# Radiation Lecture 12

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\subsection*{ Synchrotron (Magneto-Bremsstrahlung)}

Synchrotron radiation is radiation caused by a magnetic field accelerating a charge. It is generally relativistic (the Lorentz factor $\gamma\gg1$). We'll discuss this later, but first will do Cyclotrons, where $\gamma\approx1$.

\subsection*{ Cyclotron}

If we have an electron moving in circles in a magnetic field B, then the
frequency of its ``orbit* is:*
$$\omega_{cyc}={eB\over m_ec}$$
We can compute the power pattern radiated by this electron in a direction
$\theta$ (measured from the vector toward the center of the orbit) by the
acceleration of the charge:
$${dP\over d\Omega}={e^2a^2\over c^3}{\sin^2\theta\over4\pi}$$
where $a$ is the acceleration of the charge, given by:
$$a={evB\over cm_e}$$

\subsection*{ Derivation of Power Pattern}

Say that we have an $e^-$ moving in a straight line along the x axis, and
it gets accelerated, starting at $x=0$, for a duration $\Delta t$. Let
$\theta$ measure the angle of a field line emitted by the $e^-$, from the
acceleration axis. The
acceleration of $e^-$ causes a ``jog* in the electric field compared to where*
it would have been (this jog propagates outward at $c$). Say that $E_{tr}$
is the electric field carried in the transverse component
($\perp$ to $\theta$), and
$E_r$ measures the component in the radial component of the jog. Then:
$$\begin{aligned}{E_{tr}\over E_r}&={(a\Delta t)t\sin\theta\over c\Delta t}\\
&={at\sin\theta\over c}\\ \end{aligned}$$
Just to get that factor of $t$ out of there, say that $r=ct$:
$${E_{tr}\over E_r}={ar\sin\theta\over c^2}$$
We know that $E_r={e\over r^2}$, so:
$$\boxed{E_{tr}={ea\sin\theta\over c^2r}}$$
Note how the transverse field goes off as $r^{-1}$, so when
you go far enough away, it always wins out over the plain electric field.\par
What's the transverse magnetic field? Maxwell's equations tell us:
$$\begin{aligned}\dce=-{1\over c}{\partial B\over\partial t}\\
&=\begin{vmatrix}\^x&\^y&\^r\\
{\partial\over\partial x}&{\partial\over\partial y}&{\partial\over\partial r}\\
E_{tr}&0&E_r\end{vmatrix}\\
&=\^x({\partial E_r\over\partial y})+\^y({\partial E_{tr}\over\partial r}
-{\partial E_r\over\partial y})+\^r({\partial E_{tr}\over\partial y})\\
&\approx\^y{\partial E_{tr}\over\partial r}\\ \end{aligned}$$
Say we define $f$ such that:
$$E_{tr}=f(r-ct)$$
then $\vec B=f(r-ct)\^y$, so:
$$\begin{aligned}-{1\over c}{\partial f(r-ct)\^y\over \partial t}
&={-\^y\over c}{\partial f(r-ct)\over \partial(r-ct)}{\partial(r-ct)\over
\partial t}\\
&=\^y{\partial f(r-ct)\over\partial(r-ct)}\\ \end{aligned}$$
Also:
$$\begin{aligned}\dce&={\partial E_{tr}\over\partial r}\\
&=\^y{\partial f(r-ct)\over\partial(r-ct)}{\partial(r-ct)\over\partial r}\\
&=\^y{\partial f(r-ct)\over\partial(r-ct)}\\ \end{aligned}$$
Thus, $E_{tr}$ and $B_{tr}$ have the same magnitude.\par
To cut to the chase, the power radiated from the entire sphere is:
$$\begin{aligned}P_{sphere}&={e^2a^2\over4\pi r^2c^3}\int_0^\pi{\sin^2\theta
2\pi\sin\theta r^2d\theta}\\
&={2\over3}{e^2a^2\over c^3}\\ \end{aligned}$$
This is the Larmor power formula. We can also calculate the Poynting flux:
$${\vec E\times\vec B\over4\pi}c=e^2a^2{\sin^2\theta\over4\pi r^2}$$
Incidentally, you get the same result if you accelerate the charge perpendicular
to the direction it was moving.\par
The {\it axis of polarization} of this radiation is:
$$(\vec a\times\vec r)\times\vec r$$
Remember that $\vec a$ is the acceleration axis, and $\theta$ is always measured
from that axis.
\begin{itemize}
\item Returning to the cyclotron example, if we observe from above
an $e^-$ going in circles, we should see a constant flux (our angle with respect
to the $e^-$ isn't changing), and it will be {\it circularly polarized},
because the axis of polarization goes in circles as the $e^-$ goes in circles.
If we were to observe this circling edge-on, then we'd receive linearly
polarized radiation, and the flux from the accelerating $e^-$ oscillates
with a frequency twice that of the $e^-$'s orbit.
\end{itemize}

Looking at the electron cyclotron emission from solar wind particles, we see that the flux density of the aurora from most planets drops after a frequency of about 1MHz, as dictated by the cyclotron frequency: $$\omega_{cyc}={eB\over m_ec}$$ This tells us that the magnetic field for most planets is about the same. Jupiter is different because it has a very strong magnetic field.

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