<latex> \documentclass[11pt]{article} \def\inv#1Template:1 \over \def\ddtTemplate:D \over dt \def\mean#1{\left\langle {#1}\right\rangle} \def\sigot{\sigma_{12}} \def\sigto{\sigma_{21}} \def\eval#1{\big|_{#1}} \def\tr{\nabla} \def\dce{\vec\tr\times\vec E} \def\dcb{\vec\tr\times\vec B} \def\wz{\omega_0} \def\ef{\vec E} \def\ato{{A_{21}}} \def\bto{{B_{21}}} \def\bot{{B_{12}}} \def\bfieldTemplate:\vec B \def\apTemplate:A^\prime \def\xp{{x^{\prime}}} \def\yp{{y^{\prime}}} \def\zp{{z^{\prime}}} \def\tp{{t^{\prime}}} \def\upxTemplate:U x^\prime \def\upyTemplate:U y^\prime \def\e#1{\cdot10^{#1}} \def\hf{\frac12} \def\^{\hat } \usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document}

\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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