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As far as I can see, Einstein's $E=mc^2$ is most often mentioned in the context of nuclear physics, even though it is more generally applicable. I understand that this is due to the large nuclear binding energies that are involved.

In what other situations (outside of nuclear physics) is this mass-energy equivalence

  1. relevant?
  2. measurable?
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closed as too broad by Aaron Stevens, FGSUZ, John Duffield, John Rennie, Kyle Kanos May 7 at 11:46

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I'll take measurability first. Particle physics and nuclear physics you already mention. In other areas: modern mass measurements are precise enough to detect the impact of electron binding energies in atoms and molecules. Tests of the equivalence principle are sensitive to many different contributions to the total mass-energy of whatever objects are used. Precision spectroscopy in atoms can detect tiny contributions to the energy levels; many of these are, as we say, 'relativistic corrections', which amounts to saying they are closely related to $E=mc^2$.

Now on relevance. The relationship is relevant whenever speeds reach a significant fraction of the speed of light, or whenever high precision is available and needed. The former includes very hot plasmas in astrophysics and laser physics, as well as nuclear and subatomic particle physics. The latter includes precision measurements in atoms, including the determination of the fine structure constant and the gyromagnetic ratio of the electron. These present precision tests of fundamental physics.

But one could argue that the relation has a wider relevance because it is a central part of the ability of physics to be logically and mathematically consistent.

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  • $\begingroup$ Are you saying that a difference in masses of the reactants and products in a chemical reaction can be measured? $\endgroup$ – M. Enns May 7 at 0:03
  • $\begingroup$ Yes, but only in the most precise mass comparisons that are done in Penning traps. The required precision is around a few parts in $10^{11}$ which is not available in more ordinary methods. $\endgroup$ – Andrew Steane May 7 at 7:58

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