# Difference between revisions of "Synchrotron Self-Interactions"

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===Reference Material=== | ===Reference Material=== | ||

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+ | ===Need a Review?=== | ||

+ | * [[Black-Body Radiation]] | ||

+ | * [[Synchrotron Radiation]] | ||

+ | * [[Inverse Compton Scattering]] | ||

+ | * [[Energy Density]] | ||

<latex> | <latex> | ||

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peak amplitude of emission (that is, $I_\nu$ at $\nu=\nu_{sync}$). | 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 [[Black-Body Radiation|blackbody]] (perfectly absorbing |

and emitting) particles | and emitting) particles | ||

with a temperature carefully chosen so that $kT_{\nu}\sim\gamma m_e c^2$. Then: | with a temperature carefully chosen so that $kT_{\nu}\sim\gamma m_e c^2$. Then: | ||

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$$I_\nu(\nu=\nucrit)=B_\nu(T\sim{\gamma m_ec^2\over k},\nu=\nucrit)$$ | $$I_\nu(\nu=\nucrit)=B_\nu(T\sim{\gamma m_ec^2\over k},\nu=\nucrit)$$ | ||

− | 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. | + | 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 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: | If $h\nucrit\ll\gamma m_ec^2$, then: | ||

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Electrons undergoing synchrotron radiation create a photon bath which | Electrons undergoing synchrotron radiation create a photon bath which | ||

− | other electrons will then interact with via | + | 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 | 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 | the number of photons per $\gamma$ is ${dN\over d\gamma}=N_0\gamma^s$, where | ||

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Now we're going to make an approximation that we are on the Rayleigh-Jeans | Now we're going to make an approximation that we are on the Rayleigh-Jeans | ||

− | side of the blackbody curve, so that: | + | side of the [[Black-Body Radiation|blackbody curve]], so that: |

$$\begin{aligned}U_{ph}=U_{ph,sync}&\propto\nu_mI_\nu(\nu_m)\\ | $$\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{2kT_B\over\lambda_m^2}\\ | ||

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where $\nu_m$ is the frequency of peak of synchrotron emission. | where $\nu_m$ is the frequency of peak of synchrotron emission. | ||

− | Now $U_B\propto B^2$ | + | Now $U_B\propto B^2$ (see [[Energy Density]]): |

$$\nu_m\sim\gamma_m^2\nu_{cyc}\propto\gamma_m^2B$$ | $$\nu_m\sim\gamma_m^2\nu_{cyc}\propto\gamma_m^2B$$ | ||

where this $\gamma_m$ is not $\gamax$. Making the approximation that we | where this $\gamma_m$ is not $\gamax$. Making the approximation that we |

## Revision as of 12:11, 5 December 2017

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### Reference Material

### Need a Review?

<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}

\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 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.

\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\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)}}

\section*{ 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 $U_B\propto B^2$ (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>