View source for Synchrotron Self-Interactions

[[Radiative Processes in Astrophysics|Course Home]]

===Short Topical Videos===
* [https://youtu.be/HeuuX31Cyq0 Synchrotron Self-Interactions (Aaron Parsons)]

===Reference Material===

===Need to Review?===
* [[Black-Body Radiation]]
* [[Synchrotron Radiation]]
* [[Inverse Compton Scattering]]
* [[Energy Density]]

<latex>
\documentclass[11pt]{article}
\def\inv#1{{1 \over #1}}
\def\ddt{{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\bfield{{\vec B}}
\def\ap{{a^\prime}}
\def\xp{{x^{\prime}}}
\def\yp{{y^{\prime}}}
\def\zp{{z^{\prime}}}
\def\tp{{t^{\prime}}}
\def\upx{{u_x^\prime}}
\def\upy{{u_y^\prime}}
\def\e#1{\cdot10^{#1}}
\def\hf{\frac12}
\def\^{\hat }
\def\.{\dot }
\usepackage{fullpage}
\usepackage{amsmath}
\usepackage{eufrak}
\begin{document}

\section*{ Synchrotron Self-Absorption}
To discuss synchrotron self-absorption, we need to discuss what the spectrum
of an optically {\it thick} medium of relativistic electrons looks like.
If we take a bunch of $e^-$ spiraling around magnetic field lines with
$E=\gamma m_ec^2$, then the energy of the photons emitted by these
electrons is $h\nu\sim h\nu_{sync}\sim h\nu_{cyc}\gamma^2$. This emitted photon can be absorbed be a neighboring relatvistic electron in the self-absorption process.

If this gas
were optically thin, we'd just see a sharply peaked spectrum around $\nu_{sync}$
with $\nu\ll\nu_{sync}$ going as $\nu^{1\over3}$ and $\nu\gg\nu_{sync}$ going
as $\nu^\hf e^{-{\nu\over\nu_{sync}}}$. That is, when we are in the optically thin regime, more electrons in the synchrotron lead to more emission. Now let's add more electrons to this
gas.  For a while in the optically thick regime, the more electrons we add, the more emission we see.

At some point, though, self-absorption starts making a difference, and we
get less emission per electron added.

Let's examine the 
peak amplitude of emission (that is, $I_\nu$ at $\nu=\nu_{sync}$).  
To do this,
imagine we have an optically thick ball of [[Black-Body Radiation|blackbody]] (perfectly absorbing 
and emitting) particles
with a temperature carefully chosen so that $kT_{\nu}\sim\gamma m_e c^2$. Then:
\def\nucrit{{\nu_{sync}}}
$$I_\nu(\nu=\nucrit)=B_\nu(T\sim{\gamma m_ec^2\over k},\nu=\nucrit)$$

We are assuming we are in the Rayleigh-Jeans tail. Note that we are not arguing that we are operating in thermal equilibrium. Instead, we know that [[Black-Body Radiation|blackbodies]] are perfect absorbers and perfect emitters. By taking a blackbody as our model, we are defining the maximum intensity, $I_{max}$ we can observe. The actual observed intensity will be below $I_{max}$, but this gives us a good upper limit.

If $h\nucrit\ll\gamma m_ec^2$, then:
$$\begin{aligned}I_\nu(\nucrit)|_{e^-,thick}
&=B_\nu(T\sim{\gamma m_ec^2\over k},\nu=\nucrit)\\ 
&={2kT\over\lambda_{crit}^2}={2\gamma m_ec^2\over\lambda_{crit}^2}\\ 
&=2\gamma m_e\nucrit^2\\ \end{aligned}$$
Since $\nucrit\sim\nu_{cyc}\gamma^2\propto B\gamma^2$, we have that
$\gamma\sim\left({\nucrit\over B}\right)^\hf$.  Thus, for an optically
thick gas:
$$I_{\nu,max}\propto\nucrit^{5\over2}B^{-\hf}\propto\nu^{5\over2}B^{-\hf}$$
It is important to remember that each $\gamma$ has a unique corresponding
$\nucrit$.  Also note that optical thickness depends inversely on frequency,
so if we plot $I_\nu$ vs. $\nu$, we get a power law $\nu^{(1+p)\over2}$ for
high frequencies ($I_\nu$ for low optical thickness), and a $\nu^{5\over2}$ dependence
for low frequencies ($I_\nu$ for high optical thickness, under self-absorption).  This implies that there will be some turnover point at $\nu_m$ where $I_\nu$ is maximal. This turnover point can give us information about the strength of the B-field and the density of electroncs in the region.

