Is EM Radiation Present with Free vs Bound Electrons?

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The discussion centers on whether electromagnetic (EM) radiation occurs when two free electrons accelerate due to their mutual repulsion. It is established that EM radiation requires a change in the dipole moment, which does not occur in the case of two free electrons, leading to the conclusion that no radiation is emitted. When electrons are loosely bound to a nucleus, classical electrodynamics suggests they would radiate, but quantum electrodynamics asserts that radiation only occurs with energy state changes. The conversation also touches on scenarios involving collisions and bremsstrahlung, emphasizing that elastic collisions do not produce radiation. Ultimately, the consensus is that while free electrons can radiate under certain conditions, the specific scenarios discussed do not lead to radiation.
Hornbein
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Suppose there are two free electrons. Their mutual repulsion causes each to accelerate. Is there EM radiation?

Suppose those electrons are loosely bound to a nucleus, as in a metal. Is there EM radiation?
 
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Hornbein said:
Suppose there are two free electrons. Their mutual repulsion causes each to accelerate. Is there EM radiation?

Yes.

Hornbein said:
Suppose those electrons are loosely bound to a nucleus, as in a metal. Is there EM radiation?

No, assuming the electrons don't change energy states.
 
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Hornbein said:
Suppose there are two free electrons. Their mutual repulsion causes each to accelerate. Is there EM radiation?

EM radiation requires a change in the dipole moment. There is no change, so no radiation. (Sorry, Drakkith) If you had just one electron accelerate, there would be radiation, but it is canceled by the second one.
 
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Vanadium 50 said:
EM radiation requires a change in the dipole moment. There is no change, so no radiation. (Sorry, Drakkith) If you had just one electron accelerate, there would be radiation, but it is canceled by the second one.

Interesting. I just thought 'accelerated charges' and assumed there was radiation.
 
Vanadium 50 said:
EM radiation requires a change in the dipole moment. There is no change, so no radiation.
Interesting. Is that still true when the two charges are far away from each other? It would seem like the vector sum from Ampere's Law would give a non-zero component in the direction away from the axis of movement...?

https://academic.mu.edu/phys/matthysd/web004/l0220.htm
 
There are some subtleties, but as described, it's zero radiated power.
 
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Vanadium 50 said:
EM radiation requires a change in the dipole moment. There is no change, so no radiation. (Sorry, Drakkith) If you had just one electron accelerate, there would be radiation, but it is canceled by the second one

Bremsstrahlung?
 
hutchphd said:
Bremsstrahlung?

What about it?
 
dipole?
 
  • #10
Vanadium 50 said:
EM radiation requires a change in the dipole moment. There is no change, so no radiation
Isn't there a change in the dipole moment because their distance is increasing?
 
  • #11
Hornbein said:
Suppose those electrons are loosely bound to a nucleus, as in a metal. Is there EM radiation?
According to classical electrodynamics there is EM radiation even in this case, when electrons orbiting the nucleus. However according to quantum electrodynamics there is no EM radiation unless the electrons change states. The electrons orbiting the nucleus is a situation that falls into the realm of quantum electrodynamics, classical electrodynamics are inappropriate to explain well this situation.
 
  • #12
Vanadium 50 said:
EM radiation requires a change in the dipole moment. There is no change, so no radiation. (Sorry, Drakkith) If you had just one electron accelerate, there would be radiation, but it is canceled by the second one.
Delta2 said:
According to classical electrodynamics there is EM radiation even in this case, when electrons orbiting the nucleus. However according to quantum electrodynamics there is no EM radiation unless the electrons change states. The electrons orbiting the nucleus is a situation that falls into the realm of quantum electrodynamics, classical electrodynamics are inappropriate to explain well this situation.
In metals there are many electrons not orbiting a nucleus.
 
  • #13
Hornbein said:
In metals there are many electrons not orbiting a nucleus.
Hmmm so you are interested for the EM radiation from the so called free electrons in good conductors. I think according to classical electrodynamics there is EM radiation due to their random movements, however it is canceled by each other electron (i mean we have a very large number of electrons, each doing a random movement, somehow if we do some sort of statistical analysis we ll find out that the EM radiation is cancelled) so macroscopically there is no EM radiation.
 
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  • #14
Black Body?
 
  • #15
Delta2 said:
Isn't there a change in the dipole moment because their distance is increasing?

