# Synchrotron Frequency

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

### Synchrotron Characteristic Frequency

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 ${\displaystyle {2\pi \over \omega _{B}}}$, neglecting radiation reactions, so:

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

### 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. 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:

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

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

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

### Finishing Synchrotron Characteristic Frequency

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):

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

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

${\displaystyle {\omega _{sync}={3 \over 2}\gamma ^{2}\omega _{cyc}\sin \alpha }\,\!}$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 46, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}