\section*{ Synchrotron Self-Compton (SSC)}

Electrons undergoing synchrotron radiation create a photon bath which
other electrons will then interact with via [[Inverse Compton Scattering]].  For original [[Synchrotron Radiation]], that $F_\nu$, between
some minimum and maximum frequency cut-off, goes as $K\nu^\alpha$, and that
the number of photons per $\gamma$ is ${dN\over d\gamma}=N_0\gamma^s$, where
$\alpha={1+s\over2}$.  These frequency cut-offs were set by $\gamma_{min}^2
\nu_{cyc}$ and $\gamma_{max}^2\nu_{cyc}$.  After this radiation is processed
by SSC, approximately every photon is upscattered to a new energy
${4\over3}\gamma^2\nu$.  We are assuming that the relationship between
an incoming photon frequency and its final frequency are related via a
delta function.  Thus:
\def\tn{{\tilde\nu}}
$$F_{\nu,SSC}(\nu)=\tau\int_{\tn}{K\tn^\alpha d\tn\delta\left(\tn-
{\nu\over\gamma^2}\right)\int_\gamma{N_0\gamma^sd\gamma}}$$
Keep in mind that $N_0$ is normalized to so the integral comes out to 1
(it just accounts for the 
``shape'' of the energy distribution function). $\tau$ is what contains the actual
\# density of $e^-$'s.  It is the fraction scattered,
and is generally $\ll1$. $\nu\sim\tn\gamma^2$.\par
For a fixed $\nu\sim\tn\gamma^2$, we find that $\gamma\sim\left({\nu\over
\tn}\right)^\hf\propto\tn^{-\hf}$.

\def\numin{{\nu_{min}}}
\def\numax{{\nu_{max}}}
\def\gamin{\gamma_{min}}
\def\gamax{\gamma_{max}}
\def\gul{{min(\sqrt{\nu\over\numin},\gamax)}}
\def\gll{{max(\sqrt{\nu\over\numax},\gamin)}}
In deriving the interaction of synchrotron radiation with
synchrotron electrons, we derived the following formula for flux:
$$F_{\nu,SSC}(\nu)=\tau\int_{\tn}{K\tn^\alpha d\tn\delta\left(\tn-
{\nu\over\gamma^2}\right)\int_\gamma{N_0\gamma^sd\gamma}}$$
Now the integral over $\gamma$ is along ``slant paths'' through
the rectangle ($\gamin\to\gamax$, $\numin\to\numax$).
Some of these slant paths will not stretch all the way to $\gamax$
or $\gamin$ because of the boundaries imposed by $\numin$ and
$\numax$.  That is to say, photons emitted by an electron at $\gamin$ can be scattered off another electron at $\gamax$ or vice versa. So we need to be a little more precise about the bounds
on the $\gamma$ integral:
$$\begin{aligned}F_{\nu,SSC}(\nu)
&\sim\tau\int_\numin^\numax{K\tn^\alpha\delta(\tn-{\nu\over\gamma^2})d\tn
\int_\gll^\gul{N_0\gamma^sd\gamma}}\\ 
&\sim\tau K\left({\nu\over\gamma^2}\right)^\alpha
\int_\gll^\gul{N_0\gamma^sd\gamma}\\ 
&\sim\tau K\nu^\alpha N_0
\int_\gll^\gul{{\gamma^s\over\gamma^{2\alpha}}d\gamma}\\ \end{aligned}$$
But since $\alpha={1+s\over2}$, the integrand is simply $\inv{\gamma}$,
giving us:
$$\boxed{F_{\nu,SSC}(\nu)\sim\tau k\nu^\alpha N_0\ln\left({\gul\over\gll}\right)}$$
Recall that $N_0$ was normalized so that $\int{N_0\gamma^sd\gamma}=1$, so
saying that $\gamax$ is just some multiple of $\gamin$, it must be that
$$N_0\sim\gamin^{-1-s}$$
Note for $\nu\sim\gamin^2\numin$, $F_\nu$ looks like (using the above 
relationship):
\def\gllogul{\left({\gul\over\gll}\right)}
$$\begin{aligned}F_{\nu,SSC}&=\tau K\gamin^{2\alpha}\numin^\alpha N_0\ln\gllogul\\ 
&=\tau K\numin^\alpha\gamin^{2\alpha}\gamin^{-1-s}\ln\gllogul\\ 
&=\tau K \numin^\alpha\ln\gllogul\\ \end{aligned}$$

