone? Only the seconds are left!
Oh, I see, because he pulls the same trick in the next line,
Quote:
But what Ashmore found was that the quantity ‘hr/m per cubic metre’ (h is the planck constant, m is the mass of the electron and r is the classical radius of the electron) is also equal to 2.1exp(-18 ) per sec! Or, to put it another way, ‘hr/m per cubic metre’ in astronomical units is 64 km/s per Mpc – the Hubble constant.
This is plain gibberish and pure numerology, and thus hardly fit as the basis for a scientific theory. Add to this that the mechanism he describes is equally ridiculous (papageno explains this way better than I ever can), and you get an idea of why I have a problem with Ashmore's Hat of Tricks and with the way he expresses his ideas.
Again, it is not Ashmore's fault that you can not follow the math. This issue is also touched on in the first page of the first Ashmore's Paradox thread. If you read his paper or his website, you can see how all of the units are derived and converted. If you think he has done something wrong, then please explain. Don't rely on others to provide your opinion for you.
I'll certainly not rely on you to provide an opinion, rest assured. In what way have I relied on others? It's not because I say that papageno explains it better that I don't have an opinion.
Apart from that: if you cannot understand what I have said here, tough luck.
I'll repeat:
Quote:
Quote:
If we convert the Hubble constant into SI units, 64 km/s per Mpc becomes 2.1exp(-18 ) per sec.
Where have all the units gone? Only the seconds are left!
What isn't clear about this? What is the unit for the 2.1exp(18 )? Why isn't it included?

I know the answer; because he has to match it with his other quantity, which has completely different units. In the second case, he doesn't only remove units, he even adds some (where do the seconds come from?)! In what way are a mass and a distance related?
I have read his explanations, and they don't explain what he does there and why that would be allowed in physics. It's just gibberish, numerology. It has nothing to do with being able to follow the math. I have not said that his number of 2.1exp(18 ) is right or wrong, and that's the math part. It seems to me that you on the other hand cannot follow the logic needed to make or understand a coherent theory that's based in reality. (you see, it's not so hard to dismiss people's arguments by claiming that they aren't smart enough, even I can do it: but it's not very productive, so please refrain from it in the future).
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  #70 (permalink)  
Old 12-May-2005, 02:59 PM
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Quote:
Originally Posted by lyndonashmore
Quote:
TheAtomium
Like I said I don't know much about this side of physics so please understand if I'm asking questions that have obvious answers.

By what process does the photon loose energy to the electron? Does the electron just say I'll have some of that, strip the electron of some energy, and send it on its way? If this is right, then when does the electron blip out some microwave (CMB) energy?
Or is it lost in the 'rebounding' part where some momentum is transfered into the matrix of plasma electrons*? In that case how is the energy then re-emitted as microwave energy?
Basically, in the same way as the frog in my post to Sylas above. Electrons in plasma can oscillate. Th incoming photon sets an electron into oscillation. It is absorbed.
It is not a macroscopic electromagnetic wave.
What happens to energy and momentum if the electron absorbs the photon?
Sylas has shown the math: both cannot be conserved in that case.
But you did not address the math. You preferred another way to avoid the issue.

Quote:
Originally Posted by lyndonashmore
The oscillating electron then re-emits a new photon.
Why would the electron oscillate?
It is not bound to anything.

Quote:
Originally Posted by lyndonashmore
This is what happens in all transparent medium and the way light travels through them.
Wrong.
Light traveling through a dielectric medium polarizes temporarily neutral atoms.
Yo don't have to trust:l read the second volume of Feynman's
Lectures on Physics
(since you quoted him, I think you would trust Feynman).

Quote:
Originally Posted by lyndonashmore
But in plasma the electrons recoil on absorption and re-emission so some energy is transferred from the photon to the electron.
In your latest posts you insisted that the electron is affected by "restraining forces": why would it recoil?
Or are you saying it can recoil only in one direction?
Since the plasma is homogeneous and isotropic, why would that happen? Why would that be the direction coinciding with the propagation of the photon?

Quote:
Originally Posted by lyndonashmore
The frequency of the photon reduces, the wavelength increases - it is redshifted. This does not happen in glass because the elctrons are 'fixed' and so cannot recoil.
So, now the electrons are free in a plasma!
But the scattering of a photon with a free electron is correctly described by Compton effect, with frequency shift.

