In computing the time width of the pulses received from an $e^-$ 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): $$\Delta t_{21}={2\over\gamma\omega_B\sin\alpha}$$ 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 \def\tto{{t_{21}}} the emission goes from being $c\Delta\tto$ to $(c-v)\Delta\tto$ (we assume that for the duration of the emission, the $e^-$ is going approximately straight at you). Thus we can get the actual time width of the pulse: $$\Delta t={(c-v)\Delta\tto\over c}=(1-\beta)\Delta\tto$$ For $\gamma\gg1$, $\beta\to1$, so: $$\Delta t\approx\inv{\omega_{cyc}\gamma^2\sin\alpha}$$ This factor of $\inv{\gamma^2}$ comes from the following: $${\gamma_{rel\atop mass}\over \gamma_{{phot\atop chase}\atop phot}^2\gamma_{beam}}$$ Using $\Delta t$, we can deduce the synchrotron frequency. We've just done an approximate computation here. Rybicki \& Lightman do it for real and get $$\boxed{\omega_{sync}={3\over2}\gamma^2\omega_{cyc}\sin\alpha}$$
Notice that the $\gamma^2$ makes synchrotron radiation harder than cyclotron radiation. In a cyclotron, the power radiated into all solid angles is: $$P={2\over3}{e^2a^2\over c^3}$$ Let's derive this for the synchrotron. In the electron frame: $$P^\prime={2\over3}{e^2a^{\prime 2}\over c^3}$$ It turns out that power is a relativistic invariant. To see this, note $P^\prime={dU^\prime\over dt^\prime}$, and we know the following: \begin{matrix} \begin{aligned} U^\prime&=\gamma(U-vp_x)\\ t^\prime&=\gamma(t-{vx\over c^2})\\ \end{aligned}& \begin{aligned} U&=\gamma(u^\prime+vp_x^\prime)\\ t&=\gamma(t^\prime+{v\xp\over c^2})\\ \end{aligned} \end{matrix} and in the prime frame ($s^\prime$), $d\xp=dp_x^\prime=0$.\par So we've shown that once we calculate $P^\prime$, we know $P$ in all frames. What we need to do now is calculate $a^\prime$. We'll do the following: \begin{itemize} \item Get $a^\prime(a)$, the Lorentz transform of acceleration. \item Get $a^\prime$ due to $E^\prime$. \end{itemize} As an exercise in special relativity, we'll derive what $\ef^\prime$ is by \def\sigo{\sigma_0} investigating two parallel plates--the bottom one having charge density $\sigo$ and the top having $-\sigo$ We know in a motionless frame that the field between the plates is: \begin{aligned}E_{y_0}&=4\pi\sigma_0\\ B_0&=0\\ \end{aligned} In a frame ($s$) where we are moving at $v=-v_0\^x$, these become: \begin{aligned}E_y&=4\pi\sigma=4\pi\gamma_0\sigo=\gamma_0E_{y_o}\\ B_z&={4\pi J\over c}={4\pi\over c}\sigma v_0={4\pi\over c}\gamma_0\sigo v_0= E_{y_0}{v_0\over c}\gamma_0\\ \end{aligned} Jumping into one more frame ($s^\prime$) where $v=v^\prime\^x$ relative to frame $s$, we have: \begin{aligned}E_y^\prime&=4\pi\sigma^\prime\\ B_z^\prime={4\pi\over c}\sigma^\prime v^\prime\\ \end{aligned} We need to figure $\sigma^\prime$, and our first instinct might be to say $\sigma^\prime=\gamma^\prime\sigma$, but that is wrong. We have to reference