# Introduction to Synchrotron Radiation and Relativistic Beaming

### Related Topics

A synchrotron is a relativistic cyclotron. Electrons moving at relativistic speeds spiral around a static B-field, so the Lorentz factor ${\displaystyle \gamma \gg 1}$. As a result, the angular power pattern of an ${\displaystyle e^{-}}$ circling in a B field will take a new form. Instead of having a “donut” of power emitted from the accelerated charge as in a cyclotron, much more of the field is going to be thrown in the forward direction, and much less in the backward direction. This is due to two effects: (1) the length contraction experienced by moving from the electron’s rest frame and (2) the [[Doppler shift|Doppler shifting] due to the electron chasing the photons it has emitted.

In the electron’s rest frame, it is emitting a donut of power in the direction perpendicular to its acceleration. However, in the lab frame, the forward lobe (in the direction of the electron’s motion) gets stretched out, while the backward lobe gets squashed. The resultant forward lobe has a half-angle ${\displaystyle \theta =1/\gamma }$, for a total angular spread of ${\displaystyle 2/\gamma }$ (see below for further derivation details), contributing a ${\displaystyle 1/\gamma }$ effect from the Lorentz transformation.

Doppler shifting occurs because the electron is moving in the direction that it emits photons. In the forward lobe, the photons are blue-shifted by ${\displaystyle \nu _{+}\sim \nu 2\gamma ^{2}}$, while the backward lobe is red-shifted by ${\displaystyle \nu _{-}\sim \nu {\frac {1}{2\gamma ^{2}}}}$ (see below for further details). Overall, this means the forward lobe will have much more power than the backward lobe, besides being stretched out by the Lorentz transformation, creating a characteristic baseball bat-like beaming profile.

### Synchrotron Characteristic Frequency

We can derive that the characteristic cyclotron frequency is ${\displaystyle \omega _{cyc}={eB \over m_{e}c}}$, which is emitted in the characteristic Larmor profile. Because the cyclotron radiation is emitting in all directions perpendicular to the electron’s motion, an observer will observe the electric field varying as a sine wave with ${\displaystyle \omega _{cyc}}$ over time. The Fourier Transform of the observed power will be sharply peaked at the characteristic frequency. However, due to beaming effects, this is not true for an observed synchrotron.

Suppose you are observing an ${\displaystyle e^{-}}$ emitting synchrotron radiation. Since radiation is just being emitted forward by the ${\displaystyle e^{-}}$, you won’t see radiation from the electron very often. In fact, you’ll just see it once per revolution. Each time you do see it, you will see a spike in power. We’d like to figure out long in time these pulses are separated. To do this, we have the following equations for relativistic motion:

{\displaystyle {\begin{aligned}{d \over dt}{\vec {P}}&={e \over c}{\vec {v}}\times {\vec {B}}\\{d \over dt}(\gamma m{\vec {v}})&={e \over c}{\vec {v}}\times {\vec {B}}+e{\vec {E}}\\\end{aligned}}\,\!}
${\displaystyle {d \over dt}(Energy)={d \over dt}(\gamma mc^{2})=e{\vec {v}}\cdot {\vec {E}}\,\!}$

Now ${\displaystyle {\vec {B}}={\vec {B}}_{external}=B{\hat {z}}}$ and ${\displaystyle {\vec {E}}={\vec {E}}_{external}=0}$. There are also contributions of the self-interaction of the electron’s field with the electron, but we’ll neglect these as being a minor perturbation. Then:

${\displaystyle {\gamma m{d{\vec {v}} \over dt}={e \over c}{\vec {v}}\times {\vec {B}}}\,\!}$

Let’s define ${\displaystyle \alpha }$ to be the “pitch angle” between ${\displaystyle {\vec {B}}}$ and ${\displaystyle {\vec {v}}}$ (that is, the angle which makes the ${\displaystyle e^{-}}$ travel in a helix instead of a circle).

Then:

${\displaystyle \gamma m{v_{\perp }^{2} \over r_{p}}={evB\sin \alpha \over c}\,\!}$

where ${\displaystyle r_{p}}$ is the projected radius of orbit, looking down on ${\displaystyle {\vec {B}}}$. Thus:

${\displaystyle \gamma m{(v\sin \alpha )^{2} \over r_{p}}={evB\sin \alpha \over c}\,\!}$
${\displaystyle r_{p}={\gamma mc \over eB}v\sin \alpha \,\!}$

The time to make an orbit is ${\displaystyle {2\pi \over \omega _{rot}}}$, neglecting radiation reactions, so:

${\displaystyle \omega _{rot}={v_{\perp } \over r_{p}}={v\sin \alpha \over r_{p}}={eB \over \gamma mc}={\omega _{cyc} \over \gamma }\,\!}$

Here, the ${\displaystyle \gamma }$ term takes into account that we need to consider the mass of the electron as ${\displaystyle \gamma m_{e}}$ rather than just ${\displaystyle m_{e}}$ in the synchrotron case. We must also take into account the time width over which we view the pulse from the synchrotron emission.

