# Radiation Lecture 13

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\usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document} \subsection*{ Clarification on Milne Relation}

Recall that we had an expression for the rate of photoionization: $$R_{ion}={B_\nu4\pi d\nu\over h\nu}n_0\sigbf \overbrace{\left(1-{g_0\over g_+}{n_+\over n_0}\right)}^{{correction\ for\atop stimulated}\atop recombination}$$ It looks as though, for a highly ionized gas, the correction for stimulated emission (which we may alternately write as $1-e^{-h\nu\over kT}$) could become negative (which is bad). Defining G to be the ratio of the rate stimulated recombination to the rate of spontaneous recombination, so that: $$R_{ion}={4\pi B_\nu\over h\nu}d\nu\sigbf(\nu)\inv{1+G}$$ Then by analogy with line emission: $$G={n_2\bto\_J\over n_2\ato}={\_J\over {2h\nu^3\over c^2}}$$ Recall that $\_J=\int_0^\infty{J_\nu\phi(\nu)d\nu}$, and in thermal equilibrium $J_\nu=\inv{4\pi}\int{I_\nu d\Omega}$, so assuming that $\phi(\nu)$ is about a delta function, $\_J\approx B_\nu$. Thus for G, we get: $$G={B_\nu\over{2h\nu^3\over c^2}}$$ We know by conservation of energy that for stimulated emission: $$\begin{aligned}h\nu+\hf m_ev_e^2&=2h\nu-\chi\\ h\nu&=\hf m_ev_e^2+\chi>0\\ \end{aligned}$$ so it turns out that the stimulated emission term cannot be negative. In general, if you're unsure about whether to include the correction for stimulated recombination, compute G, and it will tell you.

\subsection*{ Synchrotron Radiation}

The Synchrotron is a relativistic cyclotron, so the Lorentz factor
$\gamma\gg1$. As a result, the angular power pattern of an $e^-$ circling in a B
field will take a new form. Instead of having a ``donut* of power emitted*
from the accelerated charge, much more of the field is going to be thrown in
the forward direction, and much less in the backward direction. To see this,
let's define the {\it primed frame} to be the instantaneous rest frame of an
$e^-$ which, in our frame, is moving in the $\^x$ direction and being
accelerated in the $\^y$ direction.
In the primed frame, the acceleration of the electron ($\ap$) is still pointing
in the $\^y$ direction. We can relate $\ap$ to $a$ by
noting that if $\vec\ap=\ap_y\^y^\prime+\ap_x\^x^\prime$:
$$\begin{aligned}a_x&={\ap_x\over\gamma^3(1+{vu_x^\prime\over c^2})}\\
a_y&={\ap_y\over\gamma^2(1+{vu_x^\prime\over c^2})}\\ \end{aligned}$$
where $\upx$ is the observed velocity in the primed frame:
$\upx\equiv{d\xp\over d\tp}$.
\begin{itemize}
\item Proof: The Lorentz transformation tells us:
$$\begin{aligned}\xp&=\gamma(x-vt)\\
\yp&=y\\
\zp&=z\\
\tp&=\gamma(t-{vx\over c^2})\\ \end{aligned}$$
Taking derivatives of these equations:
$$\begin{aligned}d\xp&=\gamma(dx-vdt)\\
d\yp&=dy\\
d\zp&=dz\\
d\tp&=\gamma(dt-{vdx\over c^2})\\ \end{aligned}$$
Thus:
$$\begin{aligned}{dx\over dt}&={d\xp+vd\tp\over d\tp+{vdxp\over c^2}}\\
&={\upx+v\over1+{vu_x\over c^2}}\\
{dy\over dt}&={\upy\over\gamma(1+{vu_x\over c^2})}\\ \end{aligned}$$
Defining $\theta$ to be the angle of a vector from the $\^x$ direction (and
similarly, $\theta^\prime$ from $\^x^\prime$), then:
$$\begin{aligned}\tan\theta&={u_y\over u_x}={\upy\over\gamma(\upx+v)}\\
&={\tan\theta^\prime\over\gamma(1+{v\over\upx})}\\ \end{aligned}$$
Since $\upx=c\cos\theta^\prime$:
$$\begin{aligned}\tan\theta&={\tan\theta^\prime\over\gamma(1+{v\over
c\cos\theta^\prime})}\\
&={\sin\theta^\prime\over\gamma(\cos\theta^\prime+\beta)}\\ \end{aligned}$$
\end{itemize}

