I have already pointed you at pages that explain this, but I'll have another go.

Take a wire of 1 m length with a 1 A current going through it. Now let's place another similar wire 1 m away. This corresponds to your current belief. You can calculate the force on the first wire due to the magnetic field generated by the second wire. Now let's increase the length of the second wire. By your reasoning, there is no change in the magnetic field at the first wire. This is wrong as the extra section of the second wire also generates a magnetic field, which adds vectorially to the original magnetic field, thus increasing the field at the first wire. As the field has increased, the first wire will experience a greater force on it, and yet it is still only one metre in length. Now do you see why you can't just apply the infinite solution to the case that you're considering?

At which point? Please read a decent introduction to electromagnetism, and you'll find out what is going wrong with your model.

Again, may I refer you to a decent book on EM. A tiny loop of current produces a magnetic dipole. It has nothing to do with magnetic monopoles. It is a part of elementary electromagnetism.

So all of this, presumably inifinite, negative electric charge is always there?

Not really. A moving magnetic field generates (via Faraday's law, or the appropriate bit of Maxwell's equations) an electrostatic field. It is the electrostatic field that causes static charges to move, not the magnetic field. Look up the Lorentz force to find out how charges move under combinations of electrostatic and magnetic forces.

Odd, other virtual particles have normal charges. As you have said that you don't know their charges, how do you know that they are small?

Where does the charge go? If I illuminate a positively charged black sheet of metal with light, why doesn't the sheet lose positive charge?

You can't just assume a 1 kg mass because 1 N is the force required to accelerate 1 kg by 1 ms^-2. It is also the force required to accelerate 0.1 kg by 10 ms^-2. You need to know the mass that you're accelerating otherwise it is meaningless.

If you don't know these values then you can't calculate the acceleartion due to a magnetic force either. (Note that you will also neeed to know the particle's velocity.)

A naive averaging of v and v/2 would give an answer of 3/4. The change in velocity would be v/2. This still doesn't explain your factor of 1/3?

You haven't answwered the question. You're trying to use this model to explain the redshifts. The values that you get for the redshifts must depend on the input parameters for your model, so if you know what the measured redshifts are you might be in with a chance of determining what the input parameters of your model must be in order to match the observables.

By the way, Michael, if you use the expression that I give for the magnetic force between two, non-relativistic, particles moving with a velocity v, you can very nicely see how the magnetic force is much weaker than the force due to the electric field.

Magnetic field:
F_m=e^2.v^2.(mu_0/(4.Pi))./r^2

Electric field:
F_e=e^2/(4.Pi.epsilon_0.r^2)

Dividing one by the other gives

F_m/F_e = mu_0.epsilon_0.v^2

but mu_0.epsilon_0 = 1/c^2, where c is the speed of light in a vacuum, so

F_m/F_e = v^2/c^2

The reason that we think that forces due to magnetic fields can be comparitively high, is typically because the moving charges tend to be embedded in a conductor with a net charge of zero. This is not the case for the scenario that you have proposed.

cyrek reply

These long posts take too much time to edit, so I have just included the more prevalent material in this reply.

fortis Quote: (excerpts)
In the non-relativistic regime, the equation below gives you the
magnitude magnetic field due to a point charge (with charge q) moving
with a constant velocity, v. B = (mu_0 /4Pi)*(q*vxr/r^3) (If you don't
want to derive it, it can be found in a number of places, such as,

http://cpt.phys.utk.edu/~th/Physics231/Lecture08.pdf )

The variables in bold are vectors, and the "x" is the vector cross
product. Now if we have another charge roaming around, then the force
exerted on it due to the magnetic field of the first charge is given
by F=q_2*v_2xB
If you want to put your theory on anything like solid foundations,
you should start with this. For example, if we have two electrons
moving in parallel with identical velocities, v, and seperated by a
distance r, then the magnitude of the force between them is given by
F=e^2.v^2.(mu_0/(4.Pi))./r^2 where e is the charge on an electron.
Notice how the force goes as 1/r^2, not 1/r as is implied by the
equation that you are trying to use.

fortis wrote
How did you go from your first incorrect result to a force in Newtons?
What value did you use for the second current?

cyrek reply
The Ampere Rating Formula (ARP) uses one ampere in both wires.
The formula I used was based on the (ARF) and I assumed the setup was
for one Newton on the left side. Instead, the left side is equal to
2x10^-7 Newtons which is considerably smaller (10^-7 N/A^2).

