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 $E=C*mc^2$. 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?

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    $\begingroup$ I want to challenge your thinking a bit. Does Newton's second law $F=ma$ mean that force can be converted into mass or into acceleration? $\endgroup$ – Aaron Stevens Jul 14 at 11:23
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    $\begingroup$ The factor $c^2$ is just our convention to measure mass. In fact, it is arguably more natural to work in the so-called natural units, in which $c=1$ and this question does not arise. It is common to say that the rest mass of an electron is about 511keV. We could come up with different conventions to measure the "rest energy" of a particle, and could attach other units to whatever we want to measure this in. $\endgroup$ – user178876 Jul 14 at 11:32
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    $\begingroup$ @Marmot: but when Einstein published this equation one kilogram was defined by a prototype. It seems fortunate and unlikely that this prototype was chosen so that the constant C in E=Cmc^2 should become 1? $\endgroup$ – Andy Jul 14 at 11:55
  • $\begingroup$ @Andy ??? But energy was (and is) defined in these conventions, too. $\endgroup$ – user178876 Jul 14 at 12:07
  • $\begingroup$ @AaronStevens Nice point. However, isn't it also a bit weird to say that energy can be converted into the mass or vice-versa even in the context of $E_0=mc^2$? It is clearer to simply say that the rest energy has the property of inertia. In particular, the mass of an isolated closed system never changes because since the four-momentum is conserved, its norm is also conserved. $\endgroup$ – Dvij Mankad Jul 14 at 13:41

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.

  • $\begingroup$ Thank you very much Andrew for clearing this up! Am I right in assuming that classical electromagnetism with its formula for a photon: E=pc seemed to contradict classical mechanics. Then Einstein came along and by his box thought experiment showed how these two can be reconciled if we are willing to give up conservation of mass, and assume mass that can be converted to energy. And if so this energy would follow E=mc^2? $\endgroup$ – Andy Jul 14 at 13:42
  • $\begingroup$ +1: This is a great explanation! As far as I remember, I struggled with this thought experiment while reading it from Feynman lectures because the discussion used the language of assigning a relativistic mass to every form of energy which sort of assumed the validity of what was attempted to be shown. This is much cleaner. $\endgroup$ – Dvij Mankad Jul 14 at 13:51
  • $\begingroup$ @Andy you are basically correct in this comment, except the historical sequence was a bit more complicated; the mass/energy relation was worked out from momentum arguments and then Einstein proposed his box argument with a view to finding a simple way of presenting the main idea. $\endgroup$ – Andrew Steane Jul 14 at 13:57
  • $\begingroup$ @Andrew: I understand that the box argument was proposed one year after he had already derived his equation using another thought experiment that involved radiation from an object. This derivation seems far more complicated involving Doppler shift and Lorentz transformations (?). I am assuming that this is what you are referring to(?) $\endgroup$ – Andy Jul 14 at 14:03

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.

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.”

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    $\begingroup$ The reason is that the formula for kinetic energy has the speed squared. $\endgroup$ – MarianD Jul 14 at 11:14
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    $\begingroup$ @Andy $c$ is more of "the speed limit of the universe", and photons happen to move at this speed. This speed is not unique to just light. $\endgroup$ – Aaron Stevens Jul 14 at 11:27
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    $\begingroup$ @Aaron: thank you for a great phrase. I remember now that nothing can travel faster than light, but this phrase makes it much easier to remember. $\endgroup$ – Andy Jul 14 at 12:04
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    $\begingroup$ @Andy, $c$ is the invariant speed - an object with speed $c$ in one inertial frame of reference (IRF) has speed $c$ in all IRFs (by the Lorentz transformation). A such, the invariant speed $c$ is a universal 'conversion factor' between time and length. Since the Lorentz transformations 'mix' the dimensions of time and length, it shouldn't be surprising that the conversion factor $c$ is ubiquitous in relativistic expressions. $\endgroup$ – Alfred Centauri Jul 14 at 12:30
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    $\begingroup$ @MichaelWalsby, keep in mind that if one accepts that the mass of an object increases with speed, one also accepts that the mass of the (relatively) moving object increases more in the direction of motion than in the directions orthogonal to the motion, i.e., $m_{||} \ne m_\perp$ $\endgroup$ – Alfred Centauri Jul 14 at 15:30

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