Scattering

Dru B. Renner
www.DruBryantRenner.org

2 May 2003

The difference of two forward, time-like vectors with the same invariant magnitudes is (1) space-like or (2) identically zero. This situation occurs in elastic scattering with single photon (or other vector particle) exchange. Consider two vectors, $k$ and $k^\prime$, with the same mass $m$ and positive energy, i.e. ${k}^2=m^2$, ${k^\prime}^2=m^2$, $k_0>0$, and $k^\prime_0>0$. Under these circumstances, we can boost to the rest frame for $k^\prime$. In this frame $k^\prime=(m,0,0,0)$ and $k=(E,\vec{p})$ where $E^2=\vec{p}^2+m^2$. The square of the difference is $q^2={k}^2+{k^\prime}^2-2k\cdot k^\prime=2m^2-2Em=-2m(E-m)\le 0$. The condition for $q^2=0$ is $E=m$ or $\vec{p}=0$ which implies that $k=k^\prime$. Therefore, $q^2<0$ or $q=0$; both of which satisfy $q^2\le 0$.

The sum of two forward, time-like vectors with the same invariant magnitudes is (1) time-like or (2) light-like if the two vectors are identical and have a vanishing magnitude. This situation occurs in pair annihilation to a photon (or other vector particle). Again consider two vectors, $k$ and $k^\prime$, with the same mass $m$ and positive energy, i.e. ${k}^2=m^2$, ${k^\prime}^2=m^2$, $k_0>0$, and $k^\prime_0>0$. Under these circumstances, we can boost to the center of mass frame. In this frame $k=(E,\vec{p})$ and $k^\prime=(E,-\vec{p})$ where $E^2=\vec{p}^2+m^2$. The sum is $q=k+k^\prime=(2E,0,0,0)$ with $q^2=4E^2=4m^2+4\vec{p}^2$. The condition for $q^2=0$ is $m=0$ and $\vec{p}=0$ which implies that $k=k^\prime$. Therefore, $q^2>0$ or $k=k^\prime$ and $m=0$; both of which satisfy $q^2\ge 4m^2\ge 0$.