# How to derive $E=mc^{2}$?

Is there some way to derive $E=mc^{2}$? I can understand that energy in something is proportional to its mass, but the $c^{2}$ part. I have no idea. It seems like the way the units are going it would end up as $kg*(m/s)^{2}$, which I think is the unit for Newtons, not energy. And why is there no constant of proportionality $k$? Did Einstein just set it up perfectly so that there wouldn't be one?

• The units are correct. Compare to the classical $E = \frac{1}{2} m v^2$. As for deriving it, doing this would basically be walking you through the initial derivation of Lorentz transforms in relativity, but maybe someone knows a quicker way to get there... – zeldredge Apr 29 '15 at 2:06
• Do you know calculus? If not, I will have to use a different version of the proof. – Jimmy360 Apr 29 '15 at 2:39
• Possible duplicates: physics.stackexchange.com/q/43813/2451 and links therein. – Qmechanic May 23 '15 at 23:53

A simple derivation that is accessible to lay people who can only do primary school level math, starts from the fact that a pulse of electromagnetic radiation with energy $\mathbf{E}$ has a momentum of $\dfrac{\mathbf{E}}{\mathit{c}}$. In addition, one assumes conservation of momentum. We do a thought experiment involving a closed box containing two objects of mass $\mathtt{M}$. No external forces act on the box, therefore the momentum of the box is conserved. This means that whatever happens inside the box, if the box was at rest initially, it will stay at rest. In particular, it is impossible for the center of mass of the box to move.

We then consider the following process happening inside the box. One object emits a pulse of light of energy $\mathbf{E}$ that is completely absorbed by the other object. Due to the recoil of the light, the masses will move, but the velocities will obviously be very small (for macroscopic objects). We can thus safely make the assumption that classical mechanics will be valid after the pulse of light has been absorbed. If you want to be very rigorous, you must take the limit of $\mathtt{M}$ to infinity so that classical mechanics will be exactly correct.

Due to conservation of momentum, in the original rest frame of the box, the center of mass of the box will not change. Now classical mechanics doesn't give a correct description of relativistic phenomena like light, but it can be used to describe the situation in the box both before the pulse of light was emitted and after it was absorbed.

What is clear is that the object that emits the pulse of light will move away from the other object with the velocity of $$v = \dfrac{\mathbf{E}}{\mathtt{M} \cdot \mathit{c}}$$. If the other object is a distance of $\mathit{L}$ away, then it will take a time of $T = \dfrac{\mathit{L}}{\mathit{c}}$ before the other object absorbs that pulse of light. So, it seems that the center of mass will shift by $\frac{1}{2} v\cdot T = \frac{1}{2} \dfrac{\mathbf{E} \cdot \mathit{L}}{\mathtt{M} \cdot {\mathit{c}}^2}$ because of the time lag between when the object emitting the puls of light starts to move and when the object absorbing the pulse of light starts to move.

But, of course, a closed box upon which no external forces act cannot move all by itself. The masses must have changed due to the transfer of energy. Since classical mechanics is assumed to be valid we also have conservation of mass, so the sum of the masses of the objects will not have changed. If the emitting object has lost a mass of $d\mathtt{M}$, the receiving object will have gained a mass of $d\mathtt{M}$. A transfer of a mass of $d\mathtt{M}$ from the emitting object to the receiving object will have shifted the center of mass by an amount of $\mathit{L} \dfrac{d\mathtt{M}}{2\mathtt{M}}$ in the opposite direction as the shift due to the motion. For the two effects to cancel each other out, requires that $$\mathit{L} \dfrac{d\mathtt{M}}{2\mathtt{M}} = \frac{1}{2} \dfrac{\mathbf{E} \cdot \mathit{L}}{\mathtt{M} \cdot {\mathit{c}}^2} \implies \mathbf{E} = d\mathtt{M} {\mathit{c}}^2$$.

