# Synchrotron Spectra

The Synchrotron Spectrum of a Single Electron

Our next problem is to explain how the synchrotron mechanism can yield radio radiation at frequencies much higher than $\omega_{\rm B} = \omega_{\rm G} / \gamma$. To solve it, we first calculate the angular distribution of the radiation in the observer's frame.

Relativistic aberration causes the Larmor dipole pattern in the electron frame to become beamed sharply in the direction of motion as $v$ approaches $c$. This beaming follows directly from the relativistic velocity equations implied by the differential form of the Lorentz transform. For an electron moving in the $x$ direction,
$$v_{\rm x} \equiv {d x \over d t} = {d x \over d t'} {d t' \over d t}$$ $$v_{\rm x} = \gamma \biggl({ d x' \over d t'} + v {d t' \over d t'} \biggr) \biggl( { d t \over d t'} \biggr)^{-1}$$ $$v_{\rm x} = \gamma (v_{\rm x}' + v) \biggl[\gamma \biggl( 1 + {\beta \over c} {d x' \over d t'} \biggr) \biggr]^{-1}$$ $$v_{\rm x} = ({v_{\rm x}' + v}) \biggl( 1 + {\beta v_{\rm x}' \over c} \biggr)^{-1} ~.$$
In the $y$ direction perpendicular to the electron velocity, the velocity formula gives
$$v_{\rm y} \equiv {d y \over d t} = {d y \over d t'} {d t' \over d t} = {d y' \over d t'} \biggl( {d t \over d t'} \biggr)^{-1}$$ $$v_{\rm y} = { v_{\rm y}' \over \gamma} \biggl( 1 + {\beta v_{\rm x}' \over c} \biggr)^{-1}$$

Consider the synchrotron photons emitted with speed $c$ at an angle $\theta'$ from the $x'$ axis. Let $v_{\rm x}'$ and $v_{\rm y}'$ be the projections of the photon speed onto the $x'$ and $y'$ axes. Then
$$\cos\theta' = {v_{\rm x}' \over c} \qquad \sin\theta' = {v_{\rm y}' \over c}$$ In the observer's frame we have for the same photons,
$$\cos\theta = {v_{\rm x} \over c} \qquad \sin\theta = {v_{\rm y} \over c}$$ Using our equations for velocities, we get
$$\cos\theta = \biggl({v_{\rm x}' + v \over 1 + \beta v_{\rm x}' / c }\biggr){1 \over c} = \biggl({c \cos \theta' + v \over 1 + \beta c \cos \theta' / c }\biggr){1 \over c}$$ $$\cos \theta = { \cos \theta' + \beta \over 1 + \beta \cos \theta'}$$ $$\sin\theta = {v_{\rm y}' \over c \gamma ( 1 + \beta v_{\rm x}'/ c ) }$$ $$\sin\theta = { \sin \theta' \over \gamma( 1 + \beta \cos \theta')}$$ In the frame moving with the electron, the Larmor equation implies a power pattern proportional to $\cos^2 \theta'$ with nulls at $\theta' = \pm \pi/2$. In the observer's frame, these nulls appear at much smaller angles
$$\sin\theta = {\sin \theta' \over \gamma (1 + \beta \cos \theta')} \approx { 1 \over \gamma} \approx \theta$$ since $1/\gamma \ll 1$. We see the radiation confined to a very narrow beam of width $2/\gamma$ between nulls, as shown below.

Relativistic beaming transforms the dipole pattern of Larmor radiation in the electron frame (dotted curve) into a narrow searchlight beam in the observer's frame.
The solid curve is the observed power pattern for $\gamma = 5$. The observed angle between the nulls of the forward beam $\approx 2 / \gamma$, and the peak power gain $\approx 2 \gamma$.

For example, a 10 Gev electron has $\gamma \approx 2 \times 10^4$ so $2 / \gamma \approx 10^{-4} {\rm ~rad} \approx 20$ arcsec! The observer sees a short pulse of radiation emitted during only the tiny fraction
$${2 \over 2 \pi \gamma} = {1 \over \pi \gamma }$$ of the electron orbit when the electron is moving directly toward the observer.

The duration $\Delta t$ of the observed pulse is even shorter than the time the electron needs to cover $1 / (\pi \gamma)$ of its orbit because the electron is visible only while it is moving directly toward the observer with a speed approaching $c$.  In the observer's frame, the electron nearly keeps up with the radiation that it emits.