Reducing the ARF down to the atomic level:
I use one wire which is the electron in the HA rated at one Coulomb
(fraction of an Ampere), one wire length (force length in meters) which
is the length of the electron transitional path in the atom and the
extent of the magnetic field between the wires as the half wavelength
of the pulse (extent of fielf strength) which is 4.86x10^-7 meters.

However these calculations are questionable as being rated in Newtons
since the nature of the electric field is not known.

The binary nature of the HA is given by two formulas.

The radius in ground state (GS) = h^2 / 4pi^2xm(sub)exk(sub)exe^2
The velocity in GS = h / 2pixR(sub)nxm(sub)e

h=6.626x10^-34 J/s k(sub)e=8987x10^9 N-m^2/c^2
m(sub)e=9.109x10^-31 kgs e=1.602x10^-19 C

Outer orbit radiuses are = to R(sub)n = R(sub)n x n^2 n=orb. number
Velocity = V(sub)n = V(sub)1 / n

I relied on this data for my calculations.

cyrek Quote:
The virtual particles are a theory of electric field formations. They
were introduced to me by a PhD while explaining tha electric field
patterns between the electric charges using the Maxwell equations.

fortis wrote (excerpt)
If you're talking about virtual particles and the EM field then I'm
pretty sure that he would have been referring to virtual photons being
the mediator of the EM field. These photons are neutral, i.e. they
aren't the source of the EM field, that role is reserved for particles
with electric charges.

cyrek reply
The PhD illustrated Virtual Particles with a plus inserted to show that
they had positive charges leaving the main positive charge moving
toward the main negative cherge. From his explanatio, I than presumed
these fields are composed of Virtual Charged Particles.

Quantum Chromdynamics is not relavent here. QCD is nulear, not
atomic.

fortis Quote
QED is the accepted theory of the EM field. Unfortunately it
doesn't correspond to your model

cyrek quote
These are field particles, not the self destructing space
pairs. These VNCP are responsive to magnetic forces just like
electrons are. The photon is the 'condensed congregate' of these
charged particles that would then be repelling each other to return
to their normally distributed state of tranquility. There is no
comservation problem since charged particles are just being condensed,
not added or subtracted.

fortis quote
It is the electrostatic field that causes static charges to move, not
the magnetic field. Look up the Lorentz force to find out how charges
move under combinations of electrostatic and magnetic forces.

cyrek reply
What generates the electricity in a generater?

__________________
aka Michael Cyrek

The EMF (i.e.: voltage) generated by a generator is due to the difference of magnetic flux moving past any given wire in the coil, the electrons are moved by electrostatic forces that the magnetic flux has induced upon the wire. While magnetic fields push on magnetic poles, electric fields push on electric charges. With a coil of wire near a moving magnet, the moving magnet's electric field pushes charges through the coil and eventually through the entire circuit.
If you're talking about a DC generator, you have a commutator (like a non-solid state diode ) and instead of AC current you get pulsating DC current, wich by use of a filter circuit (including an avalanche mode Diode) you can create good `ol constant DC...
Or you could use a diode rectifier bridge...
Most explanations of a generator boil it down so the "layman" can understand it. Don't rely on these distilled answers. LOOK STUFF UP! You have the sum of human knowledge at your fingertips (Internet) but you still insist on making up 'ghost stories.'

__________________

Feynman
>~~~~<
Science is a way of trying not to fool yourself. The first principle is that you must not fool yourself, and you are the easiest person to fool.

Religion is a culture of faith; science is a culture of doubt.

travisM Wrote:


Here's a good illustration showing how an AC current is induced into the wire loop.

I love elctricity. =D>

Michael,
Does your response mean that you accept that your equation was both inappropriate and incorrect? (As I have given you the force due to the magnetic field between two charged particles, it seems unlikely that you would want to hang on to your original derivation. )

I'm still unclear about the origin of your VNCP model of the photon. Are you suggesting that this is what the PhD explained to you, or is it your own creation? (If it is the former, then I believe either you misunderstood what was being explained, or your PhD person lives life at the outer fringe of physical theory.)

