special relativity - 4-momentum of photon

The 4-momentum is defined as $p=mU$ where m is the rest mass of the particle and $U$ is the 4-velocity. Now I am confused as to how this applies to a photon for which one can't define $U$ since there can be no rest frame for a photon. I'm trying to see why $p$ is still tangential to it's world line in any frame. I want to arrive at the conclusion that $p$ is a null vector. So I am not looking for an explanation which uses that equation $E^2 = (m c^2)^2 + p^2 c^2$in first place(or that photons have $zero$ rest mass). I want see how it follows from that fact that photon travels at speed $c$. Just like how we use this fact to conclude that 4-position vector is a null vector. By null vector I mean whose magnitude vanishes under Lorentz metric. This is not homework. Any help is appreciated. Thanks.

I have already gone through If photons have no mass, how can they have momentum? and it doesn't answer my question.

Answer

Because the photon definitely has energy, it must have a four-momentum vector, but it must be defined differently from mU because the proper time, $\tau$, along its worldline is zero.

$$d\tau= dt\sqrt{1-v^2/c^2}$$ The photon four-momentum vector is defined to be $$\textbf{p}=\left[\matrix{p^t\\p^x\\p^y\\p^z}\right]=\left[\matrix{E/c\\Ev_x/c^2\\Ev_y/c^2\\Ev_z/c^2}\right],$$ with $$v_x^2+v_y^2+v_z^2=c^2$$

The four-vector scalar product of this will be zero, implying that the mass of the photon is zero because $\textbf{p}\cdot\textbf{p}=m^2$ = 0 (within a sign factor depending on your sign convention for the scalar product).