Electron Bremsstrahlung

If $f_e$ is the energy spectrum of cosmic electrons in units of $particle$/$cm^2$ $sec$ $sr$ $MeV$, the photons of energy $E_{\gamma}$ produced by electrons of energy $E_e$ are:

$$dN_{\gamma}(E_e,E_{\gamma})= 4 \pi \sigma(E_{\gamma},E_e) f(E_e) n_H dE_{\gamma} dE_e =$$ (1.15)

$$= \frac{4 \pi n_H}{x_0 E_{\gamma}} f(E_e) dE_{\gamma} dE_e$$

where we used:

$$\sigma(E_{\gamma},E_e) = \frac{1}{x_0 E_{\gamma}}$$ (1.16)

with :

$$x_0 = \frac{1}{4 \alpha r_e^2 G_a}$$ (1.17)

$\alpha$ is the constant of fine structure, $r_e$ is the classical electron radius. The Gaunt factor $G_a$ for bare nuclei is:

$$G_a = ln (2\gamma) - \frac{1}{3}$$

and for neutral atoms is:

$$G_a = ln (\frac{183}{Z^{1/3}}) - \frac{1}{18}$$

The Bremsstrahlung gamma-ray emissivity of interstellar matter is therefore:

$$g_{br} = \frac{4 \pi n_H}{x_0 E_{\gamma}}$$ (1.18)

The radiation lost bu Bremsstrahlung of a relativistic electron of energy $E_e$ moving through a medium formed by Hydrogen atoms is :

$$I(\omega) = \frac{e^6 N}{16 \pi^3 \epsilon_0^3 c^4 m_e^2} G_a = 4 c \hbar \alpha r_e^2 N G_a$$ (1.19)

It shows a flat energy spectrum up to energy $\omega=E_{e}/\hbar$. The photon spectrum is then:

$$n_{\gamma}(\hbar \omega) = \frac{ 4 c \alpha r_e^2 N}{\hbar \omega } G_a$$ (1.20)

The integration of the (1.19) on $\omega$ leads to the energy emitted (therefore lost) by the electron :

$$- \frac{dE}{dt} = 4 c \alpha r_e^2 N G_a \;E$$ (1.21)

For a distribution of electrons with spectrum $N_e(E)= k E^{-a}$ we get:

$$I(\hbar \omega)= 4 Z^2 c \alpha r_e^2 N G_a k \frac{(\hbar \omega)^{-a}}{a-1}$$ (1.22)

Therefore the index of photon spectrum of electron spectrum have the same value.