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.