In relation to this question: What is potential energy truly?, I'm wondering if $E=mc^2$ has been experimentally verified to hold true for macroscopic objects with increased potential energy? I'm particularly interested in whether the following examples have been tested:

  • Does a macroscopic object at a higher position in a gravitational field have more mass due to the gravitational potential energy?
  • Does a spring weigh more when it is compressed compared with uncompressed?
  • Does a charged object weigh more when it is in an electric field?

If anyone could post links that provide more details on actual experiments that have shown these, that would be great.

EDIT: I've edited the question to try to be more specific about what I am asking. Apologies to those who have posted answers already if it makes your answer seem less relevant.

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    $\begingroup$ I take it that the spectacular demonstration of $E=mc^2$ at Hiroshima is not what you are after... you want a careful weighing experiment. I imagine the best bet is to determine the mass of nuclear fuel rods before & after operation (assuming their casing doesn't corrode). According to this Q&A the aggregate fuel in a 1 GW reactor loses about 1 kg of mass after 4 years of operation. That is measurable. $\endgroup$
    – Floris
    Jan 16, 2015 at 5:21
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    $\begingroup$ The nuclear "mass defect" (the difference in mass between, e.g. 56 moles of hydrogen-1 and 1 mole of iron-56) is evidence of relativity, but that's a bit microscopic. $\endgroup$
    – hobbs
    Jan 16, 2015 at 5:26
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    $\begingroup$ Obligatory comment: Experiments don't verify theories, they fail to falsify them. $\endgroup$
    – ACuriousMind
    Jan 16, 2015 at 18:52
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    $\begingroup$ I started writing an answer, but in a sense it comes down to a comment, so I'll put it here. Your request for a macroscopic example is reasonable on the face of it, but not really reasonable in detail. More than 99% of the mass of "ordinary" macroscopic objects is binding energy rather than the bare mass of fundamental particles, so everything around you is an example. Though I understand that this is probably not a satisfying response. $\endgroup$ Jan 16, 2015 at 21:59
  • $\begingroup$ @dmckee: Thanks for your comment - that's very interesting and I wasn't aware of it. I'm starting to get the impression that the answer to my 'macroscopic' question is probably no, which is fair enough if that's the case. I mean, it makes sense, as the amount of mass increase for compressing a spring must be incredibly small if the energy is divided by $c^2$. $\endgroup$
    – Time4Tea
    Jan 16, 2015 at 23:08

4 Answers 4


The fact that you can't compute the weight of an isotope by adding up the mass of the neutrons, protons, and electrons is because the kinetic and potential energy of the interactions affects the total energy, and hence mass. But you object because that include nuclear interactions.

OK. Look at molecules. The reason a molecule's weight isn't the sum of the weights of the isotopes of the atoms involved is because the the kinetic and potential energy of the interactions affects the total energy, and hence mass. So that counts.

Has this been measured? Yes. Firstly, we can measure the weight of molecules, and we can measure the weight of a macroscopic amount of molecules. Secondly, the energy difference is available as produced heat or required heat. When we measure the heat released or absorbed, we are measuring the energy associated with the change in mass. Just like in a nuclear reactor, except smaller amounts of energy per gram are needed/released since electromagnetic potential energies and kinetic energies are so much smaller.

  • $\begingroup$ Thanks for your answer. It's not that I'm saying that the nuclear/molecular examples don't count - I totally accept that they are valid verifications of $E=mc^2$. It's just that I'm looking specifically for macroscopic examples. $\endgroup$
    – Time4Tea
    Jan 21, 2015 at 18:37
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    $\begingroup$ @Time4Tea You might get a better answer if you made it more clear what counts. A spring is macroscopic, but the energy storied in a spring is caused by microscopic interactions, and if you take almost anything and look at its mass, the mass is much less than the sum of the masses of the leptons and quarks inside it, so almost all energy we attribute to mass (98%+) is really other forms of energy. Gravitational potential energy is another example you mentioned, but I'd have to say that is the trickiest "potential energy" to nail down precisely theoretically. $\endgroup$
    – Timaeus
    Jan 22, 2015 at 1:27

For the Spring

I know the answer for a compressed spring. The compressed spring has higher mass than the uncompressed spring. The stored energy in a compressed spring shows up as somewhat higher electric fields, overall, between the atoms of the spring.

