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What is the proof of Einsteins famous equation E = mc2 ?
stevendaryl said:There is a heuristic derivation of the relativistic expressions for energy and momentum (not a rigorous proof) that goes something like this[...]
bcrowell said:Since this is only for point particles, I think it's subject to the same criticisms that Ohanian makes of Einstein's 1905 derivation of E=mc2.
stevendaryl said:[*]Assume that the energy of a particle is proportional to the mass and is a function of the magnitude of the velocity: [itex]E = g(v) m[/itex] (By "mass", I mean rest mass).
DrStupid said:Is there a justification for this assumption? Why not [itex]E = E_0 + g(v) m[/itex]?
stevendaryl said:Right off the top of my head, I don't remember an argument for why rest energy should be proportional to mass.
DrStupid said:Is there a justification for this assumption? Why not [itex]E = E_0 + g(v) m[/itex]?
stevendaryl said:I don't remember an argument for why rest energy should be proportional to mass.
This is a bit of an overstatement. You're using that ##F=\frac{dp}{dt}## and that ##p=\gamma m v##. These are non-trivial assumptions. They can be considered definitions of the left-hand sides, but you can also take ##E^2=p^2c^2+m^2c^4## to be the definition of E, and then you don't have to do any calculations at all. This definition needs no justification other than the fact that the theory makes very accurate predictions (as already mentioned by Russ).DrStupid said:...without any assumptions apart from the basics of classical mechanics and Lorentz transformation (and without energy at all):
PeterDonis said:##E_0## is just ##m##
Fredrik said:You're using that ##F=\frac{dp}{dt}## and that ##p=\gamma m v##. These are non-trivial assumptions.
Right, but so is ##F=m\frac{d^2x}{dt^2}##, and this formula doesn't lead to the desired result. If you want the calculation to be considered a proof, or just a fairly solid argument, you have to explain why you're using that specific statement of Newton's 2nd. If it's just because it leads to the result we want, you might as well just write down the result we want and call it a definition.DrStupid said:##F=\frac{dp}{dt}## is Newton's second law
Can you justify the last equality in the first line without this?DrStupid said:and ##p=\gamma m v## the result of the derivation.
Fredrik said:Right, but so is ##F=m\frac{d^2x}{dt^2}##
Fredrik said:you have to explain why you're using that specific statement of Newton's 2nd
Fredrik said:If it's just because it leads to the result we want, you might as well just write down the result we want and call it a definition.
Is the stuff you feed into the calculation (of the work required to accelerate a massive point particle in SR) motivated by analogy with non-relativistic classical mechanics, or defined in the framework of SR? If it's the former, you certainly do have to explain why you favor dp/dt over ma. If it's the latter, then you have to explain why you consider your definitions (of momentum, work, etc) more fundamental than the end result, which can also be taken as a definition.DrStupid said:No. That results from the second law in classical mechanics but it is not identical with the second law. In SR you get another result.
The use of the original statement of Newton's second law does not need to be explained. It is the special case you mentioned above that would need a justification.
Matterwave said:He states that if you double the quantity of matter (his word for mass) you double the "motion", and if you double the velocity you also double the motion. THIS explicitly assumes ##p \propto mv## and NOT ##p=\gamma mv## (obviously he would have no idea about the second case as that arose 300 years after him).
Fredrik said:Is the stuff you feed into the calculation (of the work required to accelerate a massive point particle in SR) motivated by analogy with non-relativistic classical mechanics, or defined in the framework of SR?
DrStupid said:I can't follow your argumentation. Could you please explain why his statement excludes ##p=\gamma mv##?
Matterwave said:If you double the velocity you double the motion.
DrStupid said:I read that as "If you double the velocity [and keep quantity of matter constant] you double the motion."
Without this assumption it won't even be correct in classical mechanics. In SR you will need an open system to meet this condition.
Matterwave said:He says explicitly that if you double the mass you double the motion. If you double the velocity you double the motion. If you double both mass and velocity you quadruple the motion. This is not compatible with ##p=\gamma mv## because ##\gamma## depends on ##v## and his statements will no longer be true. If you double v, you certainly more than double ##\gamma mv## (assuming v<.5c).
Matterwave said:? Yes, keep m constant, ##\gamma m v## still more than doubles when you double v
DrStupid said:Remember: Motion is defined the product of quantity of matter and velocity. That means in SR quantity of matter is not m but ##\gamma m##.
DrStupid said:Remember: Motion is defined the product of quantity of matter and velocity. That means in SR quantity of matter is not m but ##\gamma m##.
stevendaryl said:What I was looking for with my derivation, and Dr. Stupid presumably was looking for in his derivation, was an alternate theory of energy, momentum and force that reduces to the Newtonian case in the low-velocity limit and which is invariant under Lorentz transformations.
DrStupid said:My derivation also works for relativistic velocities.
Matterwave said:Saying things such as "quantity of matter" is ##\gamma m## is certainly NOT something Newton would agree with in his time.
Matterwave said:My point is that, as formulated by Newton, the second law should read, given modern notation, ##F=\frac{dp}{dt}## but it MUST be appended with the explicit assumption that ##p=mv##.
Matterwave said:But given Newton's definition, there is no way to define quantity of matter as ##\gamma m##, both the density and volume should be measured in the rest frame.
DrStupid said:Why should we change the original definitions just to make them invalid? That doesn't make any sense.
I can't think of anywhere in Newton's writing where he analyzes the motion of an object whose mass changes with time (for example a rocket, assuming classical mechanics). So I think you are putting ideas into Newton's head with the benefit of hindsight, with no firm evidence that he ever thought about the momentum of a system with variable mass, let alone a system whose mass might be function of its velocity.Matterwave said:My point is that, as formulated by Newton, the second law should read, given modern notation, ##F=\frac{dp}{dt}## but it MUST be appended with the explicit assumption that ##p=mv##.
AlephZero said:I can't think of anywhere in Newton's writing where he analyzes the motion of an object whose mass changes with time (for example a rocket, assuming classical mechanics). So I think you are putting ideas into Newton's head with the benefit of hindsight, with no firm evidence that he ever thought about the momentum of a system with variable mass, let alone a system whose mass might be function of its velocity.