Physical interpretation of relativistic synchrotron radiation

Question

When an electron approaches the speed of light, the emission pattern is sharply collimated forward, in the direction of motion. I can see how this is mathematically true from taking the relativistic limit of the angular distribution of the radiated energy (as is done on the Wikipedia page about synchrotron radiation).

What I am wondering is how to think about this phenomenon from a physical point of view. How can both the particle and the radiation move in the forward direction? Why isn't the collimated beam pointing in the opposite, i.e. the ''backwards'', direction instead?

Answer 1

If the charge radiated backwards, it would be gaining an additional momentum-energy.

Answer 2

This collimation of light in the forwards direction is in fact also what you would expect if non-relativistically.

From the electron's perspective, light is radiated uniformly in all directions. However, the electron is itself moving with respect to an observer in the laboratory. If we didn't know anything about special relativity, we would say that to find the velocity of a given photon in the lab frame $\textbf{v}_\text{lab}$, we should add the velocity of the photon in the electron frame $\textbf{v}_{ph}$ and the velocity of the electron $\textbf{v}_e$, $$\textbf{v}_\text{lab} = \textbf{v}_{ph} + \textbf{v}_e.$$

This means that an observer in the laboratory frame would see the photon gain additional velocity $\textbf{v}_e$ in the direction that the electron is moving , i.e. it would be "collimated forwards".

Of course, special relativity teaches us that actually light does not behave like this at all, but I hope that this "galilean" analogy gives some intuition for what's going on.

Answer 3

The Bremsstrahlung diagram is used for a decelerating electron.

It could equally apply to an accelerating one, (instead of a nucleus, just the accelerating field). Feynman diagrams are in the center of mass system of the input and output particles

The collinearity comes because the whole diagram should be Lorentz transformed to the lab system where the electron has very high energy

Here u is the velocity of the particle in the CM frame. This function is always strongly peaked in the forward direction unless u≈c and cosθC≈−1.

Answer 4

One recalls that there is no explanation for the cause of the Lorentz force, the various Hall effects and the synchrotron radiation. At the latter effect, one has stated that photons are emitted at the deflection of the electrons. That these energy losses also occur with the other mentioned effects is less present, however. Even without experimental proof of the occurring radiation in all these phenomena, there is a theoretical reason and an experimental observation, which speak for the fact that this radiation occurs.

1. Particles with kinetic energy are deflected by a magnetic field although the theory says that the external magnetic field cannot do any work. (Incidentally, this is empirically confirmed by the fact that when a permanent magnet is used, its field does not weaken with time). What then is the effect of the deflection?

2. If one experiments with particles in the vacuum, then one sees that the deflection does not take place in a circular path but a spiral. The particle loses its kinetic energy and comes to a standstill in the centre of the spiral. What causes the deceleration?

I propagate already for a long time a model, where the deflection of a moving charge in a magnetic field takes place exactly by the radiation of photons (here and here and here).

In short, the spin of the electron is aligned by the external magnetic field and thereby emits a photon. Since the photon has a moment, the electron is deflected sideways and at the same time, its spin is de-aligned again. This process continues as long as the electron still has kinetic energy.

By the way, this model assumes that the synchrotron radiation does not radiate exactly tangentially, but deviates from the tangent to the outside.