## Section52.14Massless Particles

Massless particles refer to particle whose rest mass $m_0$ is zero.

$$\textrm{Massless particles: } m_0 = 0.\tag{52.14.1}$$

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

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

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,

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

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.

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

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.

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

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

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

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

$$\lambda = \dfrac{c}{f},\tag{52.14.7}$$

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

$$p\lambda = h.\tag{52.14.8}$$

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.

Hint

Use formulas in this section.

$1.13\times10^{-11}\: \textrm{J}\text{,}$ $3.77\times 10^{-20}\:\textrm{kg.m/s}\text{.}$

Solution

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*}