Introduction to Synchrotron Radiation and Relativistic Beaming

Short Topical Videos

• The Basics of Synchrotron Emission (Aaron Parsons, UCB)

Synchrotron Radiation

The Synchrotron is a relativistic cyclotron, so the Lorentz factor $\gamma \gg 1$. As a result, the angular power pattern of an $e^{-}$ circling in a B field will take a new form. Instead of having a “donut” of power emitted from the accelerated charge, much more of the field is going to be thrown in the forward direction, and much less in the backward direction. To see this, let’s define the primed frame to be the instantaneous rest frame of an $e^{-}$ which, in our frame, is moving in the ${\hat {x}}$ direction and being accelerated in the ${\hat {y}}$ direction. In the primed frame, the acceleration of the electron (${a^{\prime }}$) is still pointing in the ${\hat {y}}$ direction. We can relate ${a^{\prime }}$ to $a$ by noting that if ${\vec {a^{\prime }}}={a^{\prime }}_{y}{\hat {y}}^{\prime }+{a^{\prime }}_{x}{\hat {x}}^{\prime }$:

${\begin{aligned}a_{x}&={{a^{\prime }}_{x} \over \gamma ^{3}(1+{vu_{x}^{\prime } \over c^{2}})}\\a_{y}&={{a^{\prime }}_{y} \over \gamma ^{2}(1+{vu_{x}^{\prime } \over c^{2}})}\\\end{aligned}}\,\!$

where ${u_{x}^{\prime }}$ is the observed velocity in the primed frame: ${u_{x}^{\prime }}\equiv {d{x^{\prime }} \over d{t^{\prime }}}$.

• Proof: The Lorentz transformation tells us:

${\begin{aligned}{x^{\prime }}&=\gamma (x-vt)\\{y^{\prime }}&=y\\{z^{\prime }}&=z\\{t^{\prime }}&=\gamma (t-{vx \over c^{2}})\\\end{aligned}}\,\!$

Taking derivatives of these equations:

${\begin{aligned}d{x^{\prime }}&=\gamma (dx-vdt)\\d{y^{\prime }}&=dy\\d{z^{\prime }}&=dz\\d{t^{\prime }}&=\gamma (dt-{vdx \over c^{2}})\\\end{aligned}}\,\!$

Thus:

${\begin{aligned}{dx \over dt}&={d{x^{\prime }}+vd{t^{\prime }} \over d{t^{\prime }}+{vdxp \over c^{2}}}\\&={{u_{x}^{\prime }}+v \over 1+{vu_{x} \over c^{2}}}\\{dy \over dt}&={{u_{y}^{\prime }} \over \gamma (1+{vu_{x} \over c^{2}})}\\\end{aligned}}\,\!$

Defining $\theta$ to be the angle of a vector from the ${\hat {x}}$ direction (and similarly, $\theta ^{\prime }$ from ${\hat {x}}^{\prime }$), then:

${\begin{aligned}\tan \theta &={u_{y} \over u_{x}}={{u_{y}^{\prime }} \over \gamma ({u_{x}^{\prime }}+v)}\\&={\tan \theta ^{\prime } \over \gamma (1+{v \over {u_{x}^{\prime }}})}\\\end{aligned}}\,\!$

Since ${u_{x}^{\prime }}=c\cos \theta ^{\prime }$:

${\begin{aligned}\tan \theta &={\tan \theta ^{\prime } \over \gamma (1+{v \over c\cos \theta ^{\prime }})}\\&={\sin \theta ^{\prime } \over \gamma (\cos \theta ^{\prime }+\beta )}\\\end{aligned}}\,\!$

Here’s a chart relating $\theta$ and $\theta ^{\prime }$ for $\beta \approx 1$, $\gamma \gg 1$:

${\begin{matrix}\theta ^{\prime }&\theta \\0^{\circ }&0^{\circ }\\45^{\circ }&{{\sqrt {2}} \over 2+{\sqrt {2}}}{1 \over \gamma }\ll 1\\90^{\circ }&{1 \over \gamma }\\135^{\circ }&{{\sqrt {2}} \over 2-{\sqrt {2}}}{1 \over \gamma }\\180^{\circ }&180^{\circ }\\\end{matrix}}\,\!$

As is evident, photons emitted various angles in the $e^{-}$’s rest frame end up being beamed forward in the lab frame. This has applications in gamma ray bursts. When a star goes supernova, it’s electrons are accelerated to relativistic velocities. Because of relativistic beaming, we can only see the few electrons whose beams point at us. As the $e^{-}$’s slow down (because they are radiating power), they stop being relativistic, and we get to see radiation from a larger angle. Thus, the flux curve is “flattened” for times shortly after a star goes supernova.