29.4 Photon momentum  (Page 3/6)

Converting this to eV by multiplying by $(\text{1 eV})/(1\text{.}\text{602}\times {\text{10}}^{\text{–19}}\mathrm{J})$ yields

The photon energy $E$ is

which is about five orders of magnitude greater.

Discussion

Photon momentum is indeed small. Even if we have huge numbers of them, the total momentum they carry is small. An electron with the same momentum has a 1460 m/s velocity, which is clearly nonrelativistic. A more massive particle with the same momentum would have an even smaller velocity. This is borne out by the fact that it takes far less energy to give an electron the same momentum as a photon. But on a quantum-mechanical scale, especially for high-energy photons interacting with small masses, photon momentum is significant. Even on a large scale, photon momentum can have an effect if there are enough of them and if there is nothing to prevent the slow recoil of matter. Comet tails are one example, but there are also proposals to build space sails that use huge low-mass mirrors (made of aluminized Mylar) to reflect sunlight. In the vacuum of space, the mirrors would gradually recoil and could actually take spacecraft from place to place in the solar system. (See [link] .)

Relativistic photon momentum

There is a relationship between photon momentum $p$ and photon energy $E$ that is consistent with the relation given previously for the relativistic total energy of a particle as $E^2=(\text{pc}{)}^2+(\text{mc}{)}^2$ . We know $m$ is zero for a photon, but $p$ is not, so that $E^2=(\text{pc}{)}^2+(\text{mc}{)}^2$ becomes

or

To check the validity of this relation, note that $E=\text{hc}/\lambda$ for a photon. Substituting this into $p=E\mathrm{/}c$ yields

as determined experimentally and discussed above. Thus, $p=E\mathrm{/}c$ is equivalent to Compton’s result $p=h/\lambda$ . For a further verification of the relationship between photon energy and momentum, see [link] .