Synchrotron Radiation

The electromagnetic (EM) process are always related to photons because they are the particles to carry energy and momentum in electromagnetic interactions. Therefore, some of the EM reactions would GENERATE photons, such as pair annihilation $e^{+}e^{-}\to 2\gamma$, the inverse process of the pair creation. Electrons are also producing photons by interacting with magnetic field, which is called synchrotron radiation, the representative radiation process that can be described in the standard context of the classical electromagnetic theory.

Now let us derive the radiation intensity, i.e. the radiated energy per unit time $I\equiv d\epsilon/dt$. First we expand the treatment of the dipole radiation given by Eq. 8 to the super-relativistic condition that particles are moving almost at speed of light. The dipole ${\bf d}$ is related to acceleration ${\bf\omega}$ by $\ddot{d}=e\omega$ and Eq. 8 gives the total radiation intensity as

$$d\epsilon^{'} = {2e^2\over 3c^3}\omega^2dt$$ (37)

The radiated 4-momentum $dP^{\mu}=(d\epsilon/c, {\bf dP})$ should be equivalent to $(d\epsilon^{'}/c, {\bf0})$ when electron is at rest. We find

$$dP^{\mu} = -{2e^2\over 3c}{du^{\nu}\over ds}{du_{\nu}\over ds}dx^{\mu}$$ (38)

$$u^{\nu} \equiv {dx^{\nu}\over ds}$$

since

$${du^{\nu}\over ds}{du_{\nu}\over ds}\simeq -{\omega^2\over c^4}$$ (39)

in limit of $v\to 0$. The time component ($\mu=0$) of Eq. 39 gives the total radiation intensity, which reads

$$d\epsilon = {2e^2\over 3c^3}{w^2-{({\bf v}\times{\bf\omega})^2\over c^2} \over (1-{v^2\over c^2})^2}dt$$ (40)

The equation of motion in ${\bf B}$ and ${\bf E}$ is

$$m{\bf\omega}= e{\bf E}+ {e\over c}{\bf v}\times {\bf B},$$ (41)

then we finally obtain

$${d\epsilon\over dt}={2e^4\over 3m^2c^3} {({\bf E}-{1\over c}... ...bf B})^2 -{1\over c^2}({\bf Ev})^2\over (1-{v^2\over c^2})^2}.$$ (42)

In the uniform ${\bf B}$ field, an electron's movement is characterized by the synchrotron frequency

$$\omega_B = {ecB\over E}$$ (43)

where $E$ is energy of a charged particle. Then Eq. 42 leads to

$${d\epsilon\over dt}={2e^4\over 3m^2c^5} {B^2v^2\over (1-{v^2\over c^2})^2}\sin^2\phi.$$ (44)

Here $\phi$ is a pitch angle between electron momentum and ${\bf B}$ field. In ultra-relativistic case of $v\to c$, we obtain

$${d\epsilon\over dt}={2e^4B^2\over 3m^2c^3} ({E\over mc^2})^2\sin^2\phi.$$ (45)

Averaging the pitch angle and using the Thompson scattering cross section 12 gives the final expression:

$${d\epsilon\over dt}={4\over 3}U_B\sigma_T{m_e^2\over m^3c}({E\over mc^2})E,$$ (46)

where $U_B=B^2/8\pi$ is the energy density of magnetic field. When a particle moving in $B$ is an electron, $m$ is set to $m_e$ in the above equation.

Energies of the radiated photons are narrowly distributed around

$$\epsilon\simeq ({E\over mc^2})^2\hbar {eB\over mc}$$ (47)