Most people know the famous equation:


What were his steps of thinking for this equation that helped us discover so much about our world?


5 Answers 5


You can find the shortest and easiest derivation of this result in the paper where it was released by Einstein himself (what better reference can you find?) in 1905. It is not the main paper of Special Relativity, but a short document he added shortly afterwards.

A. Einstein,Ist die Trägheit eines Körpers von seinem Energieinhalt Abhängig?, Annalen der Physik 18 (1905) 639. A pdf file of the English translation Does the Inertia of a Body Depend upon its Energy-Content? is available here. (hattip: user53209.)

It is a delightful document to read. There is no dramatic references to huge power release nor anything similar. He simply states after the derivation "If a body gives the energy away $L$ in form of radiation, then its mass decreases in an amount $L/V^{2}$ (...) the mass of a body is a measure for its energy content (...) One can not exclude the possibility that, with the bodies whose energy content changes rapidy, for example radium salts, a proof of the theory will be found (...) If the theory adjusts to the facts, then the radiation transports inertia between emitters and absorbers."

Google for that short paper and see the derivation yourself, it is very easy. The Minkowsky four-dimensional spacetime had not yet been incorporated to special relativity, so the equations are formally very simple, easy to follow with little mathematical training.

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    $\begingroup$ @Larry Harson, and I doubt you've even read my answer, because if you had, you'd seen that I make no mention to any proof, but rather I explicitely use the word "derivation", and in the first line. Follow it with your finger, where I say "derivation of this result". $\endgroup$ Jan 29, 2013 at 2:03
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    $\begingroup$ @Larry Harson, please read the Einstein paper, it contains the original, fully correct derivation. Yes, derivation. And that "much" that has been written against 1905 Einstein papers consist on a bunch of pseudo-scientific journalism, mainly from nazi morons. $\endgroup$ Jan 29, 2013 at 14:40
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    $\begingroup$ Special Relativity can be (and will be) super-seeded by new theories, but it is self-consistent, as it is the Einstein derivation (again, derivation) of $E=mc^{2}$ of 1905. $\endgroup$ Jan 29, 2013 at 14:43
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    $\begingroup$ @Qmechanic, thanks for the nice edit with the link to the english version. I've just added the link to the original one too. What does that "hattip: userXXXX" means? $\endgroup$ Feb 28, 2013 at 3:53
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    $\begingroup$ The proper spelling is in two words: hat tip. $\endgroup$
    – Qmechanic
    Feb 28, 2013 at 7:03

Let us take a condition,

Suppose, in this picture your friend is standing with a special torch (which can emit light from either two sides) in space and you are watching your friend from a distance. Your reference frame is S and your friend is at rest in S reference frame. Now, your friend switches on the torch for once and instantly switches off the torch. So the torch emits two same light pulses from either two sides. Suppose, the frequency of light is $\nu$ in S frame.

So, the energy of each light pulse is $e = h\nu$ [$h$ is plank constant]

Total energy lost is $E = 2e$

Your friend measured the energy before emission was $E_0$

Energy remained after emission is $E_1$

So the relation is, $E_1 = E_0 - E$

Now let's take another condition,

enter image description here

Here, you are moving towards right in a rocket with velocity $v$ where your friend is standing.

Let's name this reference frame S'. In this condition, you will see your friend moving towards left with velocity $v$ where you are inside the rocket as for relative motion.

enter image description here

In this reference frame your friend does the same. He switches on the torch for once and instantly switches off that torch. Now, applying relativistic doppler effect on both light pulses we get, the frequency of the left light pulse is $\nu\cdot\sqrt{\frac {c+v }{c-v }}$, with $c$ the speed of light.

and the frequency of right light pulse is $\nu\cdot\sqrt{\frac {c-v }{c+v }}$.

Multiplying the frequency by plank constant, we get the energy of the light pulses.

So, total energy lost in S'

\begin{align} &= h\nu\cdot \frac {\sqrt{c+v} }{\sqrt{c-v} }+h\nu\cdot \frac {\sqrt{c-v} }{\sqrt{c+v} }\\ &=e\left[ \frac {\sqrt{c+v} }{\sqrt{c-v} }+\frac {\sqrt{c-v} }{\sqrt{c+v} }\right] & (\text{since, } e = h\nu)\\ &=e\left[\frac{c+v+c-v}{\sqrt{c+v}\sqrt{c-v}}\right]\\ &=e\frac{2c}{\sqrt{c^2-v^2}}\\ &=2e\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\\ &=E\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} & (\text{since, } E = 2e) \end{align}

