thus also be multiplied by the bulk Lorentz factor $\Gamma$: $$E_{\rm obs} = E_{\rm src} \Gamma / (z + 1), \label{e_obs}$$ where the characteristic energy at the source is $E_{\rm src}$. Under the most common assumptions, a limit on the bulk Lorentz factor can be determined as follows: the blast wave must be optically thin to its own photons, otherwise it would still be a fireball, due to photon-photon pair production. The emission from a relativistic blast wave is highly beamed, raising the pair-production energy threshold to 1 MeV $\Gamma / \gamma > 1$, for photons with energies of $\gamma$, in units of the electron’s rest energy. A more careful treatment would have to include details of the spectrum against which the highest energy photon is scattering [@BaringHarding97; @2006ApJ...650.1004B]. Thus, in the simplest case, if a photon with energy $\gamma$ is observed, it puts a limit on the bulk Lorentz factor $\Gamma > \gamma$. Thus, $\Gamma$, as determined by the maximum observed photon energy, is found to vary from burst to burst. The intrinsic variability of the central engine or the possibility that relativistic shock collisions accelerate particles at the expense of $\Gamma$ [@reesmesz], make it very likely that $\Gamma$ 