 Solution

With $c=1$, the total four-momentum is $P_{\mathrm{tot}}={\sum }_i{P}_i$. A sum of Lorentz four-vectors transforms as a four-vector, sois a Lorentz invariant. Its value is the squared invariant mass of the complete system, irrespective of whether its constituents interact. Isolation makes the total four-momentum constant in time.

For a photon gas, write ${\mathbf{p}}_i=W_i{\mathbf{n}}_i$ with $|{\mathbf{n}}_i|=1$. Its individual null norms do not make the total norm vanish:For an isotropic gas in its momentum rest frame, ${\sum }_i{\mathbf{p}}_i=0$ and $s^2=W_{\mathrm{tot}}^2$. Zero is possible only when all nonzero photons propagate in the same direction. Two equal-energy photons at right angles, for example, give $s^2=2W^2$, rather than zero. Without an angular distribution, the pairwise expression is the general answer; isotropy is an additional assumption.