QCD is relevant because it shares a similar property to your theory, in that the exchange particle of the strong interaction (the gluon) possesses colour charge in the same way that the photon (which is the exchange particle of the EM force) of your theory possesses electrical charge. In QCD the gluon-gluon interaction leads to some far reaching differences from QED-like behaviour. For example, look at page 7 of
http://www.hep.phy.cam.ac.uk/~thomso...pp2004_qcd.pdf
As your "photons" carry electric charge, you would also expect that the predictions for this theory would look dramatically different to those of QED. As QED works extremely well (I heartily recommend getting a copy of
QED by Richard Feynman which is a very good, non-mathematical description of QED for the layman.
[edit]Fixed "screen stretcher", though it looked fine in IE6... [/edit]


As for the dynamo question, please look at
http://hyperphysics.phy-astr.gsu.edu...magfor.html#c2
where you will see that a magnetic field only acts on charges that are moving. (Obviously this is the case when you have a wire moving through a magnetic field. The free electrons on the conductor experience a force as they move through the field.)

For the situation where you have a static conductor in a changing magnetic field, may I refer you to Faraday's law as encapsulated in Maxwell's equations.
http://hyperphysics.phy-astr.gsu.edu...maxeq2.html#c3
Hopefully you can see that a changing magnetic field leads to the creation of an electric field. It is this electric field that acts on the static charges, not the magnetic field.

I believe that I explained this in my previous post.

Please, please, please, read a decent elementary text on electromagnetism. You'll find that it will make your life much easier. [/url]

^^^^^^^^^^^^^^^^^^

Screen Stretcher Screen Stretcher

Maybe it's all those intrinsically expanding photons stretching your screen! :wink:

__________________
Microsoft is over if you want it.

The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.

cyrek reply

fortis quote
Michael,
Does your response mean that you accept that your equation was both
inappropriate and incorrect? (As I have given you the force due to the
magnetic field between two charged particles, it seems unlikely that
you would want to hang on to your original derivation. )

reply
I appreciate your help but my interest is in applying the hydrogen
atom radiations and the characteristics of the photon.
The solution I derived in the 10^-29 rnnge is correct but useless
when applied to free space.
So I have changed my objective now to Euclidean flat space and
correlation with 'angular diameters' and luminosity radiation. See
below.

fortis quote
I'm still unclear about the origin of your VNCP model of the photon.
Are you suggesting that this is what the PhD explained to you, or is
it your own creation? (If it is the former, then I believe either you
misunderstood what was being explained, or your PhD person lives life
at the outer fringe of physical theory.)

reply
His reply used the 'Virtual positive charged particles', VPCP so I
changed that to negative charged particles.
I accepted his idea because these fields are composed of something
real rather than something abstract.

If you think his reply was unscientific, than I would like to ask you
what you think is the composition of these electric and magnetic
fields? They are real and must be composed of real substance.

Fortis quote (regarding QED's

http://www.amazon.com/exec/obidos/tg...691024170/qid=
1106193904/sr=8-5/ref=pd_csp_5/104-5320640-2118324?v=glance&amp;s=books&amp;n
=507846

which is a very good, non-mathematical description of QED for the
layman.

reply
I want to concentrate now on flat space dimensions as to resolving the
cosmological redshift relation to distance with M87 as the distance
indicator.
M87 is the central elliptical giant galaxy in Virgo.
With a redshift of .004, this places it at a distance of 54x10^6 light
years with a Hubble constant of 72 km/s/mpc.
Its angular diameter at that distance is 9 minutes of arc.
Mulltiply that by 60 and you get 540 seconds times 54x10^9 =
30 billion ys
for one second of arc for the tiny specks in the HDFN. These are
SSU dimensions.
Its luminosity determined at the ratio of 540 at that distance at a
visible magnitude of about 9+, would give it a magnitude of 21+.

540^2 = 290,000 reduction ratio for its luminosity at that distance
as the inverse square.
This reduction is equal to about 12 magnitudes plus 9+ = 21+.
This does not correlate too well since those tiny specks were estimated
to be in the range of 30th magnitude.
This will require further research.

