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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?

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  • $\begingroup$ 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. $\endgroup$ May 4, 2019 at 3:49

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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.

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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.

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$\hbar \omega/c$ 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 $0$.

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  • $\begingroup$ 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. $\endgroup$ May 4, 2019 at 3:57

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