Conservation of $p_a^\mu p_b^\nu \eta_{\mu \nu}$ for relativistic collisions Two photons with 4-momenta $p_1^\mu$ and $p_2^\mu$ collide to produce 2 new photons with momenta $p_3^\mu$ and $p_4^\mu$. I need to show that $p_a^\mu p_b^\nu \eta_{\mu \nu} = p_c^\mu p_d^\nu \eta_{\mu \nu}$ for each permutation $(a,b,c,d)$ of $(1,2,3,4)$.
I have tried doing this momentum and energy conservation but don't seem to get anywhere. Even if I could solve it this way it seems tedious to check each case (i.e. checking each possible value for a/b/c/d separately) so I am wondering if there is a more elegant way.
 A: The only thing you need is actually the momentum conservation, and the fact, that the photon is massless $p_\mu p^{\mu} = 0$. Then simply write momentum conservation in some order, for example:
$$
p_1 + p_2 = p_3 + p_4
$$
And take the square of both parts. This will give:
$$
2 (p_1, p_2) = 2 (p_3, p_4)
$$
Where $(p_a, p_a)$ vanish due to the $m = 0$. And by round bracket I mean $(p_a, p_b) = \eta_{\mu \nu} p_a^{\mu} p_b^{\nu}$. Rearraging the terms in momentum conservation in all possible ways will give the result for other permutations.
A: This is a case where the Mandelstam invariants come in handy.
$$
\begin{aligned}
s&=(p_{1}+p_{2})^{2}=(p_{3}+p_{4})^{2}\,,\\
t&=(p_{1}-p_{3})^{2}=(p_{4}-p_{2})^{2}\,,\\
u&=(p_{1}-p_{4})^{2}=(p_{3}-p_{2})^{2}\,.
\end{aligned}
$$
Since you have photons $p_a^2 = 0$, so
$$
\begin{aligned}
\tfrac12 s&=p_{1}\cdot p_{2}=  p_{3}\cdot p_{4}\,,\\
-\tfrac12 t&=p_{1}\cdot p_{3}=  p_{4}\cdot p_{2}\,,\\
-\tfrac12u&=p_{1}\cdot p_{4} = p_{3}\cdot p_{2}\,.
\end{aligned}
$$
These are all the inequivalent permutations.
A: $$p_1^\mu+p_2^\mu=p_3^\mu+p_4^\mu$$
$$p_1^\mu p_{1\mu}+2p_1^\mu p_{2\mu}+ p_2^\mu p_{2\mu}=p_3^\mu p_{3\mu}+2p_3^\mu p_{4\mu}+ p_4^\mu p_{4\mu}$$
For photon $p^\mu p_\mu=0$
$$2p_1^\mu p_{2\mu}=2p_3^\mu p_{4\mu}$$
$$\eta_{\mu\nu} p_1^\mu p_{2}^\nu=\eta_{\mu\nu} p_3^\mu p_{4}^\nu$$
You can take the combinations in momentum conservation for the other permutations, likewise.
$$p_1^\mu-p_4^\mu=p_3^\mu-p_2^\mu$$
