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Synchrotron radiation results from a particle gyrating around a magnetic field line. Following full relativistic theory gives
\[\frac{d}{dt} \left[ \gamma m \vec{v}\right]= \frac{q}{c} \vec{v}\times \vec{B}\] \[\frac{d}{dt} \left[ \gamma m c^2\right]= q \vec{v}\cdot \vec{E}\]With no electric field, it is apparent that $\gamma$ and $|\vec{v} |$ are constant. Using this, the other equation implies
\[\gamma m \frac{d \vec{v}}{dt} = \frac{q}{c} \vec{v}\times \vec{B}\]Dividing into components of the velocity parallel $\parallel$ and perpendicular $\perp$ to the magnetic field, I get
\[\frac{d v_\parallel}{dt} =0\] \[\frac{d \vec{v}_\perp}{dt} = \frac{q}{mc \gamma } \vec{v}_\perp \times \vec{B}\]This drives helical motion in the plane perpendicular to $\vec{B}$ with a rotation frequency $\omega_B = \frac{qB}{\gamma m c}$, also known as the relativistic Larmor frequency.
Using the relativistic formula with the above equations, we get the radiated power from the particle’s interaction with the magnetic field:
\[a_\perp = \omega_B v_\perp = \omega_B v \sin \alpha\] \[P = \frac{2q^2 }{3c^3 } \gamma^4 \left( \frac{q B}{mc} \frac{1}{\gamma }v \sin \alpha \right)^2 = \frac{2}{3}\frac{q^4}{m^2 c^4} \frac{B^2}{c} v^2 \sin^2 \alpha \gamma^2 = \frac{2}{3} r_q^2 B^2 c \beta ^2 \gamma^2 \sin^2 \alpha\]where $\alpha$ is the pitch angle, the angle between $\vec{v}$ and $\vec{B}$, $r_q = \frac{q^2}{m c^2}$ is the particle’s classical radius, and $\beta = \frac{v}{c}$ is the particle’s speed in terms of the speed of light.
Synchrotron spectrum
For the spectrum of synchrotron raditation, we need to consider that the particle’s radiation is beamed as it rotates around the magnetic field line. As it traces out a portion of arc $\Delta s$ over an angular displacement $\Delta \theta$. Also, for these small differentials, we have
\[\Delta \theta = \frac{\|\Delta \vec{v}\|}{v}\] \[\Delta s = v \Delta t\]The acceleration is
\[a = \Omega_B v \sin \alpha =\frac{\|\Delta \vec{v}\|}{\delta t} = \frac{v^2 \Delta \theta}{\Delta s}\]giving an equation
\[\frac{\Delta \theta}{\Delta s} = \frac{1}{v} \Omega_B \sin \alpha\]Assuming the particle is radiating in a cone with angle $\Delta \theta \approx \frac{2}{\gamma}$, the change in arc is
\[\Delta s \approx \frac{2 v}{\gamma \Omega_B \sin \alpha }\] \[\Delta t_\mathrm{emit} \approx \frac{2 }{\gamma \Omega_B \sin \alpha }\]The arrival time interval is $\Delta t_\mathrm{obs}$ less than $\Delta t_\mathrm{emit}$ because the radiation travels faster than the particle:
\[\Delta t_\mathrm{emit} -\Delta t_\mathrm{obs} =\frac{\Delta s}{c} = t_\mathrm{emit} \frac{v}{c}\] \[\Delta t_\mathrm{obs} = \Delta t_\mathrm{emit} \left( 1 - \frac{v}{c} \right)\]For large $\gamma$, the extra factor is
\[\left( 1 - \frac{v}{c} \right) \approx \frac{1}{2\gamma^2}\]giving an observed time interval
\[\Delta t_\mathrm{obs} = \frac{1 }{\gamma^3 \Omega_B \sin \alpha }\]This defines a critical frequency
\[\omega_c = \frac{3}{2} \gamma^3 \omega_B \sin \alpha\]where the spectrum of synchrotron radiation falls off. I.e. this is the highest frequency emission from a particle emitting synchrotron radiation.