## FANDOM

162 Pages

In the classical Compton scattering, the probability of a photon to scatter in a certain direction depends only on the scattered photon and electron. However, since photons are bosons, the probability of scattering in a certain direction increases with the number of photons already moving in that direction. Let $N \left( \mathbf{k} \right)$ be the distribution function of photons, such that the number density of photons (photons per unit volume) is given by $n = \int \frac{d^3 k}{\left( 2 \pi \right)^3} N \left( \mathbf{k} \right)$, and $\mathbf{k}$ is the wave - number (and the momentum of the photons). Let the distribution functions of the electrons be $f \left( \mathbf{p} \right)$, where $\mathbf{p}$ is the momentum of the electron. Let $w \left(\mathbf{p}, \mathbf{k}, \mathbf{k}' \right)$ be the probability of scattering an electron with momentum $\mathbf{p}$ and a photon of momentum $\mathbf{k}$ to a state where the momentum of the photon is $\mathbf{k}'$ (and the momentum of the electron is $\mathbf{p} + \mathbf{k} - \mathbf{k}'$). The rate equation for the photons is given by

$\frac{\partial N}{\partial t} = \int d^3 p \int \frac{d^3 k}{\left( 2 \pi \right )^3} f \left( \mathbf{p} \right ) N \left( \mathbf{k}' \right ) \left[ 1 + N \left( \mathbf{k} \right ) \right ] w \left(\mathbf{p}, \mathbf{k}', \mathbf{k} \right ) -$

$\int d^3 p \int \frac{d^3 k}{\left( 2 \pi \right )^3} f \left( \mathbf{p} \right ) N \left( \mathbf{k} \right ) \left[ 1 + N \left( \mathbf{k}' \right ) \right ] w \left(\mathbf{p}, \mathbf{k}, \mathbf{k}' \right )$

If the electron does not recoil, then the incoming and outgoing photons have the momenta, so detailed balance reads

$w \left(\mathbf{p}, \mathbf{k}, \mathbf{k}' \right) = w \left(\mathbf{p}, \mathbf{k}', \mathbf{k} \right)$

Substituting into the rate equation eliminates the extra terms (due to symmetry) and only the regular scattering is left.

However, if recoil is taken into account, then detailed balance reads

$w \left(\mathbf{p}, \mathbf{k}, \mathbf{k}' \right) = w \left(\mathbf{p} - \Delta \mathbf{k}, \mathbf{k}', \mathbf{k} \right)$

where $\Delta \mathbf{k} = \mathbf{k}' - \mathbf{k}$. Substituting this into the equation, assuming $k \ll p$ and applying a Taylor expansion yields

$\frac{\partial N}{\partial t} = \int d^3 p \int \frac{d^3 k}{\left( 2 \pi \right )^3} f \left( \mathbf{p} \right ) \left[ N \left( \mathbf{k}' \right ) - N \left(\mathbf{k} \right ) \right ] w \left(\mathbf{p}, \mathbf{k}, \mathbf{k}' \right ) +$

$\int d^3 p \int \frac{d^3 k}{\left( 2 \pi \right )^3} \Delta \mathbf{k} \cdot \frac{\partial f}{\partial \mathbf{p}} N \left( \mathbf{k}' \right ) N \left(\mathbf{k} \right ) w \left(\mathbf{p}, \mathbf{k}, \mathbf{k}' \right )$

The first integral is just the classical Compton scattering. The second term is a correction due to recoil and induced emission.

## Order of Magnitude Estimate Edit

Induced Compton scattering is a positive feedback process. The more photons of one energy there are, the more likely it is for photons of a different energy to be scatter to that energy. The number of photons with the first energy increases exponentially with the optical depth. We proceed to estimate the optical depth for induced Compton scattering.

We recall that the optical depth for incoherent Compton scattering is

$\tau_{i} \approx n_e \sigma_t x$

where $n_e$ is the number density of electrons, $\sigma_t$ is the Thompson cross section and $x$ is the distance. From the discussion in the previous section, the optical depth is reduced by the relative energy difference. If the energy of the photons is much smaller than the rest mass energy of the electron, then the energy transferred in the collision is $\hbar^2 \omega^2/m_e c^2$, so the optical depth is reduced by a factor of $\hbar \omega/m_e c^2$. The optical depth also increases by a factor that is comparable to the occupation number, which can be expressed in terms of the brightness temperature $k T_b/\hbar \omega$ so overall the optical depth is roughly given by

$\tau \approx n_e \sigma_t x \frac{k T_b}{m_e c^2}$

A more detailed calculation shows that an exponential increase only occurs in frequency ranges where the spectrum increases super-linearly with the frequency.