Synchrotron Spectrum

Reference Material

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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, i.e. \begin{aligned}{dN\over dE} \propto E^p, \end{aligned} 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, we want to relate this electron energy, E, to a photon frequency $\omega$. For cyclotron emission this was straightforward, since an electron traveling with a given energy would only radiate at one frequency, a one-to-one correspondence. However, as we found in the last section, synchrotron electrons with a specific energy radiate at a whole continuum of frequencies, represented by the function F($\omega/\omega_c$). Luckily, we saw that this function is sharply peaked around $\omega \sim .29 \omega_c$, so to reasonable approximation we can use \begin{aligned}F\left(\frac{\omega}{\omega_{c}}\right) \approx \delta \left(\frac{\omega}{\omega_c}-.29\right) \end{aligned} Which means that we have recovered the 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 $P_e=P_{mag}$.
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} $${C\jn\over\pmag^{3\over4}\nu_1^{-\hf}}\sim\pmag$$ $${B^2\over8\pi}=\pmag\sim(C\jn\nu_1^\hf)^{4\over7}$$
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 $h\nu\sim h\nu_{crit}\sim h\nu_{cyc}\gamma^2$. If this gas 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 as $\nu^\hf e^{-{\nu\over\nu_{crit}}}$. Now let's add more electrons to this gas. For a while, 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.\par Let's examine the peak amplitude of emission (that is, $I_\nu$ at $\nu=\nu_{crit}$). To do this, imagine we have an optically thick ball of blackbody (perfectly absorbing and emitting) particles with a temperature carefully chosen so that $kT\sim\gamma m_e c^2$. Then: \def\nucrit{{\nu_{crit}}} $$I_\nu(\nu=\nucrit)=B_\nu(T\sim{\gamma m_ec^2\over k},\nu=\nucrit)$$ Now we'd like to argue that this blackbody particle system is emissively identical to 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: \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: $$I_\nu\propto\nucrit^{5\over2}B^{-\hf}\propto\nu^{5\over2}B^{-\hf}$$ 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 (low optical thickness), and a $\nu^{5\over2}$ dependence for low frequencies (high optical thickness). There is also some $\nu_m$ where $I_\nu$ is maximal.