Synchrotron Self-Interactions

Synchrotron Self-Absorption

To discuss synchrotron self-absorption, we need to discuss what the spectrum of an optically thick medium of relativistic electrons looks like. If we take a bunch of ${\displaystyle e^{-}}$ spiraling around magnetic field lines with ${\displaystyle E=\gamma m_{e}c^{2}}$, then the energy of the photons emitted by these electrons is ${\displaystyle h\nu \sim h\nu _{sync}\sim h\nu _{cyc}\gamma ^{2}}$. 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 ${\displaystyle \nu _{sync}}$ with ${\displaystyle \nu \ll \nu _{sync}}$ going as ${\displaystyle \nu ^{1 \over 3}}$ and ${\displaystyle \nu \gg \nu _{sync}}$ going as ${\displaystyle \nu ^{\frac {1}{2}}e^{-{\nu \over \nu _{sync}}}}$. 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, ${\displaystyle I_{\nu }}$ at ${\displaystyle \nu =\nu _{sync}}$). To do this, imagine we have an optically thick ball of blackbody (perfectly absorbing and emitting) particles with a temperature carefully chosen so that ${\displaystyle kT_{\nu }\sim \gamma m_{e}c^{2}}$. Then:

${\displaystyle I_{\nu }(\nu ={\nu _{sync}})=B_{\nu }(T\sim {\gamma m_{e}c^{2} \over k},\nu ={\nu _{sync}})\,\!}$

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 ${\displaystyle I}$ below ${\displaystyle I_{m}ax}$, but this gives us a good upper limit.

If ${\displaystyle h{\nu _{sync}}\ll \gamma m_{e}c^{2}}$, then:

{\displaystyle {\begin{aligned}I_{\nu }({\nu _{sync}})^{\cdot 10^{,}thick}&=B_{\nu }(T\sim {\gamma m_{e}c^{2} \over k},\nu ={\nu _{sync}})\\&={2kT \over \lambda _{crit}^{2}}={2\gamma m_{e}c^{2} \over \lambda _{crit}^{2}}\\&=2\gamma m_{e}{\nu _{sync}}^{2}\\\end{aligned}}\,\!}

Since ${\displaystyle {\nu _{sync}}\sim \nu _{cyc}\gamma ^{2}\propto B\gamma ^{2}}$, we have that ${\displaystyle \gamma \sim \left({{\nu _{sync}} \over B}\right)^{\frac {1}{2}}}$. Thus, for an optically thick gas:

${\displaystyle I_{\nu ,max}\propto {\nu _{sync}}^{5 \over 2}B^{-{\frac {1}{2}}}\propto \nu ^{5 \over 2}B^{-{\frac {1}{2}}}\,\!}$

It is important to remember that each ${\displaystyle \gamma }$ has a unique corresponding ${\displaystyle {\nu _{sync}}}$. Also note that optical thickness depends inversely on frequency, so if we plot ${\displaystyle I_{\nu }}$ vs. ${\displaystyle \nu }$, we get a power law ${\displaystyle \nu ^{(1+p) \over 2}}$ for high frequencies (${\displaystyle I_{\nu }}$ for low optical thickness), and a ${\displaystyle \nu ^{5 \over 2}}$ dependence for low frequencies (${\displaystyle I_{\nu }}$ for high optical thickness, under self-absorption). This implies that there will be some turnover point at ${\displaystyle \nu _{m}}$ where ${\displaystyle I_{\nu }}$ is maximal. This turnover point can give us information about the strength of the B-field and the density of electroncs in the region.

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 ${\displaystyle F_{\nu }}$, between some minimum and maximum frequency cut-off, goes as ${\displaystyle K\nu ^{\alpha }}$, and that the number of photons per ${\displaystyle \gamma }$ is ${\displaystyle {dN \over d\gamma }=N_{0}\gamma ^{s}}$, where ${\displaystyle \alpha ={1+s \over 2}}$. These frequency cut-offs were set by ${\displaystyle \gamma _{min}^{2}\nu _{cyc}}$ and ${\displaystyle \gamma _{max}^{2}\nu _{cyc}}$. After this radiation is processed by SSC, approximately every photon is upscattered to a new energy ${\displaystyle {4 \over 3}\gamma ^{2}\nu }$. We are assuming that the relationship between an incoming photon frequency and its final frequency are related via a delta function. Thus:

