4-momentum of photon

Question

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 word 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.

Comments

Voting to close as high-school level. PhysicsOverflow is a site for postgraduate-level physics.

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... for graduate+ level physics.