How to understand $E=mc^2$?

Question

I am trying to understand Einstein's mass-energy equation $E=mc^2$. Knowing about the atomic bomb, I am inclined to believe that $E$ is proportional to the mass: $E=Cm$. For this equation to be dimensionally correct $E$ must also be proportional to some speed squared. We can use the "speed-limit of the universe" $c$. So $E=C \times mc^2$. But why does $C$ turn out to be 1? Should the equation really be: $E\propto mc^2$?

The above reasoning takes a "leap of faith". If I assume that energy is proportional to mass the equation follows (except maybe for a constant).

I am looking for the simplest way to prove this. I read about Einsteins box in which a photon is emitted from one end towards the other. This seem to offer a simple way to derive this equation. If I accept that the photon has a momentum: $p=E/c$, even though it has no mass it follows from conservation of momentum that the photon has a “relativistic mass”: $m = E/c^2.$

However I also read that $E=mc^2$ only applies to a resting object (not photons) and that in general: $E^2=p^2c^2+m^2c^4$.

But does this does not mean that one cannot understand $E=mc^2$ from Einsteins box? What is the point of this thought experiment then?

Answer 1

Your question is a good one. It springs from a slightly incorrect understanding of the logic of the Einstein box argument. It is not necessary to ascribe mass to a photon in order to present the argument, and indeed that would be an incorrect way to proceed. Rather, one asserts that a pulse of electromagnetic radiation carries both energy and momentum, and the relationship between the energy and momentum of such a pulse is $$ E = p c. $$ This formula comes from classical electromagnetism (Maxwell equations etc.), not directly from relativity. (In a modern argument one would assert relativity first and then derive electromagnetism, but I won't get into that.) The rest of the argument is based on conservation laws.

Suppose the box has length $L$ and starts off centred at the origin, with its ends at $\pm L/2$. First, when the pulse is emitted by one end of the box, that end recoils with momentum $p$ and also gives up energy $E$. For example it could be thermal energy. In Newtonian physics this need have no effect on the mass of the wall of the box, but if one assumes that then one ends up in a contradiction. So let's assume instead that when the wall of the box gives up energy $E$, its mass falls a little, by an amount $m$ to be discovered. Then the recoil velocity of the wall is $$ v = \frac{p}{M-m} = \frac{E}{c(M-m)} $$ The pulse of light now propagates to the other end of the box, through a distance $L$, taking time $$ t = L/c . $$ When the other end of the box receives the energy and momentum of the pulse, its energy goes up by $E$ so its mass goes up by $m$ (the quantity we are trying to calculate). So now this end of the box is located at $L/2$ and has mass $M+m$, while the other end has moved a bit, to $-(L/2) - v t$, so the centre of mass of the whole box is now located at $$ x_{\rm cm} = (M-m) \left(-\frac{L}{2} - v t\right) + (M+m)\frac{L}{2}\\ = (M-m) \left(-\frac{L}{2} - \frac{E L}{(M-m)c^2}\right) + (M+m)\frac{L}{2}\\ = m L - \frac{EL}{c^2} . $$ But internal changes cannot shift the centre of mass, so we must have $x_{\rm cm} = 0$ and therefore $E = m c^2$.

The above is directly based on the discussion in a book called "The wonderful world of relativity" by myself (publisher Oxford University Press).

If we now look back over the derivation, we see that the mass $m$ is not associated with the pulse of light (or photon if you prefer). Rather, $m$ is the change in the mass of the wall of the box. You are right to quote the formula $$ E^2 = p^2 c^2 + M^2 c^4 $$ This formula applies equally well to bodies with zero rest mass (such as photons) as to bodies with non-zero rest mass (such as molecules). In the above argument when I said the wall gets a velocity $p/(M-m)$ I was in fact neglecting some small corrections which are negligible in the limit where this velocity is small compared to $c$. If one keeps those small corrections one still gets the answer $E = M c^2$ for a body with zero momentum.

Answer 2

It's very easy to understand $E=mc^2$.

All Einstein's very short paper on the subject, with the English translated title "Does The Inertia of a Body Depend on Its Energy Content?", sais is that $E=mc^2$ means is that the inertial mass of a body increases (or decreases) by the amount $m=E/c^2$ if its energy content increases (or decreases) by an amount $E$.

If that doesn't satisfy it, this legit video is very accessible:

https://youtu.be/Xo232kyTsO0

The Energy–momentum relation was established by Paul Dirac in 1928, but is fundamentally different than $E=mc^2$. Those are two separate things. (I don't understand why other answers involve that to be honest.)

Anyway, what's not so easy to understand is why it is what it is. There's a lot of history behind $E=mc^2$.

Einstein wasn't first to equate forms of mass to energy, nor did he definitively prove the relationship. Before him it was thought to be $m=(4⁄3)E/c^2$ (Oliver Heaviside) and later $m=(8⁄3)E/c^2$ (Fritz Hasenöhrl) for example. $E=mc^2$ is the short punch line to a long and winding scientific story.

It's all here for the interested:

https://www.scientificamerican.com/article/was-einstein-the-first-to-invent-e-mc2/

It's a nice read imo.

Now some people may wonder why $c$ is squared.

The probably unsatisfying answer is that it's just a conversion factor nothing more nothing less .. and in SI units it follows from dimensional analysis.

But it’s possible that there’s deeper explanations that I’m not considering right now. Coz, philosophically speaking, dimension analysis is a circular argument: saying that energy is like energy because it is energy.

Answer 3

In both equations, “your” $E=cm$ and Einstein's $E=mc^2$, the energy is proportional to the mass, they differ only in the coefficients: your one is $c$, while Einstein's is $c^2$.

The terms “relativistic mass” and “rest mass” were canceled by Einstein himself, and nowadays they a considered as obsolete, confusing, and inappropriate. Sadly, many textbooks still use it.

You're right, nowadays the mass is considered as a form of energy.

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Einstein himself completely stopped using the term “relativistic mass” already in 1906, and he explained, why. In the letter to L. Barnett he wrote:

“It is not good to introduce the concept of the mass $M = m/(1 − v2/c2)^{1/2}$ of a moving body for which no clear definition can be given. It is better to introduce no other mass concept than the ’rest mass’ $m$. Instead of introducing $M$ it is better to mention the expression for the momentum and energy of a body in motion.”