Dihedral Momentum Invariants | M.E. Irizarry-Gelpí

Tue 27 January 2015
Physics
#kinematics

In this post I introduce what I call dihedral momentum invariants. As usual, I will assume that every energy-momentum vector $p_{I}$ satisfies an on-shell constraint of the form

$$ p_{I}^{2} = -m_{I}^{2} $$

with the mass $m_{I}$ being real and positive.

Introduction

The Mandelstam momentum invariants pretty much tell you almost all of the kinematic information (specially if you are only working with scalar quanta). Indeed, once you have all the 2-Mandelstam invariants, you can use them to write all higher non-trivial Mandelstam invariants. So the 2-Mandelstam invariants are special. You also know about Gram momentum invariants. Since the 2-Gram invariant is quadratic in the 2-Mandelstam invariant, inverting this relation introduces complicated non-polynomial relations. The dihedral momentum invariant is linear in the 2-Mandelstam invariant. Further more, it is also dimensionless. Given two linearly-independent energy-momentum vectors, the dihedral momentum invariant is defined as

$$\xi_{IJ} \equiv - \frac{p_{I} \cdot p_{J}}{\sqrt{-p_{I}^{2}} \sqrt{-p_{J}^{2}}} = \frac{s_{IJ} - m_{I}^{2} - m_{J}^{2}}{2 m_{I} m_{J}}.$$

In the second step I have used the definition of the 2-Mandelstam invariant, and the on-shell constraint, to relate the dihedral invariant to the 2-Mandelstam invariant.

Note two special cases: when $s_{IJ} = (m_{I} + m_{J})^{2}$ you have $\xi_{IJ} = 1$, and when $s_{IJ} = (m_{I} - m_{J})^{2}$ you have $\xi_{IJ} = -1$. If the energy-momentum vectors lived in a space with Euclidean signature, then $\xi_{IJ}$ would correspond to a cosine function, which has a bounded range. However, in a space with Minkowski signature, $\xi_{IJ}$ does not has a bounded range. Indeed, $\xi_{IJ}$ only allows a cosine interpretation in the kinematic region

$$(m_{I} - m_{J})^{2} \leq s_{IJ} \leq (m_{I} + m_{J})^{2}.$$

Outside of this kinematic region is the scattering region.

2-Gram Invariant

Writting the 2-Gram invariant in terms of the dihedral invariant, you find

$$G_{IJ} = m_{I}^{2} m_{J}^{2} (\xi_{IJ} + 1) (\xi_{IJ} - 1) .$$

In this form, the geometrical meaning of $\sqrt{G_{IJ}}$ as the area of the 2-parallelotope generated by $p_{I}$ and $p_{J}$ is more transparent.

3-Gram Invariant

The expression for the 3-Gram invariant in terms of the three 2-Mandelstam invariants is not very illuminating. Using the three corresponding dihedral invariants, you find

$$G_{IJK} = m_{I}^{2} m_{J}^{2} m_{K}^{2} (1 - \xi_{IJ}^{2} - \xi_{IK}^{2} - \xi_{JK}^{2} + 2 \xi_{IJ} \xi_{IK} \xi_{JK} ) .$$

This expression might not be very useful to determine the sign properties of $G_{IJK}$, but it is a start. Again, the geometrical meaning of $\sqrt{G_{IJK}}$ as the volume of the 3-parallelotope generated by $p_{I}$, $p_{J}$, and $p_{K}$ is more transparent.