# Difference between revisions of "Synchrotron Radiation"

## Synchrotron Power

Notice that the ${\displaystyle \gamma ^{2}}$ 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:

${\displaystyle P={2 \over 3}{e^{2}a^{2} \over c^{3}}\,\!}$

Let’s derive this for the synchrotron. In the electron frame:

${\displaystyle P^{\prime }={2 \over 3}{e^{2}a^{\prime 2} \over c^{3}}\,\!}$

It turns out that power is a relativistic invariant. To see this, note ${\displaystyle P^{\prime }={dU^{\prime } \over dt^{\prime }}}$, and we know the following:

{\displaystyle {\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{x^{\prime }} \over c^{2}})\\\end{aligned}}\end{matrix}}\,\!}

and in the prime frame (${\displaystyle s^{\prime }}$), ${\displaystyle d{x^{\prime }}=dp_{x}^{\prime }=0}$.

So we’ve shown that once we calculate ${\displaystyle P^{\prime }}$, we know ${\displaystyle P}$ in all frames. What we need to do now is calculate ${\displaystyle a^{\prime }}$. We’ll do the following:

• Get ${\displaystyle a^{\prime }(a)}$, the Lorentz transform of acceleration.
• Get ${\displaystyle a^{\prime }}$ due to ${\displaystyle E^{\prime }}$.

Since ${\displaystyle e^{-}}$ is at rest in the primed frame:

${\displaystyle {\vec {a}}^{\prime }={e{\vec {E}}^{\prime } \over m_{e}}\,\!}$
${\displaystyle {\vec {E}}_{\|}^{\prime }={\vec {E}}_{\|}=0\,\!}$
${\displaystyle {\vec {E}}_{\perp }^{\prime }=\gamma {{\vec {v}}\times {\vec {B}} \over c}\,\!}$
${\displaystyle |E_{\perp }|={\gamma vB \over c}\sin \alpha \,\!}$

Therefore, the magnitude of the acceleration is:

${\displaystyle |a^{\prime }|={e\gamma vB \over m_{e}c}\sin \alpha \,\!}$

(See below for derivations up to this point.)

And so the power radiated is:

${\displaystyle {P^{\prime }={2 \over 3}{e^{4}\gamma ^{2}v^{2}B^{2} \over m_{e}^{2}c^{5}}\sin ^{2}\alpha }\,\!}$

Note that as ${\displaystyle v\to c}$,

${\displaystyle P\to {2 \over 3}{e^{4}\gamma ^{2}B^{2} \over m_{e}^{2}c^{3}}\sin ^{2}\alpha \,\!}$

Thus we get way more power (${\displaystyle \gamma _{v\to c}^{2} \over \beta _{v\ll c}^{2}}$) out of the synchrotron. How long can an ${\displaystyle e^{-}}$ hold up radiating this kind of power?

${\displaystyle t_{life}\sim {\gamma m_{e}c^{2} \over \left({e^{4}\gamma ^{2}B^{2} \over m_{e}^{2}c^{3}}\right)}\,\!}$

The time it takes an ${\displaystyle e^{-}}$ to go in the circle is just:

${\displaystyle t_{orb}\sim {\gamma m_{e}c \over eB}\,\!}$

Taking the ratio of these, we find that the critical ${\displaystyle B}$ required to make these timescales comparable is:

${\displaystyle \gamma ^{2}B\sim {c^{4}m_{e}^{2} \over e^{3}}\approx {3^{4}\cdot 10^{40}\cdot 10^{-54} \over 5^{3}\cdot 10^{-30}}\approx 10^{16}cgs\,\!}$

Getting back to P, there is a prettier way of writing it:

${\displaystyle P=\overbrace {2\sigma _{T}c\,U_{B}} ^{E\ intercepted \atop by\ e^{-}}\sin ^{2}\alpha \overbrace {\gamma ^{2}} ^{relativistic \atop enhancement}\,\!}$

where ${\displaystyle \sigma _{T}={8\pi \over 3}r_{0}^{2}}$ is the Thomson cross-section (${\displaystyle r_{0}}$ being defined by ${\displaystyle {e^{2} \over r_{0}}=m_{e}c^{2}}$), and ${\displaystyle U_{B}}$ is the magnetic field Energy Density ${\displaystyle U_{B}={B^{2} \over 8\pi }}$.