Quote:
Originally Posted by lyndonashmore
The energy transferred to the recoiling electron is emitted by a process known as bremstrahhlung (German for 'braking radiation').
Bremsstrahlung occurs in materials, where an electron interacts with many other particles.
Why would that happen in a plasma?

Quote:
Originally Posted by lyndonashmore
The recoiling electron nears other charges, decelerates and radiates the energy. I calculate the energy of this radiation and show it to be in the microwave region.
You have to provide the spectrum: where is it?
(Modern cosmology can give the whole spectrum.)


Quote:
Originally Posted by lyndonashmore
Quote:
Where does the SHM motion of plasma electrons come from? Is it because a displaced electron will want to slide back to a point of zero net electric force (terminology ops?
Plasma likes to be neutral, when a recoiling electron moves backwards it moves towards other electrons behind of it and away from electrons in front of it. This leads to electrostatic restoring forces that try to bring it back. Its like a mass between two springs - displace it and it oscillates. Electrons in dense plasma are known to perform SHM when displaced. I assume it still happens in the Sparse plasma of IG space. I work out an expression for the Hubble constant, insert known data and get the correct value. As a result I believe that the assumption is correct.
If you read the posts in this thread, you would know that this works for the charge denisty (many electrons at one time), not for single electrons.
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"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)
  #71 (permalink)  
Old 12-May-2005, 03:35 PM
lyndonashmore lyndonashmore is offline
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One of the great successes of this Tired Light theory is that it starts from first principles with a known interaction and derives an expression for the Hubble constant that is consistent with observation when published parameters are substituted.
Can the BB do that? I think not.
Papageno et al, say that this is a coincidence because they cannot bring themselves to admit that they are wrong.
Let’s look at one or two more ‘coincidences’ shall we.
Let’s start with the wavelength at which the intensity of the CMB curve peaks. This is at a wavelength of 2.1x10^-3metre.
Using my theory we can calculate the wavelength of the incoming photon that, in being redshifted, causes such a CMB photon to be emitted, it is 2.46x10^-15m and a frequency of 6.05x10^15Hz (in the UV). This has an energy (E = hf) of 3.87x10^-18J.
I wonder what the temperature of a plasma cloud would be if, on average, each electron had this much energy?
We can work it out by 3.87x10^-18J = 3kT/2 and the temperature comes out to be 1.2x10^5K.
Shock, horror, this is the temperature of the plasma of Intergalactic space. What a ‘coincidence’! That is the watershed where the CMB has its maximum intensity corresponds to the situation where the incoming photon has the same energy as the average KE of the electrons in the plasma of IG space.
But wait a minute. There is another coincidence here. People on this site go on about a ‘Compton Effect’.
When the energy of the incoming photon is greater than the average KE of the electrons in the plasma of IG space, the photons are scattered and lose energy to the electrons.
When the energy of the incoming photon is less than the average KE of the electrons in IG space then here the photons gain energy (inverse Compton scatter).
So what happens when the energy of the incoming photons is equal to the average KE of the electrons in the plasma of IG space? Nothing. There is no transfer of energy between the photons and electrons by Compton scatter! There is no Compton scatter for photons of this energy.
So the watershed where the CMB has its maximum intensity corresponds not only to the situation where the incoming photon has the same energy as the average KE of the electrons in the plasma of IG space but also to the point where there is no Compton scatter. Then there is being able to predict H of course.
And yet Papageno et al say there is nothing in all of this.
Dream on folks.
Cheers,
Lyndon
PS remember, you saw it first on the BABB
  #72 (permalink)  
Old 12-May-2005, 03:40 PM
lyndonashmore lyndonashmore is offline
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Papaggeno, not you too.
Quote:
It is not a macroscopic electromagnetic wave.
What happens to energy and momentum if the electron absorbs the photon?
Sylas has shown the math: both cannot be conserved in that case.
But you did not address the math. You preferred another way to avoid the issue.
Sylas' maths is totally incorrect. I showed it in my frog post. He has messed up the vector nature of momentum by including two interactions in opposing directions in one overall calculation.
Cheers,
Lyndon
  #73 (permalink)  
Old 12-May-2005, 03:44 PM
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Quote:
Originally Posted by lyndonashmore
One of the great successes of this Tired Light theory is that it starts from first principles with a known interaction and derives an expression for the Hubble constant that is consistent with observation when published parameters are substituted.
I see that you are in denial.
Your "first principles" are wrong, as it has been exhaustively explained on this board.

