User talk:Matteo Pinzauti

Zero mass ⟷ speed of light
Hi Matteo, I've noticed your edit in Photon. I was wondering if you knew of any good sources that discuss this causality? In a quick search I only found an article at wtamu.edu, but that might not be a good-enough source. Mind helping by adding a good reference and expanding the article where appropriate? Thanks, Ponor (talk) 12:45, 22 October 2020 (UTC)


 * Hi, I gave a read to this article wtamu.edu but it's actually wrong: photon doesn't have 0 impulse (the wrong point is that they are assuming that there is a reference frame in wich light is stationary, but moves at c (the speed of light) in every frame due to the second special relativity postulate). In fact the impulse of a photon is E/c (energy over speed of light) in the direction it is moving. So the quadrimpulse of the photon is: (E/c in time direction, E/c in the direction of propagation).


 * since the lenght (p^2) of the quadrivector (with the minkowski metric (+---)) is (m*c)^2 (mass times the speed of light) for every quadrimpulse, you get


 * p^2 = m*c = (E/c)^2 - (E/c)^2 = 0 => (m*c)^2 = 0 => m=0 for a photon.


 * This said, I do not yet know how to cite things on wikipedia, but you can refer to any special relativity article that talks about the quadrimpulse of light. — Preceding unsigned comment added by 151.38.183.135 (talk) 10:56, 23 October 2020 (UTC)


 * The article does say p=E/c if m=0. I believe the four-vector-squared invariance is how we got the p=E/c formula (am I right? it can't be a classical result because there are no photons in classical electrodynamics). I was more interested in this part: if mass is zero, speed has to be c. The article is saying: if there were a zero-mass particle (imagine it's not a photon) moving at v<c, we could travel alongside that particle, and in that reference frame its p would be 0, thus total energy 0 too. And they are asking what kind of particle would a p=0, E=0 particle be. On the other hand, if that zero-mass particle is moving at c, it would move at c in every other frame (according to Lorentz and co.) and we would not be able to say that somewhere p equals 0. In short, if m=0, v v=c (gamma = (1-(v/c)^2)^(-1/2))

on the other hand if v=c (so gamma=infinity), the only way to get finite energy is m=0 (you need E = 0 * infinity indeterminate form)

so m=0 <=> v=c — Preceding unsigned comment added by 151.68.89.98 (talk) 13:31, 23 October 2020 (UTC)


 * All clear. It'd just be nice to have a reliable source (a book or peer-reviewed article) that says this (m=0 -> v=c). Wikipedia can't rely on our original research, and we could not put it in the lead section anyway. Ponor (talk) 13:50, 23 October 2020 (UTC)