### Synchrotron Cooling Time

To estimate the synchrotron cooling time, we’ll set up our standard expression of self-energy over power radiated:

${\displaystyle t_{cool}\sim {\gamma m_{e}c^{2} \over u_{B}c\sigma _{T}\gamma ^{2}2\cdot \sin ^{2}\alpha }\,\!}$

Instead of doing anything fancy with ${\displaystyle \sin ^{2}\alpha }$, we’ll just use that ${\displaystyle \left\langle {\sin ^{2}\alpha }\right\rangle ={2 \over 3}}$. ${\displaystyle U_{B}}$ we can estimate as ${\displaystyle \left\langle {B^{2}}\right\rangle \over 8\pi }$, giving us:

{\displaystyle {\begin{aligned}t_{cool}&\sim {m_{e}c^{2} \over {4 \over 3}\sigma _{T}c}{1 \over \gamma U_{B}}\\&\sim 16yr\left({1\,Gauss \over B}\right)^{2}\left({1 \over \gamma }\right)\\\end{aligned}}\,\!}
• Let’s examine the cooling time for radio jets, where ${\displaystyle B\sim 1mGauss}$, and ${\displaystyle \gamma \sim 10^{3}}$. Plugging this in, we get ${\displaystyle t_{cool}\sim 10^{4}yr}$. Compare this to ${\displaystyle t_{dyn}\sim {l \over c}\sim {1kpc \over c}\sim 10^{4}yr}$.
• We’ll also estimate the cooling time of the Crab Nebula. To set an upper bound on ${\displaystyle t_{cool}}$, we’ll use the most energetic X-rays. We can do this because ${\displaystyle \gamma }$ 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: ${\displaystyle E=4keV}$. Then:
{\displaystyle {\begin{aligned}\omega &\sim \omega _{crit}\sim \gamma ^{2}\omega _{cyc}\\&\sim {E \over \hbar }\sim \gamma ^{2}{eB \over m_{e}c}\\\end{aligned}}\,\!}

${\displaystyle B\sim }$ mG gives us that ${\displaystyle t_{cool}\sim 2yrs}$. 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.

## 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(${\displaystyle \omega }$), the Fourier Transform is taken of the sine wave which gives a delta function at the characteristic frequency, ${\displaystyle \omega _{cyc}}$.

For synchrotron radiation, however, the radiation will be emitted in a narrow beam of angular width ${\displaystyle \sim }$ ${\displaystyle {\frac {1}{\gamma }}}$, 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:

{\displaystyle {\begin{aligned}{\frac {dP}{d\nu }}&={\frac {{\sqrt {3}}e^{3}Bsin(\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}Bsin(\alpha )}{m_{e}c^{2}}}F\left({\frac {\nu }{\nu _{sync}}}\right)\end{aligned}}\,\!}

Where ${\displaystyle \nu _{sync}}$ is the critical photon frequency found for synchrotron radiation and ${\displaystyle K_{5/3}}$ 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:

{\displaystyle {\begin{aligned}F\left({\frac {\nu }{\nu _{sync}}}\right)\sim \left({\frac {\nu }{\nu _{sync}}}\right)^{1/3}\end{aligned}}\,\!}

for small ${\displaystyle \nu /\nu _{sync}}$ and

{\displaystyle {\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 ${\displaystyle \nu /\nu _{sync}}$. Furthermore, it is highly peaked around ${\displaystyle \sim }$ .3 ${\displaystyle \nu _{sync}}$, 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 ${\displaystyle \nu }$, giving