Suppose, in S' reference frame energy before emission was $H_0$

and energy after emission is $H_1$

So the relation is, $H_1 = H_0 - E\frac {1}{\sqrt{1-\frac {v^2}{c^2}}}$

We can expand this expression using a Taylor series, $$(1-x)^{-1/2}\approx 1 + \frac {1}{2}x+\frac 3 8 x^2+\frac 5 {16} x^3+\dots$$

So, $$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\approx 1 + \frac {1}{2}\frac{v^2}{c^2}+\frac 3 8 \frac{v^4}{c^4}+\frac 5 {16} \frac{v^6}{c^6}+\dots$$

If $v\ll c$ we can neglect terms of order higher than $v^2/c^2$.

And so, we can write $H_1 = H_0 - E\frac {1}{\sqrt{1-\frac {v^2}{c^2}}}$ —this equation as $H_1 \approx H_0 - E\left(1 + \frac {1}{2}\frac {v^2}{c^2}\right)$

Now, let's find out the difference of energy of that torch between S and S' reference frame.

The only difference of energy between the frames is kinetic energy. In reference frame S the torch is at rest and in S' frame the torch has kinetic energy. Let's assume the kinetic energy before emission in S' frame was $K_0$

So, $H_0 - E_0 = K_0 + B$ [$B$ is constant] — (i)

We know after emission the torch loses energy. Because kinetic energy is changing, we assume the kinetic energy after emission is $K_1$

So, $H_1 - E_1 = K_1 + B$(ii)

As velocity isn't changing in S' frame, we can say mass is changing. Because, kinetic energy depends on mass and velocity. Suppose, the mass before emission was $m_0$ and mass after emission is $m_1$.

Subtracting eq.(i) from eq.(ii) we get,

$H_0 - E_0 - H_1 + E_1 = K_0 + B - K_1 - B$

Or, $H_0 - E_0 - H_0 + E\left(1 + \frac {1}{2}\frac {v^2}{c^2}\right) + E_0 - E = K_0 - K_1$

Or, $E\left(1 + \frac {1}{2}\frac {v^2}{c^2}\right) - E =\frac {1}{2}m_0v^2- \frac {1}{2}m_1v^2$

Or, $E\left(1 + \frac {1}{2}\frac {v^2}{c^2} - 1\right) =\frac {1}{2}v^2(m_0 - m_1) $

Or, $\frac {1}{2}v^{2} \frac {E}{c^{2}}= \frac {1}{2} v^{2}m\ $ [let's take $m_0 - m_1 \equiv m$]

Or, $\frac {E}{c^{2}} = m\ $

Or, $\boxed{E = mc^2}$

This is how Einstein derived $E = mc^2$

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    $\begingroup$ Very nice answer! I improved your formatting, please have a look at what I did if you want to improve formatting in the future $\endgroup$ May 10, 2022 at 13:48
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    $\begingroup$ Look at the equation. You can find out that 'E' is the lost energy when the torch is at rest. This means the equation is only applicable when an object is at rest. Normally, when an object is moving it gets some extra energy for its momentum. $\endgroup$ May 10, 2022 at 16:14

Einstein's equation doesn't have a "proof" because it's not a mathematical theorem. It's a physical theory that is overwhelmingly supported by experimental data. So you could say that the "proof" is in the mountains of experimental results that agree with the theory.

To understand Einstein's motivation for developing the theory of relativity, as well as mass-energy equivalence, Wikipedia has an excellent article on the history of relativity.

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    $\begingroup$ Einstein gave an argument which is summarized on Wikipedia, and also regurgitated on Terrence Tao's blog. This answer is not reasonable, physical statements have physical arguments, and these are what people normally mean by "proof" in this context. $\endgroup$
    – Ron Maimon
    Nov 9, 2012 at 21:20
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    $\begingroup$ Ron is out of his mind. It does not matter how beautiful a theory may be nor how neat the derivation, the proof is only in the agreement with the physical world. $\endgroup$ Feb 28, 2013 at 5:02
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    $\begingroup$ I will support @dmckee , since this came up again. A theory in physics cannot be proven, only disproven. It can only be validated if experiments agree with it. Even one solid disagreement disproves a theory. The questioner assumes that physical models are the same as mathematical models which end with the QED, but this is not true. The title is misleading, the content of the question is OK and is answerd by Eduardo. $\endgroup$
    – anna v
    Feb 28, 2013 at 5:26
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    $\begingroup$ Yes, Ron is probably out of his mind but he still knows more physics than ten anna vs and four dmckees put together. $\endgroup$ Feb 28, 2013 at 5:28
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    $\begingroup$ @MartyGreen Anna V is quite correct. Science doesn't deal with proof. Proof only exists in mathematics and in courtrooms. Even then, it doesn't mean the same thing in both places. Supporting evidence doesn't constitute proof. Evidence accumulates indefinitely while proof connotes finality. $\endgroup$
    – user11266
    Feb 28, 2013 at 15:11

I agree with Ron completely and wouldn't recommend "studying" Einstein's original papers about it. It's old fashioned and is explained much more accessible nowadays. Also there's a whole history behind the famous equation, where I gave an answer about, or rather a link in the answer here:

How to understand $E=mc^2$?

