Photons

The general energy-momentum relation, (3.179), implies that a particle with zero rest mass has the simplified energy-momentum relation

$$E = p\,c.$$ (3.196)

Consider a photon. The photoelectric effect demonstrates that the energy of a photon is related to its angular frequency, $\omega$, according to

$$E =\hbar\,\omega,$$ (3.197)

where $\hbar$ is Planck's constant divided by $2\pi$. (See Section 4.1.2.) However, we also know that a photon travels at the speed of light in all inertial reference frames. Thus, the relativistic energy, (3.177), of a photon can only be finite if the photon is a massless particle. In other words, a photon's rest mass must be zero. Hence, the previous two equations suggest that the momentum of a photon has the magnitude

$$p=\frac{\hbar\,\omega}{c}.$$ (3.198)

But, the dispersion relation of electromagnetic radiation in a vacuum, and, hence, of a photon moving through a vacuum, is

$$\omega=k\,c.$$ (3.199)

Here, ${\bf k}$ is the wavevector of the radiation, and, hence, of the photon. Note that the direction of ${\bf k}$ corresponds to the direction of motion of the photon. It is, thus, plausible that the momentum of our photon is written

$${\bf p} = \hbar\,{\bf k}.$$ (3.200)

Consider the two inertial reference frames, $S$ and $S'$, discussed in the Section 3.3.6. Let $\omega$ and ${\bf k}$ be the angular frequency and wavevector, respectively, of our photon in $S$. Likewise, let $\omega'$ and ${\bf k}'$ be the angular frequency and wavevector, respectively, of our photon in $S'$. Equations (3.180)–(3.183), (3.197), and (3.200) suggest that

$$k_x' = \gamma\left(k_x-\frac{v\,\omega}{c^2}\right),$$ (3.201)

$$k_y' = k_y,$$ (3.202)

$$k_z' =k_z,$$ (3.203)

$$\omega' =\gamma\,(\omega-v\,k_x).$$ (3.204)