# Why is the magnitude of a photons 4 momentum vector 0 if it has momentum?

I've just started studying 4 vectors. I understand because a photon has 0 rest mass and travels at c, you can not define a 4 velocity for it, and as momentum 4 vector = mass x velocity 4 vector, it should be 0 for a photon. But we know a photon as momentum equal to $$\frac{\hbar w}{c}$$. So how come we still say that the magnitude of the 4 momentum vector is 0 and not this?

• The $4$ momentum is not the $4$ velocity times mass. Energy of a photon carries momentum - which in $3$ space, is $\frac{\hbar\omega}{c}$ - but you did not determine the momentum in $3$ space by taking the $3$ velocity times mass either. It turns out the magnitude of the momentum in $3$ space is the same as in $4$ space. May 4, 2019 at 3:49

A massless particle does have 4-momentum, it just has the peculiar property of being a null vector due to the Minkowski metric signature. The momentum 4-vector is:

$$\vec{P} = (E,p_x,p_y,p_z)$$

and the length (using Minkowski metric) of this 4-vector is:

$$E^2-p_x^2-p_y^2-p_z^2 = E^2-p^2 = m^2$$

which is zero for a massless particle ($$m=0$$). But $$E$$ is still non-zero, and $$(p_x,p_y,p_z)$$ is non-zero and has a spatial direction.

There is no contradiction between the two statements. As you said yourself in the question $$\frac{\hbar \omega}{c}$$ is the momentum, not the four-momentum. The question is: momentum of what? It's the momentum of the associated wave in the context of the wave-particle duality, that is $$p=\hbar.k$$. This does not contradict the fact that photon has a zero rest mass, because you can use Einstein's equation and write $$E^2=p^2c^2+m_0^2.c^4$$, which, when you replace with $$m_0=0$$ and $$p=\hbar.k$$, gives $$E=\hbar\omega$$ as expected, that is, a non-zero energy for a zero-rest mass particle.

$$\hbar \omega/c$$ is the (magnitude of the) $$3$$-momentum vector, not the $$4$$-momentum. In particular, the $$4$$-momentum $$p^\mu=(E/c, \vec p)$$ satisfies $$p^\mu p_\mu=-m^2c^2$$ so if $$m=0$$ for photons one has the length-squared of the $$4$$-momentum as $$0$$ even if the $$4$$-momentum itself is not $$0$$.

• You don't even have to square the $4$ momentum to see that it's momentum is not zero. Setting $\vec p$ to zero gives you $p^\mu=(E/c,0)$ which is non zero $4$ momentum. Taking the square and then the square root won't make it zero. May 4, 2019 at 3:57