Mandelstam variables for 2 to 3 particle scattering I'm trying to work out the mandelstam variables for 2 particles scattering to produce 3 particles. Also each particle is massless.
I think there must be 5 because all possible scalar products of the incoming and outgoing momentum must be expressible in terms of the mandelstam variables, and there are 6 possible scalar products. But only 5 are linearly independent. Thus there are 5 Mandelstam variables.
But I don't know how to actually calculate them? Are they equal to the 5 linearly independent scalar products? Any help in the right direction would be appreciated. 
 A: There are potentially two different questions: how many different non trivial mandelstam variables and how many independent mandelstam variables there are. The claim is that there are 10 different mandelstam variables for 5 particle process, and there are 5 independent mandelstam variables. 
First, all possible mandelstam variables can be written as $s_{i_1\dots i_l}=-(p_{i_1}+\dots+p_{i_l})^2$ with $l\leq5$.But due to the momentum conservation, $\sum_{i=1}^5p_i=0$, they are not all distinct. More specifically, we see $l=1,4$ or $5$ give trivial ones. $l=2$ gives the same variables as $l=3$ because of momentum conservation.Thus we may just consider the case with $l=2$, and it gives ${5 \choose 2}=10$ different non-trivial mandelstam variables, and $s_{ij}=-2p_ip_j$.
Now for linearly independent ones, we see that the following relations hold for all $i$: 
$$\sum_{j\neq i}s_{ij}=0,$$
as we can directly compute using the momentum conservation: 
$$\sum_{j\neq i}s_{ij}=-2p_i\sum_{j\neq i}p_j=2p_i^2=0.$$
Thus we have at most 5 linearly independent mandelstam variables. It takes a little bit more to show the other direction to conclude that for example, $s_{12},s_{13},s_{14},s_{23},s_{24}$ is a set of generators.