No, because the dipole moment is zero at all times.

I am not sure what @hutchphd is trying to get at with the one word replies, but I presume it's the subtlety I tried to avoid. In the problem described, there is no radiation. If I repeat the analysis in the frame of electron #1, it doesn't radiate (it can't), but electron #2 can and does, because in that case the dipole moment is growing with time. But if I repeat the analysis in the frame of electron #2, it doesn't radiate (now it's the one that can't), but electron #1 does, because in that case the dipole moment is growing with time.

So what's up? How can both electrons both radiate and not radiate?

The answer is that the B and E fields are the same (up to a Lorentz transformation) in all three descriptions, but the portion of them attributed to near field and far field (radiation) is different. And in the frame described, there is no radiation.
 
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  • #16
As an aside, neither of the old canards "SR can't handle accelerations" and "photons are hard little balls of energy emitted by radiating objects" works here, and both lead you astray.
 
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  • #17
@Vanadium 50 would like to insist about dipole moment, where is that result from (that when dipole moment is not changing we have no radiation). Is it anywhere in Jackson's (I have 1st edition) or Griffiths (2nd edition).
 
  • #18
Vanadium 50 said:
No, because the dipole moment is zero at all times.
I used this definition of dipole moment for a system of point charges $$\mathbf{p(r)}=\sum q_i(\mathbf{r_i-r})$$ if apply it for a system of two electrons with ##q_1=q_2=e## i get $$\mathbf{p(r)}=e(\mathbf{r_1}+\mathbf{r_2})-2e\mathbf{r}$$ which obviously is not identically zero. Furthermore it will change with time as the position of the two electrons ##r_i## will be changing.
 
  • #19
Drakkith said:
No, assuming the electrons don't change energy states.
I think the Feynman diagram puts it nicely. A photon is 'exchanged' between them but no energy need be lost. (There's no "why" answer here, of course.)
Image from Wikipedia.
1607173275901.png
 
  • #20
I don't have a copy of Jackson at hand because of Covid. But you can find the expression for power radiation from a multipole and see that it depends on the derivatives.
 
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  • #21
Drakkith said:
Interesting. I just thought 'accelerated charges' and assumed there was radiation.
That's correct, but in a bound state (i.e., an energy eigenstate) nothing moves and thus nothing is accelerated, which solves the paradox of the Bohr ad hoc model, in which the classical electrons are simply forbidden to radiate though accelerated only because Bohr demanded it ;-)).
 
  • #22
Vanadium 50 said:
No, because the dipole moment is zero at all times.

I am not sure what @hutchphd is trying to get at with the one word replies, but I presume it's the subtlety I tried to avoid. In the problem described, there is no radiation. If I repeat the analysis in the frame of electron #1, it doesn't radiate (it can't), but electron #2 can and does, because in that case the dipole moment is growing with time. But if I repeat the analysis in the frame of electron #2, it doesn't radiate (now it's the one that can't), but electron #1 does, because in that case the dipole moment is growing with time.

So what's up? How can both electrons both radiate and not radiate?

The answer is that the B and E fields are the same (up to a Lorentz transformation) in all three descriptions, but the portion of them attributed to near field and far field (radiation) is different. And in the frame described, there is no radiation.
Wait, now it's confusing. Are we discussing the bound state or the scattering of two electrons? For the bound state there's no radiation because an energy eigenstate doesn't radiate (modulo you are in an excited state and there's some probability for spontaneous emission of a photon).

Of course in the scattering of two electrons there's radiation, because the electrons are accelerated (classical picture). From a QT point of view there is radiation, because the cross section for bremsstrahlung, i.e., ##\text{e}+\text{e} \rightarrow \text{e} +\text{e} + \gamma## is non-zero in QED.
 
  • #23
vanhees71 said:
Are we discussing the bound state or the scattering of two electrons?

I'm pretty sure there isn't a bound state of two electrons. :wink:

I am imagining the situation as the two electrons begin at rest and then are free to move.
 
  • #24
I didn't claim this. The problem with this thread now is that in the OP two questions are asked at once. I referred to the 2nd case (two electrons bound to a nucleus, i.e., a helium atom or helium-like ion).

If the electrons begin at rest and then are let free to move they are accelerated and thus they radiate (classical picture).

So the short answer is: In the first scenario the electrons radiate, in the 2nd scenario nothing radiates.
 
