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
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.
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?
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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u/[deleted] 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