Four-momentum of photons

Required math: calculus

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 3; Problem 3.4.

The four-momentum is defined for a particle with a rest mass {m} as

\displaystyle  \mathbf{p}=\gamma m\left[1,v_{x},v_{y},v_{z}\right] \ \ \ \ \ (1)

where {\gamma=1/\sqrt{1-v^{2}}}.

For particles such as photons that have no rest mass, this formula obviously doesn’t work. However, the first component of {\mathbf{p}} is taken as the particle’s energy, so if we make that identification for photons, we get

\displaystyle  \mathbf{p}=E\left[1,v_{x},v_{y},v_{z}\right] \ \ \ \ \ (2)

For any massless particle, {v=\sqrt{v_{x}^{2}+v_{y}^{2}+v_{z}^{2}}=1}, so {\mathbf{p}\cdot\mathbf{p}=E^{2}\left(v_{x}^{2}+v_{y}^{2}+v_{z}^{2}-1\right)=0}, making {\mathbf{p}} a null vector.

As an example, suppose we have a positive pion {\pi^{+}} (rest mass 140 MeV) at rest that decays into an antimuon {\mu^{+}} (rest mass 106 MeV) and a neutrino. Avoiding any controversy over whether or not the neutrino has mass, we’ll just assume it’s massless so its momentum is (assuming it travels along the {+x} axis):

\displaystyle  \mathbf{p}_{\nu}=E_{v}\left[1,1,0,0\right] \ \ \ \ \ (3)

For the muon we have

\displaystyle  \mathbf{p}_{\mu}=E_{\mu}\left[1,v,0,0\right] \ \ \ \ \ (4)

where {E_{\mu}=106\gamma\;\mbox{MeV}}.

From conservation of momentum we get from the energy component

\displaystyle   E_{\nu}+E_{\mu} \displaystyle  = \displaystyle  E_{\pi}\ \ \ \ \ (5)
\displaystyle  E_{\nu}+106\gamma \displaystyle  = \displaystyle  140 \ \ \ \ \ (6)

From the second component we get

\displaystyle   E_{\nu}+vE_{\mu} \displaystyle  = \displaystyle  0\ \ \ \ \ (7)
\displaystyle  E_{\nu}+106v\gamma \displaystyle  = \displaystyle  0 \ \ \ \ \ (8)

Subtracting these two equations we get

\displaystyle   106\gamma\left(1-v\right) \displaystyle  = \displaystyle  140\ \ \ \ \ (9)
\displaystyle  \frac{1-v}{\sqrt{1-v^{2}}} \displaystyle  = \displaystyle  1.32\ \ \ \ \ (10)
\displaystyle  1-2v+v^{2} \displaystyle  = \displaystyle  1.744-1.744v^{2} \ \ \ \ \ (11)

This quadratic has only one acceptable solution (there is also {v=1} which is spurious since it makes {\gamma} infinite), which is {v=-0.271}. This gives {\gamma=1.0389}, from which we get the energies:

\displaystyle   E_{\nu} \displaystyle  = \displaystyle  140-106\gamma\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  29.88\;\mbox{MeV}\ \ \ \ \ (13)
\displaystyle  E_{\mu} \displaystyle  = \displaystyle  110.12\;\mbox{MeV} \ \ \ \ \ (14)

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