How can photon have energy, but no mass if “m=E/c^2” (E= mc^2)
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Because E=mc² is not the complete thing. We get these results form using special relativity (and the Lagrangian formalism) to calculate the motion of a charged particle.
I will show a few interesting results and add some context if you are interested. So the stuff in brackets are not that important just some extra.
The “full” result is: E=mc² / (1-v²/c²)^0.5
It comes from this equation:
d/dt mc²/ (1-v²/c²)^0.5 = qEv
E here is the electric field and q is charge. Now qEv is the power of the electric field. And power is the time derivitive of energy. (Change in energy over time is power by definition.) So that thing after d/dt is energy.
This 1/sqrt(1-v²/c²) shows up all the time it’s the gamma factor. I’m going to use y to label it.
So if we plug in 0 for v we get E=mc². As y = 1 here.
We can define the momentum vector not as p=mv but, p=ymv. (This again would require more context so: When we derive the stuff we get something that looks almost like F=ma for the electromagnetic case where F=q(E+v×B) The Coulomb and Lorenz forces. And the other side is the time derivitive of this new modified momentum. So F is still dp/dt it’s just that p also gets gamma factored.)
In special relativity we like the four momentum. (vectors with space and time like components, 1 time and 3 space of course.) p is the spacial momentum and we also need a time like thing into it. (ymc, p) = (E/c, p). As E=ymc².
#Energy is c× the first component of four momentum.
Lets calculate its lenght squared (We call the scalar product of a vector with itself its lenght squared. Here we have two kinds of vectors, and upper and lower index variant and the rule is that you multiply the upper and the lower together. The difference boils down to the lower index variant having -p not +p. So thats where the minus sign comes from.) I’m gonna use p_4 for the four momentum but this is the only time I’m using it:
p_4 p_4 = (E/c)² – p² = (m²c²-m²v²)/(1-v²/c²) = m²c²
#Mass is the lenght of four momentum. (per c)
Another rather useful way of writing things is:
E²=(mc²)²+p²c² (The square is important cause momentum alone is a vector and we can only talk about its length in terms of energy.)
If m=0, E=|p|c. (This is quite useful when you calculate stuff like Compton scattering, where you can treat light as little balls and calculate their momentum using this equation. And for light E=hf is also true and momentum is what we really care about. Or you can also use it to calculate radiation pressure. )
And as you can see E/c = |p|. So (E/c)²=p². So for light if we plug this into the lenght calculation we did, which defines mass, (thats the big deal) p_4 p_4 = (E/c)²-p² = p² – p² = 0. Mass squared is that per c² so m²=0 for light.
This might be a bit complicated but at least now we understand why saying stuff like mass depends on velocity is conceptually wrong and this is how formally you can introduce momentum for light and now we defined what mass is.
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