{\displaystyle {\begin{aligned}P={\frac {{\sqrt {3}}e^{3}Bsin(\alpha )}{m_{e}c^{2}}}\int \limits _{0}^{\infty }F\left({\frac {\nu }{\nu _{sync}}}\right)d\nu \end{aligned}}\,\!}

which, making the substitution ${\displaystyle x\equiv \nu /\nu _{sync}}$, gives

{\displaystyle {\begin{aligned}P&={\frac {{\sqrt {3}}e^{3}B\sin(\alpha )}{m_{e}c^{2}}}\nu _{sync}\int \limits _{0}^{\infty }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 _{0}^{\infty }F\left(x\right)dx\end{aligned}}\,\!}

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

The power spectrum of a single ${\displaystyle e^{-}}$ undergoing synchrotron radiation peaks at ${\displaystyle \omega _{sync}\sim \gamma ^{2}\omega _{cyc}}$. For small ${\displaystyle \omega }$, ${\displaystyle P}$ goes as ${\displaystyle \omega ^{1 \over 3}}$, and for ${\displaystyle \omega \gg \omega _{sync}}$, ${\displaystyle P}$ goes as ${\displaystyle \omega ^{\frac {1}{2}}e^{-\omega }}$. In general, recall that ${\displaystyle P\propto B^{2}\gamma ^{2}}$.

We would like to calculate the power spectrum of an ensemble of ${\displaystyle e^{-}}$. To do this, we need to describe how many electrons there are per energy. We’ll assume a power law distribution of ${\displaystyle e^{-}}$ energies, i.e.

{\displaystyle {\begin{aligned}{dN \over dE}\propto E^{p},\end{aligned}}\,\!}

where ${\displaystyle p}$ is the 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 ${\displaystyle E}$ and ${\displaystyle E+dE}$:

{\displaystyle {\begin{aligned}dP&=dN\times P{\big |}_{single\ e^{-}}\\&\propto \left({dN \over dE}dE\right)\gamma ^{2}B^{2}\propto (E^{p}dE)E^{2}B^{2}\\&\propto E^{2+p}B^{2}dE\\\end{aligned}}\,\!}

Thus we have:

${\displaystyle {dP \over dE}\propto E^{2+p}B^{2}\,\!}$

Now, we want to relate this electron energy, E, to a photon frequency ${\displaystyle \omega }$. 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(${\displaystyle \omega /\omega _{sync}}$). Luckily, we saw that this function is sharply peaked around ${\displaystyle \omega \sim .29\omega _{sync}}$, so to reasonable approximation we can use

{\displaystyle {\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 ${\displaystyle E}$ and ${\displaystyle \omega }$. That is, we can say that electrons of a specific energy ${\displaystyle E}$ are responsible for generating photons of frequency ${\displaystyle \omega _{sync}\sim \gamma ^{2}\omega _{cyc}}$. Note also that ${\displaystyle \omega _{sync}\propto \gamma ^{2}B\propto E^{2}B}$, so ${\displaystyle E\propto \left({\omega _{sync} \over B}\right)^{\frac {1}{2}}}$.

Therefore:

{\displaystyle {\begin{aligned}{dP \over d\omega }&={dP \over dE}{dE \over d\omega }\\&\propto (E^{2+p}B^{2})B^{-{\frac {1}{2}}}\omega ^{-{\frac {1}{2}}}\\&\propto \left({\omega \over B}\right)^{2+p \over 2}B^{3 \over 2}\omega ^{-{\frac {1}{2}}}\\\end{aligned}}\,\!}
${\displaystyle {{dP \over d\omega }\propto B^{1-p \over 2}\omega ^{1+p \over 2}}\,\!}$

Usually, ${\displaystyle p<0}$. Note that our assumption of one-to-one correspondence was not necessary to get this power law dependence on ${\displaystyle \omega }$. 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 ${\displaystyle \omega }$. Also note that this expression relies on an influx of ${\displaystyle e^{-}}$ to replace old ones which cooled down. If we cut off this influx, we’ll see that since ${\displaystyle t_{cool}\sim {1 \over \gamma B^{2}}}$, the ${\displaystyle e^{-}}$ emitting higher ${\displaystyle \omega }$ photons decay first, and so we see turn-offs from a power law distribution for increasingly low ${\displaystyle \omega }$ as time goes by.

