Difference between revisions of "Synchrotron Radiation"

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* [[Lorentz transformations]]
* [[Lorentz transformations]]
* [[Fourier Transform]]
* [[Fourier Transform]]
* [[Energy Density]]
===Related Topics===
===Related Topics===

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<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 Power}

Notice that the $\gamma^2$ factor in the critical frequency makes synchrotron radiation ``harder than cyclotron radiation. In a cyclotron, the power radiated into all solid angles is given by the Larmor Forumla: $$P={2\over3}{e^2a^2\over c^3}$$ Let's derive this for the synchrotron. In the electron frame: $$P^\prime={2\over3}{e^2a^{\prime 2}\over c^3}$$ It turns out that power is a relativistic invariant. To see this, note $P^\prime={dU^\prime\over dt^\prime}$, and we know the following: $$\begin{matrix} \begin{aligned} U^\prime&=\gamma(U-vp_x)\\ t^\prime&=\gamma(t-{vx\over c^2})\\ \end{aligned}& \begin{aligned} U&=\gamma(u^\prime+vp_x^\prime)\\ t&=\gamma(t^\prime+{v\xp\over c^2})\\ \end{aligned} \end{matrix}$$ and in the prime frame ($s^\prime$), $d\xp=dp_x^\prime=0$.\par So we've shown that once we calculate $P^\prime$, we know $P$ in all frames. What we need to do now is calculate $a^\prime$. We'll do the following: \begin{itemize} \item Get $a^\prime(a)$, the Lorentz transform of acceleration. \item Get $a^\prime$ due to $E^\prime$. \end{itemize}

Since $e^-$ is at rest in the primed frame: $$\vec a^\prime={e\ef^\prime\over m_e}$$ $$\ef_\|^\prime=\ef_\|=0$$ $$\ef_\perp^\prime=\gamma{\vec v\times\bfield\over c}$$ $$|E_\perp|={\gamma vB\over c}\sin\alpha$$ Therefore, the magnitude of the acceleration is: $$|a^\prime|={e\gamma vB\over m_ec}\sin\alpha$$

(See below for derivations up to this point.)

And so the power radiated is: $$\boxed{P^\prime={2\over3}{e^4\gamma^2v^2B^2\over m_e^2c^5}\sin^2\alpha}$$

Note that as $v\to c$, $$P\to{2\over3}{e^4\gamma^2B^2\over m_e^2c^3}\sin^2\alpha$$ Thus we get {\it way} more power ($\gamma_{v\to c}^2\over\beta_{v\ll c}^2$) out of the synchrotron. How long can an $e^-$ hold up radiating this kind of power?

$$t_{life}\sim{\gamma m_ec^2\over\left({e^4\gamma^2B^2\over m_e^2c^3}\right)}$$ The time it takes an $e^-$ to go in the circle is just: $$t_{orb}\sim{\gamma m_ec\over eB}$$ Taking the ratio of these, we find that the critical $B$ required to make these timescales comparable is: $$\gamma^2B\sim{c^4m_e^2\over e^3}\approx{3^4\e{40}\e{-54}\over 5^3\e{-30}}\approx10^{16}cgs$$

Getting back to P, there is a prettier way of writing it: $$P=\overbrace{2\sigma_Tc\,U_B}^{E\ intercepted\atop by\ e^-} \sin^2\alpha\overbrace{\gamma^2}^{relativistic\atop enhancement}$$ where $\sigma_T={8\pi\over3}r_0^2$ is the Thomson cross-section ($r_0$ being defined by ${e^2\over r_0}=m_ec^2$), and $U_B$ is the magnetic field Energy Density $U_B={B^2\over8\pi}$.

\subsection*{ Synchrotron Cooling Time}

To estimate the synchrotron cooling time, we'll set up our standard expression of self-energy over power radiated: $$t_{cool}\sim{\gamma m_ec^2\over u_Bc\sigma_T\gamma^22\cdot\sin^2\alpha}$$ Instead of doing anything fancy with $\sin^2\alpha$, we'll just use that $\mean{\sin^2\alpha}={2\over3}$. $U_B$ we can estimate as $\mean{B^2}\over 8\pi$, giving us:

$$\begin{aligned}t_{cool}&\sim{m_ec^2\over{4\over3}\sigma_Tc}\inv{\gamma U_B}\\ &\sim16yr\left({1\,Gauss\over B}\right)^2\left(\inv{\gamma}\right)\\ \end{aligned}$$

\begin{itemize} \item Let's examine the cooling time for radio jets, where $B\sim1mGauss$, and $\gamma\sim10^3$. Plugging this in, we get $t_{cool}\sim10^4yr$. Compare this to $t_{dyn}\sim{l\over c}\sim{1kpc\over c}\sim10^4yr$.