Quote:
Originally Posted by lyndonashmore
Can the BB do that? I think not.
Papageno et al, say that this is a coincidence because they cannot bring themselves to admit that they are wrong.
[Dr. Evil] Riiiiiiight![/Dr. Evil]
There is not much we can do if you simply ignore criticism.

Quote:
Originally Posted by lyndonashmore
Let’s look at one or two more ‘coincidences’ shall we.
Let’s start with the wavelength at which the intensity of the CMB curve peaks. This is at a wavelength of 2.1x10^-3metre.
Modern cosmology can explain the whole spectrum. Can you?

Quote:
Originally Posted by lyndonashmore
Using my theory we can calculate the wavelength of the incoming photon that, in being redshifted, causes such a CMB photon to be emitted, it is 2.46x10^-15m and a frequency of 6.05x10^15Hz (in the UV). This has an energy (E = hf) of 3.87x10^-18J.
I wonder what the temperature of a plasma cloud would be if, on average, each electron had this much energy?
We can work it out by 3.87x10^-18J = 3kT/2 and the temperature comes out to be 1.2x10^5K.
Shock, horror, this is the temperature of the plasma of Intergalactic space. What a ‘coincidence’! That is the watershed where the CMB has its maximum intensity corresponds to the situation where the incoming photon has the same energy as the average KE of the electrons in the plasma of IG space.
Ah yes, your "watershed" idea.
Unfortunately for you, it has been shown to be non-sense.


Quote:
Originally Posted by lyndonashmore
But wait a minute. There is another coincidence here. People on this site go on about a ‘Compton Effect’.
When the energy of the incoming photon is greater than the average KE of the electrons in the plasma of IG space, the photons are scattered and lose energy to the electrons.
When the energy of the incoming photon is less than the average KE of the electrons in IG space then here the photons gain energy (inverse Compton scatter).
Again, the momentum (and kinetic energy) of the electron does not affect the Compton scattering: if the direction of the photon does not change, the wavelength does not change, whatever the initial momentum of the electron was before scattering.
There is no "watershed".

Quote:
Originally Posted by lyndonashmore
So what happens when the energy of the incoming photons is equal to the average KE of the electrons in the plasma of IG space? Nothing. There is no transfer of energy between the photons and electrons by Compton scatter! There is no Compton scatter for photons of this energy.
Wrong. It is only Compton scattering, there is no other "mechanism".

Quote:
Originally Posted by lyndonashmore
So the watershed where the CMB has its maximum intensity corresponds not only to the situation where the incoming photon has the same energy as the average KE of the electrons in the plasma of IG space but also to the point where there is no Compton scatter. Then there is being able to predict H of course.
And yet Papageno et al say there is nothing in all of this.
Dream on folks.
You are the only one here living in a dream.
It is a pity that refuse reality.
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  #74 (permalink)  
Old 12-May-2005, 03:46 PM
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Quote:
Originally Posted by papageno
You are the only one here living in a dream.
It is a pity that refuse reality.
Wrong Papageno, there are at least two of them. Oh no, it's contagious
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  #75 (permalink)  
Old 12-May-2005, 03:47 PM
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Quote:
Originally Posted by lyndonashmore
Papaggeno, not you too.
Quote:
It is not a macroscopic electromagnetic wave.
What happens to energy and momentum if the electron absorbs the photon?
Sylas has shown the math: both cannot be conserved in that case.
But you did not address the math. You preferred another way to avoid the issue.
Sylas' maths is totally incorrect. I showed it in my frog post.
No, you called him a half-wit, you did not provide evidence that he is wrong.
I strongly suspect that you actually do not understand his argument, hence you had to resort to some other way to avoid the issue.