${\displaystyle F_{\nu ,SSC}(\nu )=\tau \int _{\tilde {\nu }}{K{\tilde {\nu }}^{\alpha }d{\tilde {\nu }}\delta \left({\tilde {\nu }}-{\nu \over \gamma ^{2}}\right)\int _{\gamma }{N_{0}\gamma ^{s}d\gamma }}\,\!}$

Keep in mind that ${\displaystyle N_{0}}$ is normalized to so the integral comes out to 1 (it just accounts for the “shape” of the energy distribution function). ${\displaystyle \tau }$ is what contains the actual # density of ${\displaystyle e^{-}}$’s. It is the fraction scattered, and is generally ${\displaystyle \ll 1}$. ${\displaystyle \nu \sim {\tilde {\nu }}\gamma ^{2}}$.

For a fixed ${\displaystyle \nu \sim {\tilde {\nu }}\gamma ^{2}}$, we find that ${\displaystyle \gamma \sim \left({\nu \over {\tilde {\nu }}}\right)^{\frac {1}{2}}\propto {\tilde {\nu }}^{-{\frac {1}{2}}}}$.

In deriving the interaction of synchrotron radiation with synchrotron electrons, we derived the following formula for flux:

${\displaystyle F_{\nu ,SSC}(\nu )=\tau \int _{\tilde {\nu }}{K{\tilde {\nu }}^{\alpha }d{\tilde {\nu }}\delta \left({\tilde {\nu }}-{\nu \over \gamma ^{2}}\right)\int _{\gamma }{N_{0}\gamma ^{s}d\gamma }}\,\!}$

Now the integral over ${\displaystyle \gamma }$ is along “slant paths” through the rectangle (${\displaystyle \gamma _{min}\to \gamma _{max}}$, ${\displaystyle {\nu _{min}}\to {\nu _{max}}}$). Some of these slant paths will not stretch all the way to ${\displaystyle \gamma _{max}}$ or ${\displaystyle \gamma _{min}}$ because of the boundaries imposed by ${\displaystyle {\nu _{min}}}$ and ${\displaystyle {\nu _{max}}}$. That is to say, photons emitted by an electron at ${\displaystyle \gamma _{min}}$ can be scattered off another electron at ${\displaystyle \gamma _{max}}$ or vice versa. So we need to be a little more precise about the bounds on the ${\displaystyle \gamma }$ integral:

{\displaystyle {\begin{aligned}F_{\nu ,SSC}(\nu )&\sim \tau \int _{\nu _{min}}^{\nu _{max}}{K{\tilde {\nu }}^{\alpha }\delta ({\tilde {\nu }}-{\nu \over \gamma ^{2}})d{\tilde {\nu }}\int _{max({\sqrt {\nu \over {\nu _{max}}}},\gamma _{min})}^{min({\sqrt {\nu \over {\nu _{min}}}},\gamma _{max})}{N_{0}\gamma ^{s}d\gamma }}\\&\sim \tau K\left({\nu \over \gamma ^{2}}\right)^{\alpha }\int _{max({\sqrt {\nu \over {\nu _{max}}}},\gamma _{min})}^{min({\sqrt {\nu \over {\nu _{min}}}},\gamma _{max})}{N_{0}\gamma ^{s}d\gamma }\\&\sim \tau K\nu ^{\alpha }N_{0}\int _{max({\sqrt {\nu \over {\nu _{max}}}},\gamma _{min})}^{min({\sqrt {\nu \over {\nu _{min}}}},\gamma _{max})}{{\gamma ^{s} \over \gamma ^{2\alpha }}d\gamma }\\\end{aligned}}\,\!}

But since ${\displaystyle \alpha ={1+s \over 2}}$, the integrand is simply ${\displaystyle {1 \over \gamma }}$, giving us:

${\displaystyle {F_{\nu ,SSC}(\nu )\sim \tau k\nu ^{\alpha }N_{0}\ln \left({{min({\sqrt {\nu \over {\nu _{min}}}},\gamma _{max})} \over {max({\sqrt {\nu \over {\nu _{max}}}},\gamma _{min})}}\right)}\,\!}$

