Photons emitted by an accelerated charge

In summary, the sentence "accelerated charges radiate em waves" is a well-known concept, but deriving it from Maxwell's equations can be difficult. When considering the quantum-mechanical problem of the em field coupled to a 4-current, the cross section for photons being radiated can be found, although it is only a probability. The physics of elementary particles and their self energy is still not fully understood, but boundary conditions may play a key role. However, the radiation pattern does not depend on the history of the motion of the charge, and the retarded time equation is the only boundary condition needed. A paper by H. A. Haus presents an alternative derivation of the radiation field of a point charge, which shows that a
  • #1
Dickfore
2,987
5
Almost everyone is familiar with the sentence "accelerated charges radiate em waves".

Nevertheless, if you are asked to derive this starting from Maxwell's equations, you might find it difficult. Surely the radiation pattern depends on the history of the motion of the charge.

Then, there comes an even harder problem: Consider the quantum-mechanical problem of the em field coupled to a 4-current corresponding to the accelerated motion of the charge. Treat this as a classical source term, without considering the dynamics of the charge. Can we find the differential cross-section of photons with a particular frequency and polarization being radiated in a particular direction?

Finally, this cross section is some sort of probability. Suppose a photon is detected. Will this alter the field in some way, similar to the disturbance of quantum particles by measuring their position/momentum, and, if so, how?
 
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  • #2
This is too good a question to disappear without response. One place to start would be the QM savy chapter 16 of Jackson's "Classical Electrodynamics" which takes a fairly thorough study of radiation reaction. But Jackson begins the chapter with the caveat that a fully satisfactory understanding cannot be found at this point because that would entail a better understanding of elementary particles and what their self energy is and how that affects radiation.

I'd interpret that remark to mean that the physics of elementary particles is developed around the idea of symmetries in spatially extended fields that have different qualities in different types of particles. The creation and annihilation of all types of particles, even with fractional charges, is well described but I've never seen an attempt made at determining boundary conditions for the fields comprising the particles. Boundary conditions would enclose the energy of the particle in a way that could give it qualities of self energy.

De Broglie, in the context of Wave Mechanics, has expressed that problem in the fact that, with no boundary conditions, all waves associated with particles would very quickly dissipate. The overlooked importance of Mossoti's definition of charge is brought to bear on this especially well in that his definition expresses charge in terms of a boundary condition.
 
  • #3
Nevertheless, if you are asked to derive this starting from Maxwell's equations, you might find it difficult. Surely the radiation pattern depends on the history of the motion of the charge.
Actually no, it does not depend at all on the history. And the only boundary condition needed is the retarded one. If you take a look in Jackson, Chap 14, the Lienard-Weichert potentials for an accelerating charge are given, and they are quite simple. They depend only on the instantaneous location and velocity of the charge at the retarded time.
 
  • #4
But, the retarded time in those equations is found from the condition:
[tex]
\vert \mathbf{x} - \mathbf{r}(t_{ret}) \vert = c \, (t - t_{ret})
[/tex]
where [itex](\mathbf{x}, t)[/itex] are the space-time coordinates where the potential is calculated, and [itex]\mathbf{r}(t)[/itex] is the equation of motion for the charge.

As you can see, this is an implicit equation for [itex]t_{ret}(\mathbf{x}, t)[/itex] that strongly depends on the motion of the charge (its history). For complicated motions, it is impossible to solve exactly.
 
  • #5
You may find this paper interesting: "On the radiation from point charges" by H. A. Haus from American Journal of Physics, Volume 54, Issue 12, pp. 1126-1129 (1986)

http://adsabs.harvard.edu/abs/1986AmJPh..54.1126H

An alternative derivation of the radiation field of a point charge is presented. It starts with the Fourier components of the current produced by the moving charge. The electric field is found from the vector wave equation. Each step in the integration permits physical interpretation. The retarded time appears very naturally in this derivation. The interpretation of the present derivation is that a charge at constant velocity v(||v||<c) does not radiate, not because it is unaccelerated, but because it has no Fourier components synchronous with waves traveling at the speed of light. Of course, Cherenkov radiation in a medium, in which the velocity of electromagnetic propagation is less than c, is the classic example of radiation by a charge moving at constant velocity.
 
  • #6
As you can see, this is an implicit equation for tret(x,t) that strongly depends on the motion of the charge (its history). For complicated motions, it is impossible to solve exactly.
At the observation point P, construct the past null cone. This intersects the world line of the source at exactly one point Q, which is the retarded location of the particle. No "history", only one instant. The fact that you may not be able to write down a simple analytic solution for Q is irrelevant.
 
  • #7
Bill_K said:
At the observation point P, construct the past null cone. This intersects the world line of the source at exactly one point Q, which is the retarded location of the particle. No "history", only one instant. The fact that you may not be able to write down a simple analytic solution for Q is irrelevant.

OK, thanks for your help. Have a nice day.
 

Related to Photons emitted by an accelerated charge

What is a photon?

A photon is a fundamental particle of light that has zero mass and carries energy in the form of electromagnetic radiation.

How are photons emitted by an accelerated charge?

When an electrically charged particle, such as an electron, is accelerated, it creates a changing electric field that propagates through space as electromagnetic radiation, which is composed of photons.

Why do photons emitted by an accelerated charge have a specific energy?

The energy of a photon is directly related to its frequency, which is determined by the acceleration of the charged particle. As the acceleration changes, so does the frequency and therefore the energy of the emitted photons.

Do photons emitted by an accelerated charge have a constant speed?

Yes, photons travel at the speed of light in a vacuum, which is a constant value of approximately 299,792,458 meters per second. This speed is not affected by the acceleration of the charged particle.

What are some practical applications of understanding photons emitted by an accelerated charge?

Understanding the behavior of photons emitted by an accelerated charge is crucial in fields such as optics, telecommunications, and particle physics. It also plays a significant role in technologies such as lasers, solar panels, and medical imaging devices.

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