I have been reviewing your past posts and have womdered about this
statement regerding magnetic influence on virtual negative particles:

fortis quote from past post:
Not really. A moving magnetic field generates (via Faraday's law, or
the appropriate bit of Maxwell's equations) an electrostatic field.
It is the electrostatic field that causes static charges to move,
not the magnetic field. Look up the Lorentz force to find out how
charges move under combinations of electrostatic and magnetic forces.

cyrek reply
Does this reply violate the 'conservation of electric charges'?
The moving MF has generated new electric charges, right?

__________________
aka Michael Cyrek

Just a very quick one, but
Riddle me this, "What are electrons made of?"

(In QED, the EM field is due to the coupling of virtual photons with electric charges. Have a read of QED, you may enjoy it. )

Just buy me a virtual beer sometime...

It isn't correct in the context that you're trying to apply it in even if you are looking for a force between two moving charges for the reasons that I previously gave.

The only way that virtual positively charged particles come into QED is in the context of things like the little electron-positron loops (or particle-antiparticle loops). Charge is always conserved at the verticies of the Feynam diagrams, so for a positive charged particle to appear, an equivalent negatively charged particle (obeying the other conservation laws) must also appear. I suspect that you either misunderstood what he was saying, or he was way beyond the fringe.

The moving MF has not generated new electric charges. In a material such as a metal, there exist a whole hoard of conduction electrons, whose charges are balanced by the positive charges of the lattice (a full description would require a discussion of QM as it relates to solid state physics, but this should suffice for the current explanation.) When an electric field is applied, it causes these electrons to move (the lattice, that isn't free to move, doesn't.)

Take an even easier example, let's consider an electron at rest (relative to the observer.) A changing magnetic field induces an electric field. This field causes the electon to experience a force, and if it is free, then it will accelerate. I don't understand how you are concerned about conservation of charge in this instance.

Notice that these are quite different scenarios to your VNCP model, so you still have a charge conservation problem.

cyrek reply

fortis wrote
Riddle me this, "What are electrons made of?"

reply
You are changing the subject but the answer is matter.
If you want a further beakdown, then the inventor of quarks could answer that.

Fortis wrote
(In QED, the EM field is due to the coupling of virtual photons with electric charges. Have a read of QED, you may enjoy it. )

reply
Virtual photons? What wavelength are they?
And why do you repudiate the 'virtual charged photon congregates' that I proposed exist? They have specific wavelengths too.

Ha Ha

__________________
aka Michael Cyrek

cyrek reply

fortis wrote (excerpt)
Take an even easier example, let's consider an electron at rest (relative to the observer.) A changing magnetic field induces an electric field. This field causes the electon to experience a force, and if it is free, then it will accelerate. I don't understand how you are concerned about conservation of charge in this instance.

reply
According to QM, new particles have to be in pairs. You have introduced a new negative field into use. Besides, how do we know that this negative field is moving the electron rather than the MF?

fortis
Notice that these are quite different scenarios to your VNCP model, so you still have a charge conservation problem.

reply
Not true. I already explained that the virtual field particles compose the field so they are already there.
The magnetic photon just condenses them to a congregation of condensed particles. No charge violation.

__________________
aka Michael Cyrek

Same problem as before as mentioned by Fortis. Ok, ok, lemme keep up here. Now they're "Positively" charged virtual particles. If you shine that same theoretical light does that same theoretical plate become positively charged? You claim no charge violations?

[edit to add]

Just how do these virtual particles get to be positively charged? Are they anti-virtual photons/electrons i.e.: virtual photinos/positrons?

__________________

Feynman
>~~~~<
Science is a way of trying not to fool yourself. The first principle is that you must not fool yourself, and you are the easiest person to fool.

Religion is a culture of faith; science is a culture of doubt.

And "matter" is made of?
My point was that at some point you have to just say something to the effect of "an electron is an electron" (or "a photon is a photon") and it interacts with its surroundings according to a set of rules. (There is some finessing of this statement when you start talking about "bare" and "dressed" photons, but the jist of it still stands.)

By the way, electrons aren't made of quarks, and as far as anyone can tell, experimentally an electron is just an electron, with no structure beneath it.

What wavelength are "real" photons?
Your VCPCs can have any "wavelength" that you like, but they carry electric charge which means that your photons will interact with each other in the same way that other charged particles interact, except the photon is also the particle that mediates the EM interaction, so the EM interaction now looks very strange. (c.f. gluons in the strong interaction.)