In computing the time width of the pulses received from an ${\displaystyle e^{-}}$ emitting synchrotron radiation, we first measure the time during which the electron emits radiation toward the observer (starting at some initial point 1 and finishing at point 2):

${\displaystyle \Delta t_{21}={2 \over \gamma \omega _{rot}\sin \alpha }\,\!}$

However, during the time the electron was emitting this radiation, it was also moving toward us, so we receive these photons in a shorter burst than they were emitted at. The difference in the actual physical length of the emission goes from being ${\displaystyle c\Delta {t_{21}}}$ to ${\displaystyle (c-v)\Delta {t_{21}}}$ (we assume that for the duration of the emission, the ${\displaystyle e^{-}}$ is going approximately straight at you). Thus we can get the actual time width of the pulse:

${\displaystyle \Delta t={(c-v)\Delta {t_{21}} \over c}=(1-\beta )\Delta {t_{21}}\,\!}$

For ${\displaystyle \gamma \gg 1}$, ${\displaystyle \beta \to 1}$, so:

${\displaystyle \Delta t\approx {1 \over \omega _{cyc}\gamma ^{2}\sin \alpha }\,\!}$

This factor of ${\displaystyle {1 \over \gamma ^{2}}}$ comes from the following:

${\displaystyle {\gamma _{rel \atop mass} \over \gamma _{{phot \atop chase} \atop phot}^{2}\gamma _{beam}}\,\!}$

Using ${\displaystyle \Delta t}$, we can deduce the synchrotron frequency.

That is, if we also take into account the beaming effect (from Lorentz) and the Doppler shift, we get that the peak synchrotron frequency is:

${\displaystyle \omega _{sync}\sim {\frac {\omega _{cyc}}{\gamma _{relmass}}}*(\gamma _{beaming})*(\gamma _{Doppler}^{2})\sim \omega _{cyc}\gamma ^{2}\,\!}$

We’ve just done an approximate computation here. Rybicki & Lightman do it for real and get

${\displaystyle {\omega _{sync}={3 \over 2}\gamma ^{2}\omega _{cyc}\sin \alpha }\,\!}$

## Synchrotron Polarization

(See Polarization for a review of polarization.)

### Cyclotron Polarization

Recall for cyclotron radiation, the emitted power pattern as a function of angle looks like a torus centered around the acceleration vector (right), that is if the center hole’s size is infinitely small. This behavior is due to the ${\displaystyle \sin(\theta )^{2}}$ term in the Larmor Formula. The polarization of Larmor (or cyclotron) radiation is linear and directed along the “kink” in the electric field created by the acceleration of the charge. For Larmor radiation, the polarization is therefore parallel to the acceleration vector (i.e. along the red line in the figure to the right). One way to remember this is that polarization of Larmor radiation is linear and follows a “donut cut” along the emitted torus-shaped power profile (red line), as opposed to a bagel cut (blue line).

Now imagine one non-relativistic electron spiralling around a magnetic field. If we were to look at this electron along the magnetic field direction we would see the electron carve out a circular path around the magnetic field lines (i.e. face-on). Now try to picture the behavior of the donut-shaped power pattern as the electron spirals around the magnetic field. If the electron’s orbit radius is small enough, it’s almost as if we took the donut and spun it on its side, like a nickel on a table top. If the red-line in the figure above represents the linear polarization of Larmor radiation, then we can see that Cyclotron radiation viewed face-on is circularly polarized. Remember that observing face-on refers to being not face-on with respect to the donut, but being face-on with respect to the circular path the electron traces out in its trajectory (which are perpendicular to each other!). This is similar to how a spinning nickel viewed from above has an observed edge that rotates over time.

Now image we view the same non-relativistic electron spiralling around a magnetic field but view it from the side, such that it traces out a trajectory that resembles a dipole (edge-on). You can picture the Larmor radiation pattern and see that the observed polarization is always the same throughout one cycle of the electron: it has linear polarization that is constant over time. Another way to state this is that the polarization of Larmor radiation is linear along the direction of acceleration, and because the projected acceleration vector always lies along a single line for cyclotron emission viewed edge-on, the polarization is also linear and constant over time.

If we don’t view the electron directly edge-on we get elliptical polarization that depends on the inclination angle of our observation.

### Synchrotron Polarization

Synchrotron emission due to highly relativistic electrons focuses the donut-shaped emission pattern into highly collimated beams. While we typically work with diagrams that show the beaming to occur in the plane of the orbit, in practice highly relativistic electrons also have large velocities along the magnetic field lines, which means their beamed radiation is viewed neither directly face-on nor edge-on (see Figure 6.5 in Rybicki and Lightman). This means that synchrotron polarization, for a single electron, is generally elliptically polarized.

