# Is $E=mc^2$ artificially defined?

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

## 1 Answer

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.

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