it from $\sigo$: $$\sigma^\prime=\gamma^\prime\sigo$$ Now we do some algebra: \def\gampTemplate:\gamma^\prime \def\gamoTemplate:\gamma 0 \begin{aligned}E_y^\prime&=4\pi\gamp\sigo\\ &=4\pi\gamma\gamo(1-{vv_0\over c^2})\sigo\\ &=\underbrace{4\pi\gamo\sigo}_{E_y}\gamma-\underbrace{4\pi\gamo v_0\sigo\over c}_{B_z}{\gamma v\over c}\\ \end{aligned} Thus we have: $$\begin{matrix} \boxed{E_y^\prime=\gamma E_y-{\gamma v\over c}B_z}& \boxed{E_z^\prime=\gamma E_z+{\gamma v\over c}B_y}& \boxed{B_y^\prime=\gamma B_y+{\gamma v\over c}E_z} \end{matrix}$$ And, of course, $E_x^\prime=E_x$. The only thing we need to get now is $B_x^\prime$. For this we'll talk about a solenoid aligned with the $\^x$ direction with $n$ turns per unit length. The field of this solenoid in the rest frame is: $$B_x={4\pi\over c}nI$$ Jumping to a frame where $\vec v=v\^x$, $B_x^\prime={4\pi n^\prime I^\prime\over c}$. Using that: \begin{aligned}n^\prime&=\gamma n\\ I^\prime&={dQ^\prime\over dt^\prime}={dQ\over\gamma dt}={I\over\gamma}\\ \end{aligned} We find that: $$\boxed{B_x^\prime=B_x}$$ Recall that we've derived all of this for boosts in the $\^x$ direction. To be completely general, we'll write them for any direction: \begin{matrix}\begin{aligned}E_\|^\prime&=E_\|\\ B_\|^\prime&=B_\|\\ \end{aligned}& \begin{aligned}\vec E_\perp^\prime &=\gamma(\ef_\perp+{\vec v\over c}\times\bfield)\\ \vec B_\perp^\prime&=\gamma(\bfield_\perp-{\vec v\over c}\times\ef)\\ \end{aligned} \end{matrix} So finally, back to the synchrotron. Since $e^-$ is at rest in the primed frame: $$\vec a^\prime={e\ef^\prime\over m_e}$$ $$\ef_\|^\prime=\ef_\|=0$$ $$\ef_\perp^\prime=\gamma{\vec v\times\bfield\over c}$$ $$|E_\perp|={\gamma vB\over c}\sin\alpha$$ Therefore, the magnitude of the acceleration is: $$|a^\prime|={e\gamma vB\over m_ec}\sin\alpha$$ And so the power radiated is: $$\boxed{P^\prime={2\over3}{e^4\gamma^2v^2B^2\over m_e^2c^5}\sin^2\alpha}$$ Note that as $v\to c$, $$P\to{2\over3}{e^4\gamma^2B^2\over m_e^2c^3}\sin^2\alpha$$ Thus we get {\it way} more power ($\gamma_{v\to c}^2\over\beta_{v\ll c}^2$) out of the synchrotron. How long can an $e^-$ hold up radiating this kind of power? $$t_{life}\sim{\gamma m_ec^2\over\left({e^4\gamma^2B^2\over m_e^2c^3}\right)}$$ The time it takes an $e^-$ to go in the circle is just: $$t_{orb}\sim{\gamma m_ec\over eB}$$ Taking the ratio of these, we find that the critical $B$ required to make these timescales comparable is: $$\gamma^2B\sim{c^4m_e^2\over e^3}\approx{3^4\e{40}\e{-54}\over 5^3\e{-30}}\approx10^{16}cgs$$ Getting back to P, there is a prettier way of writing it: $$P=\overbrace{2\sigma_Tc\,U_B}^{E\ intercepted\atop by\ e^-} \sin^2\alpha\overbrace{\gamma^2}^{relativistic\atop enhancement}$$ where $\sigma_T={8\pi\over3}r_0^2$ is the Thomson cross-section ($r_0$ being defined by ${e^2\over r_0}=m_ec^2$), and $U_B$ is the energy stored in the magnetic field $U_B={B^2\over8\pi}$.