**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 as

where .

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

For any massless particle, , so , making a null vector.

As an example, suppose we have a positive pion (rest mass 140 MeV) at rest that decays into an antimuon (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 axis):

For the muon we have

where .

From conservation of momentum we get from the energy component

From the second component we get

Subtracting these two equations we get

This quadratic has only one acceptable solution (there is also which is spurious since it makes infinite), which is . This gives , from which we get the energies:

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