\item We'll also estimate the cooling time of the Crab Nebula. To set an upper bound on $t_{cool}$, we'll use the most energetic X-rays. We can do this because $\gamma$ uniquely determines the electron energy and as a result, it uniquely determines a photon energy. For the Crab Nebula, we'll pick a photon energy: $E=4keV$. Then: $$\begin{aligned}\omega&\sim\omega_{crit}\sim\gamma^2\omega_{cyc}\\ &\sim{E\over\hbar}\sim\gamma^2{eB\over m_ec}\\ \end{aligned}$$ $B \sim$ mG gives us that $t_{cool}\sim 2 yrs$. Clearly, since the Crab Nebula was created some 1000 years ago, there must be a source of fresh electrons. This source is the pulsar, sitting in the middle of the nebula. \end{itemize}

\section*{Single Electron Power Distribution} For a non-relativistic cyclotron emission, an observer will see a power spectrum with time that oscillates sinusoidally with a single characteristic frequency. To obtain the power spectrum with frequency, P($\omega$), the Fourier Transform is taken of the sine wave which gives a delta function at the characteristic frequency, $\omega_{cyc}$.

For synchrotron radiation, however, the radiation will be emitted in a narrow beam of angular width $\sim$ $\frac{1}{\gamma}$, so P(t) will be a series of sharp peaks. Taking the Fourier transform of this distribution is not so straightforward. Jumping to the result, the spectrum for a single electron can be found:


\frac{dP}{d\nu} &= \frac{\sqrt{3}e^3 B sin(\alpha)}{m_e c^2}\frac{\nu}{\nu_{sync}} \int_{-\infty}^{\infty}{\frac{\nu}{\nu_{sync}}K_{5/3}(x)dx}\\

&\equiv \frac{\sqrt{3}e^3 B sin(\alpha)}{m_e c^2} F\left(\frac{\nu}{\nu_{sync}}\right) 


Where $\nu_{sync}$ is the critical photon frequency found for synchrotron radiation and $K_{5/3}$ is a modified Bessel function. In general this function is hard to work with, however it has some very nice properties. Note in the plot how: $$\begin{aligned}F\left(\frac{\nu}{\nu_{sync}}\right) \sim \left(\frac{\nu}{\nu_{sync}}\right)^{1/3} \end{aligned}$$ for small $\nu/\nu_{sync}$ and

$$\begin{aligned} F\left(\frac{\nu}{\nu_{sync}}\right) \sim \left(\frac{\nu}{\nu_{sync}}\right)^{1/2}e^{- \left(\frac{\nu}{\nu_{sync}}\right)}

\end{aligned}$$ for large $\nu/\nu_{sync}$. Furthermore, it is highly peaked around $\sim$ .3 $\nu_{sync}$, which allows us to approximate the power as being radiated at that single frequency when doing further calculations.

To find the total power radiated we can then integrate with respect to $\nu$, giving
$$\begin{aligned}P = \frac{\sqrt{3}e^3 B sin(\alpha)}{m_e c^2} \int \limits ^{\infty}_{0} F\left(\frac{\nu}{\nu_{sync}}\right) d\nu

\end{aligned}$$ which, making the substitution $x \equiv \nu/\nu_{sync}$, gives

$$\begin{aligned}P &= \frac{\sqrt{3}e^3 B \sin(\alpha)}{m_e c^2} \nu_{sync} \int \limits ^{\infty}_{0} F\left(x\right) dx \\
& = \frac{3\sqrt{3}e^4 \gamma^2 B^2 \sin^2(\alpha)}{4 \pi m_e^2 c^3} \int \limits ^{\infty}_{0} F\left(x\right) dx


The integral over F is just some constant factor, so we see that the important scalings are P $\propto \gamma^2 B^2$, which is what we found earlier using the Larmor Formula.

\section*{ Spectra of Synchrotron Radiation}

The power spectrum of a single $e^-$ undergoing synchrotron radiation peaks at $\omega_{sync}\sim\gamma^2\omega_{cyc}$. For small $\omega$, $P$ goes as $\omega^{1\over3}$, and for $\omega\gg\omega_{sync}$, $P$ 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($\omega/\omega_{sync}$). Luckily, we saw that this function is sharply peaked around $\omega \sim .29 \omega_{sync}$, so to reasonable approximation we can use $$\begin{aligned}F\left(\frac{\omega}{\omega_{sync}}\right) \approx \delta \left(\frac{\omega}{\omega_{sync}}-.29\right) \end{aligned}$$ Which means that we have recovered the one-to-one relationship between $E$ and $\omega$. That is, we can say that electrons of a specific energy $E$ are responsible for generating photons of frequency $\omega_{sync}\sim\gamma^2\omega_{cyc}$. Note also that $\omega_{sync}\propto\gamma^2B\propto E^2B$, so $E\propto\left({\omega_{sync}\over B} \right)^\hf$.

