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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?

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This is not a complete answer. The amplitude of the scattering process with 4 particles $(1) + (2) \to (3) + (4)$ must be lorentz invariant (i.e. cannot depend on the frame) and thus, can only depend on product of 4-momenta (scalars). There are 10 such products ($p1.p1, p1.p2, p1.p3$ etc). Because of momentum conservation (4 constraints) and the on-mass shell constraint (4 constraints), only 2 independent combinations remain. We have 3 Mandelstam variables $s,t,u$. But they are not independent since the relation $s+t+u = \sum_{i=1}^4 m_i^2$ is satisfied. And so, again, only 2 variables are needed.

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  • $\begingroup$ What you're saying makes sense, but to prove the relation between the Mandlestam variables we have to use momentum conservation, so it can't be an independent constraint to momentum conservation. What does that tell us about the number of degrees of freedom? $\endgroup$ – Joe Jun 4 '15 at 16:25
  • $\begingroup$ yes, both momentum conservation and on-shell condition are needed to establish the Mandelstam relation. The number of degrees of freedom is thus 2. $\endgroup$ – Paganini Jun 4 '15 at 16:48
  • $\begingroup$ Oh, whoops! I obviously wasn't paying attention there, thanks $\endgroup$ – Joe Jun 4 '15 at 17:38
  • $\begingroup$ So what that means is that the thing I read about phase space symmetry was wrong, and we have $3N-10$ Mandlestam variables in an $N$ particle process? $\endgroup$ – Joe Jun 4 '15 at 17:40
  • $\begingroup$ No, now that I think, this isn't exactly what I mean. We have 3N-10 degrees of freedom after using the equivalent relations to $s+t+u=4M^2$, but what I want to know is how many scattering channels we have before we use that relation? I guess you can't calculate that just by counting degrees of freedom $\endgroup$ – Joe Jun 4 '15 at 17:46

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