To minimize confusion, consider a small bar of metal made of a single crystal of aluminum atoms. In the crystal, the nuclei and electron clouds of the atoms configure themselves to a lowest energy configuration. The details of this are complex, but for this example we just need to accept that the atoms wind up at a spacing from each other where if they were closer together they would be repelling each other because (primarily) of the positive charges on the nuclei pushing against each other, and if they were further apart they would be pulling towards each other, primarily because the electron cloud of each atom is distorted to attract the positive nucleus of the neighboring atom.

SO now consider compression. We push on the ends of the bar, forcing the atoms of metal to be somewhat closer to each other then in their relaxed state. The atoms push back, and they do it by the extra electrostatic repulsion of the positively charged nuclei. But when you push those nuclei closer to each other you are increasing the net electric field between them.

Electric fields store energy. The energy density of an electric field is proportional to field-strength squared. So the compressed metal, with somewhat higher electric fields because its positively charged ions are closer together than in relaxed state, has its compression energy stored in the higher electric fields!

BUt of course by relativity, that energy density of an electric field means the electric field has mass density $m=E/c^2$.

So the answer for the compressed spring is that in compression, interatomic electric fields are slightly higher than when the spring is relaxed, and the energy of that slightly higher field is the energy that can be recovered by decompressing that spring, and that higher electric field has a mass from its energy density due to relativity. So a compressed spring weighs more than an uncompressed spring because of the mass of its slightly higher interatomic electric fields.

Unfortunately, I can find no reference to someone measuring the mass difference of a compressed spring and an uncompressed spring. The mass of electric and magnetic fields is indirectly verified by the gravitational red-shift or blue-shift of radio waves. This is verified, oscillators on satellites are measured at higher frequency on the earth's surface, and the reason is that as the radio energy goes down to earth it is increased in energy by the earth's gravitational pull on the mass of its electric and magnetic fields. The energy of a quantum of radiation, a photon, is $hf$ where $h$ is Planck's constant and $f$ is the frequency of the radio. The increase in the frequency of the radio wave as it gets to the earth's surface is exactly consistent with the $mgh$ energy of a "falling" photon from satellite height $h$, photon mass of $hf/c^2$, in gravitational pull $g$.

For the Gravitational Potential

Just as the compressed spring has higher electrostatic field in compression than relaxed, moving a small mass away from the earth increases the net gravitational field found by summing the gravitational field of the earth with the gravitational field of the object being raised. Fields add linearly: linear superposition is the solution of the net gravitational field of two objects. But the energy density in the field proportional to field strength squared. So the field configuration with the objects slightly further apart has more energy in it, found by integrating the energy density over space.

Without actually proving it or having been told it authoritatively by someone who did the math, I will assert by analogy with the spring case that this gravitational field energy density must balance, must account for, the higher potential energy of the mass lifted higher above the earth. This would then suggest that the weight of the object when it is at higher height is the same as when it is on the ground, because the extra energy, the potential energy, is not physically located in the object lifted, but in the gravitational field around and in between the object and the earth.

As far as I know, the energy density of a gravitational field is a lot less obvious than the energy density of a electric and magnetic fields. Gravitational radiation is much less a part of our daily lives than the nearly omnipresent electromagnetic radiation, and so experimental evidence that gravitational field has mass when it is in gravitational radiation may not be available.


The energy stored in a spring is associated with higher electric fields inside the lattice of the spring. Those electric fields have a known energy density which by $E=mc^2$ has a known mass density. The mass of electric and magnetic fields is known to be consistent with gravitational "acceleration" of radio waves known as blue shift.

The energy stored in height above a planetary mass it would seem should, by analogy, be stored in the gravitational field. But since Gravity is weak compared to electromagnetic forces, and gravitational radiation is not something we can do much with experimentally, the actual experimental evidence that gravitational potential energy is stored in the gravitational field between and around the objects is harder to come by.

