# Circularly-polarized plane wave and electron

1. Aug 3, 2008

### fizzle

Can a classical circularly-polarized plane electromagnetic wave, that's bounded in time, transfer net energy/momentum to an electron?

2. Aug 3, 2008

### cesiumfrog

So you're excluding Compton scattering..

A point charge in a circularly polarized wave will be driven around, by the direction of the local field, in a little circle (so some energy must be transferred at least temporarily). And since it is accelerated into such motion, the point charge must radiate its own EM wave. This is very similar to the way light interacts with a medium; in one direction the little EM wave will slightly cancel (and retard) the driving wave, and in the opposite direction this little EM wave can be thought of as reflection of the driving wave. By global argument, the electron must be pushed forward (tracing a slackening helix) to complement the reflected component of the wave momentum.

Last edited: Aug 3, 2008
3. Aug 3, 2008

### Andy Resnick

Are you asking if electromagnetic radiation can transfer (angular) momentum to a free particle? Circularly polarized light incident on a linearly birefringent object will transfer angular momentum to the object.

Or are you asking if the electromagnetic field can drive angular momentum transitions for a bound electron? Yes- but circularly polarized light will not transfer angular momentum. Instead, so-called optical vortices are associated with orbital angular momentum- circular polarization states are associated with spin transitions. However, the literature is a little unclear (to me) if an optical vortex is formally identical to the angular momentum operator.

4. Aug 3, 2008

### fizzle

I need to be more specific! I'm talking about a single free electron initially at rest and a single CP plane wave in a completely classical world.

The CP wave forces the electron in a helical path in the direction the CP wave is travelling, while under the influence of the wave. I'm interested in the end result where the CP wave's amplitude increases from zero to some max value then decreases back to zero. When I do a simple computer simulation of this the result is no net transfer of energy/momentum to the electron. It always returns to rest when the CP wave's amplitude is a Gaussian.

Last edited: Aug 3, 2008
5. Aug 3, 2008

### cesiumfrog

By "bounded" (Gaussian) you must mean a (half-) wave "pulse"?

I used global argument because the mechanism is not obvious. There have been published papers on the topic of how a transverse wave on a string can transfer longitudinal momentum (that case can be answered with Pythagorean longitudinal stretching of the string). It is well known that a mirror recoils from a light pulse, but the mechanism is obscure since ideal mirrors have no longitudinal extent and since the EM fields appear perfectly transverse.

I suspect your simulation would have to be extremely precise (this is a small effect) and take into account the finite extent of the charge as well as its self-interaction. There is no effect if the driving field is static or if the field of the charge itself is insignificant. There is reasonable basis against treating the electron as a point particle in such context as this. By the way, did you have a specific reference mentioning the effect?

Last edited: Aug 3, 2008
6. Aug 3, 2008

### fizzle

This is all standard electromagnetic theory so there's no need to get into self-interaction, etc. The E field in the CP wave drives the electron in a circle and that circular motion results in v x B forcing the electron in the direction of the wave's motion. The non-zero amplitude could be hundreds, thousands, millions, etc. of cycles of the wave.

Why am I asking this question? Because it would have significant implications in any classical description of Compton Scattering. It means that an electron would never get "ejected" from the target block simply because it interacted with an incident CP wave. The electron must interact with something else (eg. an atom, another electron, etc.) while interacting with the incident CP wave in order to retain any energy. This would explain one of the problems in a classical description of Compton Scattering => the speed of the electron required to produce the correct angular-dependence of the scattered radiation due to the relativistic Doppler Effect is much less than the speed of an ejected electron. When I calculate the orbital component of the electron's velocity when interacting with a CP wave (when the electron has effectively absorbed a "photon" of energy), I find that the orbital speed is equal to the maximum possible ejection velocity.

I think this gives a much more complete classical description of the Compton Effect, including the seemingly random electron ejection direction. Basically, the incident CP sets the electron on its helical path where it then has to interact with a third party to retain any energy.