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 $-3\le p\le -2$. 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 $\nu_1<\nu_2$ (where we can define energies $E_1$ and $E_2$ that directly correspond to $\nu_1$ and $\nu_2$). 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} $${C\jn\over\pmag^{3\over4}\nu_1^{-\hf}}\sim\pmag$$ $${B^2\over8\pi}=\pmag\sim(C\jn\nu_1^\hf)^{4\over7}$$

\subsection*{Supernova Remnants & Effective Source Sizes} Imagine that we had a continuous, power law source of highly relativistic electrons that are subsequently carried away by a non-relativistic magnetized flow at speed v. An example of this situation would be the shock of a supernova remnant. Typical speeds are of order v $\sim$ 1250 $km/s$. We want to see if the finite cooling time for synchrotron radiation will effect what we observe.

The distance such an electron will travel will before cooling will be simply: $$\begin{aligned}vt_{cool}&\sim.02 pc\left({1\,Gauss\over B}\right)^2\left(\inv{\gamma}\right)\\ &\sim 10^{7} pc\left({40\,\mu G\over B}\right)^2\left(\inv{\gamma}\right)\\ \end{aligned}$$

where 40 $\mu$G is a more appropriate magnetic field for the post shock environment in a supernova remnant. We can convert the electron energy, $\gamma$ to a photon energy using the critical frequency found in past lectures: $$\begin{aligned}\nu_c = \frac{3\gamma^{2}eB}{2 m_e c}\\ \Rightarrow \gamma = \sqrt{\frac{2\nu_c m_e c}{3 e B}}\end{aligned}$$ so $$\begin{aligned} vt_{cool} \sim .9 pc\left({40\,\mu G\over B}\right)^{3/2}\left({1 keV \over h\nu}\right)^{1/2}\end{aligned}$$

Which is smaller than the typical supernova remnant radius around keV energies. As a result, when viewing supernova remnants at higher energies ($\sim$ keV and above), we see synchrotron emission concentrated around very thin nonthermal filaments that should decrease in size with energy. However, electrons also have the ability to diffuse through media, which is characterized by a diffusion coefficient, $\kappa$ with dimensions of [length][velocity] and can be written as : $$\begin{aligned} \kappa = \lambda_{mfp} v. \end{aligned}$$ For very relativistic electrons, $v\sim c$ and $\lambda_{mfp}$ is taken as some constant multiple of the gyroradius $\left(\frac{\gamma m_e c^2}{e B}\right)$, so that $$\begin{aligned} \kappa \sim \frac{\gamma m_e c^3}{e B} \end{aligned}$$ Quick dimensional analysis tells us the the characteristic length scale of diffusion is


\sqrt{\kappa t_{cool}} &\sim 5 \times 10^{-8} pc\left(\frac{1 Gauss}{B}\right)^{3/2} \\ &\sim 30 pc \left(\frac{40 \mu G}{B}\right)^{3/2} \end{aligned}$$

Chandra X-ray Image of SN1006. The nonthermal emission is concentrated inside the thin, white filaments Credit: http://chandra.harvard.edu/photo/2013/sn1006/

which is independent of the emitting frequency and comparable to $vt_{cool}$ at keV energies. At higher energies, then, diffusion will be the limiting factor for effective source size while for lower energies convection will be the limiting factor. Since both results have depend on the magnetic field as $B^{-3/2}$, we can use effective source sizes to infer local values of the magnetic field, a parameter that is otherwise hard to get at.

\subsection*{Spectrum} The cooling time also has an impact on the shape of the spectra. To calculate the total $F_\nu$ from our source, we would integrate the specific intensity produced by these electrons over the solid angle subtended by the source: $$\begin{aligned} F_\nu &= \int I_{\nu} d \Omega


Where D is the distance to the source. The integration over A is in principle of the area of the entire source, but as we have shown above this is limited to a region of width $vt_{cool}$ or $\sqrt{\kappa t_{cool}}$ before the radiation will effectively shut off. For the case of a spherical shock and constant specific intensity, then, the integral becomes