Quote:
Originally Posted by lyndonashmore
He has messed up the vector nature of momentum by including two interactions in opposing directions in one overall calculation.
Where is the math to support your point? Show us.
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  #76 (permalink)  
Old 12-May-2005, 03:49 PM
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Quote:
Originally Posted by Fram
Quote:
Originally Posted by papageno
You are the only one here living in a dream.
It is a pity that refuse reality.
Wrong Papageno, there are at least two of them. Oh no, it's contagious
Unless akirabakabaka is just an alter-ego of Ashmore, I still have some hope that he will try to learn.
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Old 12-May-2005, 03:56 PM
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Quote:
Originally Posted by Fram
I'll repeat:
Quote:
Quote:
If we convert the Hubble constant into SI units, 64 km/s per Mpc becomes 2.1exp(-18 ) per sec.
Where have all the units gone? Only the seconds are left!
What isn't clear about this? What is the unit for the 2.1exp(18 )? Why isn't it included?
Have you ever looked at Hubble's Constant and thought "what the heck is km/s/Mpc?" These strange units are only used in the context of Hubble's Constant and are derived from the assumption that space is expanding. They have no meaning without this assumption.

Let's look at the FAQ:
Quote:
Originally Posted by FAQ
2) Why the 'per cubic metre of space?

For two quantities to be equal to each other, they must have the same magnitude and the same units. The Hubble constant is a 'per sec' (s-1) and so our function of the electron must also be a 'per sec' (s-1). For this to be so we must look at 'hr/m in each cubic metre of space' as this has the same units i.e. 'per sec' (s-1). What we are saying in Tired Light is that the Hubble constant is equal to 'this much of an electron in each cubic metre of space'. In actual fact when we work it out from first principles, we find that H = 2nhr/m where 'n' is the number of electrons in each cubic metre of Intergalactic space. This is where the 'per cubic metre of space' comes from - it is the number density of electrons, the more electrons we have the more collisions and so (up to a point), the greater ther value of H.
Ashmore's model does not use these units, because it reinterprets Hubble's redshifts. Now we have a different assumption, instead of expanding space we say that light loses energy as it hits electrons. His model relies on electron densities in IG space, so now we are looking at number of electrons per cubic meter (instead of the rate at which space expands).

Quote:
Originally Posted by Fram
I know the answer; because he has to match it with his other quantity, which has completely different units. In the second case, he doesn't only remove units, he even adds some (where do the seconds come from?)! In what way are a mass and a distance related?
I have read his explanations, and they don't explain what he does there and why that would be allowed in physics. It's just gibberish, numerology.
I have tackled this once before in another thread. The values 72km/s/Mpc and 0.6Ne/m^3 are derived from mutually exclusive theories (actually they are both measured observationally). If we convert both values to SI units, we discover they have matching dimensions, so we can make a comparison between them. To convert from one to the other is actually meaningless, except to show that both theories give the same answer. But if you do the math both sides have the same units, in the same way that 1m/s = 100cm/s.

Quote:
Originally Posted by Fram
It has nothing to do with being able to follow the math. I have not said that his number of 2.1exp(18 ) is right or wrong, and that's the math part. It seems to me that you on the other hand cannot follow the logic needed to make or understand a coherent theory that's based in reality. (you see, it's not so hard to dismiss people's arguments by claiming that they aren't smart enough, even I can do it: but it's not very productive, so please refrain from it in the future).
I didn't mean to offend you, but you are asking trivial questions that are answered in the first section of Ashmore's paper, and claiming he is wrong because you don't understand it. I humbly suggest reading the material more thoroughly. Please ask questions if you need to, that's what this thread is for, but let's not all assume we have the answers before we've seen the material.
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  #78 (permalink)  
Old 12-May-2005, 04:03 PM
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Quote:
Originally Posted by akirabakabaka
I didn't mean to offend you, but you are asking trivial questions that are answered in the first section of Ashmore's paper,...
These trivial questions raise the issue of whether Ashmore has a basic understanding of physics.
He has been found to be severly lacking.

Quote:
Originally Posted by akirabakabaka
... and claiming he is wrong because you don't understand it.
Claims supported by evidence.

Quote:
Originally Posted by akirabakabaka
I humbly suggest reading the material more thoroughly.
Do us the same courtesy.
You could start with Feynman's Lectures on Physics.
These should make you realize how wrong Ashmore is.