7. Aug 3, 2008

### cesiumfrog

I had forgotten to consider the magnetic field, thank you.

(FWIW, self-interaction is relevant to some problems in a completely classical world, such as the inertia of a charged object/"electron".) Why did your simulation fail?

8. Aug 4, 2008

### Andy Resnick

I don't follow you here. Why do you think a circularly polarized plane wave will "force the (free) electron (to move) in a helical path"? Under what conditions?

A circularly polarized beam incident onto a conductor should generate synchotron radiation, if your idea is correct.

9. Aug 4, 2008

### fizzle

Standard em theory. The electric field forces the electron to travel in a circle while v x B forces the electron to move in the direction of travel of the wave. The overall result is a helical path. This is all well known.

A CP wave incident on a conductor results in a reflected CP wave.

10. Aug 4, 2008

### Andy Resnick

Not to me.... can you provide a reference?

Erm, yes... but if the conduction electrons are induced to move in little circles, as you claim is well-known, they are undergoing an acceleration, which produces a field. Has anyone measured or detected this field?

11. Aug 4, 2008

### fizzle

Reference? This is very old stuff. I guess you could look up Thomson Scattering and then not ignore v x B. More modern references are in papers on laser acceleration.

The field radiated by electrons is responsible for the reflection off a conductor. You measure and see it all the time. How do you think a mirror works? The electric field in the incident em wave causes the electrons to accelerate and emit a wave whose electric field is the inverse of that in the incident wave. You typically ignore qv x B in the incident wave because it's force is neglible compared to qE.

I'm interested in the regime where v x B is not neglible => very high field strengths in the incident wave.

12. Aug 4, 2008

### Andy Resnick

Oh- plasmas. I don't know much about them. From what I just read, an interesting experiment would be to use dusty plasmas- the particle size is larger, so the optical wavelength would be larger as well- one could probably use visible light and track individual particles.

I don't know the relevant literature- maybe someone has already done it.

13. Aug 4, 2008

### cesiumfrog

Why are you specifically interested in circularly polarised light?

Why are you so interested in describing Compton scattering classically (whereas it seems like evidence for the quantisation of light)?

Are you trying to describe bound electrons as classical charged particles orbiting a nucleus? If so, how do you explain that they do not radiate and fall in?

Do you agree that a free electron exposed to (circularly polarised) light, with a finite envelope, will indeed experience a net transfer of energy and momentum (rather than being restored to its original velocity)? Do you have any contrary evidence/argument?

Why do you say (classically) that an electron would need to interact with some other object in order to "be ejected" by (sufficiently intense) light?

Since you did mention the photon, could you clarify what you mean by "classical description"?

14. Aug 5, 2008

### fizzle

I'm interested in CP waves because they're the easiest to analyze.

I'm interested in a classical description of Compton Scattering because I want to see how far I can get with it. Does it provide any physical insight into quantum theory? Where does it go wrong? In the abstract of his 1925 paper, "A Quantum Theory of the Scattering of X-Rays by Light Elements", Compton explicitly states that the angular-dependence of the wavelength shift can be explained by classical theory if you assume an electron moving in the direction of propagation (the "drift velocity").

I'm not trying to describe bound electrons, just the most simple electromagnetic interaction possible: a single free electron and a single plane wave.

I do not agree that a free electron will receive net energy/momentum from a bounded CP em wave. By "net" I mean that the energy/momentum of the free electron will be the same before and after the CP wave has passed it by. Here's a paper by McDonald with more details and references:

http://www.hep.princeton.edu/~mcdonald/accel/dressing.pdf

I say that a free electron, initially at rest, must interact with a third party in order to be "ejected" (aka. gain net energy/momentum) from a bounded CP wave ... otherwise it would have returned to rest after the wave had passed by. It will be displaced by the wave but will not net any energy/momentum.

By "classical description" I mean classical electromagnetic theory plus special relativity.