So, transferring the energy via the pulse of light led to mass being transferred according to $\mathbf{E} = \mathtt{M} {\mathit{c}}^2$. Due to conservation of energy and the ability of energy to transform from one form to another (e.g. the pulse of light will become heat, chemical energy or whatever), one can then argue that any transfer of energy implies a transfer of the equivalent mass according to $\mathbf{E} = \mathtt{M} {\mathit{c}}^2$. This in turn then leads to the conclusion that the mass of an object is simply the total energy content of the object divided by ${\mathit{c}}^2$.

• I'm lost here. The center of mass is also determined by the position of the EM pulse which, if you take it into account, means the center of mass of the whole system doesn't shift at all. Are the masses attached to the ends of the box, or free to move inside it? – Larry Harson Feb 15 '18 at 21:19
• @Count Iblis, isn't the formula P=E/c a result we get by playing with mc^2 already? It does not look right to use it for proof of mc^2 to me.. – physicsguy19 Sep 9 '18 at 12:29
• @Larry Harson, does the EM pulse have a definite mass? I believe I read somewhere that photons are massless and therefore don’t contribute to the center of mass. – frogeyedpeas Aug 30 '19 at 15:26

First a note not to forget that $E=m\,c^2$ is only a special case of the more general formula $\frac{E^2}{c^2} - p^2 = m^2 c^2$ for the Lorentz-invariant square "length" of the momentum four-vector in terms of the rest mass $m$.

That matter out of the way, there are two methods springing to mind:

Method 1: A neat way to do this is to imagine light in a massless box with perfectly reflecting mirrors in it and to analyse what happens when you shove the box.

Suppose this box is at first at rest with respect to your inertial frame. At equilibrium, the bouncing light exerts equal pressure on all mirrors.

However, now we wish to make the box move at some small, constant speed $v\ll c$ relative to our frame. As the box begins to move, the light reflecting off the hinder end of the box (in the sense defined by the motion) is blue shifted as it meets the mirror at the end of the box and that reflected from the forward end of the box is red shifted. So the pressures now differ, with the result that you need to supply impulse in the direction of motion to allow the light to reach a steady state moving at constant velocity relative to your frame. When you do the calculation, you find that this impulse is equal to $E\,v/c^2$. So you therefore ascribe a "resistance" to shove measured by an "inertia" $E/c^2$. I show how to do this calculation with photons (easy) here. One can also do a fully classical using the Lorentz transformation (much harder and messier) and the results are the same. Some day I must work this full analysis into a Physics SE answer, but have as yet not found occasion to do so: the technique is very like that in my answer here which describes "squashing light" in a cavity with moving mirrors.

Note that, although we are calculating, using classical, nonrelativistic mechanics, the force needed on the box, we are actually working out the limiting behavior as $v\to0$: we are considering a system moving at velocities as small as we please. So this method shows that we must ascribe a "mass" $E/c^2$ to a quantity of energy if special relativity and classical mechanics are to be mutually consistent, i.e. equal in the limit as $v\to0$. Thus we avoid problems associated with discussion of so called "relativistic" mass, which can yield misleading results, although it yields by chance the right answer in Jimmy 360's answer.

Method 2: This is the method initially used by Einstein in his first paper after his big special relativity paper of 1905:

This is very like the method of Count Iblis's Answer

The answer really depends on how much you assume. You could start with the relativistic action, $$S=-mc\int d\tau$$ where $c^2d\tau^2=dx^\mu dx_\mu$, such that the Lagrangian is $$L=-mc\sqrt{1-(v/c)^2}$$ whence the momentum follows from $p=\partial L/\partial v$ and the total energy is $E=pv-L$--this leads directly to \begin{align} E&=\gamma mc^2=\left(1-v^2/c^2\right)^{-1/2}mc^2\tag{1}\\ p&=\gamma mv=\left(1-v^2/c^2\right)^{-1/2}mv\tag{2} \end{align} Where you can then use the first to solve for $v$ and insert into the latter to get the expected relation for $p=0$ (inserted after solving).

