Difference between revisions of "Synchrotron Self-Interactions"

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m (moved Synchrotron Self-Absorption to Synchrotron Self-Interactions: Consolidating self-absorption, self-Compton, and Compton catastrophe into one page.)
(Consolidated 3 pages into 1.)
Line 37: Line 37:
 
\usepackage{eufrak}
 
\usepackage{eufrak}
 
\begin{document}
 
\begin{document}
Recall that last time we derived for an optically thin synchrotron gas that:
 
\def\pmag{{P_{mag}}}
 
$$j_\nu(\nu_1)\propto P_eP_{mag}^{3\over4}\nu_1^{-\hf}$$
 
$$\begin{aligned}P_{tot}&=P_e+P_{mag}\\
 
&={Cj_\nu\over\pmag^{3\over4}\nu_1^{-\hf}}+\pmag\\ \end{aligned}$$
 
Thus, the minimum total power occurs near the equipartition point between
 
$\pmag$ and $P_e$.  This gives us:
 
\def\jn{j_\nu}
 
$${C\jn\over\pmag^{3\over4}\nu_1^{-\hf}}\sim\pmag$$
 
$${B^2\over8\pi}=\pmag\sim(C\jn\nu_1^\hf)^{4\over7}$$
 
  
 
\subsection*{ Synchrotron Self-Absorption}
 
\subsection*{ Synchrotron Self-Absorption}
 
 
To discuss synchrotron self-absorption, we need to discuss what the spectrum
 
To discuss synchrotron self-absorption, we need to discuss what the spectrum
 
of an optically {\it thick} medium of relativistic electrons looks like.
 
of an optically {\it thick} medium of relativistic electrons looks like.
 
If we take a bunch of $e^-$ spiraling around magnetic field lines with
 
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
 
$\_E=\gamma m_ec^2$, then the energy of the photons emitted by these
electrons is $h\nu\sim h\nu_{crit}\sim h\nu_{cyc}\gamma^2$. If this gas
+
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.
were optically thin, we'd just see a sharply peaked spectrum around $\nu_{crit}$
+
 
with $\nu\ll\nu_{crit}$ going as $\nu^{1\over3}$ and $\nu\gg\nu_{crit}$ going
+
If this gas
as $\nu^\hf e^{-{\nu\over\nu_{crit}}}$. Now let's add more electrons to this
+
were optically thin, we'd just see a sharply peaked spectrum around $\nu_{sync}$
gas.  For a while, the more electrons we add, the more emission we see.
+
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
 
At some point, though, self-absorption starts making a difference, and we
get less emission per electron added.\par
+
get less emission per electron added.
 +
 
 
Let's examine the  
 
Let's examine the  
peak amplitude of emission (that is, $I_\nu$ at $\nu=\nu_{crit}$).   
+
peak amplitude of emission (that is, $I_\nu$ at $\nu=\nu_{sync}$).   
 
To do this,
 
To do this,
 
imagine we have an optically thick ball of blackbody (perfectly absorbing  
 
imagine we have an optically thick ball of blackbody (perfectly absorbing  
 
and emitting) particles
 
and emitting) particles
with a temperature carefully chosen so that $kT\sim\gamma m_e c^2$. Then:
+
with a temperature carefully chosen so that $kT_{\nu}\sim\gamma m_e c^2$. Then:
\def\nucrit{{\nu_{crit}}}
+
\def\nucrit{{\nu_{sync}}}
 
$$I_\nu(\nu=\nucrit)=B_\nu(T\sim{\gamma m_ec^2\over k},\nu=\nucrit)$$
 
$$I_\nu(\nu=\nucrit)=B_\nu(T\sim{\gamma m_ec^2\over k},\nu=\nucrit)$$
Now we'd like to argue that this blackbody particle system is emissively
+
 
identical to
+
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 blackbodies are perfect absorbers and perfect emitters. By taking a blackbody as our model, we are defining the maximum intensity we can observe. The actual observed intensity will be some $I$ below $I_max$, but this gives us a good upper limit.
the optically thick synchrotron gas. To see why this is true, imagine we
+
 
overlaid our sphere of blackbody particles on our sphere of synchrotron gas.
 
Since all particles in it have the same energy, then energy cannot be
 
transfered between the two system by collisions. The only option for exchanging
 
energy is through photons of energy $h\nucrit$.  However, this cannot create
 
a net flow of energy because (I don't know).\par
 
 
If $h\nucrit\ll\gamma m_ec^2$, then:
 
If $h\nucrit\ll\gamma m_ec^2$, then:
 
$$\begin{aligned}I_\nu(\nucrit)^{\e,thick}
 
$$\begin{aligned}I_\nu(\nucrit)^{\e,thick}
Line 85: Line 73:
 
$\gamma\sim\left({\nucrit\over B}\right)^\hf$.  Thus, for an optically
 
$\gamma\sim\left({\nucrit\over B}\right)^\hf$.  Thus, for an optically
 
thick gas:
 
thick gas:
$$I_\nu\propto\nucrit^{5\over2}B^{-\hf}\propto\nu^{5\over2}B^{-\hf}$$
+
$$I_{\nu,max}\propto\nucrit^{5\over2}B^{-\hf}\propto\nu^{5\over2}B^{-\hf}$$
 
It is important to remember that each $\gamma$ has a unique corresponding
 
It is important to remember that each $\gamma$ has a unique corresponding
 
$\nucrit$.  Also note that optical thickness depends inversely on frequency,
 
$\nucrit$.  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
 
so if we plot $I_\nu$ vs. $\nu$, we get a power law $\nu^{(1+p)\over2}$ for
high frequencies (low optical thickness), and a $\nu^{5\over2}$ dependence
+
high frequencies ($I_\nu$ for low optical thickness), and a $\nu^{5\over2}$ dependence
for low frequencies (high optical thickness).  There is also some $\nu_m$ where
+
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.
$I_\nu$ is maximal.
+
 
 +
\title{ 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)}}
 +
 
 +
\subsection*{ Compton Catastrophe}
 +
 
 +
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 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 $U_B\propto B^2$ is pretty obvious: 
 +
$$\nu_m\sim\gamma_m^2\nu_{cyc}\propto\gamma_m^2B$$
 +
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}
 
\end{document}
  
 
<\latex>
 
<\latex>

Revision as of 15:17, 13 December 2016

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Reference Material

<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 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 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 blackbodies are perfect absorbers and perfect emitters. By taking a blackbody as our model, we are defining the maximum intensity we can observe. The actual observed intensity will be some $I$ 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.

\title{ 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\tnTemplate:\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)}}

\subsection*{ Compton Catastrophe}

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 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 $U_B\propto B^2$ is pretty obvious: $$\nu_m\sim\gamma_m^2\nu_{cyc}\propto\gamma_m^2B$$ 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>