I also noticed that someone who asked a question what the physical reason is of why $c$ is squared in the most famous equation of all. Which is not fully known, but I try to explain there as well.

But so a more modern view about how Einstein realized that $E$ equals $mc^2$ is as follows:

Einstein looked at the case of an object emitting two rays of light in opposite directions. As a result, the object's motion is unchanged, since the light pressure reaction from the two rays canceled each other. However, the object loses the energy that is carried by those rays of light.

Looking at the same object from a moving coordinate system, the situation appears different. The rays of light now carry a certain amount of momentum, as measured in that coordinate system. Yet the velocity of the object itself is unchanged. Since momentum is conserved, this is only possible if the object's inertial mass has changed, and indeed, Einstein found that the change in inertial mass is equal to $ΔE/c^2$, where $ΔE$ is the amount of energy emitted in the form of those light rays.

In Einsteins paper there's no mention of (conservation of) momentum, though. Instead, he argues that the kinetic energy of the object changes relative to the moving frame, without a change in its speed. This is accounted for by a loss of inertial mass of the object in the amount of $L/c^2$, where $L$ is the combined energy of the two identical but oppositely directed light pulses.

As I understand he later used the first argument, which is more thoroughly explained in a classic book "Physics for the Inquiring Mind", in a nutshell note:

enter image description here

PS. The style of scientific publishing at that time was somewhat rougher than it is today (where nowadays there must first be an introduction, an elaboration and then the conclusion). Einstein started right in the middle of the story.


It is incorrect to say that $E=m$ cannot be proven, it is a mathematical theorem and can be proven within the axiomatic system of special relativity. The point of experiment is to verify this axiomatic system or theory (or more precisely, to alter our confidence in this axiomatic system), but "it's all experiment" is an unsatisfying and useless answer when it comes to understanding the math. It's particularly bad in this case, because it doesn't teach you what energy is, so it's silly to say you can experimentally prove things about it.

The assumptions you need to prove this (and other relativistic dynamics) are: (1) the postulates of special relativity (2) de Broglie's relation (3) conservation laws (this along with 1 also imply the Newtonian approximation at low speeds, but it's better to list it directly, since it's where the definitions of energy and momentum come from). I mean, in principle you could use some other set of results as your axioms, but that's useless and boring, don't do it.

The proof is fairly simple. An object releases light $E$ in opposite directions, and you consider the rest frame and a frame moving at $v$. You consider a low-velocity approximation, where the Doppler factors $\sqrt{\frac{1+v}{1-v}}$ and its inverse approach $1+v$ and $1-v$. The total energy is clearly conserved anyway, so we look at momentum conservation instead -- the momentum is equal to the energy in magnitude (by a factor of $c$, but that's just 1), so we argue that the change in momentum due to the light in the moving frame of reference must be $(1+v)E/2-(1-v)E/2=vE$. So this must equal the change in momentum from the decrease in mass $m$ (the mass is moving at $v$ in the opposite direction in the moving reference frame), and we have $vE=mv$, or $E=m$.

You can also use this to prove other dynamical results, see e.g.

This is Einstein's derivation -- there's a lot of crap online, such as from minutephysics, claiming to be Einstein's original derivation, and if you believe them, I have a mass-loss bridge to sell you.

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    $\begingroup$ In this derivation you assume Einstein's relation $E=h\nu$ for the light pulses, which is unnecessarily special assumption. The formula at question is valid for any energy loss via EM radiation, not just those that conform to that idea of quantization. What is really needed are the transformation properties of momentum and energy of a light pulse (the relation $p=vE$ you have used above). These follow directly from EM theory and Lorentz transformations, no quantum theoretical assumptions are needed. $\endgroup$ Jul 5, 2018 at 22:13
  • $\begingroup$ @JánLalinský You're right -- we just need $E=pc$ for light. Deriving it from $p=vE$ just seems un-motivated (and also trivial). $\endgroup$ Jul 7, 2018 at 4:23

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