Quote:
Originally Posted by akirabakabaka
Please ask questions if you need to, that's what this thread is for,...
We did, but Ashmore prefers avoiding the issues and mindlessly repeating his claims.

Quote:
Originally Posted by akirabakabaka
...but let's not all assume we have the answers before we've seen the material.
You assume that we did not read Ashmore's writings.
Unfortunately for him we did, and we can see what is wrong with them.
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  #79 (permalink)  
Old 12-May-2005, 04:05 PM
akirabakabaka akirabakabaka is offline
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Quote:
Originally Posted by papageno
Quote:
Originally Posted by akirabakabaka
Take a look at the website already.
Do you think that I have not?
You asked a question that was specifically addressed by Ashmore on his website, so no, I figured you may not have seen it.

Quote:
Originally Posted by papageno
Since you seem convinced that Ashmore is right, you should not have a problem explaining his "theory" without resorting to quotes.
I am sorry that you feel you are being attacked or that I have sided with lyndonashmore because I made you aware of material that you asked about.

Quote:
Originally Posted by papageno
Quote:
Originally Posted by Ashmore's Tired Light Theory
This is clearly not the case. The reason being that the collision cross-section is proportional to the wavelength and each time a photon interacts with an electron it is redshifted and undergoes an increase in wavelength. That is, as the photon travels along and is absorbed and re-emitted by the electrons in the plasma of intergalactic space, the wavelength will get bigger and bigger and so the chances of it making a collision will also become bigger and bigger. The distance travelled by the photon between collisions becomes less and less. The rate at which the photon is redshifted becomes greater and greater as the photon travels further and further.
Still the same problem: Ashmore still wrongly assumes that he can apply that formula to plasma.
He's not even talking about the mechanism here! He is saying that if we assume that the mechanism is correct, then this is what would happen. We know you have decided he is fundamentally wrong. That is completely besides the point. We can still discuss the theory, can't we? This explanation by Ashmore is internally consistent with his theory, that is all that matters.


Quote:
Originally Posted by papageno
Did you actually read the following posts, or even just this thread?
Do you actually understand my criticism to Ashmore's ideas?
How many times do we need to explain that the "mechanism" describes does not happen?
Why don't you address this post?
We all understand wonderfully that you have decided the theory is wrong. If you think that continually stating your position is going to brainwash anyone into believing you.. well.. I think it's worked actually, but not on me.

If anyone can show a fault in Ashmore's math, then please do so. So far the only objection is that he is misapplying theorems. This may be technically true, but by doing so he is able to predict quite a few things to impressive degrees of accuracy. So maybe this means he is not misapplying them afterall.
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Old 12-May-2005, 04:12 PM
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Quote:
Originally Posted by papageno
Quote:
Originally Posted by lyndonashmore
So what happens when the energy of the incoming photons is equal to the average KE of the electrons in the plasma of IG space? Nothing. There is no transfer of energy between the photons and electrons by Compton scatter! There is no Compton scatter for photons of this energy.
Wrong. It is only Compton scattering, there is no other "mechanism".
Compton scattering only applies to bound electrons, and implies a change of direction. Our IG plasma photon-electron collision has neither of these conditions, so it can't be Compton. Have you thought of Raman scattering? I wonder how much of Ashmore's theory overlaps Jerry's CREIL.
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  #81 (permalink)  
Old 12-May-2005, 04:12 PM
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Quote:
Originally Posted by akirabakabaka
We all understand wonderfully that you have decided the theory is wrong.
Yes I did, and I explained in length where it is wrong.

Quote:
Originally Posted by akirabakabaka
If you think that continually stating your position is going to brainwash anyone into believing you.. well.. I think it's worked actually, but not on me.
You are projecting Ashmore's behavior on me.

Quote:
Originally Posted by akirabakabaka
If anyone can show a fault in Ashmore's math, then please do so.
What math? His unbalanced units in his first "paradox"?

Quote:
Originally Posted by akirabakabaka
So far the only objection is that he is misapplying theorems.
And equating atoms to free electrons.

Quote:
Originally Posted by akirabakabaka
This may be technically true, but by doing so he is able to predict quite a few things to impressive degrees of accuracy.
He still has not provided the spectrum of the CMB.

