Can we derive the momentum of photon from special relativity?

Question

I don't really have strong backgrounds studying quantum physics, but I did learn special and general relativity, and I have now a question how to get the momentum of photon.

For my understanding, Einstein "predicted" energy and momentum of photon through his research on photoelectric effects (1905) and Compton scattering (1916) by assuming photon as a "quantum" which contains the wave-particle duality. Those Einstein's predictions were actually "verified" by experiments.

However, some people are explaining the momentum of photon can be "derived" analytically from the special relativity by using $E^2 = (mc^2)^2 + (pc)^2$ with $m=0$.

But I can't derive the same conclusion because that famous equation comes from the "invariants" in space-time which is "interval (or displacement)" in Minkowski 4D space. In special relativity, if the "particle" is moving with the speed of light, the invariant (interval) must be zero. Considering four-velocity (derivative of "proper length" with respect to "proper time"), the invariant becomes the square of speed of light, and then the four-momentum is still defined by four-velocity multiplied by invariant mass $m_{inv}$. This should eventually give $0 = 0$ because $m_{inv} = m_{rest}$ in special realativity, not tell us $E^2 = (pc)^2$ when $m_{inv} = 0 $ .

I could not find any resources telling why/how we can use special relativity for "quantum" or "massless particle."

Can we really derive the momentum of photon from the special relativity? Do we have to use quantum physics, not special relativity to get the momentum of "massless quantum (or particle)?"

Answer 1

The four-momentum of any particle must be tangent to its world line, including photons. Since $ds^2=0$ for a photon, its four-momentum must also satisfy $P\cdot P=0$ to ensure that it is tangent along the photon's world line.

We must also note that the four-velocity of a photon cannot be defined, because the proper time experienced along a photon's world line is always zero. Thus, our definition for massive particles $P=mU$ will not work for light. However, the most basic definition of four-momentum is that it's a vector whose components contain the energy and momentum of the particle, like so:

$$P=\begin{bmatrix}E/c\\p_x\\p_y\\p_z\end{bmatrix}$$

For a photon moving in the $x-$direction, $p_y=p_z=0$. The inner product is:

$$P\cdot P=0$$ $$E^2-c^2p_x^2=0$$ $$p_x=E/c$$

Note that we have not imposed $m=0$ yet. In fact, we deduce that photons have zero mass because their world lines and four-momenta are null.

$$P\cdot P=m^2c^2=0$$

Answer 2

The answer to the question in the title is: yes, we can. Special relativity states the energy momentum relation as $E^2=m^2+p^2$ in units where c=1. Setting m=0 gives $E=p$. The four momentum of a photon is $\hbar(\omega,\vec k)$.

In fact, $E=p$ can be derived from Maxwell's equations. It does not even require photons.

Answer 3

When the particle is going at the speed of light, its line interval $ds^2 = 0$, but that does not mean that any relativistic invariant is zero.

Considering, as you did, the four-velocity $\underline{U}^2 = c^2$. The four-momentum is defined by $\underline{P} = m\underline{U} = 0$ if $m=0$, so $\underline{P}^2 = 0$. Since $P^\mu = \{E/c, \vec{p}\}$, you obtain $E^2 = c^2p^2$.