0
$\begingroup$

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?
$\endgroup$
7
  • $\begingroup$ As usual, the answer to this depends on what you're willing to accept as definitions for 'energy', 'momentum' as well as 'mass'. $\endgroup$
    – jacob1729
    Commented Oct 27, 2019 at 22:43
  • $\begingroup$ @jacob1729 Can we define them simutaneously? They depend on each other, so in the end it is still circular. $\endgroup$
    – Ma Joad
    Commented Oct 27, 2019 at 22:49
  • 1
    $\begingroup$ 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. $\endgroup$
    – jacob1729
    Commented Oct 27, 2019 at 23:09
  • 1
    $\begingroup$ 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. $\endgroup$
    – knzhou
    Commented Oct 28, 2019 at 0:54
  • 1
    $\begingroup$ 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. $\endgroup$
    – knzhou
    Commented Oct 28, 2019 at 0:55

1 Answer 1

8
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Thank you. So could you please tell me why $E=p/v$? That is also part of my question. $\endgroup$
    – Ma Joad
    Commented Oct 28, 2019 at 2:10
  • $\begingroup$ 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. $\endgroup$
    – jacob1729
    Commented Oct 28, 2019 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.