Quote:
Originally Posted by akirabakabaka
So maybe this means he is not misapplying them afterall.
I guess you won't bother to read anything besides Ashmore's writings, or maybe do some research on your own.
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  #82 (permalink)  
Old 12-May-2005, 04:14 PM
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Quote:
papageno
No, you called him a half-wit, you did not provide evidence that he is wrong.
I strongly suspect that you actually do not understand his argument, hence you had to resort to some other way to avoid the issue.
I called him no such thing.
I aked him to 'imagine one stood at the side of the pond' - there was no actual reason to edit the post.
I showed where his maths is incorrect in the analogy to the frog.
Cheers,
lyndon
  #83 (permalink)  
Old 12-May-2005, 04:18 PM
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Quote:
Originally Posted by akirabakabaka
Quote:
Originally Posted by papageno
Quote:
Originally Posted by lyndonashmore
So what happens when the energy of the incoming photons is equal to the average KE of the electrons in the plasma of IG space? Nothing. There is no transfer of energy between the photons and electrons by Compton scatter! There is no Compton scatter for photons of this energy.
Wrong. It is only Compton scattering, there is no other "mechanism".
Compton scattering only applies to bound electrons, and implies a change of direction.
Wrong.
If you look it up, Compton scattering is observed in materials with X-rays. I already explained in this very thread why.
The theory explaining the Compton effect (simply conservation of momentum and energy) deals with a free electron, and the case where the photon does not change direction is considered (no loss of energy, hence no change in frequency)

Quote:
Originally Posted by akirabakabaka
Our IG plasma photon-electron collision has neither of these conditions, so it can't be Compton.
So, now the electrons in a plasma are no longer affected by "restraining forces".

Quote:
Originally Posted by akirabakabaka
Have you thought of Raman scattering? I wonder how much of Ashmore's theory overlaps Jerry's CREIL.
So, even you cannot distinguish between free electrons and atoms.
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  #84 (permalink)  
Old 12-May-2005, 04:24 PM
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Quote:
Originally Posted by lyndonashmore
Quote:
papageno
No, you called him a half-wit, you did not provide evidence that he is wrong.
I strongly suspect that you actually do not understand his argument, hence you had to resort to some other way to avoid the issue.
I called him no such thing.
You were not very subtle, don't think you can fool us.

Quote:
Originally Posted by lyndonashmore
I aked him to 'imagine one stood at the side of the pond' - there was no actual reason to edit the post.
I showed where his maths is incorrect in the analogy to the frog.
You did not.
He wrote down the equations for conservation of energy and momentum, and showed that an "absorption" of the photon by the electron leads to an unphysical situation.
You replied with an analogy (no, no equations), which does not work for a plasma.


EDIT to fix quote attribution.
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  #85 (permalink)  
Old 12-May-2005, 04:35 PM
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Quote:
Originally Posted by akirabakabaka
He's not even talking about the mechanism here! He is saying that if we assume that the mechanism is correct, then this is what would happen. We know you have decided he is fundamentally wrong. That is completely besides the point. We can still discuss the theory, can't we? This explanation by Ashmore is internally consistent with his theory, that is all that matters.
First problem with this is that there is a huge fudge factor—n, the density of the electrons in IG space. The value of n is not well known—and the methods we use to calculate it now derive from the big bang model and expansion. Even assuming that the mechanism exists, he’s right about n, and he’s right about the rate of collisions based on his cross section equation, Ashmore has not proven that the “bremsstrahlung” radiation produces the spectrum of the CMB, merely that it is in the correct neighborhood for some frequencies of light. Without the spectrum, it is numerology. I ran a few simulations myself, and didn’t get anything resembling the CMB, but I’m not convinced I’ve taken everything into account.

Whether or not Ashmore’s equations accurately model observed redshift depend entirely on the selection of n, regardless of the correctness of his mechanism—and an error in his collision cross section equation would completely change the result. It is again numerology.

