# Single Electron Power Spectrum

### Single Electron Power Distribution

For the case of non-relativistic cyclotron emission, an observer will see a power spectrum with time that oscillates sinusoidally with a single characteristic frequency. To obtain the power spectrum with frequency, P(${\displaystyle \omega }$), the Fourier transform is taken of the sine wave which gives a delta function at a the characteristic frequency. For synchrotron radiation, however,we have found in earlier lectures that the radiation will be emitted in a narrow beam of angular width ${\displaystyle \sim }$ ${\displaystyle {\frac {1}{\gamma }}}$, so P(t) will be a series of sharp peaks. Taking the Fourier transform of this distribution is not so straight forward. The result is derived in many textbooks, but it is not very instructive, so we will simply state the result:

{\displaystyle {\begin{aligned}{\frac {dP}{d\nu }}&={\frac {{\sqrt {3}}e^{3}Bsin(\alpha )}{m_{e}c^{2}}}{\frac {\nu }{\nu _{c}}}\int \limits _{\frac {\nu }{\nu _{c}}}^{\infty }K_{5/3}(x)dx\\&\equiv {\frac {{\sqrt {3}}e^{3}Bsin(\alpha )}{m_{e}c^{2}}}F\left({\frac {\nu }{\nu _{c}}}\right)\end{aligned}}\,\!}

Where ${\displaystyle \nu _{c}}$ is the critical photon frequency found for synchrotron radiation and ${\displaystyle K_{5/3}}$ is a modified Bessel function. In general this function F is hard to work with, however it has some very nice properties. Note in the plot how:

{\displaystyle {\begin{aligned}F\left({\frac {\nu }{\nu _{c}}}\right)\sim \left({\frac {\nu }{\nu _{c}}}\right)^{1/3}\end{aligned}}\,\!}

for small ${\displaystyle \nu /\nu _{c}}$ and

{\displaystyle {\begin{aligned}F\left({\frac {\nu }{\nu _{c}}}\right)\sim \left({\frac {\nu }{\nu _{c}}}\right)^{1/2}e^{-\left({\frac {\nu }{\nu _{c}}}\right)}\end{aligned}}\,\!}

for large ${\displaystyle \nu /\nu _{c}}$. Furthermore, it is highly peaked around ${\displaystyle \sim }$ .3 ${\displaystyle \nu _{c}}$, which allows us to approximate the power as being radiated at that single frequency when doing further calculations.

To find the total power radiated we can then integrate with respect to ${\displaystyle \nu }$, giving

{\displaystyle {\begin{aligned}P={\frac {{\sqrt {3}}e^{3}Bsin(\alpha )}{m_{e}c^{2}}}\int \limits _{0}^{\infty }F\left({\frac {\nu }{\nu _{c}}}\right)d\nu \end{aligned}}\,\!}

which, making the substitution ${\displaystyle x\equiv \nu /\nu _{c}}$, gives

{\displaystyle {\begin{aligned}P&={\frac {{\sqrt {3}}e^{3}B\sin(\alpha )}{m_{e}c^{2}}}\nu _{c}\int \limits _{0}^{\infty }F\left(x\right)dx\\&={\frac {3{\sqrt {3}}e^{4}\gamma ^{2}B^{2}\sin ^{2}(\alpha )}{4\pi m_{e}^{2}c^{3}}}\int \limits _{0}^{\infty }F\left(x\right)dx\end{aligned}}\,\!}

The integral over F is just some constant factor, so we see that the important scalings are P ${\displaystyle \propto \gamma ^{2}B^{2}}$, which were what we found earlier using the Larmor formula.