## FANDOM

167 Pages

Charged particles accelerated by a magnetic field will radiate due to Larmour's formula. For non-relativistic electrons the frequency of radiation is simply the gyration frequency:

$\nu_{cyc}=\frac{eB}{2 \pi m_e c}=3\times 10^{3} \frac{B}{\mbox{mGauss}}\mbox{Hz}$

and the radiation is cyclotron which is weak (in comparison to synchrotron).

For relativistic electrons, the radiation is Synchrotron, which spans a wide range of frequencies and is more powerful. Three different effects modify the cyclotron typical frequency to give the synchrotron one:

1. For relativistic particles, the effective mass is $\gamma$ times larger, causing the gyration frequency to be smaller by $\gamma$.
2. Due to relativistic beaming, the particle’s radiation can only be observed so long as it is within an angle of the line of sight to the observer, so the effective time of emission is shorter by $\gamma$ as compared to the full orbit.
3. As the electron is relativistic it almost “catches up” with the first photon emitted towards the observer, by the time it emits the last photon of the same cycle to reach the observer. Therefore, the delay between detection of first and last photon (of each cycle) is shorter by $\frac{1}{1-\beta} \approx 2\gamma^2$ as compared with the delay between their emission.

Finally, the typical synchrotron frequency is: $\nu_{syn} \approx \frac{\nu_{cyc} \gamma \gamma^2}{\gamma}=\frac{eB\gamma^2}{2 \pi m_e c}$.

Note that as the cut-off due to the last two processes is not sharp, synchrotron will also be emitted at smaller frequencies, but with less power. We therefore obtain a broad spectrum instead of delta function around the typical frequency as in cyclotron. It turns out that the synchrotron spectrum produced by a single electron can be roughly approximated as: $F_{\nu} \propto \nu^{1/3}$ for $\nu<\nu_{syn}$ (where $F_{\nu}$ is the flux per unit frequency) and it falls exponentially above this frequency.

The total power produced by synchrotron emission from a single electron is given by: $P=\frac{4}{3}\sigma_T c \beta^2 \gamma^2 \frac{B^2}{8\pi}$. This can be understood in the following way:

1. In a unit time the electron effectively covers a volume: $c \sigma_T$
2. The magnetic energy in that volume is: $c \sigma_T \frac{B^2}{8\pi}$
3. The electron scatters the magnetic field, amplifying its energy by $\gamma^2$. This is because synchrotron can be viewed as up-scattering by Inverse-Compton of the virtual photons composing the magnetic field. In each Inverse-Compton collision the energy of the outgoing photon is $\gamma^2$ times its initial energy. This can be understood by moving to the electron's rest frame, where the virtual photon is heading towards the electron with an energy $\gamma h\nu$. Assuming that $m_e c^2 \gg \gamma h \nu$, the collision simply causes the photon to bounce backwards from the electron with the same energy $\gamma h\nu$. Now, moving back to the lab frame, the photon's energy is increased by an additional factor of $\gamma$ and we obtain the desired $\gamma^2 h \nu$.

As a side remark, note that the Thomson scattering cross section goes as $m_e ^{-2}$ and as a result the synchrotron power for protons with the same Lorentz factor, will be smaller by: $\bigg(\frac{m_e}{m_p}\bigg)^2\approx 3 \times 10^{-7}$. For this reason it is usually possible to ignore synchrotron radiation from protons in most astrophysical situations.

### Spectrum Edit

Let us consider a single particle rotating at the speed of light. The vector potential is given by

$\vec{A} \left(\vec{r},t \right ) = \frac{1}{c} \int \frac{\vec{J} \left( \vec{r}',t'\right )}{|\vec{r} - \vec{r}'|} \delta \left(t-t'-\frac{|\vec{r}-\vec{r}'|}{c} \right ) d^{3} r' dt'$

The current density is given by

$\vec{J} \left(\vec{r}', t' \right ) = q \dot{\vec{R}} \left(t' \right ) \delta \left(\vec{r}'-\vec{R}\left(t' \right ) \right)$

where $\vec{R}$ is the trajectory of the particle. Substituting into the expression for the vector potential yields

$\vec{A} \left(\vec{r},t \right ) = \frac{q}{c} \int \frac{\dot{\vec{R}} \left(t' \right )}{|\vec{r} - \vec{R} \left(t' \right )|} \delta \left(t-t'-\frac{|\vec{r}-\vec{R} \left(t' \right )|}{c} \right ) dt'$

This expression can be further simplified using the far field assumption

$\vec{A} \left(\vec{r},t \right ) \approx \frac{q}{cr} \int \dot{\vec{R}} \left(t' \right ) \delta \left(t-t'-\frac{r}{c} +\frac{\hat{r} \cdot \vec{R} \left(t' \right )}{c} \right ) dt'$

Taking the Fourier transform

$\tilde{A} \propto \int dt' e^{i \omega \left(t'-r/c-\hat{r} \cdot \vec{R}/c \right )} \dot{R}_{\perp} \left(t' \right )$

Since the particle moves in a circle

$\hat{r} \cdot \vec{R} \left(t' \right ) = R \sin \left(\Omega t' \right ) \approx R \left(\Omega t - \frac{1}{6} \Omega^3 t^3 \right)$

and the component of the velocity perpendicular to the line of sight

$\dot{R}_{\perp} = R \sin \left(\Omega t\right) \approx R \Omega t$

where $\Omega R = c$. The Fourier transform of the vector potential therefore has the form

$\tilde{A} \propto \int t' dt' e^{-i \omega \Omega^2 t^3/6}$

Introducing a new variable $x = t \Omega^{2/3} \omega^{1/3}$ yields

$\tilde{A} \propto \omega^{-2/3}$

The electric field is given by

$\tilde{E} \propto \omega \tilde{A} \propto \omega^{1/3}$

and the energy spectrum therefore scales as $I \propto \tilde{E}^2 \propto \omega^{2/3}$.

This spectrum extends indefinitely only for the case of edge on motion. In the case where the radiation comes from an oblique motion, the spectrum is truncated above some critical frequency. Suppose that the particle moves at an angle $\alpha$ relative to the line of sight. In this case photons emitted from that particle will be to the observer while it is moving on an arc whose angle is also $\alpha$. The time it take the particle to cover this distance is $\alpha R / c$, but the observer sees a pulse that only lasts $R \left(\alpha-\sin \alpha \right)/c \approx \alpha^3 R/c$. Therefore, the maximum frequency above which the spectrum is truncated is $\omega_{m} \propto \alpha^{-3}$. The number of particles that have pitch angle $\alpha$ scales linearly with the angle $N \propto \alpha$. If we divide the particles to different groups according to pitch angle, then each contributes a different peak to the spectrum, and the sum of all peaks yields the spectrum

$I \propto \omega^{2/3} N \propto \omega^{2/3} \alpha \propto \omega^{1/3}$