# Why does E=mc²?

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I cannot find a decent proof that E=mc². Can someone help?
Sean Carroll says that in SR the time component of the 4-momentum of a particle is its energy. It is of course also ##mc^2dt/d\tau##. He uses that to prove that ##E=mc^2##. Which begs the question why does ##E=p^0##?

Misner, Thorne, Wheeler do roughly the same thing.

I find these 'proofs' unsatisfying but neither of the above books are on special relativity so I look elsewhere. I have found nothing useful on the web!

I checked two videos that claim to prove it (here and here). The first immediately picks a definition of relativistic doppler shift: "Transverse Doppler effect, geometric closest approach" is the second of eight different equations for special relativistic doppler shift given in this Wikipedia article. Why should I choose that one?
The second is long and proves after much calculus (about 23 minutes in) that a particle's kinetic energy is$$KE=mc^2\gamma-mc^2$$##\gamma## is the usual thing related to the particle's velocity. It then turns that round to say$$mc^2\gamma=KE+mc^2$$and that ##mc^2\gamma## is the total energy of the particle (WHY?) and therefore when the particle is at rest ##KE=0## and ##E=mc^2##. I am in despair!

I'm sure Einstein did better than this. Can somebody point me to a decent proof of ##E=mc^2##? Or even quickly put the proof here?

• Delta2

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• George Keeling
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• George Keeling
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The first thing you need to do is to establish what is meant by the relation. Typically, this is not always clear. Some will tell you that it is incomplete and just a special case of ##E^2 = m^2 c^4 + p^2 c^2##. However, that just tells you that ##E = m c^2## is the rest energy of an object and that the mass is defined as the modulus of the 4-momentum. This in itself is not very enlightening.

The more interesting interpretation is that in terms of the mass-energy equivalence and how the classical inertial mass can be identified with the rest energy of an object. In order to do this, what you need to do is to look at the non-relativistic limit and find out essentially how the components of the 4-momentum behave in that limit and identify them with the classical counterparts.

• George Keeling
Staff Emeritus
The second is long and proves after much calculus (about 23 minutes in) that a particle's kinetic energy is$$KE=mc^2\gamma-mc^2$$##\gamma## is the usual thing related to the particle's velocity. It then turns that round to say$$mc^2\gamma=KE+mc^2$$and that ##mc^2\gamma## is the total energy of the particle (WHY?) and therefore when the particle is at rest ##KE=0## and ##E=mc^2##. I am in despair!

I don't understand what is still puzzling to you. You can just take it as just terminology.

Suppose you have two identical wads of chewing gum, each has rest mass ##m##, with one traveling to the left at speed ##v## and the other traveling to the right at speed ##v##. After they collide, they form a bigger wad of chewing gum at rest. The new wad of chewing gum must have mass ##M = 2 \gamma m##. If energy is conserved, then there has to be energy associated with the big wad, even though it is at rest.

The total kinetic energy before the collision is

##KE = 2 (\gamma - 1) mc^2## (the 2 is because there are two particles)

Given that ##M = 2 \gamma m##, this can be written as:

##KE = Mc^2 - 2mc^2##

or

##Mc^2 = 2 mc^2 + KE##

So the natural interpretation of this is a conservation of energy equation: The rest energy of the large wad is the sum of the rest energies of the two smaller wads plus the sum of their kinetic energies.

It's just terminology to call it "total energy", but that terminology allows us to assert a conservation of energy law.

• George Keeling
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Thanks for all the replies. I will resume Monday. More replies welcome.

Gold Member
Thanks for all the replies. I will resume Monday. More replies welcome.
You can get it from the relativistic momentum equation doing some "cheating." This is my own little derivation, and I'm just a student with some undergrad math/physics classes, so read it with a grain of salt, but at the end, you get the right equation. Basically just using the work-energy theorem to get the relativistic kinetic energy equation, and from that you can see the rest energy equation. This is high school or 1st or second year college level calculus, going through step by step:

## p = γmu##

Force:

##\frac{dp}{dt} = \frac{d}{dt}(γmu)##
##\frac{dp}{dt} = m \frac{d}{dt}(γu)##

Work:

##W = \int_{0}^{u}m \frac{d}{dt}(γu) ~dx##
##W = m\int_{0}^{u} \frac{dx}{dt}~d(γu) ##
##W = m\int_{0}^{u} u~d(γu) ##

Then some multiplication rule and chain rule stuff:

##W = m\int_{0}^{u} u~(γdu+udγ) ##
##W = m\int_{0}^{u} u~\left([1 - \frac{u^2}{c^2}]^{-\frac{1}{2}}du+u\left(-\frac{1}{2}[1 - \frac{u^2}{c^2}]^{-\frac{3}{2}} \left[- \frac{2u}{c^2}\right] \right)~du \right) ##