\def\gul{{min(\sqrt{\nu\over\numin},\gamax)}}
\def\gll{{max(\sqrt{\nu\over\numax},\gamin)}}


[[File:Synchrotron_Self_Interaction.png|500px|thumb|center|Spectra for synchrotron radiation, with and without self-interactions. The model used to generate this figure assumes a rectangular slab as the emitting/absorbing region and that the initially emitted photons only interact with other electrons once before leaving the radiating slab. The radiating region is 1 pc thick, has an electron number density of 1 cm$^{-3}$, and is immersed in a uniform magnetic field of 100 $\mu$G. The electrons obey a power law distribution in energy, with a minimum Lorentz factor of 10, a maximum Lorentz factor of 10,000, and a power-law index of -2.5. Note that $I_\nu$ has a slope of 5/2 for low frequencies, and a slope of -3/4 after the turnover, as expected from the above discussion about self-absorption. Also note the bounds on the range of frequencies for the SSC spectrum\textemdash the lower bound is $\gamma_\mathrm{min}^2$ times greater than the minimum frequency of the original synchrotron spectrum, and the upper bound is $\gamma_\mathrm{max}^2$ times greater than the upper bound of the original spectrum.}]]

\section*{ Compton Catastrophe}

\def\gamax{\gamma_{max}}

If you keep scattering the same electrons via Synchrotron Self-Compton and things are dense enough,
there is a danger of a runaway amplification
of radiation energy density, or a ``Compton Cooling Catastrophe''.  However,
we've never seen anything with a brightness temperature of $10^{12}K$.  
What sets the ``inverse Compton limit'' at this temperature?  Comparing,
for a single electron,
the luminosity of inverse Compton scattering to synchrotron scattering:
$${L_{IC}\over L_{sync}}={{4\over3}\beta^2\gamma^2\sigma_TcU_{ph}\over
{4\over3}\beta^2\gamma^2\sigma_TcU_B}={U_{ph}\over U_B}
\begin{cases} >1&catastrophe\\ <1 &no\ catastrophe\end{cases}$$

Now we're going to make an approximation that we are on the Rayleigh-Jeans
side of the [[Black-Body Radiation|blackbody curve]], so that:
$$\begin{aligned}U_{ph}=U_{ph,sync}&\propto\nu_mI_\nu(\nu_m)\\ 
&\propto\nu_m{2kT_B\over\lambda_m^2}\\ 
&\propto\nu_m^3T_B\\ \end{aligned}$$
where $\nu_m$ is the frequency of peak of synchrotron emission. 

Now $U_B\propto B^2$ (see [[Energy Density]]):  
$$\nu_m\sim\gamma_m^2\nu_{cyc}\propto\gamma_m^2B$$
where this $\gamma_m$ is not $\gamax$.  Making the approximation that we
are in the optically thick synchrotron spectrum, so that $\gamma m_ec^2\sim
kT$, then we get $\nu_m\sim T_B^2B$.  We can say that the kinetic temperature
is the brightness temperature because we are talking about the average kinetic
energy of the electrons generating the synchrotron radiation with a particular
brightness temperature (i.e. another frequency of synchrotron radiation will
have another brightness temperature, and another set of electrons moving
with a different amount of kinetic energy). Thus,
$${U_{ph}\over U_B}=C{\nu_m^3T_B\over\nu_m^2}T_B^4
=\left({\nu_m\over10^9Hz}\right)\left({T_B\over10^{12}K}\right)^5=1$$

A way of think about this is that, in order to avoid having infinite energy
in this gas of electrons, there has to be a limit on the brightness 
temperature (which is determined by the density of electrons).  This is a
self-regulating process--if the brightness temperature goes too high, an
infinite energy demand is set up, knocking it back down.

\end{document}

</latex>