The beamed radiation from a relativistic electron is visible only while the electron's velocity points within $\pm 1/\gamma$ of the line-of-sight ($\Delta \theta \approx 2 / \gamma$).  During that time $\Delta t$ the electron moves a distance $\Delta x = v \Delta t$ toward the observer, almost keeping up with the radiation that travels a distance $c \Delta t$.  As a result, the observed pulse duration is shortened by a factor $(1 - v/c)$.

$$\Delta t_{\rm p} = t{\rm (end~of~observed~pulse)} - t{\rm (start~of~observed~pulse)}$$ $$\Delta t_{\rm p} = {\Delta x \over v} + {( x - \Delta x) \over c} - {x \over c}$$ The first term in this equation represents the time taken by the electron to cover the distance $\Delta x$, the second is the light travel time from the electron position at the end of the pulse to the observer, and the third is the light travel time from the electron position at the beginning of the pulse to the observer. Note that the observed pulse duration
$$\Delta t_{\rm p} = {\Delta x \over v} - {\Delta x \over c} = {\Delta x \over v} \biggl(1 - {v \over c} \biggr) \ll {\Delta x \over v}$$ is much less than the time $\Delta t$ the electron needs to move a distance $\Delta x$ because, in the observer's frame, the electron nearly keeps up with its radiation. In the limit $v \rightarrow c$, $$\biggl( 1 - {v \over c} \biggr) = \biggl( 1 - {v \over c} \biggr) { 1 + v/c \over 1 + v/c} = {1 - v^2/c^2 \over 1 + v/ c} \approx { \gamma^{-2} \over 2} = { 1 \over 2 \gamma^2 }$$ so
$$\Delta t_{\rm p} = { \Delta x \over v} { 1 \over 2 \gamma^2} ={ \Delta \theta \over \omega_{\rm B} } { 1 \over 2 \gamma^2}~.$$ Recall that $\Delta \theta \approx 2/\gamma$ so
$$\Delta t_{\rm p} = { 2 \over \gamma \omega_{\rm B} 2 \gamma^2} = {1 \over \gamma^3 \omega_{\rm B} } = {1 \over \gamma^2 \omega_{\rm G}}$$ is the full observed duration of the pulse.

Example: Just how short is the observed duration of one synchrotron pulse? In the typical interstellar magnetic field of our Galaxy, $B \approx 5 \times 10^{-6}$ G so $\omega_{\rm G} \approx 2 \pi \times 14$ rad s$^{-1}$. For an electron having $\gamma \approx 10^4$,
$$\Delta t_{\rm p} \approx {1 \over \gamma^2 \omega_{\rm G}} \approx {1 \over (10^4)^2 \times 2 \pi \times 14} \approx 10 ^{-10}{\rm ~s}$$

Allowing for the motion of the electron parallel to the magnetic field, we replace the total magnetic field by its perpendicular component $B \sin \alpha$, yielding
$$\Delta t_{\rm p} = {1 \over \gamma^2 \omega_{\rm G} \sin \alpha}~,$$ where $\alpha$ is the pitch angle of the electron.  Thus a plot of the power received as a function of time is very spiky.

Synchrotron radiation is a very spiky series of widely spaced narrow pulses.  The numerical values indicated on the plot of power versus time correspond to an electron with $\gamma \approx 10^4$ in a magnetic field $B \approx 5 \mu$G typical of the interstellar medium in our Galaxy.

If $\gamma \approx 10^4$, the duration of each pulse is $\Delta t_{\rm p} \sim 10^{- 10}$ s and if $B \approx 5\mu$G, the spacing between pulses is $\gamma / \nu_{\rm G} \sim 10^3$ s.

The observed synchrotron power spectrum is the Fourier transform of this time series of pulses.  Instead of calculating the Fourier transform directly, we note first that the pulse train is the convolution of the profile $p(t)$ of an individual pulse with the Shah function (see Appendix B of Rohlfs & Wilson or Bracewell's valuable reference book The Fourier Transform and Its Applications):
$$I I I (t / \Delta t) \equiv \sum_{n = -\infty}^\infty \delta [(t / \Delta t)] - n)$$ where each delta function $\delta$ is an infinitesimally narrow spike at integer $t / \Delta t= n$ whose integral is unity. Then we use the convolution theorem to show that the Fourier transform of the pulses is the product of the Fourier transform of one pulse times the Fourier transform of the shah function.

The Fourier transform of the shah function is also the shah function, so by the similarity theorem, the Fourier transform of
$$III \biggl( {t \nu_{\rm G} \over \gamma} \biggr)$$ is proportional to
$$III \biggl( { \nu \gamma \over \nu_{\rm G} } \biggr)~,$$ a nearly continuous series of spikes in the frequency domain separated in frequency by only
$$\Delta\nu = { \nu_{\rm G} \over \gamma} \sim 10 ^{-3}{\rm ~Hz}~.$$ Although this is not formally a continuous spectrum, the distortions caused by even tiny fluctuations in electron energy, magnetic field strength, or pitch angle cause frequency shifts larger than $\Delta \nu$, so the spectrum of synchrotron radiation is effectively continuous.

