What experiments have been done that confirm $E=mc^2$?

Are there experimental results that contradict $E=mc^2$?

Or are experimental results consistently showing this famous formula to be true?

  • $\begingroup$ Does E=mc^2 represent anything real without specifying "at rest" or adding the momentum term? $\endgroup$ – Jitter Sep 30 '15 at 12:56
  • $\begingroup$ The formula is part of the special relativity conditions and special relativity has been validated innumerable number of times in particle physics experiments. $\endgroup$ – anna v Sep 30 '15 at 15:04
  • $\begingroup$ @annav Which "particle physics experiments"? $\endgroup$ – user2800708 Sep 30 '15 at 15:06
  • $\begingroup$ All the experiments that give the data in the particle data book pdg.lbl.gov are based on the validity of the Lorenz transformations , from the decay of the muon to the discovery of the Higgs. $\endgroup$ – anna v Sep 30 '15 at 15:29

There is a huge mass of experimental evidence that confirms the mass-energy equivalence. The clearest example that it happens is, of course, nuclear power, both in its explosive and civilian forms. If you want a detailed breakdown of all the experiments that corroborate it, I would recommend the entire archive of Physical Review C or a similar nuclear-physics journal.

If you want a clear-cut example, I think the best way to see things is via something called the mass defect. In essence, you weigh all the different nuclei and you divide their mass by the number of nucleons they contain. Somewhat surprisingly, the resulting mass average is not constant: a helium atom is less massive (by about $7\:\mathrm{MeV}·c^2$) than two protons and two neutrons out by themselves, so we say that helium has a mass defect of $7\:\mathrm{MeV}·c^2$, or a binding energy of $7\:\mathrm{MeV}$.

Binding energy curve for known nuclides.

Image source

In a huge number of nuclear reaction experiments, we induce nuclear changes so that e.g. one cobalt nucleus turns into something else. We weigh all the ingredients and all the outputs, which are typically lighter than the ingredients, and we also measure the energy of all the outgoing particles. And, indeed, the extra energy in the output has always matched the mass difference between input and output - there are no observed deviations from this rule.


Emilio's answer was also the first that came to my mind, but I was not quick enough to post. However, even more precise experiments come from particle accelerators, and similar devices.


The power of the magnets in the LHC is determined by the relativistic mass of the particles going through it, which is over 7000 times the mass these particles have at rest. If $e=mc^2$ was wrong, and they underestimated the mass, the particles would hit the outer wall of the accelerator.


To answer the second question:

I am not aware of any experiment refuting $e=mc^2$. There may be problems near or in a black hole, where relativity slams into quantum stuff.


  • 2
    $\begingroup$ Great wikipedia page $\endgroup$ – Gremlin Sep 30 '15 at 11:45
  • 2
    $\begingroup$ The needs of the bending magnets for relativistic beams are more consistently stated in terms of momentum. This has the added advantage of letting you stop using the unnecessarily confusing "relativisitc mass". Serious scientist don't put the $\gamma$ with the $m$ and give that a name anymore. Not that you can't do science with the term and the idea, but it doesn't add anything useful and generally leads beginners astray. For most particle physicists, "mass" always means the invariant mass (AKA the rest mass). $\endgroup$ – dmckee Sep 30 '15 at 15:26

Another set of experiments which support $E=mc^2$ are Compton scattering experiments. The mass-energy of the electron is an important quantity in analyzing these events, and the results are consistent across a wide range of energies for the primary photon and scattering angles. The energy of the secondary photon is given by $$E_{\gamma '}= \frac{E_{\gamma}}{1+\left(\frac{E_{\gamma}}{m_ec^2} \right)\left(1-\cos\theta\right)} $$

EDIT: Here's a link to the Wikipedia version of the derivation, although any modern physics or nuclear physics text will have it as well.

  • $\begingroup$ Might be useful to include a link to a page showing the calculuation, if you don't want to provide it here. $\endgroup$ – Kyle Kanos Sep 30 '15 at 17:59
  • $\begingroup$ @KyleKanos Done. $\endgroup$ – Bill N Sep 30 '15 at 19:49

protected by Qmechanic Sep 30 '15 at 19:59

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