# When Does EM Radiation Occur?

• Hornbein
In summary, the presence of two free electrons does not necessarily result in EM radiation, as it requires a change in the dipole moment which is not present in this scenario. However, if the electrons are loosely bound to a nucleus, there may be EM radiation according to classical electrodynamics, but not according to quantum electrodynamics unless there is a change in energy states. In the case of a metal, the EM radiation from the free electrons is cancelled out by each other, resulting in no macroscopic radiation. The concept of dipole moment and its change over time is crucial in determining whether or not there will be EM radiation.

#### Hornbein

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?

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.

vanhees71
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.

berkeman
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.

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...?

There are some subtleties, but as described, it's zero radiated power.

berkeman
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?

dipole?

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?

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.

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.

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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Black Body?

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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davenn and Delta2
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.

davenn, vanhees71 and sophiecentaur
@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).

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.

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.

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.

vanhees71 and Delta2
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 ;-)).

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.

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.

I am imagining the situation as the two electrons begin at rest and then are free to move.

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.

Whatever the details, if there's an elastic collision, there can't be any radiation.

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.

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?

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.

sophiecentaur
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?

sophiecentaur
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.

sophiecentaur
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.

hutchphd
@cmb, this is not the situation in the original question, nor is it classical physics, which is the section this is in.

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.

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...

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.