  • $\begingroup$ A conceptually accessible view on gravitational potential energy is that it uses up energy to separate gravitational sources, the potential energy keeps track of the fact that some of that work has already been done. But the separation between two gravitational sources does not for instance show up in the stress-energy tensor, so in that sense it is not like the energy density of the electromagnetic field. $\endgroup$
    – Timaeus
    Jan 22, 2015 at 5:03
  • $\begingroup$ @mwengler: I think your example of the radio waves falling shows that GPE doesn't contribute to mass, because if it did then the frequency wouldn't change as the waves fall (as GPE is converted into KE). $\endgroup$
    – Time4Tea
    Jan 23, 2015 at 3:50
  • $\begingroup$ That would be consistent with what you say in the next section, that the mass from GPE is contained in the gravitational field and not in the object that is raised. $\endgroup$
    – Time4Tea
    Jan 23, 2015 at 3:54
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    $\begingroup$ This doesn't answer the question. The question asks for experimental evidence. $\endgroup$
    – user4552
    Sep 2, 2018 at 13:34

The special theory of relativity has been validated innumerable times in particle physics experiments and nuclear physics experiments. Macroscopic verification of the special theory of relativity also exist.

I will quote from this article, "Einstein's relativity and everyday life"

But what about Einstein's theories of special and general relativity? One could hardly imagine a branch of fundamental physics less likely to have practical consequences. But strangely enough, relativity plays a key role in a multi-billion dollar growth industry centered around the Global Positioning System (GPS).

The system is based on an array of 24 satellites orbiting the earth, each carrying a precise atomic clock


The satellite clocks are moving at 14,000 km/hr in orbits that circle the Earth twice per day, much faster than clocks on the surface of the Earth, and Einstein's theory of special relativity says that rapidly moving clocks tick more slowly, by about seven microseconds (millionths of a second) per day.

The Lorenz transformations are what defines special relativity, and have to be used.

Also, the orbiting clocks are 20,000 km above the Earth, and experience gravity that is four times weaker than that on the ground. Einstein's general relativity theory says that gravity curves space and time, resulting in a tendency for the orbiting clocks to tick slightly faster, by about 45 microseconds per day. The net result is that time on a GPS satellite clock advances faster than a clock on the ground by about 38 microseconds per day.

But at 38 microseconds per day, the relativistic offset in the rates of the satellite clocks is so large that, if left uncompensated, it would cause navigational errors that accumulate faster than 10 km per day! GPS accounts for relativity by electronically adjusting the rates of the satellite clocks, and by building mathematical corrections into the computer chips which solve for the user's location. Without the proper application of relativity, GPS would fail in its navigational functions within about 2 minutes.

The fact that the theory works to this accuracy validates it, and this includes the E=mc^2 which is part of special and general relativity.

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    $\begingroup$ Thanks for your answer anna v, although I'm really looking for something more direct. You're saying that special relativity has been validated, therefore $E=mc^2$ has been validated (which I totally accept). However, what I'm looking for are experiments that have been done on macroscopic objects that demonstrate that if they have more energy, they have more mass. $\endgroup$
    – Time4Tea
    Jan 16, 2015 at 19:16
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    $\begingroup$ If they did not have the mass expected from the special relativity, their velocity would not fit the final equations that give the corrections for the GPS. In this sense they are measured. $\endgroup$
    – anna v
    Jan 16, 2015 at 19:26
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    $\begingroup$ the whole article is describing how they cannot neglect special relativity for GPS. the theory is a whole, not a la carte $\endgroup$
    – anna v
    Jan 17, 2015 at 15:24
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    $\begingroup$ The way one calculates relativistic effects one uses the rest mass of the involved bodies, which does not change. To find the famous E=m*c^2 one has to use the Lorentz transformatios, which evidently they do as they compute the time delay. The same Lorenz transformations will give the m in the famous formula. It is not a useful variable for relativistic calculations. $\endgroup$
    – anna v
    Jan 17, 2015 at 16:54
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    $\begingroup$ I would really appreciate if the people down voting would explain the downvote. As an experimentalist using continuously the algebra of relativity and trusting that since it has been validated on one part of an experiment the Lorenz transformation describes the kinematics of the rest of the constituents I cannot see where my answer is so wrong to deserve three downvotes. $\endgroup$
    – anna v
    Jan 21, 2015 at 18:56

One can offer a simple logic to clarify the relation between rest mass, relativistc mass, energy and momentum and connecton to E=m c^2. The essence of it has been told by many, but not all in one place unfortuanely.see http://geocalc.clas.asu.edu/pdf/ZBW_I_QM.pdf. https://arxiv.org/pdf/1602.07534.pdf and I have many others.

