# Radiation Lecture 15

<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 } \def\.{\dot } \usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document} \subsection*{ Synchrotron Cooling Time}

To estimate the Synchrotron cooling time, we'll set up our standard expression of self-energy over power radiated: $$t_{cool}\sim{\gamma m_ec^2\over u_Bc\sigma_T\gamma^22\cdot\sin^2\alpha}$$ Instead of doing anything fancy with $\sin^2\alpha$, we'll just use that $\mean{\sin^2\alpha}={2\over3}$. $U_B$ we can estimate as $\mean{B^2}\over 8\pi$, giving us: $$\begin{aligned}t_{cool}&\sim{m_ec^2\over{4\over3}\sigma_Tc}\inv{\gamma U_B}\\ &\sim16yr\left({1\,Gauss\over B}\right)^2\left(\inv{\gamma}\right)\\ \end{aligned}$$ \begin{itemize} \item Let's examine the cooling time for radio jets, where $B\sim1mGauss$, and $\gamma\sim10^3$. Plugging this in, we get $t_{cool}\sim10^4yr$. Compare this to $t_{dyn}\sim{l\over c}\sim{1kpc\over c}\sim10^4yr$.

\item We'll also estimate the cooling time of the Crab Nebula. To set an upper bound on $t_{cool}$, we'll use the most energetic X-rays. We can do this because $\gamma$ uniquely determines the electron energy and as a result, it uniquely determines a photon energy. For the Crab Nebula, we'll pick a photon energy: $E=4keV$. Then: $$\begin{aligned}\omega&\sim\omega_{crit}\sim\gamma^2\omega_{cyc}\\ &\sim{E\over\hbar}\sim\gamma^2{eB\over m_ec}\\ \end{aligned}$$ $B\sim mG$ gives us that $t_{cool}\sim2yrs$. Clearly, since the Crab Nebula was created some 1000 years ago, there must be a source of fresh electrons. This source is the pulsar, sitting in the middle of the nebula. \end{itemize}

\subsection*{ Polarization}

Returning to the non-relativistic cyclotron, recall that when we observe an $e^-$ in a circular orbit from the plane of that orbit, we see linearly polarized photons, and when we observe it from above the plane of the orbit, we see circularly polarized photons. In the synchrotron, the total emission of the $e^-$ (both the linearly polarized photons in the plane of the orbit and the circularly polarized photons perpendicular to the orbit) are swept into the forward direction. Thus, for a single electron, we'd see the top and bottom fringes of the beam are circularly polarized, and the center of the beam is linearly polarized. If we have several electrons, the bottom (circularly polarized) fringe of its beam will overlap the top (oppositely circularly polarized) fringe of the next electrons beam. These will tend to cancel, and we find that synchrotron radiation is generally dominated by linearly polarized photons. The upper bound for how much of the radiation is linearly polarized is set by a perfectly uniform $B$ field, where 75\% is linearly polarized.