Recall that ${\displaystyle N_{0}}$ was normalized so that ${\displaystyle \int {N_{0}\gamma ^{s}d\gamma }=1}$, so saying that ${\displaystyle \gamma _{max}}$ is just some multiple of ${\displaystyle \gamma _{min}}$, it must be that

${\displaystyle N_{0}\sim \gamma _{min}^{-1-s}\,\!}$

Note for ${\displaystyle \nu \sim \gamma _{min}^{2}{\nu _{min}}}$, ${\displaystyle F_{\nu }}$ looks like (using the above relationship):

{\displaystyle {\begin{aligned}F_{\nu ,SSC}&=\tau K\gamma _{min}^{2\alpha }{\nu _{min}}^{\alpha }N_{0}\ln \left({{min({\sqrt {\nu \over {\nu _{min}}}},\gamma _{max})} \over {max({\sqrt {\nu \over {\nu _{max}}}},\gamma _{min})}}\right)\\&=\tau K{\nu _{min}}^{\alpha }\gamma _{min}^{2\alpha }\gamma _{min}^{-1-s}\ln \left({{min({\sqrt {\nu \over {\nu _{min}}}},\gamma _{max})} \over {max({\sqrt {\nu \over {\nu _{max}}}},\gamma _{min})}}\right)\\&=\tau K{\nu _{min}}^{\alpha }\ln \left({{min({\sqrt {\nu \over {\nu _{min}}}},\gamma _{max})} \over {max({\sqrt {\nu \over {\nu _{max}}}},\gamma _{min})}}\right)\\\end{aligned}}\,\!}
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${\displaystyle ^{-3}}$, and is immersed in a uniform magnetic field of 100 ${\displaystyle \mu }$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 ${\displaystyle I_{\nu }}$ 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—the lower bound is ${\displaystyle \gamma _{\mathrm {min} }^{2}}$ times greater than the minimum frequency of the original synchrotron spectrum, and the upper bound is ${\displaystyle \gamma _{\mathrm {max} }^{2}}$ times greater than the upper bound of the original spectrum.

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 ${\displaystyle 10^{12}K}$. What sets the “inverse Compton limit” at this temperature? Comparing, for a single electron, the luminosity of inverse Compton scattering to synchrotron scattering:

${\displaystyle {L_{IC} \over L_{sync}}={{4 \over 3}\beta ^{2}\gamma ^{2}\sigma _{T}cU_{ph} \over {4 \over 3}\beta ^{2}\gamma ^{2}\sigma _{T}cU_{B}}={U_{ph} \over U_{B}}{\begin{cases}>1&catastrophe\\<1&no\ catastrophe\end{cases}}\,\!}$

Now we’re going to make an approximation that we are on the Rayleigh-Jeans side of the blackbody curve, so that:

{\displaystyle {\begin{aligned}U_{ph}=U_{ph,sync}&\propto \nu _{m}I_{\nu }(\nu _{m})\\&\propto \nu _{m}{2kT_{B} \over \lambda _{m}^{2}}\\&\propto \nu _{m}^{3}T_{B}\\\end{aligned}}\,\!}

where ${\displaystyle \nu _{m}}$ is the frequency of peak of synchrotron emission.

Now ${\displaystyle U_{B}\propto B^{2}}$ (see Energy Density):

${\displaystyle \nu _{m}\sim \gamma _{m}^{2}\nu _{cyc}\propto \gamma _{m}^{2}B\,\!}$

where this ${\displaystyle \gamma _{m}}$ is not ${\displaystyle \gamma _{max}}$. Making the approximation that we are in the optically thick synchrotron spectrum, so that ${\displaystyle \gamma m_{e}c^{2}\sim kT}$, then we get ${\displaystyle \nu _{m}\sim T_{B}^{2}B}$. 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,

${\displaystyle {U_{ph} \over U_{B}}=C{\nu _{m}^{3}T_{B} \over \nu _{m}^{2}}T_{B}^{4}=\left({\nu _{m} \over 10^{9}Hz}\right)\left({T_{B} \over 10^{12}K}\right)^{5}=1\,\!}$

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.