The origin of everything is the EM field. This field has mechanical attributes, in the form of momentum, as you can rotate a small turbine placed in an evacuated jar using radiation. And it has energy, as it can go on and evaporate the blades by cranking up the intensity. In any case, energy is derived from momentum. Radiation has an electric attributes as we know, in the form of an electric field, and a magnetic field that is a direct consequence of it.

An energy beam does not have a rest mass, but a trapped beam between two mirrors- as in lasers, has a rest-mass. Hence one can conclude that rest mass is nothing more than a trapped EM field. The trapped momentum part gives rise to mass- as momentum is conserved, leading to gravitation and the inverse square forces (as shown in Bertrand theorem). This is also supported by pair creation and annihilation events.

The trapping of a field can be between two mirrors, but can also be in going round infinitely in circles- as in an electron. It is self trapping in this case. The resulting circulating field produces spin, magnetic dipole moment, and it produces no hard core, again as in the case for an electron. But you can have double trapping too- as in the case of the proton, wherein masses(trapped energy) have a very large kinetic energy, and the lot are further trapped in one structure forming the proton or the neutron. All composite particles follow the same logic.

The momentum for radiation is E=pc, and for a particle E=m ingeral(v.dv)=.5 m v^2. Put v=c, and get E=m c^2, with the factor 2 disappearing, as we need two opposite radiation beams to produce a trapped radiation, and a resulting rest, orzero momentum of the product. The total energy of a composite/sum particle would be that of radiation and mass giving by the more general Einstein formula; E^2=(pc)^2 + (mc^2)^2. The momentum direction inside a circulating wave is tangential and thus normal to the direction of motion of the particle. This is the reason for adding the energy vectorially and the square on the terms.. because the momenta from which the energies are obtained is orhogonal too.

Therefore, we can say that all measurement of pair production and annihilation processes and all those for particle creation and disintigration, are proofs that the formula E=m c^2 is exact and accurate.

  • $\begingroup$ Why are you using the non-relativistic formula for kinetic energy, $.5mv^2$? Also, you have some mistakes in the total energy formula. It should be $E^2=(pc)^2+(mc^2)^2$ $\endgroup$
    – PM 2Ring
    Sep 3, 2018 at 12:28
  • $\begingroup$ you can rotate a small turbine placed in an evacuated jar using radiation Do you mean the Crookes radiometer? That does not demonstrate the momentum of light. $\endgroup$
    – PM 2Ring
    Sep 3, 2018 at 23:31
  • $\begingroup$ PM; numerous experiments show light has momentum. see this; physlink.com/education/askexperts/ae694.cfm '' In their paper, La Porta and Wang use the angular momentum created by circular polarized light to torque a quartz bead. Not only can they apply torque to the bead, but, by conservation of angular momentum, the bead can apply a torque to the light, which is detected by a change in the circular polarization. '' $\endgroup$
    – Riad
    Sep 5, 2018 at 14:02
  • $\begingroup$ PM; I was writing from memory regarding the turbine blades and the momentum of light. You are right, See this quote; en.m.wikipedia.org/wiki/Nichols_radiometer '' the experimenters were able to obtain an agreement between observed and computed radiation pressures within about 0.6%. The original apparatus is at the Smithsonian Institution.[1] This apparatus is sometimes confused with the Crookes radiometer of 1873. '' $\endgroup$
    – Riad
    Sep 5, 2018 at 14:46
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    $\begingroup$ You can get tiny spheres to rotate very fast using lasers, eg nature.com/articles/ncomms3374 and google.com/amp/s/phys.org/news/… $\endgroup$
    – PM 2Ring
    Sep 6, 2018 at 14:12

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