Derivation of E=mc2

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I have looked at the Einsteins Derivation of the Relation E=mc2.
http://www.adamauton.com/warp/emc2.html

But I don't if this shorthand derivation is O.k. or not.
like the derivation, I use maxwells law and say the momentum p is given by
p = E / c ---> 1

Now from newtonian mechanics, we have
p= m* v
or, p=mc (here v=c) --->2

Combining 1 and 2
mc = E / c
therfore, E = mc2

What's wrong here?
If its wrong, why is the result correct?
If its correct, why didn't Einstein use it?
 
938
9
What's wrong here?
If its wrong, why is the result correct?
If its correct, why didn't Einstein use it?

Well, for one, photon mass is zero!

Einstein's point was to figure out how much a photon would shift the center of mass of the box and then ask how big a massive object would have to be to have a similar effect.
 

Fredrik

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Science Advisor
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But I don't if this shorthand derivation is O.k. or not.
like the derivation, I use maxwells law and say the momentum p is given by
p = E / c ---> 1
Here you're using the formula [tex]E^2=\vec p^2c^2+m^2c^4[/tex] with m=0. Nothing wrong with that.

Now from newtonian mechanics, we have
p= m* v
or, p=mc (here v=c) --->2
As Clamtrox mentioned, m=0, so this formula doesn't work. The correct formula for a massive particle (which has [itex]|\vec v|<c[/itex]) is [itex]\vec p=\gamma m \vec v[/itex]. If you square both sides and do a little algebra, you recover the formula I posted above, which actually holds for massless particles too. So instead of p=mc, you have [tex]\vec p^2=E^2/c^2-m^2c^2=E^2/c^2[/tex].

Combining 1 and 2
mc = E / c
therfore, E = mc2
If you combine 1 and the correct version of 2 (which is actually the same formula as 1), all you get is

[tex]E^2=\frac{E^2}{c^2}c^2[/tex]

which tells you exactly nothing.

See this post for a better derivation.
 
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I question if this can really be rigorously derived. In relativity, mass and rest energy are exactly the same thing, the reason why there is a c^2 is simply because we choose to use inconsistent units for the same physical quantity.

The nontrivial statement is really that (E, p) transforms under Lorentz transformations as a four-vector.
 
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Actually, the fact that dx^{mu}/dtau transforms as a four-vector is trivial. So, the non-trivial statement is that four-momentum is conserved.
 
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Four-momentum is conserved because the laws of physics are invariant under arbitrary space-time translations. So, the whole matter is trivial.

Then, given the fact that the heuristic derivation (that use intuitive concepts from classical mechanics in some ways) looks non-trivial, suggests that the derivation of classical mechanics from relativity is less trivial than is presented in most textbooks.

In classical mechanics, mass is an independent physical quantity from energy, while in relativity mass and (rest) energy are the same thing. A derivation of classical equations like
E_kin = 1/2 m v^2 is then not just a simple matter of doing a Taylor expansion around v = 0.

The non-relativistic limit involves rescaling physical variables in a certain way and then studying the infinite scaling limit. In this limit, the (trivial) relation between mass and energy breaks down (it becomes singular). So, you then have a new independent physical quantity.
 

Fredrik

Staff Emeritus
Science Advisor
Gold Member
10,730
406
I question if this can really be rigorously derived. In relativity, mass and rest energy are exactly the same thing, the reason why there is a c^2 is simply because we choose to use inconsistent units for the same physical quantity.

The nontrivial statement is really that (E, p) transforms under Lorentz transformations as a four-vector.
I agree. It isn't possible to derive [itex]E=mc^2[/itex] without making some assumptions. Some of the details about this didn't become 100% clear to me until I wrote about it the Science Advisor forum (hidden for normal users). I quoted myself from that thread here. (Start reading at "How are you going to...").
 
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