Here's a chart relating $\theta$ and $\theta^\prime$ for $\beta\approx1$,
$\gamma\gg1$:
$$\begin{matrix}
\theta^\prime&\theta\\
0^\circ&0^\circ\\
45^\circ&{\sqrt{2}\over2+\sqrt{2}}{1\over\gamma}\ll1\\
90^\circ&\inv{\gamma}\\
135^\circ&{\sqrt{2}\over2-\sqrt{2}}\inv{\gamma}\\
180^\circ&180^\circ\\
\end{matrix}$$
As is evident, photons emitted various angles in the $e^-$'s rest frame
end up being beamed forward in the lab frame.
This has applications in gamma ray bursts. When a star goes supernova,
it's electrons are accelerated to relativistic velocities. Because of
relativistic beaming, we can only see the few electrons whose beams
point at us. As the $e^-$'s slow down (because they are radiating power),
they stop being relativistic, and we get to see radiation from a larger
angle. Thus, the flux curve is ``flattened* for times shortly after*
a star goes supernova.

\subsection*{ Synchrotron Characteristic Frequency}

Suppose you are observing an $e^-$ emitting synchrotron radiation. Since
radiation is just being emitted forward by the $e^-$, you won't see radiation
from the electron very often. In fact, you'll just see it once per revolution.
Each time you do see it, you will see a spike in power. We'd like to figure
out long in time these pulses are separated. To do this, we have the
following equations for relativistic motion:
$$\begin{aligned}\ddt\vec P&={e\over c}\vec v\times\vec B\\
\ddt(\gamma m\vec v)&={e\over c}\vec v\times\vec B+e\vec E\\ \end{aligned}$$
$$\ddt(Energy)=\ddt(\gamma mc^2)=e\vec v\cdot\vec E$$
Now $\vec B=\vec B_{external}=B\^z$ and $\vec E=\vec E_{external}=0$. There
are also contributions of the self-interaction of the electron's field with
the electron, but we'll neglect these as being a minor perturbation. Then:
$$\boxed{\gamma m{d\vec v\over dt}=e\vec v\times\vec B}$$
Let's define $\alpha$ to be the ``pitch angle* between $\vec B$ and $\vec v$*
(that is, the angle which makes the $e^-$ travel in a helix instead of a
circle).
Then:
$$\gamma m{v_\perp^2\over r_p}={evB\sin\alpha\over c}$$
where $r_p$ is the projected radius of orbit, looking down on $\vec B$. Thus:
$$\gamma m{(v\sin\alpha)^2\over r_p}={evB\sin\alpha\over c}$$
$$r_p={\gamma mc\over eB}v\sin\alpha$$
The time to make an orbit is ${2\pi\over\omega_B}$, neglecting radiation
reactions, so:
$$\omega_B={v_\perp\over r_p}={v\sin\alpha\over r_p}={eB\over\gamma mc}
={\omega_{cyc}\over\gamma}$$

\subsection*{ Photons Chasing Photons}

The photons emitted by the $e^-$ are being squished together by the fact that the $e^-$ is itself moving close to the speed of light. If some $e^-$ is spiraling around, there is only a tiny arc over which the $e^-$ emits photons that we can see. We would like to calculate this arc in order to figure the width of the pulse of radiation an observer sees from synchrotron radiation. The width of the arc over which the $e^-$ emits radiation that we see (as the electron sweeps its beam past us) is just the width of the beam that the $e^-$ emits, which is $2\over\gamma$. The time interval over which this arc is swept out is determined by the time it takes the $e^-$ to travel an angle of $2\over\gamma$ around the circle. However, we can also calculate this by noting that the change in v over the interval must come from the acceleration of the $\vec B$ field: $$\gamma{\Delta v\over\Delta t_{21}}={evB\sin\alpha\over c}$$ $${\Delta v\over\Delta t_{21}}=\omega_Bv\sin\alpha$$ $$\Delta t={\Delta v\over \omega_B v\sin\alpha}$$ Now $\Delta v\sim v{2\over\gamma}$, so: $$\boxed{\Delta t_{21}\sim{2\over\gamma\omega_B\sin\alpha}}$$

\end{document}
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