  • #25
Whatever the details, if there's an elastic collision, there can't be any radiation.
 
  • #26
vanhees71 said:
If the electrons begin at rest and then are let free to move they are accelerated and thus they radiate (classical picture).

I argued that they don't, because the power radiated is zero. The power is proportional to the dipole moment (actually, its square), which is zero in this case.

One way to look at radiation is that the thing that actually produces radiation is a time-varying multipole moment. Accelerated charges radiate (in most cases) because they produce a time-varying multipole moment (in most cases). This is neither more right nor more wrong than the more conventional description, as it is more akin to changing coordinate systems, but often symmetries that are hard to see in the more conventional description are apparent when thinking in multipoles. Like this case.

If you really want to think in terms of the individual electrons, you can calculate the radiation from each and will discover that it is 180 degrees out of phase. In the radiation zone (r >> separation between the electrons) the fields (other than the overall E field from -2 units of charge) are zero.
 
  • #27
If electrons are bound ('loosely' or otherwise) to a nucleus then would I be right in saying they would emit the same EM radiation as they would absorb it. As such atoms/ions 'can' absorb some form of EM radiation, so they 'can' emit it.

If two free electrons scatter off each other, I thought this was, indeed, a case for emission of bremsstrahlung, is it not?
 
  • #28
cmb said:
If two free electrons scatter off each other, I thought this was, indeed, a case for emission of bremsstrahlung, is it not?
Well, if the collision causes some vacuum-fields, like the photon field, to become exited, then it's not a perfectly elastic collision, and there is some radiation.

So I am guessing that classical physics, which knows nothing about exited vacuum fields, says, incorrectly, that there is no radiation in this case.

Classical physics is almost correct if the collision is a low energy collision and takes a long time.
 
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  • #29
jartsa said:
Well, if the collision causes some vacuum-fields, like the photon field, to become exited, then it's not a perfectly elastic collision, and there is some radiation.

So I am guessing that classical physics, which knows nothing about exited vacuum fields, says, incorrectly, that there is no radiation in this case.

Classical physics is almost correct if the collision is a low energy collision and takes a long time.
OK, can I be a bit more specific then with a few of scenario questions I have?

Two electrons on a collision path, each at 1MeV lab-frame/collision-frame energy, at a closing angle of 45 deg.

From the instant of closest approach, is this then an example of two free electrons accelerating each other away (from their collision CoM) and if so is this EM-emitting? What is the photon energy?

Prior to the instant of closest approach, is this then an example of two free electrons decelerating each other (wrt their collision CoM) and if so is this EM-emitting? What is the photon energy?

Does the angle of incidence make a difference to whether EM is emitted or not?
 
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  • #30
sophiecentaur said:
Whatever the details, if there's an elastic collision, there can't be any radiation.
By definition an elastic collision is of the form ##A+B \rightarrow A+B## and there is no radiation for such processes by definition. That doesn't mean that it's the only possible process. Scattering of two electrons )or any electrically charged particles) always also has some probability to emit photons (bremsstrahlung). So there is radiation in scattering processes involving charged particles.
 
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  • #31
Vanadium 50 said:
I argued that they don't, because the power radiated is zero. The power is proportional to the dipole moment (actually, its square), which is zero in this case.

One way to look at radiation is that the thing that actually produces radiation is a time-varying multipole moment. Accelerated charges radiate (in most cases) because they produce a time-varying multipole moment (in most cases). This is neither more right nor more wrong than the more conventional description, as it is more akin to changing coordinate systems, but often symmetries that are hard to see in the more conventional description are apparent when thinking in multipoles. Like this case.

If you really want to think in terms of the individual electrons, you can calculate the radiation from each and will discover that it is 180 degrees out of phase. In the radiation zone (r >> separation between the electrons) the fields (other than the overall E field from -2 units of charge) are zero.
I'm very surprised that you think that two electrons in interaction don't radiate. Of course the bremsstrahlung cross section is not 0 in QED!

Of course you are right, if you consider the dipole approximation there are the well-known dipole selection rules for one-photon emission.
 
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  • #32
@cmb, this is not the situation in the original question, nor is it classical physics, which is the section this is in.
 
  • #33
vanhees71 said:
I'm very surprised that you think that two electrons in interaction don't radiate.

I didn't say that. I said that there is no radiation in the situation described, not that two electrons can never radiate. See message #15.

vanhees71 said:
...in QED!