\subsection*{ Spectra of Synchrotron Radiation}

The power spectrum of a single $e^-$ undergoing synchrotron radiation peaks at
$\omega_{crit}\sim\gamma^2\omega_{cyc}$. For small $\omega$, $P$ goes as
$\omega^{1\over3}$, and for $\omega\gg\omega_{crit}$, $P$ goes as
$\omega^\hf e^{-\omega}$. In general, recall that $P\propto B^2\gamma^2$.
We would like to calculate the power spectrum of an ensemble of $e^-$. To do
this, we need to describe how many electrons there are per energy. We'll
assume a power law distribution of $e^-$ energies (${dN\over dE}$ goes as
$E^p$, where $p$ is the {\it differential energy spectrum index}). We make this
assumption simply because this coincides with our observations (see Nilsen and
Zager). Let's consider the power radiated by electrons with energies between
$E$ and $E+dE$:
$$\begin{aligned}dP&=dN\times P\eval{single\ e^-}\\
&\propto\left({dN\over dE}dE\right)\gamma^2B^2\propto(E^pdE)E^2B^2\\
&\propto E^{2+p}B^2dE\\ \end{aligned}$$
Thus we have:
$${dP\over dE}\propto E^{2+p}B^2$$
Now let's say that we have a one-to-one relationship between $E$ and $\omega$.
That is, electrons of a specific energy $E$ are solely responsible for
generating photons of frequency $\omega\sim\gamma^2\omega_{cyc}$. Note also
that $\omega\propto\gamma^2B\propto E^2B$, so $E\propto\left({\omega\over B}
\right)^\hf$. Therefore:
$$\begin{aligned}{dP\over d\omega}&={dP\over dE}{dE\over d\omega}\\
&\propto(E^{2+p}B^2)B^{-\hf}\omega^{-\hf}\\
&\propto\left({\omega\over B}\right)^{2+p\over2}B^{3\over2}\omega^{-\hf}\\ \end{aligned}$$
$$\boxed{{dP\over d\omega}\propto B^{1-p\over2}\omega^{1+p\over2}}$$
Usually, $p<0$. Note that our assumption of one-to-one correspondence was
not necessary to get this power law dependence on $\omega$. If hadn't made
that assumption, we would have found that the sum of the contributions of
electrons in nearby energies would have yielded the same result we got, except
near the edges of $\omega$. Also note that this expression relies on an
influx of $e^-$ to replace old ones which cooled down. If we cut off this
influx, we'll see that since $t_{cool}\sim\inv{\gamma B^2}$, the $e^-$ emitting
higher $\omega$ photons decay first, and so we see turn-offs from a power law
distribution for increasingly low $\omega$ as time goes by.\par
For measuring $p$, let's define $\alpha\equiv{1+p\over2}$.
Observations of extended radio sources have measured $\alpha\approx-0.7
\Rightarrow p=-2.4$. In general, we find that $-0.75\le\alpha\le-0.5$, or
$-3\le p\le -2$.
Now we've made an assumption of a constant magnetic field.
Suppose we have an optically thin synchrotron emitting gas with a power law
emissivity
$j_{\nu_1}>j_{\nu_2}$ for $\nu_1<\nu_2$. We will use observations through
this thin gas to infer a ``minimum pressure* or ``minimum energy density*.
The electron pressure (or electric energy density) is given by:
$$\begin{aligned}P_e&\propto\int_{E_1\leftrightarrow\nu_1}^{E_2\leftrightarrow\nu_2}
{{dN \over dE}dE\,E}\\
&\propto E^{2+p}\eval{E_1}^{E_2}\\ \end{aligned}$$
For $p\le-2$,
$$P_e\propto E_1^{2+p}$$
In reality, spectra might cut off outside of the range of our observations.
To correct for this, we'll carry around a correction factor of $\left({E_{min}
\over E_1}\right)^{2+p}$. Measuring $j_\nu$, we get:
$$\begin{aligned}j_\nu(\nu_1\leftrightarrow E_1)&\propto\left({dN\over dE}dE\right){P
\eval{single\ e^-}\over\nu_1}\\
&\propto\left({dN\over dE}dE\right){B^2\gamma^2\over\gamma^2 B}
\propto{dN\over dE}dE{E_1^2B\over E_1^2}\\
&\propto\overbrace{{dN\over dE}dE\,E_1}^{P_e}{E_1B\over E_1^2}\\
&\propto P_e{B\over E_1}\propto P_eB\left({B\over\nu_1}\right)^\hf\\ \end{aligned}$$
$$j_\nu\propto P_eP_{mag}^{3\over4}\nu_1^{-\hf}$$
where $P_{mag}$ is the magnetic energy density.
Now since $P_{tot}=P_e+P_{mag}$:
$$\boxed{P_{tot}={C\,j_\nu\over P_{mag}^{3\over4}\nu^{-\hf}}+P_{mag}}$$
This expression has a minimum for a unique $P_{mag}$. This $P_{mag}$ tells
us the way energy is partitioned in the system between $P_e$ and $P_{mag}$.
The minimum should occur when $P_e=P_{mag}$.

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