The Thomson scattering optical depth for a photon ar radius r I am looking to understand some more about the physics of gamma ray bursts. In particular I am looking at the origin of the "prompt emission". Some of the energy associated with this prompt emission is thought to come from the photosphere of the expanding fireball at the centre of the GRB, which expands at relativistic speeds.
The Thomson scattering optical depth for a photon at radius $r$ is give n by
$$\tau = \int \frac{dr}{c} (c-v) \sigma_T n_e  $$
where $c$ is the speed of light, $n_e$ the electron number density and $\sigma_T$ the scattering cross section and $v$ the velocity of the expanding fireball.
Can anyone give me a physical explanation behind this equation? Also, at the photospheric radius, the optical depth becomes 1. Why should this be so? Thanks 
 A: The origin of this equation is reasonably well explained in Abramowicz (1991).
If you take a relativistically expanding enevelope and only consider Thomson scattering, then as the electron scattering cross-section in the co-moving frame $\sigma_T$ is independent of frequency, then the mean free path of a photon in the co-moving frame is independent of the envelope velocity as seen in the "stationary" (observers) frame.
Abramowicz et al. shows that in the case of photons moving "downstream", i.e. with the flow 
$$ d\tau = \gamma (1 - \beta) d\tau_0,$$
where $\beta = v/c$, $\gamma = (1-\beta^2)^{-1/2}$ is the Lorentz factor, $\tau_0$ is the optical depth in the co-moving frame and $\tau$ is the optical depth in the observer's frame.
Now $\tau_0(r) = \int_{r}^{\infty} n_{e,0} \sigma_T\ dr$, so
$$  \tau(r) = \int^{\infty}_{r} \gamma (1 - \beta)  n_{e,0} \sigma_T\ dr,$$
where $n_{e,0}$ is the electron density in the co-moving frame.
But if $ v\ll c$ then $\gamma (1 - \beta) \simeq (1- \beta)$ and $n_{e,0} \simeq n_e$ thus
$$  \tau(r) \simeq  \int_{r}^{\infty} (1 - \beta) n_e \sigma_T\ dr = \int_{r}^{\infty} \frac{(c-v)}{c}  n_e \sigma_T\ dr$$
The integral is carried out from some physical depth in the wind out to infinity. The Thomson scattering photosphere can be defined as where this optical depth is approximately unity. This is because this will be roughly where the radiation decouples from the flow and photons can escape.