For measuring ${\displaystyle p}$, let’s define ${\displaystyle \alpha \equiv {1+p \over 2}}$. Observations of extended radio sources have measured ${\displaystyle \alpha \approx -0.7\Rightarrow p=-2.4}$. In general, we find that ${\displaystyle -0.75\leq \alpha \leq -0.5}$, or ${\displaystyle -3\leq p\leq -2}$. We’ve made an assumption of a constant magnetic field. Suppose we have an optically thin synchrotron emitting gas with a power law emissivity ${\displaystyle j_{\nu _{1}}>j_{\nu _{2}}}$ for ${\displaystyle \nu _{1}<\nu _{2}}$ (where we can define energies ${\displaystyle E_{1}}$ and ${\displaystyle E_{2}}$ that directly correspond to ${\displaystyle \nu _{1}}$ and ${\displaystyle \nu _{2}}$). 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:

{\displaystyle {\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}{\big |}_{E_{1}}^{E_{2}}\\\end{aligned}}\,\!}

For ${\displaystyle p\leq -2}$,

${\displaystyle P_{e}\propto E_{1}^{2+p}\,\!}$

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 ${\displaystyle \left({E_{min} \over E_{1}}\right)^{2+p}}$. Measuring ${\displaystyle j_{\nu }}$, we get:

{\displaystyle {\begin{aligned}j_{\nu }(\nu _{1}\leftrightarrow E_{1})&\propto \left({dN \over dE}dE\right){P{\big |}_{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}^{2}B \over E_{1}^{2}}\\&\propto \overbrace {{dN \over dE}dE\,E_{1}} ^{P_{e}}{E_{1}B \over E_{1}^{2}}\\&\propto P_{e}{B \over E_{1}}\propto P_{e}B\left({B \over \nu _{1}}\right)^{\frac {1}{2}}\\\end{aligned}}\,\!}
${\displaystyle j_{\nu }\propto P_{e}P_{mag}^{3 \over 4}\nu _{1}^{-{\frac {1}{2}}}\,\!}$

where ${\displaystyle P_{mag}}$ is the magnetic energy density. Now since ${\displaystyle P_{tot}=P_{e}+P_{mag}}$:

${\displaystyle {P_{tot}={C\,j_{\nu } \over P_{mag}^{3 \over 4}\nu ^{-{\frac {1}{2}}}}+P_{mag}}\,\!}$

This expression has a minimum for a unique ${\displaystyle P_{mag}}$. This ${\displaystyle P_{mag}}$ tells us the way energy is partitioned in the system between ${\displaystyle P_{e}}$ and ${\displaystyle P_{mag}}$. The minimum should occur when ${\displaystyle P_{e}=P_{mag}}$.

Recall that last time we derived for an optically thin synchrotron gas that:

${\displaystyle j_{\nu }(\nu _{1})\propto P_{e}P_{mag}^{3 \over 4}\nu _{1}^{-{\frac {1}{2}}}\,\!}$
{\displaystyle {\begin{aligned}P_{tot}&=P_{e}+P_{mag}\\&={Cj_{\nu } \over {P_{mag}}^{3 \over 4}\nu _{1}^{-{\frac {1}{2}}}}+{P_{mag}}\\\end{aligned}}\,\!}

Thus, the minimum total power occurs near the equipartition point between ${\displaystyle {P_{mag}}}$ and ${\displaystyle P_{e}}$. This gives us:

${\displaystyle {Cj_{\nu } \over {P_{mag}}^{3 \over 4}\nu _{1}^{-{\frac {1}{2}}}}\sim {P_{mag}}\,\!}$
${\displaystyle {B^{2} \over 8\pi }={P_{mag}}\sim (Cj_{\nu }\nu _{1}^{\frac {1}{2}})^{4 \over 7}\,\!}$

