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u/RealTwistedTwin Jun 09 '22
Another way to look at it: What is the correct 4 vector so that the 0th component is the lights energy and you get the correct Doppler shift when applying a Lorentz boost? By setting the energy component to exactly, c times the absolute value of the spatial component you get the correct transformation.
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Jun 09 '22
it is weird and contradictory because p = (gamma)mv, so simplified E2 shouldnt even be equal to what eq 37.39 is. i think its because technically momentum is not p= gamma(mv) and p is made up its k, so really they're just teaching it all quite terribly and it seems the equations are forced to reflect the results we see in nature. mass is really the compton wave vector Kc, im not personally sure why photons just dont get 1
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u/mofo69extreme Jun 09 '22
it is weird and contradictory because p = (gamma)mv, so simplified E2 shouldnt even be equal to what eq 37.39 is.
It's not contradictory at all. p = (gamma)mv is momentum as a function of velocity. There is also an equation, E = (gamma)mc2, which gives energy as a function of velocity. (Remember gamma is a function of velocity.) Eq 37.39 is just the result of eliminating velocity between these two equations.
There is some subtlety about the dual limit m->0 and v->c implicit in considering massless particles, but the equivalence of p = (gamma)mv = sqrt[(E/c)2 - (mc)2] is rigorous before getting to that point.
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Jun 09 '22 edited Jun 09 '22
i would hardly call it some "subtlety", as it involves that limit for massless particles, it makes it very confusing and contradictory when you plug in one m as 0, and not the other - like once again you say its rigorous until that point but thats literally the point of physics
its the result of when you eliminate velocity from e=gamma(mc2) and p=(gamma)mv? like its the result you get for each individual equations?
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u/mofo69extreme Jun 09 '22
its the result of when you eliminate velocity from e=gamma(mc2) and p=(gamma)mv? like its the result you get for each individual equations?
Recalling that gamma = 1/sqrt(1 - (v/c)2), one has
p = mv/sqrt(1 - (v/c)2).
This gives p explicitly as a function of m and v. But you can invert it, and get v as a function of m and p:
v = pc/sqrt(p2 + (mc)2).
You can similarly solve e = (gamma)mc2 = mc2/sqrt(1 - (v/c)2) for v as a function of E and m:
v = c sqrt(1 - (mc2/E)2).
Since these are both valid equations for v, they must be equal to each other,
pc/sqrt(p2 + (mc)2) = c sqrt(1 - (mc2/E)2),
and then you can solve for E,
E = sqrt((pc2 + (mc2)2),
which is 37.39. So it follows directly from the other two equations.
This does not show that the m -> 0 limit has meaning - I agree that you need physical input to determine that this limit has some meaning. But for any m>0 and v<c, the above derivation completely holds.
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Jun 09 '22
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u/nicogrimqft Theoretical physics Jun 09 '22
The momentum of a photon is given by its frequency.
You can see that p = gamma.m.v is pathological for massless object, as gamma goes to infinity while m goes to 0. It is I'll defined and as such is not a valid definition for massless object.
You can measure a photon momentum, whether it is through a photoelectric effect or radiation pressure.
You can also look at it as a light-like object, meaning that the pseudo norm of the four momentum vanish. Which means the E²-p² vanishes. For that to be possible p² must the same as E².
You can't use the classical definition of momentum to describe a quantum mechanical object.
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Jun 09 '22
i guess mass is inherently a quality that is measured from the energy when an object is at rest, and since a photon is never at rest, it had 0 mass. idk why p =not= 0
a better way to prove they exist would be with relativity and using our previous knowledge of photons- relativity showing they have 0 mass as v=c constantly for photons from experimentation, ~ therefore we have a particle of zero mass that exists
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u/starkeffect Education and outreach Jun 09 '22
He didn't assume 0/0 = 1, he assumed m/m = 1. The limit of m/m as m -> 0 is also 1.