9.2: Invariant Mass

One of the key numbers we can extract from mass and momentum is the invariant mass, a number independent of the Lorentz frame we are in

$$W^2c^4 = (\sum_i E_i)^2 - (\sum_i {\vec{p}}_i)^2 c^2. \nonumber$$

This quantity takes it most transparent form in the center-of-mass, where $\sum_i {\vec{p}}_i = 0$. In that case

$$W = E_{\mathrm{CM}}/c^2, \nonumber$$

and is thus, apart from the factor $1/c^2$, nothing but the energy in the CM frame. For a single particle $W=m_0$, the rest mass.

Most considerations about processes in high energy physics are greatly simplified by concentrating on the invariant mass. This removes the Lorentz-frame dependence of writing four momenta. I

As an example we look at the collision of a proton and an antiproton at rest, where we produce two quanta of electromagnetic radiation ($\gamma$’s), see Figure $\PageIndex{1}$, where the anti proton has three-momentum $(p,0,0)$, and the proton is at rest.

Figure $\PageIndex{1}$: A sketch of a collision between a proton with velocity $v$ and an antiproton at rest producing two $\gamma$ quanta.

The four-momenta are

$$\begin{aligned} p_{\mathrm{p}} &=& (p_{\mathrm{lab}},0,0,\sqrt{m_p^2 c^4+ p_{\mathrm{lab}}^2 c^2})\nonumber\\ p_{\bar{\mathrm{p}}} &=& (0,0,0,m_p c^2).\end{aligned} \nonumber$$

From this we find the invariant mass

$$W = \sqrt{2 m_p^2 + 2m_p\sqrt{m_p^2+ p_{\mathrm{lab}}^2 /c^2}} \nonumber$$

If the initial momentum is much larger than $m_p$, more accurately

$$p_{\mathrm{lab}} \gg m_p c, \nonumber$$

we find that

$$W \approx \sqrt{2 m_p p_{\mathrm{lab}}/c}, \nonumber$$

which energy needs to be shared between the two photons, in equal parts. We could also have chosen to work in the CM frame, where the calculations get a lot easier.