Energy, Mass, and Momentum of Photon

A photon is an elementary particle, the quantum of light. It has no rest mass and has no electric charge. The modern photon concept was developed gradually by Albert Einstein to explain experimental observations of the photoelectric effect, which did not fit the classical wave model of light. In particular, the photon model accounted for the frequency dependence of light's energy. Max Planck explained black body radiation using semiclassical models, in which light is still described by Maxwell's equations, but the material objects that emit and absorb light, do so in amounts of energy that are quantized.

Photons are emitted in many natural processes. They are emitted from light sources such as floor lamps or lasers . For example, when a charge is accelerated it emits photons, a phenomenon known as synchrotron radiation. During a molecular, atomic or nuclear transition to a lower or higher energy level, photons of various energy will be emitted or absorbed respectively. A photon can also be emitted when a particle and its corresponding antiparticle are annihilated. During all these processes, photons will carry energy and momentum.

laser

Photons emitted in a coherent beam from a laser.

Energy of photon: From the studies of photoelectric effects, energy of a photon is directly proportional to its frequency with the Planck constant being the proportionality factor. Therefore, we already know that $E = h \nu$ (Eq. 1), where E is the energy and $\nu$ is the frequency.

Momentum of photon: According to the theory of Special Relativity, energy and momentum (p) of a particle with rest mass m has the following relationship: $E^2 = (mc^2)^2+p^2c^2$, where c is the speed of light. In the case of a photon with zero rest mass, we get $E = pc$. Combining this with Eq. 1, we get $p = \frac{h\nu}{c}=\frac{h}{\lambda}$. Here, $\lambda$ is the wavelength of the light. Since momentum is a vector quantity and p points in the direction of the photon's propagation, we can write $\bf{p} = \hbar \bf{k}$ , where $\hbar = \frac{h}{2\pi}$ and is $k$ a wave vector.

You may wonder how an object with zero rest mass can have nonzero momentum. This confusion often arises because of the commonly used form of momentum ($mv$ in nonrelativistic mechanics and $\gamma mv$ in relativisitc mechanics, where v is velocity and $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$. ) This formula, obviously, shouldn't be used in the case v=c.