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I try very hard to find a satisfying "derivation" of the relation $E=mc^2$, but it turns out that everything is circular. For example, under this answer, my2cts points out that if you assume $p=m\gamma v$ then classically we have $E=p/v$, which inevitably leads to $E=\gamma m$. (Here $c=1$.) Of course $E=mc^2$ can be backed up by a physical argument - see Einstein's original paper - but if we want to derive it mathematically, it turns out we need to assume something like $p=m\gamma v$, and I don't know why we should have such definition. (It is the spacial part of 4 momenta, of course, but why should it be 3-momenta as well.) Here are my questions:

  1. Why my2cts say that classically we have $E=p/v$? I do not completely understand it.
  2. Is $E=mc^2$ just a sensible definition made according to de Broglie's interpretation of momentum of electron $p=\hbar k$, and cannot essentially be derived?
Qmechanic
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Ma Joad
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  • As usual, the answer to this depends on what you're willing to accept as definitions for 'energy', 'momentum' as well as 'mass'. – jacob1729 Oct 27 '19 at 22:43
  • @jacob1729 Can we define them simutaneously? They depend on each other, so in the end it is still circular. – Ma Joad Oct 27 '19 at 22:49
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    You need to define all of them. Definitions don't have an idea of time, they just are. I can define the energy to be whatever I like, but its on me to show that it satisfies any property you might expect energy to have. What Einstein did was show $E=\gamma m$ does satisfy these properties and gave a physical argument why the rest mass of a body can be considered as energy. – jacob1729 Oct 27 '19 at 23:09
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    It's not circular. Often if you assume $X$ you can prove $Y$, and if you assume $Y$ you can prove $X$. That is not circular reasoning. That is just two separate proofs. The way we actually tell in reality that $X$ and $Y$ are true are by experiment. – knzhou Oct 28 '19 at 00:54
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    You can derive $E=mc^2$ in many ways (Einstein published nearly $10$ derivations of it throughout his life), if you change what you allow the starting assumptions to be. The only question is what assumptions you like best. – knzhou Oct 28 '19 at 00:55
  • "The Energy-momentum relation was first established by Paul Dirac in 1928 under the form $E=\sqrt{c^2p^2+(m_o c^2)^2} + V$, where $V$ is the amount of potential energy." - https://en.wikipedia.org/wiki/Energy-momentum_relation – safesphere Oct 28 '19 at 07:47
  • See the proof of E = mc² here — https://physics.stackexchange.com/a/707910/329778 – Debanjan Biswas May 11 '22 at 12:10

1 Answers1

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There is an ambiguity as to what is meant by 'energy', 'momentum' and indeed 'mass'. One approach (that OP presumably finds unsatisfying) is to declare $p=\gamma mv$, $E=\gamma m$ where $\gamma = (1-v^2)^{-1/2}$ and thus $E^2-p^2=m^2$ by direct calculation. It then remains to show that these deserve the names we've given them. This involves two steps:

1) Showing that as $v\to 0$ we recover $p=mv$ and $E=\frac{1}{2} mv^2$. In fact we get $E=m+\frac{1}{2} mv^2$ but we can recognise that nothing in Newtonian mechanics precluded this constant since we only ever dealt with energy changes.

2) We need to show $E,p$ are conserved. This is harder and is what the various physical arguments tend to focus on.

A 'high level' way to do this would be the following:

1) Agree on a Lagrangian/Action for the system. In this case we take the proper time along a path to be the action of that path.

2) Show that spacetime translations $x^\mu \mapsto x^\mu + a^\mu$ leave this action invariant (ie shifting a curve doesn't change its arc length). This implies by Noether's theorem a conserved quantity called $P^\mu$.

3) Declare the time part of $P^\mu$ to be called 'energy' and the space part to be called 'momentum'. Declare $P^\mu P_\mu$ to be the mass squared.

4)We are now in the same situation as before, where we need to show these quantities are correct to call by those names. Here, a little classical mechanics comes in helpful since in classical mechanics $E$ is conserved due to time translation and $p$ due to space translation. So these certainly correspond. We're left with masses, however we can write out the mass squared relation as:

$$ E^2 - p^2 = m^2$$

which looks a lot like $\cosh^2 \eta -\sinh^2 \eta = 1^2$ so we further define the variable $\eta$ by $E=m\cosh \eta, p=m\sinh \eta$. Whatever this $\eta$ is, it's true that as $\eta \to 0$ we have:

$$ E \to m(1 + \frac{1}{2}\eta ^2)$$ $$ p \to m\eta $$

all of which looks convincing enough to me to just go ahead and say for small $\eta$, $\eta$ must be the velocity, and this has recovered the usual formulas with $m$ where the mass would be, so this agrees with our intuition of what mass should mean as well.

jacob1729
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  • Thank you. So could you please tell me why $E=p/v$? That is also part of my question. – Ma Joad Oct 28 '19 at 02:10
  • In the language of my post we have $p/E = \tanh \eta$ by direct calculation. It will turn out that $\tanh \eta = v$ but I haven't formally set up the relation between $\eta$ and $v$ - to do so would I think require more work in the Nother's theorem section of this answer which should show $P^\mu \propto U^\mu$ with $m$ being the name given to the proportionality constant. – jacob1729 Oct 28 '19 at 14:20