How many relativistically invariant degrees of freedom in $n$-particle scattering?

Question

Suppose we have a scattering process with $n$ external legs with four-momenta $p_1, \cdots, p_n$. Naively there are $4n$ degrees of freedom, however most of these putative degrees of freedom are not relativistically invariant. With four external legs we get the Mandelstam variables that completely characterize the relativistically invariant degrees of freedom in our scattering problem. My question is: how many relativistically invariant degrees of freedom are associated to the more general $n$-particle scattering process outlined above?

Answer 1

For $n \geq 4$ the number is $3n-10$, because:

• You start with $4n$ degrees of freedom, from the $n$ particles' $4$-momenta.
• You lose $n$ degrees of freedom because each particle has to be on-shell, $E_i^2 = p_i^2 + m_i^2$.
• You lose $4$ degrees of freedom by overall $4$-momentum conservation.
• You lose $3$ degrees of freedom because you can always boost to the center of mass frame.
• You lose $3$ degrees of freedom because you can perform rotations.

Note that for $n = 4$ there are $2$ degrees of freedom left over, which is the expected answer. (There are $3$ Mandelstam variables, but they are redundant because $s + t + u$ is fixed to the sum of the squares of the masses.) One example of a non-redundant parametrization for $2 \to 2$ scattering is the total energy and the scattering angle in the center-of-mass frame. For $1 \to 3$ decay, one can use the "squared mass" variables on the axes of a Dalitz plot.

For $n = 3$ the answer is different because not all of the rotations actually do anything. If you have a $1 \to 2$ decay process, then the decay products have to come out back-to-back in the center of mass frame, so rotating about that axis does nothing. So you have $3n-9 = 0$ degrees of freedom in this case, also as expected.