By the way, in your theory photons will strongly repel each other. Can you think of any experimental evidence for this.

Hee Hee! I'm the laughing gnome and you can't catch me.

Where did these "new particles" come into the picture? The electron was already there. The new field was created by the changing magnetic field. As it doesn't possess any charge, etc., it isn't breaking any conservation laws.
How do we know that it is the electric field, rather than the magnetic field, that is doing the moving? You could have a look at the Lorentz force for starters and see that the magnetic term requires the charged particle to have a velocity, in order for it to experience a force due to a magnetic field. Please read an elementary text on EM. I've said it once and I'll say it again, it will make your life a lot easier if you do.

Let's look at the weak interaction, which shares the property with your theory in that some of the virtual particles involved possess electric charge, i.e. the W+ and W-. If you look at the Feynamn diagrams corresponding to various weak processes
(see http://hyperphysics.phy-astr.gsu.edu...funfor.html#c5 )
you can see that charge is conserved, e.g. the beta decay process
a neutron emits a W- and is converted inpto a proton. The W- then decays into a negatively charged electron and an electron anti-neutrino.

Notice how charge is conserved at every point of this, even though the W- is only a virtual particle?

This looks nothing like your theory.

cyrek reply

fortis wrote
cyrek1 quote:
You are changing the subject but the answer is matter.
Added reply (regarding electron structure)

fortis
And "matter" is made of?
My point was that at some point you have to just say something to
the effect of "an electron is an electron" (or "a photon is a photon")

reply
electrons are one of the most studied substances in science. It has
mass (9.11x10^-31 kgs), charge and even physical size.

fortis (excerpt)
experimentally an electron is just an electron, with no
structure beneath it.

cyrek reply
See above.

cyrek Quote:
Virtual photons? What wavelength are they?
And why do you repudiate the 'virtual charged photon congregates'
that I proposed exist? They have specific wavelengths too.

fortis quote
What wavelength are "real" photons?
Your VCPCs can have any "wavelength" that you like, but they carry
electric charge which means that your photons will interact with
each other in the same way that other charged particles interact,
except the photon is also the particle that mediates the EM
interaction, so the EM interaction now looks very strange.
(c.f. gluons in the strong interaction.)

cyrek reply
The last portion of the last paragraph above is Finemans theory that
does not have anything to do with my idea of photons.
You also say the photon is a particle which it is not.

fortis quote
By the way, in your theory photons will strongly repel each other.
Can you think of any experimental evidence for this.

cyrek reply
Photons are not that dense in space and move at the velocity of light
so coincidences of their interactions would be rare.

fortis quote
cyrek1 quote:
According to QM, new particles have to be in pairs. You have
introduced a new negative field into use. Besides, how do we know
that this negative field is moving the electron rather than the MF?

fortis quote
Where did these "new particles" come into the picture? The electron
was already there. The new field was created by the changing magnetic
field. As it doesn't possess any charge, etc., it isn't breaking
any conservation laws.

fortis wrote
cyrek quote
How do we know that it is the electric field, rather than the magnetic
field, that is doing the moving? (added - pertaining to the VP's)

fortis
You could have a look at the Lorentz force for starters and see that
the magnetic term requires the charged particle to have a velocity,
in order for it to experience a force due to a magnetic field.
Please read an elementary text on EM. I've said it once and I'll say
it again, it will make your life a lot easier if you do.

cyrek reply
The electron in the HA is always moving so its electric field is
fluctuating with its motion.
This is called the 'standing wave'.

__________________
aka Michael Cyrek

*Gasp* could it be moving in a 'cloud' around the atom? Could it be moving so fast (light speed) that you couldn't accurately pin down it's position at any given moment? *gasp* Sounding a little like the currently accepted theories... =D> I'm impressed... kinda...

__________________

Feynman
>~~~~<
Science is a way of trying not to fool yourself. The first principle is that you must not fool yourself, and you are the easiest person to fool.

Religion is a culture of faith; science is a culture of doubt.

cyrey reply

Travis
Are you kidding? If you are, then at least put a ha ha at the end.

I posted back in one of my replies that the electron orbital velocity in the ground state is - 2.19x10^6 m/s.

__________________
aka Michael Cyrek

And what is it's orbital angular momentum?

