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fizzle
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Can a classical circularly-polarized plane electromagnetic wave, that's bounded in time, transfer net energy/momentum to an electron?
fizzle said: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.
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.Andy Resnick said: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 CP wave incident on a conductor results in a reflected CP wave.A circularly polarized beam incident onto a conductor should generate synchotron radiation, if your idea is correct.
fizzle said: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.
fizzle said:A CP wave incident on a conductor results in a reflected CP wave.
fizzle said:Because [whether at CP wave transfers net energy/momentum to a free electron] 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.
fizzle said: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.
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.cesiumfrog said: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?
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!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?
Doubt McDonald's analysis at your own peril. Better yet, derive the results yourself (as I did before finding his). He is correct.(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" reference...)
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.fizzle said: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.
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.fizzle said:Do you have a link to Kibble's paper? [..]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.
A circularly-polarized plane wave is an electromagnetic wave that oscillates in a circular motion as it propagates through space. This type of wave has two components, a magnetic field that rotates in a circular pattern and an electric field that oscillates perpendicular to the magnetic field. It is a special type of plane wave that can be produced by combining two perpendicular linearly-polarized waves with a specific phase difference.
A circularly-polarized plane wave can be created by combining two perpendicular linearly-polarized waves with a specific phase difference. This can be achieved using specialized optical components such as wave plates or by using electronically-controlled devices such as antennas or lasers.
Circularly-polarized plane waves have a variety of applications in fields such as telecommunications, radar, and astronomy. They are used to transmit and receive signals with improved efficiency and to manipulate the polarization of light for various optical experiments and measurements. They are also important in the study of the behavior of electrons in magnetic fields.
When an electron is exposed to a circularly-polarized plane wave, it experiences a force due to the magnetic field component of the wave. This force can cause the electron to spiral in a circular motion, which can be used to manipulate and control the electron's motion in various applications such as particle accelerators and electron microscopy.
Circularly-polarized plane waves can have either left- or right-handed polarization, depending on the direction of rotation of the electric and magnetic fields. Left-handed polarization means that the electric field rotates counterclockwise and the magnetic field rotates clockwise, while right-handed polarization means the opposite. The handedness of circularly-polarized plane waves is important in various applications, such as in determining the chirality of molecules in chemistry and biology.