And QED is not classical physics.
 
  • #34
I think I simply don't understand your argument then. Why shouldn't two classical electrons accelerated in back-to-back motion due to their repulsion radiate?

Maybe I have to do the calculation to understand this. From a QT point of view it seems plausible though, because of the dipole selection rule. That's why I thought you argue within QT...
 
  • #35
vanhees71 said:
Why shouldn't two classical electrons accelerated in back-to-back motion due to their repulsion radiate?
This has already been discussed above. If you treat them a two classical charged particles in the absence of any other charges then you would have an elastic collision. The net EM energy that's transferred between will be zero - or they wouldn't be following Classical laws.
But electrons are not classical particles so the possibility of radiation does exist.
 
  • #36
vanhees71 said:
Why shouldn't two classical electrons accelerated in back-to-back motion due to their repulsion radiate?

Because the power radiated is proportional to the square of the change of their dipole moment, and the back to back electrons have a constant dipole moment (of zero). For this setup. Other setups would of course have different answers.
 
  • #37
I have to do the calculation to see this. Both electrons are accelerated. So why don't they radiate within classical physics? Only because they are moving on a straight line? Do you know a textbook, where this calculation is done?
 
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  • #38
This exact problem? No. But the power radiated by a multipole? Lots - pretty much all of them. If you look at one, you will see a factor of p (two, actually), the dipole moment. This is always zero for this problem.
 
  • #39
Of course, I know the expression of the radiation power in terms of the multipole expansion. Each multipole mode adds to the radiation power
$$P_{\ell,m}=\frac{c}{8 \pi k^2} |a_{\ell m}|^2.$$
See Jackson, Eq. (16.78), using Gaussian units, and indeed the ##|a_{\ell m}|^2## are proportional to the corresponding multipoles of the charge-current distribution.

So how do I see that all these vanish only because a charge moves (accelerated) along a straight line?

There are radiation losses in linear colliders though they are negligible even for electrons for all practical purposes. See, e.g.,

http://farside.ph.utexas.edu/teaching/em/lectures/node131.html

Though this doesn't make use of the multipole expansion but just of the Larmor formula for an accelerated point charge.
 
  • #40
sophiecentaur said:
Whatever the details, if there's an elastic collision, there can't be any radiation.
There cannot be any emitted radiation unless there was incident radiation. [Which, since arranging for the right radiation to be incident would be difficult, amounts to pretty much the same thing].
 
  • #41
cmb said:
OK, can I be a bit more specific then with a few of scenario questions I have?

Sorry, I can't answer any of those questions.
@Vanadium 50 What if two identical charged balls far away from each other are given simultaneous, sharp, equal pushes towards each other? Surely they would radiate? I mean pushed by a hand or something, not by an electric field.
 
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  • #42
This has been kind of annoying - responses to "the radiation in this case is zero" seem to fall into two categories: "I haven't done the calculation myself, but I just don't believe it" and "yes, but this other system has radiation!". (I wish my professors would haven fallen for this: "The answer was A. You wrote D." "Yeah, but D is the answer to another problem! I should get full credit")

@vanhees71, I am unable to get at my copy of Jackson, but somewhere in chapter 9 there is an expression for the power radiated from a dipole that looks something like dP/d\Omega \sim k^4 {\bf p}^2 \sin^2{\theta}, where p is the dipole moment vector (at maximum extent). (I am sure Jackson has all the c's and 1/12π's in the right spot, but there is no way I am going to get the leading constants right.) Since p = 0, dP/d\Omega = 0. No radiation.

One could say, "yes, but this system does not have a constant wavenumber k" which is true, but if I sum this up wavenumber by wavenumber, it's just summing up zeros.

Also, intgrating over angles to get P won't fix anything: if the power per unit angle is zero everywhere, integrating won't help.

Another way to look at this (which I suspect you will really hate) is if you have two electrons separated by a distance a (or 2a, if you prefer), then a's can only appear in radiation terms accompanied by q's. In the far field zone, the length a can be neglected because r >> a (this is the definition of the far field or radiation zone) but the dipole moment (qa) cannot, because it describes the strength of the radiation source.

Since all that matters in the radiation zone is (qa), which is how p is defined (up to a factor of 2 or 4), I can replace one of the electrons by an equal positive charge moving in the opposite direction and it will not - indeed, cannot - change what is happening in the radiation zone. But in that case, the field is obviously zero at large r.