### 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 ${\displaystyle \sim }$ 1250 ${\displaystyle km/s}$. 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:

{\displaystyle {\begin{aligned}vt_{cool}&\sim .02pc\left({1\,Gauss \over B}\right)^{2}\left({1 \over \gamma }\right)\\&\sim 10^{7}pc\left({40\,\mu G \over B}\right)^{2}\left({1 \over \gamma }\right)\\\end{aligned}}\,\!}

where 40 ${\displaystyle \mu }$G is a more appropriate magnetic field for the post shock environment in a supernova remnant. We can convert the electron energy, ${\displaystyle \gamma }$ to a photon energy using the critical frequency found in past lectures:

{\displaystyle {\begin{aligned}\nu _{c}={\frac {3\gamma ^{2}eB}{2m_{e}c}}\\\Rightarrow \gamma ={\sqrt {\frac {2\nu _{c}m_{e}c}{3eB}}}\end{aligned}}\,\!}

so

{\displaystyle {\begin{aligned}vt_{cool}\sim .9pc\left({40\,\mu G \over B}\right)^{3/2}\left({1keV \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 (${\displaystyle \sim }$ 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, ${\displaystyle \kappa }$ with dimensions of [length][velocity] and can be written as :

{\displaystyle {\begin{aligned}\kappa =\lambda _{mfp}v.\end{aligned}}\,\!}

For very relativistic electrons, ${\displaystyle v\sim c}$ and ${\displaystyle \lambda _{mfp}}$ is taken as some constant multiple of the gyroradius ${\displaystyle \left({\frac {\gamma m_{e}c^{2}}{eB}}\right)}$, so that

{\displaystyle {\begin{aligned}\kappa \sim {\frac {\gamma m_{e}c^{3}}{eB}}\end{aligned}}\,\!}

Quick dimensional analysis tells us the the characteristic length scale of diffusion is

{\displaystyle {\begin{aligned}{\sqrt {\kappa t_{cool}}}&\sim 5\times 10^{-8}pc\left({\frac {1Gauss}{B}}\right)^{3/2}\\&\sim 30pc\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 ${\displaystyle vt_{cool}}$ 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 ${\displaystyle B^{-3/2}}$, we can use effective source sizes to infer local values of the magnetic field, a parameter that is otherwise hard to get at.

### Spectrum

The cooling time also has an impact on the shape of the spectra. To calculate the total ${\displaystyle F_{\nu }}$ from our source, we would integrate the specific intensity produced by these electrons over the solid angle subtended by the source:

{\displaystyle {\begin{aligned}F_{\nu }&=\int I_{\nu }d\Omega \\\end{aligned}}\,\!}

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 ${\displaystyle vt_{cool}}$ or ${\displaystyle {\sqrt {\kappa t_{cool}}}}$ before the radiation will effectively shut off. For the case of a spherical shock and constant specific intensity, then, the integral becomes

{\displaystyle {\begin{aligned}F_{\nu }&=\int I_{\nu }{\frac {dA}{D^{2}}}\\&\sim {1 \over D^{2}}\int \limits _{0}^{vt_{cool}}\int \limits _{0}^{2\pi }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

{\displaystyle {\begin{aligned}F_{\nu }&=\int I_{\nu }{\frac {dA}{D^{2}}}\\&\sim {1 \over D^{2}}\int \limits _{0}^{\sqrt {\kappa t_{cool}}}\int \limits _{0}^{2\pi }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 ${\displaystyle \nu ^{-1}}$ 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.

