# 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.

• possible duplicate of If photons have no mass, how can they have momentum? Commented May 26, 2015 at 2:46
• This is not a duplicate, my question has to do with null vectors and the definitions of 4-velocity and 4-momentum. Commented May 26, 2015 at 2:53

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).

• Sorry for the late and possibly stupid question. I'm also reading a book to better understand GR, here the zero mass is also proven by the statement pp = -m^2. But the latter (in the book) is derived by using U so I don't understand why this must hold in general. Can you explain why pp = -m^2 for light-like vectors? Commented Apr 26 at 19:58

let speed of light c=1(by choosing the right units

We do not neccessarily need the proper time to define the four velocity and hence, the four momentum. Any scalar λ which is independent of the frame you choose is a appropriate choice to replace the proper time τ. The reason is that,four vector is nothing but a tangent vector of the particle's world line.And we can have a plenty of parameter choices for a curve(world line our case).

One can esily check out that the space component of the four velocity equals the usual veloctiy times its time component.(by chain rule)

As for a photon, we choose the parameter λ so that the tangent vector of its world line is just its four momentum.This can be done simply because four momentum parallels to four velocity.Since the square of a photon's usual velocity always equals to c=1,we get that the Lorentz metric of its momentum is zero.