F_\nu &= \int I_\nu \frac{dA}{D^2} \\ &\sim \inv{D^2} \int \limits ^{vt_{cool}}_{0} \int \limits ^{2\pi}_{0} I_\nu rdrd\phi\\ &\sim \frac{I_\nu}{D^2}(vt_{cool})^2 \\ &\propto I_\nu \nu^{-1}B^{-3}\end{aligned}$$

for convection-limited electrons and


F_\nu &= \int I_\nu \frac{dA}{D^2} \\ &\sim \inv{D^2} \int \limits ^{\sqrt{\kappa t_{cool}}}_{0} \int \limits ^{2\pi}_{0} I_\nu rdrd\phi\\ &\sim \frac{I_\nu}{D^2}(\sqrt{\kappa t_{cool}})^2 \\ &\propto I_\nu B^{-3}\end{aligned}$$ for diffusion-limited electrons.

Thus, the two transport mechanism have qualitatively different effects on the shape of the spectrum. In one case there is an extra steeping by a power of $\nu^{-1}$ while for the other the spectrum remains unchanged. We can use these differences in many situations to infer properties of diffusion and convection in the plasma. This effect is also important in other astronomical objects, such as relativistic jets and pulsar wind nebulae, but these will have different geometries which will change the degree of steepening when integrating over the area of the source.

\section*{Special Relativity Derivations}

As an exercise in special relativity, we'll derive what $\ef^\prime$ is by \def\sigo{\sigma_0} investigating two parallel plates--the bottom one having charge density $\sigo$ and the top having $-\sigo$ We know in a motionless frame that the field between the plates is: $$\begin{aligned}E_{y_0}&=4\pi\sigma_0\\ B_0&=0\\ \end{aligned}$$ In a frame ($s$) where we are moving at $v=-v_0\^x$, these become: $$\begin{aligned}E_y&=4\pi\sigma=4\pi\gamma_0\sigo=\gamma_0E_{y_o}\\ B_z&={4\pi J\over c}={4\pi\over c}\sigma v_0={4\pi\over c}\gamma_0\sigo v_0= E_{y_0}{v_0\over c}\gamma_0\\ \end{aligned}$$ Jumping into one more frame ($s^\prime$) where $v=v^\prime\^x$ relative to frame $s$, we have: $$\begin{aligned}E_y^\prime&=4\pi\sigma^\prime\\ B_z^\prime={4\pi\over c}\sigma^\prime v^\prime\\ \end{aligned}$$ We need to figure $\sigma^\prime$, and our first instinct might be to say $\sigma^\prime=\gamma^\prime\sigma$, but that is wrong. We have to reference it from $\sigo$: $$\sigma^\prime=\gamma^\prime\sigo$$ Now we do some algebra: \def\gampTemplate:\gamma^\prime \def\gamoTemplate:\gamma 0 $$\begin{aligned}E_y^\prime&=4\pi\gamp\sigo\\ &=4\pi\gamma\gamo(1-{vv_0\over c^2})\sigo\\ &=\underbrace{4\pi\gamo\sigo}_{E_y}\gamma-\underbrace{4\pi\gamo v_0\sigo\over c}_{B_z}{\gamma v\over c}\\ \end{aligned}$$ Thus we have: $$\begin{matrix} \boxed{E_y^\prime=\gamma E_y-{\gamma v\over c}B_z}& \boxed{E_z^\prime=\gamma E_z+{\gamma v\over c}B_y}& \boxed{B_y^\prime=\gamma B_y+{\gamma v\over c}E_z} \end{matrix}$$ And, of course, $E_x^\prime=E_x$. The only thing we need to get now is $B_x^\prime$. For this we'll talk about a solenoid aligned with the $\^x$ direction with $n$ turns per unit length. The field of this solenoid in the rest frame is: $$B_x={4\pi\over c}nI$$ Jumping to a frame where $\vec v=v\^x$, $B_x^\prime={4\pi n^\prime I^\prime\over c}$. Using that: $$\begin{aligned}n^\prime&=\gamma n\\ I^\prime&={dQ^\prime\over dt^\prime}={dQ\over\gamma dt}={I\over\gamma}\\ \end{aligned}$$ We find that: $$\boxed{B_x^\prime=B_x}$$ Recall that we've derived all of this for boosts in the $\^x$ direction. To be completely general, we'll write them for any direction: $$\begin{matrix}\begin{aligned}E_\|^\prime&=E_\|\\ B_\|^\prime&=B_\|\\ \end{aligned}& \begin{aligned}\vec E_\perp^\prime &=\gamma(\ef_\perp+{\vec v\over c}\times\bfield)\\ \vec B_\perp^\prime&=\gamma(\bfield_\perp-{\vec v\over c}\times\ef)\\ \end{aligned} \end{matrix}$$

\end{document} <\latex>