You could also start with (1) and assume $v\ll c$ such that you can use the approximation, $$(1-x)^n\approx1-nx$$ such that $$E\approx\left(1-\frac{v^2}{2c^2}\right)mc^2=mc^2+\frac12mv^2$$ where the last term equals 0 for a stationary particle, leading again to the famous $E=mc^2$.

which I think is the unit for Newtons, not energy.

The SI unit of energy is the Joule, defined as $$\rm Joule=1\,kg\,\frac{m^2}{s^2}=1\,N\,m$$ so $mc^2$ is indeed an energy, not force.

And why is there no constant of proportionality ? Did Einstein just set it up perfectly so that there wouldn't be one?

Einstein didn't "set it up" (which would imply that he made it up), he derived the equation. There is indeed a constant of proportionality, it is 1.

• This has now gained 3 downvotes; I'm curious what these three people think is wrong with this post? Especially so since this is the only undeleted answer with downvotes... – Kyle Kanos Oct 24 '18 at 20:05

This isn't a derivation, but you might be interested in the fact that there is experimental evidence for the relation $E=mc^2$, and it is very simple. This is due to what is known, in nuclear physics, as the “mass defect” (cfr. ).

Using methods to determine the masses of the nuclei (the so called "mass spectroscopy") one finds that the sum of the masses of the constituents of a nucleus (neutrons and protons) is always greater than the mass of the nucleus itself. That is: $$\Delta(Z,A):=ZM_H + (A-Z)M_n - M(A,Z)>0,$$ where $M_H,\, M_n,\, M(A,Z)$ are the masses of the hydrogen atom, the neutron and the atom with $A$ nucleons and $Z$ electrons (the masses of the $Z$ electrons are summed and subtracted in the formula, since the accessible quantities are the masses of the atoms).

On the other hand, it's possible to independently determine the rest energies $E(A,Z)$ of the nuclei. For instance, suppose that a nucleus $A,Z$ decays at rest into two lighter nuclei $A',Z'$ and $A'',Z''$, with $Z'+Z''=Z$ and $A'+A''=A$. The kinetic energies of the daughter nuclei can be measured, and the total kinetic energy is, by conservation of energy, the difference beetween the binding energies. The mass defect formula allows us to predict the energy realeased in the process, since: $$E(A,Z)-E(A',Z')-E(A'',Z'')=^!\\ [M(A,Z)-M(A',Z')-M(A'',Z'')] c^2\\=[\Delta(A',Z')+\Delta(A'',Z'')-\Delta(A,Z)]c^2.$$ The exclamation mark is to emphasize that the first equality is a guess for the binding energies, under the assumption that $E=mc^2$. The fact that this guess is consistent with the experimentally determined energies, is an evidence for the mass-energy equivalence.

 “Particles and Nuclei: An Introduction to the Physical Concepts”, B. Povh, K. Rith, C. Scholz, F. Zetsche - et. al - http://www.amazon.it/Particles-Nuclei-Introduction-Physical-Concepts/dp/3540793674

• Good point, sad to see such a lack of upvotes on an empirical and historically important perspective. +1 – electronpusher May 8 '17 at 5:35
• @electronpusher Yep +1, one can't beat empirical arguments in physics. – WetSavannaAnimal Jul 15 '17 at 2:41

The energy-momentum 4-vector is simply mass times the velocity 4-vector. This gives the correct momentum in the non-relativistic limit. In the rest frame it correctly gives the (three)-momentum as being zero. And gives $E = mc^2$ for the time-component of the vector.

This is clearly not going to make too much sense if you don't know relativity, but that's a long but rather interesting story in itself!

The equation $E = mc^2$ is only true for an object at rest. None of the other answers prove that's the case because they deduce it from assumptions they did not prove so I wrote an answer that actually proves that it's the case according to a certain theory in the absence of a long range force.