What do you think the size of an electron is? In the context of what you asked, a photon possesses a frequency, an energy, a momentum. What else should it have in order that it is as real as an electron?
QED gives results that are in extremely good agreement with experiment. It is interesting that you seem to be suggesting that your photons are not the exchange particle of the EM field, which obviously contradicts the picture from QED. The reason that I say this is that presumably your EM field is just the classical Maxwellian EM field. Don't you think it strange (or coincidental) that this field supports EM waves that propagate at exactly the same velocity as light?
If you're not quantising the EM field (which results in QED), what exactly do your photons do?
Are they massless?
Do the propagate with a velocity c, that is invariant between inertial reference frames?
As they possess negative charge, are they deflected by an electric field?
Does a magnetic field cause them to go round in circles (as the magnetic force on a charged particle is perpendicular to its velocity vector)?
Do we see any of this sort of behaviour on the Sun, given that it is a good source of light that also possess pretty strong magnetic fields?
If you put a lightbulb in an electric field, do you see the light redshifted on one side, and blueshifted on the other?

As for the particles bit, it is you that is claiming that photons consist of negatively charged particles.

We can make some pretty bright sources on the earth, and focus those beams pretty well. Are you saying that the photon density would still be too small?

You don't seem to have responded to this bit.

Your response doesn't seem to have anything to do with what we are talking about. We weren't talking about the hydrogen atom. In fact in my previous post (if you check the quote) you'll see that I said,
"Take an even easier example, let's consider an electron at rest (relative to the observer."

You then asked how you could tell that it was an electric rather than a magnetic force that moved the electron, and I pointed out the behaviour of the Lorentz force. Why bring in the hydrogen atom when it isn't involved?

cyrek reply

Travis wrote
Same problem as before as mentioned by Fortis. Ok, ok, lemme keep up here. Now they're "Positively" charged virtual particles. If you shine that same theoretical light does that same theoretical plate become positively charged? You claim no charge violations?

Just how do these virtual particles get to be positively charged? Are they anti-virtual photons/electrons i.e.: virtual photinos/positrons?

reply
The positive charged virtual particles were introduced when a PhD explained the nature of the expanxion in the electric field patterns. You must have seen those in physics books. He used these to show how they complied to the Maxwell equations.

So I intriduced 'negative virtual charged particles' NVCP that would surround the electrons field.
Then the electron transitions create magnetic pulses that create the 'compacted congregate' of these particles that would constitute the photon.

fortis then used his experiment to refute my hypothesis by introducing the positive charges with his questioning.
These NCVP do not have anything to do with positive charges.
He obviously confused them with the 'self annihalating virtual pairs' that spontaneously appear and disappear in space.

To answer the question of what happens when these light pulses are absorbed by an object, these NCVP's disperse as heat by transferring their energy to the electrons in the object.
Does that answer your question Travis?

Also, you have to remember that all this is a hypothetical situation.

__________________
aka Michael Cyrek

cyrek reply

fortis wrote
cyrek1 wrote:
I posted back in one of my replies that the electron orbital velocity in the ground state is - 2.19x10^6 m/s.

And what is it's orbital angular momentum?

reply
Why do you need the orbital momentum?

On page 3, and dated Jan. 19, I posted all the pertinent data for my calculations.

If tou want the OAM, then use the velocity in the GS and the mass of the electron and you can have the OAM. Check out that data.

__________________
aka Michael Cyrek

The reason for asking about the orbital angular momentum is that your model suggests that the ground state possesses a finite orbital angular momentum (i.e. you have an electron with mas m_e, orbiting the nucleus in a circular orbit at a velocity v, and an orbital radius r_e, giving an orbital angular momentum of m_e x v x r_e.)

Sadly for this model, the ground state of the hydrogen atom is an eigenstate of the angular momentum operator (strictly it is the operator for the the square of the angular momentum, L^2) with the eigenvalue of zero, i.e. the ground state has an angular momentum of zero.

cyrek reply

fortis wrote
The reason for asking about the orbital angular momentum is that your model suggests that the ground state possesses a finite orbital angular momentum (i.e. you have an electron with mas m_e, orbiting the nucleus in a circular orbit at a velocity v, and an orbital radius r_e, giving an orbital angular momentum of m_e x v x r_e.)