Then canceling the -2's out

##W = m\int_{0}^{u} u~\left([1 - \frac{u^2}{c^2}]^{-\frac{1}{2}}du+\frac{u^2}{c^2}[1 - \frac{u^2}{c^2}]^{-\frac{3}{2}} ~du \right) ##

It's easier if you break this up into two integrals:

##W = m\int_{0}^{u} u~\left(1 - \frac{u^2}{c^2}\right)^{-\frac{1}{2}} du+m\int_{0}^{u} u~\left(\frac{u^2}{c^2}[1 - \frac{u^2}{c^2}]^{-\frac{3}{2}} ~du \right) ##

Solve these with some trig substitution.

Let ##u = c~sinθ##
Then ## du = c~cosθ ~dθ## and ##\frac{u^2}{c^2} = ~sin^2θ##

So, using the trig identity that ##~cos^2θ = 1 - ~sin^2θ##:

##W = m\int_{0}^{θ} c~sinθ ~\left(~cos^2θ\right)^{-\frac{1}{2}}c~cosθ ~dθ +m\int_{0}^{θ} c~sinθ ~sin^2θ \left(~cos^2θ\right)^{-\frac{3}{2}}~c~cosθ ~dθ ##

Simplify just a little bit:

##W = mc^2 \int_{0}^{θ} ~sinθ ~\left(~cos^2θ\right)^{-\frac{1}{2}}~cosθ ~dθ +mc^2 \int_{0}^{θ} ~sinθ ~sin^2θ \left(~cos^2θ\right)^{-\frac{3}{2}}~cosθ ~dθ ##

##W = mc^2 \int_{0}^{θ} \frac{~sinθ}{~cosθ}~cosθ ~dθ +mc^2 \int_{0}^{θ} ~sinθ \left(\frac{~sin^2θ}{~cos^2θ}\right) ~dθ ##

##W = mc^2 \int_{0}^{θ} ~sinθ~dθ +mc^2 \int_{0}^{θ} ~sinθ \left(\frac{~sin^2θ}{~cos^2θ}\right) ~dθ ##

The left one is super easy but the right needs work. It can be broken into two more integrals using the identity ## ~sin^2θ =1 - ~cos^2θ##:

##W = mc^2 \int_{0}^{θ} ~sinθ~dθ +mc^2 \int_{0}^{θ} ~sinθ \left(\frac{1 - ~cos^2θ}{~cos^2θ}\right) ~dθ ##

##W = mc^2 \int_{0}^{θ} ~sinθ~dθ +mc^2 \int_{0}^{θ} \frac{~sinθ}{~cos^2θ} ~dθ - mc^2 \int_{0}^{θ} ~sinθ \frac{~cos^2θ}{~cos^2θ} ~dθ ##

##W = mc^2 \int_{0}^{θ} ~sinθ~dθ +mc^2 \int_{0}^{θ} \frac{~sinθ}{~cos^2θ} ~dθ - mc^2 \int_{0}^{θ} ~sinθ ~dθ ##

The first and the last integral cancel, leaving:

##W = mc^2 \int_{0}^{θ} \frac{~sinθ}{~cos^2θ} ~dθ ##

Using another substitution, this time Let ##w = ~cosθ## and so, ##dw = - ~sinθ dθ ##

##W = - mc^2 \int_{0}^{w} \frac{1}{w^2} ~dw ##

which is:

##W = - mc^2 \left. \left(-\frac{1}{w} \right) \right|_0^w ##

negatives cancel

##W = mc^2 \left. \left(\frac{1}{w} \right) \right|_0^w ##

Substitute our w's out:

##W = mc^2 \left. \left(\frac{1}{~cosθ} \right) \right|_0^θ ##

Recall that since ##1 = ~cos^2θ + ~sin^2θ##, ##~cosθ = \sqrt{1 - ~sin^2θ}##

##W = mc^2 \left. \left(\frac{1}{\sqrt{1 - ~sin^2θ}} \right) \right|_0^θ ##

And then we can re-substitute our u's back in, remembering that we defined ##~sin^2θ = \frac{u^2}{c^2}##:

##W = mc^2 \left. \left(\frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \right) \right|_0^u ##

And finally, when you evaluate the integral, you have:

##W = mc^2 \left(\frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \right) - mc^2 \left(\frac{1}{\sqrt{1 - \frac{0^2}{c^2}}} \right) ##

which is:

##W = mc^2 \left(\frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \right) - mc^2##

So in that equation for work is the term ##E = mc^2##. You can see it by noting that work is the same as the change in kinetic energy, and in this case, they are the same.