Thus the synchrotron spectrum of a single electron is fairly flat at low frequencies and tapers off at frequencies above
$$\nu_{\rm max} \approx { 1 \over 2 \Delta t_{\rm p}} \approx \pi \gamma^2 \nu_{\rm G} \sin \alpha \propto \gamma^2 B_\bot~.$$

It isn't really necessary to calculate the Fourier transform of the pulse shape precisely because astrophysical sources don't contain electrons with just one energy and one pitch angle in a uniform magnetic field. The actual energy distribution of cosmic rays in our Galaxy is a very broad power law, and this smears out the details of the spectrum from each energy range. Just for the record, the synchrotron power spectrum of a single electron is
$$\bbox[border:3px blue solid,7pt]{P(\nu) = {\sqrt{3} e^3 B \sin \alpha \over m c^2} \biggl( { \nu \over \nu_{ \rm c}} \biggr) \int_{\nu/\nu_{\rm c}}^\infty K_{5/3} (\eta) d \eta}\rlap{\quad \rm {(5C1)}}$$ where $K_{ 5/3}$ is a modified Bessel function and $\nu_{\rm c}$ is the critical frequency whose value is
$$\bbox[border:3px blue solid,7pt]{\nu_{\rm c} = {3 \over 2} \gamma^2 \nu_{\rm G} \sin \alpha}\rlap{\quad \rm {(5C2)}}$$ For the full mathematical derivation of these equations, see Pacholczyk's Radio Astrophysics, a valuable reference for details of radiation processes.

This figure shows four different ways to plot the synchrotron spectrum of a single electron in terms of
$$F(x) \equiv x \int_x^\infty K_{5/3} (\eta) d \eta~,$$ where $x \equiv \nu / \nu_{\rm c}$ is the frequency in units of the critical frequency $\nu_{\rm c}$.  Although they all plot the same spectrum, they look quite different and convey or suppress information in different ways. (1) Simply plotting $F(x)$ versus $x$ on linear axes (lower left panel) completely obscures the spectrum below the peak of $F(x)$ at $x \approx 0.29$. (2) Replotting on logarithmic axes (upper left panel) shows that the low-frequency spectrum has a slope of $1/3$, but it obscures the fact that most of the power is emitted at frequencies near $x \sim 1$ because $F(x)$ is the spectral power per unit frequency, not per unit $\log$(frequency). (3) The power per unit $\log(x)$ is $F[\log(x)] = \ln(10) x F(x)$, which is plotted on logarithmic axes in the upper right panel. It has a slope of $4/3$ at low frequencies, making it clearer that most of the power is emitted near $x \sim 1$, as required to justify the approximation (used in the next section) that all of the power is emitted at $x = 1$. (4) The lower right panel plots $F(\log(x)]$ with a linear ordinate but a logarithmic abscissa to expand the low-frequency spectrum lost in the lower right panel.  It is clearly consistent with the approximation that all emission is near $x = 1$ but doesn't clearly show that the low-frequency spectrum is a power law. Note also that the peak of $F(\log(x)]$ is at $x \approx 1.3$, not $x \approx 0.29$.  Areas under the curves in the two lower panels are proportional to the power radiated in given frequency ranges.  For example, both lower panels show clearly that about half of the power is emitted at frequencies below the critical frequency and half at higher frequencies.

The spectrum of a single electron has a logarithmic slope $$d \log P(\nu) / d \log \nu \approx {1 \over 3}$$
at low frequencies, a broad peak near the critical frequency $\nu_{\rm c}$, and falls off sharply at higher frequencies. One way to look at $\nu_{\rm c}$ is
$$\nu_{\rm c} = \biggl( { 3 \over 2} \sin \alpha \biggr) \biggl({E \over m c^2}\biggr)^2 {e B \over 2 \pi m_{\rm e} c} \propto E^2 B_\bot~.$$ That is, the frequency at which an electron emits most strongly is proportional to the square of the electron energy multiplied by the strength of the perpendicular component of the magnetic field.

Spectra of Optically Thin Synchrotron Sources

If a synchrotron source containing any arbitrary distribution of electron energies is optically thin ($\tau \ll 1$), then its low-frequency spectrum is the superposition of the spectra from individual electrons and can never rise more rapidly than the $1/3$ power of frequency. In other words, the [negative] spectral index $\alpha \equiv - d \log P_\nu / d \log \nu$ (be careful not to confuse this $\alpha$ with the electron pitch angle $\alpha$) must always be greater than $-1/3$. Most astrophysical sources of synchrotron radiation have spectral indices near $\alpha \approx 0.7$ at high frequencies where they are optically thin, and as we shall soon see, their overall spectral indices primarily reflect their electron energy distributions. The energy spectrum of cosmic-ray electrons in the local interstellar medium (Casadei, D., & Bindi, V. 2004, ApJ, 612,262).  In the energy range above a few GeV, N(E) is a power law with slope $\delta \approx 2.4$.