This, by the way, is the same argument as saying that the radiation from electron 2 is 180 degrees out of phase with the radiation from electron 1. One might argue that this is not exact, which is true. It is only true to order (a/r). But in the radiation zone we neglect such terms because they are part of the definition for what it means to be near field. (They will also fall off as 1/r2 and not 1/r)

I should point out that I am treating an "electron" as a classical object: an infinitesemally small charged ball. A physical electron also has spin, and a spin flip would of course radiate.
 
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  • #43
But we don't have a dipole here. It's much simpler to argue with the Larmor formula for accelerated point particles and there I don't see, why the particles shouldn't radiate when being accelerated by their mutual repulsive interaction forces. Of course you can solve this only approximately, i.e., full radiation reaction cannot be taken into account. If I find the time, I try to do the calculation, using the Larmor formula and see whether the fields are such that the Poynting vector vanishes for some reason at infinity (radiation part).
 
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  • #44
vanhees71 said:
But we don't have a dipole here.

Exactly! It's zero.
 
  • #45
Vanadium 50 said:
Exactly! It's zero.
No the dipole moment is not zero, check post #18 please. However it is not time varying as i claim it to be because after more careful consideration ##r_1+r_2## remains constant if the electrons are moving only due to mutual repulsion.
 
  • #46
vanhees71 said:
But we don't have a dipole here. It's much simpler to argue with the Larmor formula for accelerated point particles and there I don't see, why the particles shouldn't radiate when being accelerated by their mutual repulsive interaction forces. Of course you can solve this only approximately, i.e., full radiation reaction cannot be taken into account. If I find the time, I try to do the calculation, using the Larmor formula and see whether the fields are such that the Poynting vector vanishes for some reason at infinity (radiation part).
Vanadium 50 said:
Exactly! It's zero.
If I remember correctly the radiation from the Larmor formula for a point charge is preferentially emitted in the "forward" direction. How can this possibly lead to exact cancellation for equal and opposite velocities? Can you provide the highlights of your calculation ? Or a reference?
 
  • #47
Vanadium 50 said:
Exactly! It's zero.
I don't know, how you are so sure that in this situation the classical electrons don't radiate. Here is a paper, proving the opposite (though it's not the interaction of two point particles but the radial motion (as well as other solutions) for a charged particle in an external Coulomb potential):

https://journals.aps.org/pr/abstract/10.1103/PhysRev.124.616

I don't claim to know the definitive answer, but I don't understand your argument with the dipole. Why should I use a multipole expansion in this case to begin with? The currents are not restricted to a finite region. So why should the dipole approximation be correct?

The solution of the retarded potential for the motion given by the accelerated charges when solving the radial Coulomb problem neglecting the radiation loss and then plugging this solution into the formula for the retarded (Lienart-Wiechert) potential is for sure tedious to find. I'm not sure, whether it's doable in closed form.

It's anyway a subtle point whether charges in unbound trajectories radiate or not. E.g., for decades people (including Pauli in his famous encyclopedia article on relativity) thought that a particle in hyperbolic motion doesn't radiate, but that's well known to be errorneous, and it was already known for a long time by a calculation by Sommerfeld, which however had been forgotten.
 
  • #48
hutchphd said:
If I remember correctly the radiation from the Larmor formula for a point charge is preferentially emitted in the "forward" direction. How can this possibly lead to exact cancellation for equal and opposite velocities? Can you provide the highlights of your calculation ? Or a reference?
As I said in my previous posting, I don't want to claim anything without having done the calculation. I'm not sure whether it's doable at all. Here's an alternative argument for radiation, using the Abraham-Lorentz-Dirac equation, but not for exactly the case discussed here (see my previous posting):

https://journals.aps.org/pr/abstract/10.1103/PhysRev.124.616
 
  • #49
Delta2 said:
No the dipole moment is not zero, check post #18 please.

I disagree with that definition, which leads to a position dependence of dipole moment (we don't wave or want one with monopole moment, usually called "charge" do we). I am used to seeing it without the r vector. But...

Delta2 said:
it is not time varying

So the power will be zero. The k4 comes about through time derivatives.
 
  • #50
vanhees71 said:
but not for exactly the case discussed here

I wish I had you as a professor. You'd give me credit for answering C instead of A because C was the answer to some other question, right?
 

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