Justification of $P_{\text{photon}}=E/c$ in derivation of $E=mc^2$ I recently was reading up on the derivation of $E=mc^2$. Now, I came across this derivation at this link. I noticed that several lines into the derivation they throw in the equation
$$P_{\text{photon}}=E/c.$$
How did they come to this equation so easily? Did they use special relativity equation for $\gamma$ to get it, or did they just have some super easy way to get it?
 A: The general Energy equation is
$$ E^2  = (mc^2)^2 + p^2c^2 $$
where m is rest mass. Since in case of photon rest mass is 0.
So we will get
$$ E = pc $$
$$ p = E/c $$
A: No, the cited equation is not justified by the relativistic energy formula in the derivation the OP asks about. The corresponding text of the derivation is as follows:

For the momentum of our photon, we will use Maxwell’s expression for the momentum of an electromagnetic wave having a given energy. If the energy of the photon is E and the speed of light is c, then the momentum of the photon is given by:
  $$
p_{photon} = \frac{E}{c}
$$

Maxwell's expression actually relates the electromagnetic energy density $u = \frac{1}{2}\left( \epsilon_0 \vec{E}^2 + \frac{1}{\mu_0}\vec{B}^2\right)$ to the linear momentum density of the electromagnetic field for the case of a plane wave. The linear momentum is defined in terms of the Poyinting vector $\vec{S} = (1/\mu_0)\left(\vec{E} \times \vec{B} \right)$
as
$$
\vec{p} = \frac{1}{c^2}\vec{S}
$$
For a plane wave $\vec{E} = \vec{E_0} cos\left( \vec{k}\cdot\vec{r} - \omega t \right)$, $\vec{B} = (1/c)(\vec{k} \times \vec{E_0}) \cos\left(\vec{k}\cdot\vec{r} - \omega t \right)$, it can be checked that 
$$
\vec{S} = cu\frac{\vec{k}}{|\vec{k}|} 
$$
which implies
$$
|\vec{p}| = \frac{u}{c}
$$
This is the relation invoked in the derivation up to a relabeling of the energy.
A: Well, if you accept energy-momentum conservation, then the equation you referred to can be easily obtained from the energy-momentum relation of E^2=(p c)^2+(mc^2)^2, where E is the energy of the particle, p is its momentum, m is its rest mass, and c is the speed of light (see Energy–momentum relation). For photons, the rest mass is zero, so this equation reduces to the equation that you presented.