I've done the full analysis for Compton's original experiment (incident photon of 17.49 KeV) and the results are as follows:

1) An electron reaching Compton's drift velocity, where the angular-dependence of scattered em radiation matches his experiment with an incident wavelength of 7.09 x 10^-11m, is moving with the wave at 3.31% of the speed of light.

2) The electron's orbital component, remember that it's under the influence of a circularly-polarized incident wave, is 25.3% of the speed of light. Its orbital radius is 2.95 x 10^-12 m.

3) The electric field in the incident CP wave has to be 1.18 x 10^16 V/m.

#3 is the killer here. That's an extraordinarily high electric field strength. Furthermore, the electric field energy varies as the cube of the frequency -- a "Compton Catastrophe", if you will.

Just for laughs, I plugged in some other interesting numbers to see what fell out of the equations:

a) For an electron absorbing a photon of energy equal to an electron's rest mass (511 KeV), the drift velocity is exactly 50% of the speed of light. I was surprised by this initially but it makes sense when you look at it from a momentum viewpoint.

b) For an electron absorbing a photon of energy required for pair production (1.022 KeV), the drift velocity is equal to 66.67% of the speed of light, as expected after seeing (a). However, an interesting number falls out of the equations: the electron's orbital radius is equal to its Compton Wavelength (3.86 x 10^-13 m).

A final observation from the analysis is the value of the recoil electron's velocity. In Compton Scattering, the recoil/ejected electron's velocity is a certain value that depends on the recoil angle, which Compton calculated assuming a billiard ball-like interaction between a photon and an electron. For the most part, this ejection velocity is much higher than the drift velocity in the classical analysis. However, there is a one distinct match with the classical analysis: the maximum value of the ejected electron's velocity is equal to the orbital velocity in the classical analysis.

The bottom line of this exercise is that classical em theory can offer an explanation of Compton Scattering. It requires ultra-high field strengths and adds in the requirement of an interaction with a third party (which, interestingly enough, adds randomness/probability to the process).

15. Aug 5, 2008

### cesiumfrog

We agreed that the free electron will be longitudinally accelerated (in a "slackening helix") for the duration of the external pulse; by what mechanism do you propose it will be decelerated after the pulse?

Do you concede that the electron will scatter part of the EM wave? That the EM wave must therefore end up with less momentum in its original direction? Would not the principle of momentum conservation then be violated if the responsible electron finished up with the same momentum as it started with?

(I found the paper you cited to be unclear, incomplete and non-refereed, but I'll look into its http://prola.aps.org/abstract/PR/v138/i3B/pB740_1" [Broken] reference...)

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16. Aug 5, 2008

### fizzle

The electron isn't decelerated after the pulse, it's decelerated as the pulse's amplitude decreases. Anyway, I figured it out. You have to think in terms of unstable versus stable states between the electron's velocity and the incident wave. Initially, the electron is at rest and inertia keeps it from instantly adjusting its velocity to the incident wave. The wave's E and B fields "lead" the electron's velocity. This accelerates the electron orbitally and longitudinally, and is an unstable state.

Eventually, the electron's velocity becomes perfectly in phase with the wave's E and B fields, where E is perpendicular to the velocity and B is parallel. With E perpendicular to v, there's no orbital acceleration and since B is parallel to v, there's no more longitudinal acceleration. This is a stable state and I imagine with more analysis you'd find that any small deviations around this state would lead back to it.

Finally, when the amplitude of the incident wave begins to decrease, E and B in the wave begin to "lag" the electron's velocity because of inertia. E has a small component in the opposite direction of v and that reduces the orbital speed. Correspondingly, B is no longer perfectly parallel to v so v x B results in longitudinal deceleration.

What's interesting is that the time derivative of the magnitude of E and B is critical here. Obviously, you think about the spatial and temporal derivatives in em theory all the time but here the time derivative of the magnitude is the most important part.

No concession necessary. The scattering is an integral part of the whole process and carries away the energy and momentum removed from the incident plane wave. Be careful when accounting for energy and momentum in classical em fields!

