QED

$$\eta = \hbar \omega / mc^2 \sqrt{(\varepsilon E + p \times B)^2 -(p\cdot E)^2}$$

if we set the laser photon energy in the unit of $mc^2$, and frequency in the unit of $mc^2 / \hbar$.

$$\eta_e = \xi\sqrt{(\varepsilon E + p \times B)^2 -(p\cdot E)^2}$$

for laser with wave length $\lambda = 1 \mu m$, the corresponding energy is $1.024eV$, and $\xi \approx 1.024 / 0.511 \times 10^6$.

The spectrum of emission

$$\frac{\partial I}{\partial \omega} = \frac{\sqrt{3}}{2\pi} \frac{e^3 H_{eff}}{mc^2}(1-\delta)\bigg[ F_1(z_{q}) + \frac{3}{2}\delta \chi z_q F_2(z_q)\bigg]$$

with $z_q = \frac{2}{3} \frac{\delta}{\chi(1-\delta)}$, $H_{eff} = \chi E_S/\gamma$. and the schwinger field strength $E_S = m^2 c^3/ e\hbar \approx 10^{18} V/m$.

Here $\omega$ is the emitted photon frequency with $\hbar \omega = \delta \gamma m c^2$, with probability (during $\Delta t$) in cgs

$$P(\delta) = \frac{\partial I_q}{\partial \omega} \frac{\partial \omega}{\partial \delta} \frac{1}{\hbar \omega} \Delta t \\ = \bigg[\Delta t \frac{e^2 mc}{\hbar^2}\bigg] \frac{\sqrt{3} \chi (1-\delta)}{2\pi \gamma \delta} \bigg[ F_1(z_q) + \frac{3}{2} \delta \chi z_q F_2(z_q) \bigg]$$

when we change into the normalization representation:

$$P(\delta) = \bigg[ \frac{ \alpha_f \Delta t}{ \xi_0} \bigg] \frac{\sqrt{3} \chi (1-\delta)}{2\pi \gamma \delta} \bigg[ F_1(z_q) + \frac{3}{2} \delta \chi z_q F_2(z_q) \bigg]$$

here $\xi_0 = \hbar \omega_0 / mc^2$. $\omega_0$ is the frequecy we used to normalize every variable.

Similarly, the pair production per $\Delta t$ in cgs and has been normalized:

$$P(\delta_e) = \bigg[ \frac{\alpha_f \Delta t}{\xi_0} \bigg] \frac{\sqrt{3}}{2\pi} \chi \frac{1}{\varepsilon} (\delta_e - 1) \delta_e \bigg[F_1(z_p) -\frac{3}{2}\chi_\gamma z_p F_2(z_p) \bigg]$$

with $z_p = \frac{2}{3 \chi_\gamma (1 - \delta_e) \delta_e }$, electron energy: $\delta_e \varepsilon$, positron energy $(1-\delta_e) \varepsilon$. their momentum is parallel to the photon momentum with : $\delta_e \varepsilon / c$ and $(1-\delta_e) \varepsilon / c$&&

$$\chi_\gamma = \frac{\hbar \omega}{mc^2} \frac{H_{eff}^\gamma}{E_s}\\ = \xi_0 \varepsilon \sqrt{(E+ \frac{c}{\omega} k\times B )^2 - (\frac{k}{|k|}\cdot E)^2} \\ =\xi_0 \sqrt{(\varepsilon E + p \times B)^2 - (p \cdot E)^2}$$

Thus $\chi_\gamma = \chi_e$ if the field strength is the same and the rest energy is the same for electron and photon.

$$\varepsilon = \hbar \omega_\gamma / mc^2$$

Here,

$$F_1 (x) = x\int_x^\infty dt K_{5/3}(t)$$

$$F_2(x) = x K_{2/3}(x)$$

by using $2 \frac{d}{dz} K_{2/3}(z) + K_{1/3}(z) = -K_{5/3}(z)$, the photon emission probability can be simplified as

$$P(\delta) = \frac{\alpha_f \Delta t}{\xi_0 \sqrt{3}\pi \gamma} \bigg[ \bigg(2+ \frac{\delta^2}{1-\delta} \bigg) K_{2/3}(z) - \int_{z}^\infty K_{1/3}(t)dt \bigg]$$

with $z = \frac{2\delta}{3\chi(1-\delta)}$.

Also, for the pair production probability we use :

$$P(\delta_e) = \frac{\alpha_f \Delta t}{\xi_0 \sqrt{3}\pi \varepsilon} \bigg[ \bigg(2 - \frac{1}{\delta(1-\delta)} \bigg) K_{2/3}(z) - \int_{z}^\infty K_{1/3}(t)dt \bigg]$$

With $z= \frac{2}{3\chi \delta (1- \delta)}$.

while for the 2nd kind modified bessel function $K_{2/3}(x)$, this can be calculated from the gsl lib. and the $1/3$ integration can be done via a numerical interpolation.

Numerical Tricks for stochastic radiation

since the radiation spectrum is divergent in the infrared limit, a numerical tricks has been used to avoid this problem. we have borrowed this method from Ginsokov's :

let random $r_1, r_2 \in [0, 1]$. a photon with energy $r_1 \gamma mc^2$ will be emitted if $r_2 < P(r_1)$.

Tricks:

let $\delta = r_1^3$, and

$$P_m(r_1) \equiv \frac{\partial f(r)}{\partial r} P(f(r))$$

where $f(r) = r^3$.

at last, if two random $r_1, r_2$ obeys $r_2 < P_m(r_1)$. then a photon with energy $r_1 \gamma mc^2$ will be emitted.