# 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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