Doubt McDonald's analysis at your own peril. Better yet, derive the results yourself (as I did before finding his). He is correct.

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17. Aug 6, 2008

### cesiumfrog

That is incorrect, but after reading Kibble I understand what you are trying to articulate. Since the acceleration corresponds with the peaks of E, the simple harmonic motion of the charge must have velocity lagging by a quarter of a cycle (where a=E=0), which coincides upon the zero of B. Since the slope of the wave envelope can drive the charge velocity to lag by just slightly more/less than a quarter cycle (i.e., no longer exactly centred upon B=0), the net Lorentz force provides longitudinal acceleration then deceleration.

This basic derivation neglects back-reaction of any wave produced (read: back-scattered) by the charge. In this toy-context energy-momentum is conserved, and Kibble writes that the electron will no longer be returned to rest if it "scatters a photon".

18. Aug 6, 2008

### fizzle

No, it is correct. Do you have a link to Kibble's paper? It sounds like you (and he?) are referring to a linearly-polarized incident wave. I'm talking about a circularly-polarized incident wave, where |E| never goes to zero in a cycle. In that case, the electron reaches a stable state where its tangential velocity is perpendicular to E and parallel to B. I call this being perfectly in phase with the wave. If the wave's amplitude decreases a bit, the wave "lags" behind a bit because the electron is moving too fast. This lag results in E having a small component in the opposite direction of the electron's velocity. This is what slows the electron's orbital speed. Simultaneously, the lag causes B to be slightly non-parallel to the electron's velocity and results in a force that decelerates it's longitudinal speed.

Note that the electron is not "scattering a photon" when it cycles from rest to interacting to rest again. As best as I can tell, that would be equivalent to the electron absorbing then re-emitting a photon unchanged. What's called "scattering a photon" happens when the electron interacts with a third party while interacting with the incident wave and doesn't return to rest. This causes a discontinuity in the electron's radiated em wave. If this paragraph appears to be vague and "hand waving", that's because it is! I reduced it to the most simple case possible. I don't know what happens when the incident wave has different polarizations. Circular is the easiest and produces the strongest effect since the longitudinal acceleration is the greatest.

19. Aug 6, 2008

### cesiumfrog

I linked Kibble's paper in post #15, but it's properly cited inside your own reference. For the stable state I do understand broadly saying "the electron is in phase with the wave" but note that the velocity lags E by 90 degrees (and may be perfectly in phase or exactly out of phase with B depending on polarisation chirality). Sorry if, earlier, my only describing one transverse component (independent of polarisation) caused confusion.

I agree that, if the electron does not scatter any of the external wave, the electron will be longitudinally accelerated as the wave increases in intensity then decelerated back to its initial momentum as the wave decays past. (I hadn't previously been aware of this mechanism, which applies a net longitudinal force whenever the orbit of the charge is mismatched to the intensity of the wave.)

However, consider the neglected (synchrotron) radiation produced by the rapid transverse acceleration of the electron. Obviously (since the position of the charge is 180 degrees out of phase with the applied E field) this radiation will be 180 degrees out of phase from the wave (hence the component propagating in the same direction will destructively interfere with the energy of the original wave) and we can call it a reflection (or continuous classical scattering) of part of the original wave, off from the electron.

Since part of the wave is scattered, the wave momentum decreases. But the emission of radiation has a back-reaction (e.g., the Abram-Lorentz force) which will damp the transverse motion of the electron. This will maintain such a mismatch (between the external wave and the transverse orbit of the charge) as to re-invoke the earlier mechanism to also maintain a longitudinal acceleration (even now that the intensity of the incident wave is constant). However, when the incident wave intensity finally decreases away, the amount of associated longitudinal deceleration will be no more than before (despite extra intervening acceleration) therefore the free electron will ultimately retain some momentum from the wave (just as was required by global conservation of momentum to balance for the reflected part of the wave).

Last edited: Aug 6, 2008