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4-momentum of photon

+ 0 like - 1 dislike
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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.

Closed as per community consensus as the post is high-school level; thus not up to this site's level
asked May 26, 2015 in Closed Questions by Levitt (-5 points) [ revision history ]
recategorized May 26, 2015 by dimension10

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

... for graduate+ level physics.





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