Section 52.14 Massless Particles
Massless particles refer to particle whose rest mass \(m_0 \) is zero.
The momentum of a massless particle will be zero unless it has speed equal to the speed of light.
When \(v=c\text{,}\) the denominator becomes zero and momentum becomes undefined. In that case, we use Eq. (52.13.11) to find momentum of a massless particle. Thus,
Therefore, momentum of a massless particle moving at speed \(v=c\text{,}\) such as light particle photon, is not zero. Rather, it depends on energy of the particle.
The energy of each photon depends on the frequency of the light. Each photon of light of frequency \(f\) has energy equal to \(hf\text{,}\) where \(h\) is Planck constant, whose value is equal to \(6.627\times 10^{-34}\) J.s.
Therefore, momentum a photon will be related to its frequency as
Writing the ratio of \(c\) to \(f\) as the wavelength \(\lambda\) of light,
we obtain an interesting result relating the momentum and wavelength of light.
Checkpoint 52.14.1. Decay of a Neutral Pion in Photons.
A neutral pion, \(\pi^{0}\text{,}\) decays into two light particles, called photons. The rest mass of neutral pion is \(135\:\textrm{MeV/c}^2\text{.}\) That is, if you multiply \(135\:\textrm{MeV/c}^2\) by \(c^2\) you will get the rest energy in MeV unit. Find the energy and momenta of the two photons released when a neutral pion at rest decays into two photons.
Use formulas in this section.
\(1.13\times10^{-11}\: \textrm{J}\text{,}\) \(3.77\times 10^{-20}\:\textrm{kg.m/s}\text{.}\)
Let \(E\) be the energy of one of the photons. Balancing the energy in the decay reaction we find
Therefore, the energy of each photon is
The momentum of each photon