So this is your kinetic energy equation. Since total energy is kinetic energy minus rest energy, if we set ##u= 0##, and if we disregard terms like gravitational potential energy (and any energy related to the position of the object with mass m), then there is an additional rest energy term remaining: ##mc^2##:

##KE = mc^2 \left(\frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \right) - mc^2##

##KE + mc^2 = mc^2 \left(\frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \right)##

And since total energy is kinetic energy plus rest energy:

##E_{total} = mc^2 \left(\frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \right)##

And when the object is at rest, that is, at ##u = 0##:

##E_{total_{u = 0}} = mc^2 \left(\frac{1}{\sqrt{1 - \frac{0^2}{c^2}}} \right)##

##E_{total_{u = 0}} = mc^2##

• George Keeling
Gold Member
I did a version of the work-energy thing here a while back that may be of interest:

Is that the binomial theorem you used to expand the gamma factor there?

Edit: I also just noticed you have a gamma cubed factor in your integrals, which is kind of like the one I did above (hidden in the negative 3/2 power). I got to say I find that interesting. Your derivation is very concise.

SiennaTheGr8
Is that the binomial theorem you used to expand the gamma factor there?
Yes.

SiennaTheGr8
• Grasshopper
Which begs the question why does ##E=p^0##?
Instead of starting with Newton's definition of the 3-momentum, you can directly define the four-momentum (in units with ##c=1##) as
##\mathbf P = E \frac{d}{dt}\begin{pmatrix}
t \\
x \\
y \\
z
\end{pmatrix}##
For it's conservation exists experimental evidence, for example in particle storage rings.

and that ##mc^2\gamma## is the total energy of the particle (WHY?)
Reason: because the squared Minkowski-norm of the four-momentum is invariant.
##\mathbf P \cdot \mathbf P = E^2 (1 -v^2) = E_0^2##

It is helpful to understand, that the word "mass" is redundant and therefore needless in SR. However, it is still used for historical reasons, and because the word "mass" is shorter than the equivalent expression "Minkowski-norm of the four-momentum, divided by ##c##".

I'm sure Einstein did better than this. Can somebody point me to a decent proof of ##E=mc^2##? Or even quickly put the proof here?
Einstein's ##E_0=mc^2## proof of 1935:

• George Keeling
Gold Member
First as I'm English, I must congratulate the Italians 😭and apologise for the shameful behaviour of some of our fans . We came second! Next the replies are overwhelming. Two Einstein papers separated by 30 years. The first one is short, simple and has the mass being converting to emitted radiation. That's how it usually happens.
Einstein suggested the theory could be tested on 'radium salts'. As far as I can see it was first tested in 1933. (energy released from the reaction of lithium-7 plus protons giving rise to two alpha particles)

Then there was
##Mc^2 = 2 mc^2 + KE##
just with 2 bits of chewing gum, no calculus and in 10 lines! Brilliant! (although I'm not sure it answered my WHY).

Having established ##E=mc^2## it's fairly obvious why ##p^0## is energy.

Thanks everybody!

ergospherical
1) ##E = \mathbf{p} \cdot \mathbf{v} - L##
2) For a free particle, ##S = -m \int \sqrt{g_{\mu \nu} dx^{\mu} dx^{\nu} }##, therefore $$\delta S = -m\int \dfrac{g_{\mu \nu} dx^{\mu} \delta d x^{\nu}}{\sqrt{g_{\rho \sigma} dx^{\rho} dx^{\sigma} }} = -m \int u_{\mu} \delta dx^{\mu} = \left[ -m u_{\mu} \delta x^{\mu} \right]_a^b + m\int \delta x^{\mu} du_{\mu}$$Consider admissible trajectories (those for which ##\delta S = 0## if ##\delta x \big{|}_a = \delta x \big{|}_b = 0##) where the second integral vanishes. Now keeping ##\delta x \big{|}_a = 0## but varying the coordinates ##\delta x \big{|}_{b} \equiv \delta x## of the endpoint, ##p_{\mu} := mu_{\mu} = -\dfrac{\delta S}{\delta x^{\mu}}## is a four-vector with zeroth component ##p_0 = E## in analogy with classical mechanics.

Last edited:
Mentor
(although I'm not sure it answered my WHY).
These "Why?" questions seldom have good answers, in the sense that you're looking for. We can start with assumptions that we believe are good desriptions of how the universe works, restate them as mathematical formulas, and then we algebraically manipulate these formulas to derive other formulas that aren't so intuitively obvious like the ##E=mc^2## this thread is about... but that's not really an answer to "Why", it's more a discovery about the implications of our initial assumptions.

• Grasshopper