Sadly for this model, the ground state of the hydrogen atom is an eigenstate of the angular momentum operator (strictly it is the operator for the the square of the angular momentum, L^2) with the eigenvalue of zero, i.e. the ground state has an angular momentum of zero.


reply
Does that mean that the electron is standing still?
Thats what zero implies.

__________________
aka Michael Cyrek

The hydrogen atom ground state is not an eigenstate of either the linear velocity of the electron, or the electron position operator, which means that you cannot neither of these quantities are well defined. If you make a measurement of either of them, then the hydrogen atom ceases to be in an eigenstate of the Hamiltonian, i.e. it can no longer be considered to be in the ground state. On the other hand, the ground state is an eigenstate of the L^2 operator, and hence it does possess a well defined magnitude of the orbital angular momentum. (That value being zero.)

So it is not meaningful to ask what the electron velocity is in the ground state. Hopefully this will answer your question, but it emphasises the difference between the Bohr model and full QM.

As an aside, do you have any comments relating to my earlier queries about your model of the photon?

No. It doesn't answer my questions Mike, in fact, I'm more confused now about your 'hypothetical situation' than ever. :-?

What force particles carry this 'heat radiation' away?

__________________

Feynman
>~~~~<
Science is a way of trying not to fool yourself. The first principle is that you must not fool yourself, and you are the easiest person to fool.

Religion is a culture of faith; science is a culture of doubt.

Just for fun, I thought I'd calculate what the relevant expectation value (i.e. the mean value if you repeated lots of equivalent measurements) was for the hydrogen ground state using "proper" QM rather than the Bohr approach.

The normalised ground state wavefunction for the 1s orbital in hydrogen is given in spherical polars by

Psi(r)=(1/(Pi.a_0^3))^(1/2) . exp(-r/a_0)

where r is the radial distance. One thing to notice is that this this wavefunction is spherically symmetric, i.e. it only depends on the radius, and not any of the two angular variables.

Now we need an appropriate operator. We could use the linear momentum operator to give us a vectorial value for the expectation value of the momentum, but you might expect that to be zero from the symmetry of the situation. Instead we'll look at the "dot product" of the momentum with itself, as this value should always be positive, and we can always square root it at the end of the day, to give us something like what we want.

This operator looks like

p~^2 = -h_bar^2.Del^2

where Del^2 is the Laplacian operator, d=given by

Del^2 = (d^2/dx^2)+(d^2/dy^2)+(d^2/dz^2)

in Cartesian coordinates.

In spherical polars it looks a lot more complicated, but due to the symmetries of the wavefunction two of the three terms are zero, so the only part of the Laplacian that acts upon Psi(r) to give us a non-zero term is given by

Del^2_r = (1/r^2)(d/dr(r^2.d/dr))

All we need now is to stick a bra and ket around either side and away we go.

After a bit of maths you find that the expectation value, &lt;p^2>, is given by

&lt;p^2>=(h_bar/a_0)^2

or taking the square root

&lt;p^2>^(1/2) = h_bar/a_0

where a_0 is the Bohr radius, roughly equal to 0.0529 nm.

If we divide the final answer by the mass, we get the square root of the expectation value for v^2, i.e. I get

&lt;v^2>^(1/2) to be roughly 2.19x10^6 m/s. (Though notice that I didn't use the reduced mass.)

Now you claim that this would mean that the electron possessed an orbital angular momentum. As I'm sure you're aware, the angular momentum contribution comes from the components of momentum that are perpendicular to the radial vector. As the ground state is spherically symmetric, perhaps you can see why it will possess an orbital angular momentum of zero.

Cyrek reply

Travis
I said that the electron field composition that is composed of 'negative charged virtual particles', constitute the photon condensed congregate. When this photon hits a surface that absorbs it, these virtual particles 'rattle' the electrons in that surface to generate the heat waves.
Remember, these are 'negative charged particles'. This is proven because these photons can bounce electrons into outer orbits or in space (Compton effect).

fortis
After all that work with Schroedingers equations(?), you derive the same answer for the GS velocity?
OK, so you create a cloud around the proton called an orbital or a wave function?.
Now what is the diameter of that cloud?
Will it be the same diameter as I have given for the GS radius times two?

__________________
aka Michael Cyrek