# Why is the opening angle of synchrotron radiation less than $1/\gamma$?

I am currently studying free-electron laser which accelerate electrons and use undulators to create synchrotron radiation. In a variety of graphics and diagrams I see an opening angle of $$\pm 1/\gamma$$ for the radiation.

Where does this value come from and is there an intuitive explanation for it?

This kind of thing always occurs when you have radiation from a moving object. In the rest frame of the object, the radiation comes out reasonably isotropically. But when you add the velocity of the object, and the object is moving at nearly the speed of the radiation itself, that makes almost all the radiation comes out the forward direction. This is the "headlight effect" (or "Lorentz focusing" or "relativistic beaming", as pointed out in the comments), and it's common in discussions of relativistic radiation, though it also occurs for nonrelativistic waves to a lesser degree.

Explicitly, consider some radiation that comes out with velocity $$\mathbf{v} = (0, c)$$ in the rest frame of the radiator. If the radiator moves with speed $$v \hat{\mathbf{x}}$$ in the lab frame, then the velocity of that radiation in the lab frame is $$\mathbf{v}_{\text{lab}} = \left(v, \sqrt{c^2 - v^2} \right)$$ by relativistic velocity addition. If $$v$$ is near $$c$$, this is almost perfectly directed along $$\hat{\mathbf{x}}$$. Specifically, the angle to the $$\hat{\mathbf{x}}$$ axis is about $$1/\gamma$$. More generally, for most initial directions $$\mathbf{v}$$, the velocity in the lab frame is going to be within that angle or around it.

• The more formal search term is "Lorentz focusing", though "headlight effect" is so evocative that it is very common. – dmckee --- ex-moderator kitten Nov 15 '19 at 23:22
• Actually, the most common term (of the ones I know) is, according to Google "relativistic beaming." – Buzz Nov 16 '19 at 0:04
• Thank you for your answer - could you maybe give me a further reference (e.g. to a derivation of the $1/\gamma$ factor)? – Fabian Schn. Nov 16 '19 at 0:19
• @FabianSchn. This answer contains the whole derivation! But if you want more detail, see chapter 14 of Jackson's Electrodynamics. – knzhou Nov 16 '19 at 0:31

It comes from the Lorentz transformations of the wave vectors of the electromagnetic waves.

Imagine a stationary observer and a source moving towards them, emitting isotropically in the source frame of reference.

It is easy to show that for $$\gamma \gg 1$$

(I) Waves emitted towards the observer are appropriately blueshifted and the Poynting vector is boosted by $$\sim 4\gamma^2$$.

(II) Waves emitted at right angles to the observer-source line have their Poynting vector boosted by $$\sim \gamma^2$$, but importantly, their wave vector $${\bf k}$$, is transformed, such that it makes an angle $$\tan^{-1} (1/\gamma) \simeq 1/\gamma$$ to the observer-source line.

In this way, the radiation is both boosted and beamed in the direction of the observer. The cone opening angle is defined roughly by rays that are emitted at right angles to the source-observer line in the source frame of reference. Hence the opening angle of $$\pm 1/\gamma$$.