The observed energy distribution of cosmic-ray electrons in our Galaxy is roughly a power law:
$$\bbox[border:3px blue solid,7pt]{N(E) dE \approx K E^{-\delta} dE}\rlap{\quad \rm {(5C3)}}$$
where $N(E) dE$ is the number of electrons per unit volume with energies $E$ to $E + dE$. The energy range around $\gamma \sim 10^4$ is relevant to the production of radio radiation, and there the power-law slope is $\delta \sim +2.4$.  Because $N(E)$ is nearly a power law over more than two decades of energy and the critical frequency $\nu_{\rm c}$ is proportional to $E^2$, we expect the synchrotron spectrum to reflect this power law over a frequency range of at least $(10^2)^2 = 10^4$. Consequently, we can ignore the detailed spectra of individual electrons, which are smeared out in the observed spectrum by this broad power-law energy distribution. We make the very simple and crude approximation that each electron radiates all of its power
$$P = - {d E \over d t} = {4 \over 3} \sigma_{\rm T} \beta^2 \gamma^2 c U_{\rm B}$$ at the single frequency $$\nu \approx \gamma^2 \nu_{\rm G}~,$$ which is very close to the critical frequency.  Then the emission coefficient of synchrotron radiation by an ensemble of electrons is
$$\epsilon_\nu d \nu = -{dE \over dt} N(E) dE$$ where
$$E = \gamma m_{\rm e} c^2 \approx \biggl( {\nu \over \nu_{\rm G}} \biggr)^{1/2} m_{\rm e} c^2~.$$ Differentiating $E$ gives
$${d E \over d \nu} \approx {m_{\rm e} c^2 \nu^{-1/2} \over 2 \nu_{\rm G}^{1/2}}$$ so
$$\epsilon_\nu \approx \biggl( { 4 \over 3} \sigma_{\rm T} \beta^2 \gamma^2 c U_{\rm B} \biggr) ( K E^{-\delta}) \biggl( { m_{\rm e} c^2 \nu^{-1/2} \over 2 \nu_{\rm G}^{1/2}} \biggr)$$ Eliminating $E$ in favor of $\nu / \nu_{\rm G}$ and ignoring the physical constants in this equation for $\epsilon_\nu$ results in the proportionality $$\epsilon_\nu \propto \biggl( {\nu \over \nu_{\rm G}} \biggr) B^2 \biggl( {\nu \over \nu_{\rm G}} \biggr)^{-\delta /2} (\nu \,\nu_{\rm G})^{-1/2}$$ $$\epsilon_\nu \propto \biggl( {\nu \over B} \biggr) B^2 \biggl( { \nu \over B} \biggr)^{-\delta /2} (\nu\,B)^{-1/2}$$ since $\nu_{\rm G} \propto B$. We finally get:
$$\bbox[border:3px blue solid,7pt]{\epsilon_\nu \propto B^{(\delta + 1) /2} \nu^{(1 - \delta)/2}}\rlap{\quad \rm {(5C4)}}$$ Since $\nu^{-\alpha} \propto \nu^{(1 - \delta)/2}$,
$$\bbox[border:3px blue solid,7pt]{\alpha = {\delta - 1 \over 2}}\rlap{\quad \rm {(5C5)}}$$ That is, the synchrotron spectrum of a power-law energy distribution is itself a power law, and Equation 5C5 relates the slopes of these two power laws.

Example: In our Galaxy $\delta \approx 2.4$, so we expect
$$\epsilon_\nu \propto B^{1.7} \nu^{-0.7}~$$ and hence the (negative) spectral index should be
$$\alpha \approx 0.7~,$$ which is in agreement with observation.  This is also the typical spectral index of most optically thin extragalactic radio sources, even radio galaxies and quasars.  It reflects the power-law energy distribution of cosmic rays accelerated in shocks, the shocks produced by supernova remnants expanding into the ambient interstellar medium for example. Synchrotron radiation (dot-dash line) from cosmic-ray electrons accelerated by the supernova remnants of relatively massive ($M > 8M_\odot$) and short-lived ($T < 3 \times 10^7$ yr) stars dominates the radio continuum emission of the nearby starburst galaxy M82 at frequencies $\nu < 30$ GHz.  Thermal emission (dashed line) from HII regions ionized primarily by even more massive ($M > 15M_\odot$) and shorter-lived stars is strongest between about 30 and 200 GHz.  At frequencies well below 1 GHz, free-free absorption flattens the overall spectrum.