So in 4D we have three Mandlestam variables for a 4-particle scattering process. This corresponds to $p_i^\mu$ giving us 16 degrees of freedom. Momentum conservation reduces this by 4, and we have 4 mass shell constraints, leaving us with 8. Then using Lorentz invariance, we expect to be able to reduce this by a further 6 by choosing our axes carefully. I have read that we have a phase space symmetry which means we only reduce by 5, can someone explain this to me? Normally when we have symmetries it makes things easier (ie. reducing degrees of freedom, not increasing them) Clearly 16-4-4-5 = 3, in agreement with our number of Mandlestam variables
Now I want to extend to $N$ particles. My calculation gives $4N$ degrees of freedom, $N$ mass shell constraints, 4 momentum conservation conditions and 5 reductions by Lorentz invariance. $4N - N - 4 - 5 = 3N-9$ But I'm not sure about the phase space symmetry argument from above, does this extend to $N$ particles? If it doesn't I might have to change the -5 to something else.
It's also quite obvious to see that we have three Mandlestam variables in a four-particle scattering by simply drawing the Feynmann diagrams, permuting legs and seeing what happens. Is there some kind of permutational calculation we can do to give us the number of Mandlestam variables in an $N$-particle process using Feynmann diagrams? And would it depend on the types of verticles (cubic, quartic) that we have in the theory?