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Section 52.14 Massless Particles

Massless particles refer to particle whose rest mass \(m_0 \) is zero.

\begin{equation} \textrm{Massless particles: } m_0 = 0.\tag{52.14.1} \end{equation}

The momentum of a massless particle will be zero unless it has speed equal to the speed of light.

\begin{equation} \vec p = \lim_{m_0\rightarrow 0} \dfrac{m_0 \vec v}{\sqrt{1-v^2/c^2}} = 0\ \ \ (v \ne c).\tag{52.14.2} \end{equation}

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,

\begin{equation} \lim_{m_0\rightarrow 0}\left[ E^2 = p^2 c^2 + m_0^2 c^4 \right]\ \Longrightarrow\ E^2 = p^2 c^2.\tag{52.14.3} \end{equation}

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.

\begin{equation} p = \frac{E}{c}. \label{eq-sq-rel-energy-3}\tag{52.14.4} \end{equation}

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.

\begin{equation} E = h f. \label{eq-photon-energy-1}\tag{52.14.5} \end{equation}

Therefore, momentum a photon will be related to its frequency as

\begin{equation} p = \dfrac{E}{c} = \dfrac{hf}{c}. \label{eq-photon-momentu-freq}\tag{52.14.6} \end{equation}

Writing the ratio of \(c\) to \(f\) as the wavelength \(\lambda\) of light,

\begin{equation} \lambda = \dfrac{c}{f},\tag{52.14.7} \end{equation}

we obtain an interesting result relating the momentum and wavelength of light.

\begin{equation} p\lambda = h.\tag{52.14.8} \end{equation}

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

\begin{equation*} 2E = 135\:\textrm{MeV}. \end{equation*}

Therefore, the energy of each photon is

\begin{align*} E \amp = 67.5 \:\textrm{MeV} = 67.5 \:\textrm{MeV} \times 10^6\times 1.67\times 10^{-19}\: \textrm{J/MeV}\\ \amp = 1.13\times10^{-11}\: \textrm{J}. \end{align*}

The momentum of each photon

\begin{equation*} p = \frac{E}{c} = \frac{1.13\times10^{-11}\: \textrm{J}}{3.0\times 10^8\:\textrm{m/s}} = 3.77\times 10^{-20}\:\textrm{kg.m/s}. \end{equation*}