Quote:
If anyone can show a fault in Ashmore's math, then please do so. So far the only objection is that he is misapplying theorems. This may be technically true, but by doing so he is able to predict quite a few things to impressive degrees of accuracy. So maybe this means he is not misapplying them afterall.
Er, math can be intrinsically correct and still have nothing to do with the physical world. The theories that Ashmore misapply are the only link between the math and reality—misapplying them makes the correctness of the math irrelevant.
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Old 12-May-2005, 04:40 PM
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papageno
He wrote down the equations for conservation of energy and momentum, and showed that an "absorption" of the photon by the electron leads to an unphysical situation.
He showed no such thing. I will not waste my time going through bad math, but for one thing, he appears to have difffering solutions for newtonian and relativistic calculations when they should both agree.
Thye maths is wrong because he has neglected to take into account that there are two recoils here in opposite directions. Sylas is looking at the whole thing and is surprised when they cancelled.
No Sylas is wrong in basic maths.
Chers,
Lyndon
  #87 (permalink)  
Old 12-May-2005, 04:47 PM
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Quote:
Originally Posted by lyndonashmore
Quote:
Originally Posted by papageno
He wrote down the equations for conservation of energy and momentum, and showed that an "absorption" of the photon by the electron leads to an unphysical situation.
He showed no such thing.
He showed that you cannot have conservation of both momentum and energy, didn't he?
So, where exactly is he wrong?

Quote:
Originally Posted by lyndonashmore
I will not waste my time going through bad math, but for one thing, he appears to have difffering solutions for newtonian and relativistic calculations when they should both agree.
Show us that they should agree.

Quote:
Originally Posted by lyndonashmore
Thye maths is wrong because he has neglected to take into account that there are two recoils here in opposite directions.
You just showed that you do not understand conservation of momentum.

Quote:
Originally Posted by lyndonashmore
Sylas is looking at the whole thing and is surprised when they cancelled.
No Sylas is wrong in basic maths.
Provide the evidence.
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Old 12-May-2005, 04:51 PM
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Quote:
Originally Posted by lyndonashmore
He showed no such thing. I will not waste my time going through bad math,
Why not. If it's so bad it should be easy to show, in the equations, where it's wrong. Or is it you can't show where Sylas' math is wrong, and that's why you have to keep resorting to analogies hoping nobody notices you haven't or can't show where Sylas' equations are wrong?

Quote:
Originally Posted by lyndonashmore
he appears to have difffering solutions for newtonian and relativistic calculations when they should both agree.
Just appears? You can't show it?

Quote:
Originally Posted by lyndonashmore
Thye maths is wrong because he has neglected to take into account that there are two recoils here in opposite directions.
So..... where is your QM equations explaining the two recoils.

How about your full CMB spectrum that you were asked to provide more than six months ago?
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Old 12-May-2005, 04:58 PM
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Quote:
Originally Posted by lyndonashmore
I will not waste my time going through bad math, ...
That explains a lot.
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Old 12-May-2005, 05:09 PM
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Thanks for jumping in with some fresh thoughts, Demigrog

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Originally Posted by Demigrog
First problem with this is that there is a huge fudge factor—n, the density of the electrons in IG space. The value of n is not well known—and the methods we use to calculate it now derive from the big bang model and expansion.
Are you sure about that last point? I havn't come across anything that suggests this, but I'm not sure. However I have seen plenty of estimates of n and they are all between 0.1 and 10 per cubic meter, and this is just one measurement that is needed, compared to the multitude of isotope ratios required for BBN (systematic errors), so this isn't a strong objection to the model.

Quote:
Originally Posted by Demigrog
Even assuming that the mechanism exists, he’s right about n, and he’s right about the rate of collisions based on his cross section equation, Ashmore has not proven that the “bremsstrahlung” radiation produces the spectrum of the CMB, merely that it is in the correct neighborhood for some frequencies of light. Without the spectrum, it is numerology. I ran a few simulations myself, and didn’t get anything resembling the CMB, but I’m not convinced I’ve taken everything into account.
I'm not sure why this is a killer either. Even if it turns out the microwave radiation from Ashmore's model doesn't explain the CMB 100%, this has little effect on the rest of the theory. The claim that the CMB is entirely leftover from the BB itself seems kind of short sighted.

Quote:
Originally Posted by Demigrog
Er, math can be intrinsically correct and still have nothing to do with the physical world. The theories that Ashmore misapply are the only link between the math and reality—misapplying them makes the correctness of the math irrelevant.
This is a great point, but all of the same objections apply to expanding space as well.
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