Mandelstam Momentum Invariants | M.E. Irizarry-Gelpí

Tue 26 August 2014
Physics
#kinematics

In this post I would like to gather some facts about Mandelstam momentum invariants. For simplicity, I am going to consider a scattering process with $N$ external quanta. Without loss of generality, I will assume that each external quantum is incoming. I will also restrict to external scalar quanta, so all the information required to specify the state of an external quantum is contained in the energy-momentum vector $p_{a}$ carried by that quantum. I will label each of the external quanta with an integer from $1$ to $N$. Thus, there are $N$ energy-momentum vectors but the conservation constraint

$$p_{1} + p_{2} + \ldots p_{N} = 0$$

restricts the number of linearly-independent vectors to be $N - 1$.

Introduction

The Mandelstam momentum invariants are Lorentz scalar quantities that are made from linear combinations of energy-momentum vectors of different external quanta. Given $n$ distinct energy-momentum vectors, the $n$ quanta Mandelstam momentum invariant is defined by

$$s_{I_{1} I_{2} \ldots I_{n}} \equiv - (p_{I_{1}} + p_{I_{2}} + \ldots + p_{I_{n}})^{2}.$$

Since I have assumed that each external quantum is incoming, the linear combination only involves positive signs. For simplicity, I will refer to such an invariant as an $n$-Mandelstam invariant. If some of the external quanta are incoming, and some are outgoing, then the $n$-Mandelstam invariant is given by

$$s_{I_{1} I_{2} \ldots I_{n}} \equiv - (p_{I_{1}} + \sigma_{I_{1}I_{2}} p_{I_{2}} + \ldots + \sigma_{I_{1} I_{n}} p_{I_{n}})^{2}$$

where $\sigma_{IJ} = +1$ if either $p_{I}$ and $p_{J}$ are both incoming or both outgoing, and $\sigma_{IJ} = -1$ otherwise.

Two Quanta

Given two linearly-independent energy-momentum vectors, the 2-Mandelstam invariant is given by

$$s_{IJ} = - (p_{I} + p_{J})^{2}.$$

With $N$ energy-momentum vectors you can make a total of

$${N \choose 2} = \frac{N (N - 1)}{2}$$

possible 2-Mandelstam invariants. However, the conservation constraint allows you to write $N$ of these in terms of the remaining others, so in total you have

$$\mathfrak{M}_{2}(N) = \frac{N (N - 1)}{2} - N = \frac{N (N - 3)}{2}, \qquad N \geq 5$$

independent 2-Mandelstam invariants. Some special cases that I will explore later are $\mathfrak{M}_{2}(6) = 9$ and $\mathfrak{M}_{2}(8) = 20$.

Three Quanta

Given three linearly-independent energy-momentum vectors, the 3-Mandelstam invariant is given by

$$s_{IJK} = - (p_{I} + p_{J} + p_{K})^{2}.$$

A 3-Mandelstam invariant can always be written in terms of 2-Mandelstam invariants and masses:

$$s_{IJK} = s_{IJ} + s_{IK} + s_{JK} - m_{I}^{2} - m_{J}^{2} - m_{K}^{2}$$

where I have assumed that $p_{I}^{2} = -m_{I}^{2}$ with $m_{I}$ being real.

With $N$ energy-momentum vectors, you can make a total of

$$\mathfrak{M}_{3}(N) = {N \choose 3} = \frac{N (N - 1) (N - 2)}{6}, \qquad N \geq 7$$

possible 3-Mandelstam invariants. Note that $\mathfrak{M}_{3}(6) = 10$ because of the conservation constraint.

Four Quanta

Given four linearly-independent energy-momentum vectors, the 4-Mandelstam invariant is given by

$$s_{IJKL} = - (p_{I} + p_{J} + p_{K} + p_{L})^{2}.$$

Just like a 3-Mandelstam invariant, a 4-Mandelstam invariant can always be written in terms of 2-Mandelstam invariants and masses:

$$s_{IJKL} = s_{IJ} + s_{IK} + s_{IL} + s_{JK} + s_{JL} + s_{KL} - 2m_{I}^{2} - 2m_{J}^{2} - 2m_{K}^{2} - 2m_{L}^{2}.$$

With $N$ energy-momentum vectors, you can make a total of

$$\mathfrak{M}_{4}(N) = {N \choose 4} = \frac{N (N - 1) (N - 2) (N - 3)}{24}, \qquad N \geq 9$$

possible 4-Mandelstam invariants. Note that $\mathfrak{M}_{4}(8) = 35$ because of the conservation constraint.