Suppose that in any frame of reference, when ever two objects of a given positive rest mass $m$ and subluminal velocity collide and combine into a single object in the absense of a long range force, they will always combine into an object with the same positive rest mass and subluminal velocity. Furthermore, let's assume that the momentum $\vec{p}$ and energy $E$ of any object with a positive rest mass and subluminal velocity $\vec{v}$ can be defined as a function of rest mass and velocity in such a way that the following conditions are satisfied:

• Energy is always positive
• For any positive rest mass, energy and magnitude of momentum $p$ don't vary between two subluminal velocities that have the same magnitude $v$
• Momentum is zero for zero velocity and goes in the direction of the velocity for nonzero subluminal velocity
• As velocity approaches zero, momentum divided by velocity approaches rest mass
• For any given rest mass, for low speeds, energy approximates as $\frac{1}{2}mv^2$ plus a constant
• For any subluminal velocity, energy and momentum vary linearly with rest mass
• When ever two objects collide and combine into one, the total energy and momentum is conserved

It can be mathematically prove that the only solution for energy and momentum as a function of positive rest mass and for 2 objects of a given rest positive mass and subluminal velocity that combine into one what rest mass and velocity is has that satisfies the assumptions I made is:

• $E = \gamma mc^2$ where $\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$
• $\vec{p} = \gamma\vec{v}$
• When two objects collide and combine into one, we compute the total energy and momentum of the original two objects and assume that's also the momentum and energy of the combined object and then recompute the rest mass and velocity using their formulae in the last two bullet points: solving them gives us $m = \sqrt{\frac{E^2}{c^4} - \frac{p^2}{c^2}}$ and $\vec{v} = \frac{\vec{p}c^2}{E}$

I still haven't proven my original assumptions are true. That problem can be solved by defining special relativity to predict that energy and momentum vary with rest mass and velocity and two objects of a given rest mass and velocity will combine into a single object of rest mass and velocity according to the last 3 bullet points in the absence of a long range force. It can be mathematically proven that according to that prediction, energy and momentum are conserved when two objects collide and combine into one in the absence of a long range force. After reading this answer, it's easy to figure out a proof that the prediction actually predicts it.

The complete theory classical special relativity probably also predicts that an object with no electric or magnetic field cannot emit or absorb any elecromagnetic radiation at all but quantum mechanics predicts that an object with no electric or magnetic field actually can absorb a photon. Let's assume that an object absorbs a photon whose wavelength is much smaller than the size of the object. Furthermore, let's assume that a photon is a particle of nonzero energy travelling at the speed of light that just like matter also satisfies $\vec{p} = \frac{E\vec{v}}{c^2}$ and $m = \sqrt{\frac{E^2}{c^4} - \frac{p^2}{c^2}}$, then we find that $\vec{p} = \frac{E}{c}$ and $m$ = 0 for a photon. Furthermore, suppose that energy and momentum are conserved when a photon gets absorbed by matter and that in any frame of reference, an object of a given rest mass and velocity absorbing a photon of a given energy going in a given direction will always produce an object with the same rest mass and velocity, then in another frame of reference, the momentum 4-vector will undergo the same Lozentz transformation as a space-time vector. Using the formula $\lambda = \frac{h}{p}$, we find that the energy of a photon varies as the reciprocal of its wavelength. Treating the photon as a classical wave, we can also predict how its wavelength and direction change with frame of reference. Using the formula that its energy varies as the inverse of its wavelength, we can also predict how its momentum 4-vector changes and it predicts the exact same change in its momentum 4-vector as was predicted earlier.

I don't know for sure that I'm correct about the definitions of energy and momentum. The old text book Nelson Physics 12 defined relativistic mass as $\gamma m$ and relativistic momentum as $\gamma\vec{v}$ and I don't know how the momentum and energy of a piece of matter are defined now.