Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I had a special relativity course at university. Now I'm trying to extract what new insight $E=mc^2$ did give us. I mean that moving mass has/is energy (kinetic) not new. The energy merely changed from classical kinetic energy to some relativistic form. And the offset with rest mass doesn't matter for energy.

I'm hoping for some in-depth answer. Given a few hours I could probably recall physics myself, but I suppose someone will have a nicer, more fundamental answer :)

share|improve this question
The main insight was that hot coffee has more mass than cold coffee, and if you squeeze a spring you make it heavier. This was not known before, and it's fundamental as all heck. –  Ron Maimon May 4 '12 at 19:48
add comment

2 Answers

And the offset with rest mass doesn't matter for energy.

Tell it to the relatives of those who were evaporated in Hiroshima and Nagasaki.

The contribution to the energy from the rest mass is the very point of $E=mc^2$. And it matters because the rest mass of objects that exist may be changed, too. In fact, only one conservation law – the total energy i.e. the total mass conservation law – holds. All other processes that are compatible with this single law (plus other laws such as the conservation of momentum, angular momentum, and electric charge) are allowed. This allows the transformation of what we used to call "energy" to things that we used to call "mass" or "matter" and vice versa.

When we use nuclear energy, whether peacefully or less peacefully, we are changing the uranium to some other elements whose rest mass is about 0.1 percent smaller. So we may extract $0.001 mc^2$ of the energy. That makes it enough for a few pounds of a material to evaporate a city or for a few tons of a material to power a nation for quite a long time. Thermonuclear fusion increases this efficiency to 1% or so, a fact that gives us all the useful energy (from the Sun) and that may be reproduced in future thermonuclear fusion plants, too. If we annihilate particles with their antiparticles, we may convert the full $E=mc^2$ where $m$ is the total rest mass into energy (e.g. the kinetic or potential energy of a rocket) and it's a lot of energy.

And on the contrary, the LHC accelerates beams of protons. The kinetic energy of the protons, 4 TeV per proton and 8 TeV for a pair, may be converted to any other form of energy, including the energy stored in the rest mass of new particles. So by the collision of these two protons, or any other particles, we may create 10 new quarks and a Higgs boson or anything else that is allowed by the total energy conservation law.

What we used to call "energy" and what we used to call "mass" is the same thing and the different forms may be transformed to each other much like mechanical work may be transformed to heat and vice versa. It's clearly one of the deepest insights of the 20th century science.

share|improve this answer
It's certainly true. Let me try to draw a conclusion which is more the "fundamental insight" part that I was looking for. That splitting nuclei emits energy isn't surprising either due to the known concept of binding energy. But maybe the actual key point is that binding energy can be measure by the gravitational force? The conclusion I see is that kinetic energy can be converted to real particles. Is that the essence? –  Gerenuk Apr 26 '12 at 11:09
Dear @Gerenuk, yes, the total mass $m$ in $E=mc^2$ is also the mass that appears in Newton's formula for the gravitational force. It's one of the equivalent ways to look at the mass. Another equivalent one is the inertial mass determining the resistance towards acceleration. The inertial and gravitational masses are equal (in sensible units) due to the equivalence principle. It's also true that kinetic energy may be used to produce new particles. All these things are aspects of $E=mc^2$. I don't understand what you mean by the question "is that the essence". –  Luboš Motl Apr 26 '12 at 16:38
Physics is about learning right facts about the physical world and about correct equations relating various observable quantities. Physics is about $E=mc^2$. Choosing one particular insight out of many related facts and calling it "essence" isn't physics, it's a rhetorical exercise and emotional self-brainwashing. Are you looking for some cheap slogans, or are you trying to understand physics? The physical importance of $E=mc^2$ is $E=mc^2$, a relation between two quantities (namely energy and mass) that were previously thought to be unrelated. 1 may and should try to understand what it means –  Luboš Motl Apr 26 '12 at 16:39
You helped me find the answer. In my opinion you just don't get to the point and the "in-depth/fundamental" part was missing. That's as much as I can say. Thanks. –  Gerenuk Apr 27 '12 at 6:02
Also, @Gerenuk: binding energy was an unknown concept before Einstein. It was previously unknown why you had slight differences in the atomic masses per nucleon of different elements/isotopes. So it's wrong to say that Einstein piggybacked on THAT. The history is the other way around--Einstein offered an explanation for binding energy. –  Jerry Schirmer May 4 '12 at 15:09
show 2 more comments

Gerenuk, i'm not sure what you meant by 'in-depth/fundamental' that Lubos' answer didn't cover but let me offer an alternative. The equation shown us that energy and mass are two forms of the same thing. Mass can be converted into energy and the exchange rate for this conversion, is c squared. If by fundamental you are looking for a derivation, Einstein 'stumbled upon' this expression while deriving an invariant formula for momentum in spacetime. When you examine the time component of the energy momentum four vector, E=mc^2+1/2mv^2 pops out

share|improve this answer
Well, not $\frac{1}{2}mv^{2}$, but the jist of your answer is right. Relativistic KE is given by ${\rm KE}=mc^{2}\left(\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}-1\right)$ –  Jerry Schirmer May 4 '12 at 15:11
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.