The handed-ness of the polarization, however, depends on whether or not us as observers lie directly above or below the center-line of the focused beam (i.e. the point of "maximal radiation") (Figure 6.5). That is, the polarization may be left-hand circular viewed from above, linear viewed straight-on, and right-hand circular viewed from below. Because synchrotron polarization is highly collimated, for any reasonable distribution of particles and pitch angles, we can assume that we will view synchrotron radiation above for the center-line for some electrons and below it for other electrons, meaning we will be seeing both left-handed and right-handed polarized emission at the same time.

These work to cancel themselves out to form mostly linearly polarized emission. For an ensemble of particles following a power-law distribution in energy to the ${\displaystyle -p}$ degree, the intensity of radiation that is linearly polarized can be shown to be about 70 percent of the total intensity: ${\displaystyle \Pi =(p+1)/(p+7/3)}$. Knowing that in practice ${\displaystyle p}$ ranges from 2 to 3, we get back that the degree of polarization is roughly 70 percent (Rybicki and Lightman pg 181). To derive this relationship yourself, see problem 6.5 in Rybicki and Lightman.

## Lorentz and Doppler Shift Derivations

### Length Contraction and Relativistic Beaming

To see the effect of the Lorentz transformation, let’s define the primed frame to be the instantaneous rest frame of an ${\displaystyle e^{-}}$ which, in our frame, is moving in the ${\displaystyle {\hat {x}}}$ direction and being accelerated in the ${\displaystyle {\hat {y}}}$ direction. In the primed frame, the acceleration of the electron (${\displaystyle {a^{\prime }}}$) is still pointing in the ${\displaystyle {\hat {y}}}$ direction. We can relate ${\displaystyle {a^{\prime }}}$ to ${\displaystyle a}$ by noting that if ${\displaystyle {\vec {a^{\prime }}}={a^{\prime }}_{y}{\hat {y}}^{\prime }+{a^{\prime }}_{x}{\hat {x}}^{\prime }}$:

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

In a cyclotron, the power observed in a certain direction depends on the charge’s acceleration perpendicular to that direction (as seen in the Larmor formula). This is true for the synchrotron case as well, though we must also account for the relatavistic beaming effects.

The Lorentz transformation tells us:

{\displaystyle {\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:

{\displaystyle {\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:

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

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

{\displaystyle {\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 ${\displaystyle \theta }$ and ${\displaystyle \theta ^{\prime }}$ for ${\displaystyle \beta \approx 1}$, ${\displaystyle \gamma \gg 1}$:

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

As is evident, photons emitted various angles in the ${\displaystyle e^{-}}$’s rest frame end up being beamed forward in the lab frame.

This has applications in gamma ray bursts. When a star goes supernova, its electrons are accelerated to relativistic velocities. Because of relativistic beaming, we can only see the few electrons whose beams point at us. As the ${\displaystyle e^{-}}$’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.

### Doppler Shifting: Photons Chasing Photons

The photons emitted by the ${\displaystyle e^{-}}$ are being squished together by the fact that the ${\displaystyle e^{-}}$ is itself moving close to the speed of light. This blue-shifts the forward lobe and red-shifts the backward lobe.

${\displaystyle \nu _{+}=\nu _{0}{1 \over 1-\beta }\,\!}$
${\displaystyle \nu _{-}=\nu _{0}(1-\beta )\,\!}$

Where ${\displaystyle \beta =v \over c}$ and in the limit where ${\displaystyle \gamma \gg 1}$, ${\displaystyle 1-\beta \approx {1 \over 2\gamma ^{2}}}$.

We can also say that if some ${\displaystyle e^{-}}$ is spiraling around, there is only a tiny arc over which the ${\displaystyle e^{-}}$ emits photons that we can see. We would like to calculate this arc in order to figure the width of the pulse of radiation an observer sees from synchrotron radiation. The width of the arc over which the ${\displaystyle e^{-}}$ emits radiation that we see (as the electron sweeps its beam past us) is just the width of the beam that the ${\displaystyle e^{-}}$ emits, which is ${\displaystyle 2 \over \gamma }$. The time interval over which this arc is swept out is determined by the time it takes the ${\displaystyle e^{-}}$ to travel an angle of ${\displaystyle 2 \over \gamma }$ around the circle. However, we can also calculate this by noting that the change in v over the interval must come from the acceleration of the ${\displaystyle {\vec {B}}}$ field:

${\displaystyle \gamma m{\Delta v \over \Delta t_{21}}={evB\sin \alpha \over c}\,\!}$
${\displaystyle {\Delta v \over \Delta t_{21}}=\omega _{rot}v\sin \alpha \,\!}$
${\displaystyle \Delta t={\Delta v \over \omega _{B}v\sin \alpha }\,\!}$

Now ${\displaystyle \Delta v\sim v{2 \over \gamma }}$, so:

${\displaystyle {\Delta t_{21}\sim {2 \over \gamma \omega _{rot}\sin \alpha }}\,\!}$