The words "What is the Momentum of a Photon" lie on the top of a gradient that starts as orange on the left growing darker as you move right, turning into dark blue and then lightening to bright blue on the far right hand side.

A young student I was tutoring some time ago asked me how a photon could have momentum when it doesn’t have any mass. That’s a good question!

Most beginning physics students are familiar with the momentum formula that looks like this:

\\ \: \\ \Large \mathrm{\Delta \vec{p} = \Delta m \vec{v}} \\ \: \\

 But this doesn’t work with photons because for a photon, the rest mass m = 0. So what do we do? We start with energy and a famous equation given to us by Einstein.

E2 = p2c2 + m2c4

(This looks a little different from the shortened version in pop culture, doesn’t it?)

 When we substitute in m=0, this shortens to:

E2 = p2c2 + 02c4

E2 = p2c2

so we can easily take the square root of both sides.

\\ \: \\ \large \mathrm{\sqrt{(E)^2} = \sqrt{{p^2}{c^2}}}  \\ \: \\ \mathrm{E = pc} \\ \: \\ Solving this for p gives us: \\ \: \\ \large \mathrm{p = \frac{E}{c}} \\ \: \\
So the momentum of a photon depends only on the energy of the photon and a constant, the speed of light in a vacuum.

Before moving on, let’s define all of those variables. (Maybe we should’ve started with that. Oh, well. 🙂 )

  • E – energy
  • p  – momentum
  • c is the speed of light in vacuum, 3 × 108 [m/s]
  • m is the rest mass. For a photon, m = 0 because photons are massless particles.
  • f (or ν in some books) – frequency [Hz] or [cycles per second] – this is equal to c/λ  because c = λf
  • λ – wavelength [m] – this is equal to c/f  because c = λf
  • h – Planck’s constant = 6.626070 × 10-34 [J·s]

So now, with these definitions, we can now find the momentum of a photon in terms of the photon’s frequency or wavelength.

Remember that light acts both as a particle (photon) and as a wave. The energy of an electromagnetic wave is given by

\Large \mathrm{E = h}f \\ \: \\ and \\ \: \\ \large f \mathrm{= \frac{c}{λ} }  \\ \: \\ so we can substitute the wavelength into the Energy equation \\ \: \\ \large \text {E = h}f \mathrm{ = \frac{hc}{λ}} \text{ = pc }  

Looking at the first part, we have

\Large \mathrm{h}f = \mathrm{pc} \\ \: \\ solving this for momentum, p, gives us \\ \: \\ \large \text {p = } \frac{\mathrm{h}f}{c}

And our last variation involves the wavelength. Looking at the next part, we have

\Large \mathrm{\frac{hc}{λ} = pc} \\ \: \\ solving this for momentum, p, gives us \\ \: \\ \large \mathrm{p = \frac{hc}{λc} = \frac{h}{λ}}

In summary, here is the momentum of a photon in terms of its energy, its frequency, and its wavelength:

\large \mathrm{p = \frac{E}{c}} \text{ or p = } \frac{\mathrm{h}f}{c} \text { or } \mathrm{p = \frac{h}{λ}}