## Special Relativity Derivations

As an exercise in special relativity, we’ll derive what ${\displaystyle {\vec {E}}^{\prime }}$ is by investigating two parallel plates–the bottom one having charge density ${\displaystyle \sigma _{0}}$ and the top having ${\displaystyle -\sigma _{0}}$ We know in a motionless frame that the field between the plates is:

{\displaystyle {\begin{aligned}E_{y_{0}}&=4\pi \sigma _{0}\\B_{0}&=0\\\end{aligned}}\,\!}

In a frame (${\displaystyle s}$) where we are moving at ${\displaystyle v=-v_{0}{\hat {x}}}$, these become:

{\displaystyle {\begin{aligned}E_{y}&=4\pi \sigma =4\pi \gamma _{0}\sigma _{0}=\gamma _{0}E_{y_{o}}\\B_{z}&={4\pi J \over c}={4\pi \over c}\sigma v_{0}={4\pi \over c}\gamma _{0}\sigma _{0}v_{0}=E_{y_{0}}{v_{0} \over c}\gamma _{0}\\\end{aligned}}\,\!}

Jumping into one more frame (${\displaystyle s^{\prime }}$) where ${\displaystyle v=v^{\prime }{\hat {x}}}$ relative to frame ${\displaystyle s}$, we have:

{\displaystyle {\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 ${\displaystyle \sigma ^{\prime }}$, and our first instinct might be to say ${\displaystyle \sigma ^{\prime }=\gamma ^{\prime }\sigma }$, but that is wrong. We have to reference it from ${\displaystyle \sigma _{0}}$:

${\displaystyle \sigma ^{\prime }=\gamma ^{\prime }\sigma _{0}\,\!}$

Now we do some algebra:

{\displaystyle {\begin{aligned}E_{y}^{\prime }&=4\pi {\gamma ^{\prime }}\sigma _{0}\\&=4\pi \gamma {\gamma _{0}}(1-{vv_{0} \over c^{2}})\sigma _{0}\\&=\underbrace {4\pi {\gamma _{0}}\sigma _{0}} _{E_{y}}\gamma -\underbrace {4\pi {\gamma _{0}}v_{0}\sigma _{0} \over c} _{B_{z}}{\gamma v \over c}\\\end{aligned}}\,\!}

Thus we have:

${\displaystyle {\begin{matrix}{E_{y}^{\prime }=\gamma E_{y}-{\gamma v \over c}B_{z}}&{E_{z}^{\prime }=\gamma E_{z}+{\gamma v \over c}B_{y}}&{B_{y}^{\prime }=\gamma B_{y}+{\gamma v \over c}E_{z}}\end{matrix}}\,\!}$

And, of course, ${\displaystyle E_{x}^{\prime }=E_{x}}$. The only thing we need to get now is ${\displaystyle B_{x}^{\prime }}$. For this we’ll talk about a solenoid aligned with the ${\displaystyle {\hat {x}}}$ direction with ${\displaystyle n}$ turns per unit length. The field of this solenoid in the rest frame is:

${\displaystyle B_{x}={4\pi \over c}nI\,\!}$

Jumping to a frame where ${\displaystyle {\vec {v}}=v{\hat {x}}}$, ${\displaystyle B_{x}^{\prime }={4\pi n^{\prime }I^{\prime } \over c}}$. Using that:

{\displaystyle {\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:

${\displaystyle {B_{x}^{\prime }=B_{x}}\,\!}$

Recall that we’ve derived all of this for boosts in the ${\displaystyle {\hat {x}}}$ direction. To be completely general, we’ll write them for any direction:

{\displaystyle {\begin{matrix}{\begin{aligned}E_{\|}^{\prime }&=E_{\|}\\B_{\|}^{\prime }&=B_{\|}\\\end{aligned}}&{\begin{aligned}{\vec {E}}_{\perp }^{\prime }&=\gamma ({\vec {E}}_{\perp }+{{\vec {v}} \over c}\times {\vec {B}})\\{\vec {B}}_{\perp }^{\prime }&=\gamma ({\vec {B}}_{\perp }-{{\vec {v}} \over c}\times {\vec {E}})\\\end{aligned}}\end{matrix}}\,\!}