I read that most of the "mass" in the proton was actually energy from the quarks and gluons, as opposed to the actual mass which was coupled to the Higgs field. This made me start thinking about objects on macroscopic scales and the effects of energy and mass in a closed macroscopic system.

I have read vigorous debates and arguments about the significance of the mass-energy equivalence, and what it means in terms of the "weight" of an object.

So my question is that if said atomic bomb were to explode inside of a completely impenetrable container of some sort, would the measured weight of the container change, assuming no radiation was able to escape the container itself?

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    $\begingroup$ Nice container. Can I get one? Anyway, the short answer is no. The weight measured would be the same. The energy content of the container remains the same, hence the coupling to gravity (the weight) remains the same as well. I'm not sure where the debates are coming from. The physics is unambiguous: it is the energy content of the box (+ the gravitational field of the Earth) that determines the weight. All energy gravitates. It matters not one jot whether you think the energy inside counts as the rest mass of some particle(s). $\endgroup$
    – Michael
    Commented Jul 2, 2013 at 8:27
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    $\begingroup$ The box would indeed be impressive! No matter how small your bomb is, the box will need to shield neutrinos, which even for the best materials you can find makes it at least a few light-years thick. Those minor complications aside, the answer is pretty straightforward. $\endgroup$ Commented Jul 2, 2013 at 9:37
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    $\begingroup$ The company you bought the container from cheated you. And so did the company you brought the atomic bomb from. $\endgroup$ Commented Jul 3, 2013 at 5:57
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    $\begingroup$ Is the box also perfectly insulated? if yes, then the weight doesn't change. If it eventually leaks out all of the heat of the explosion, it will definitely have less mass after it cools down to room temperature. $\endgroup$ Commented Aug 14, 2014 at 17:59
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    $\begingroup$ Alternative version of the question, to stop it being about the meaning of "unbreakable": if an atomic bomb explodes inside the event horizon of a black hole, does the mass of the black hole change? (As a consequence of the explosion: I don't mean the simultaneous Hawking radiation that would have happened anyway...) $\endgroup$ Commented Dec 2, 2014 at 3:09

2 Answers 2


No, it wouldn't change due to the energy conservation law or mass conservation law – these are the same laws in relativity where $E=mc^2$. The nuclear energy extracted from the nuclei by fission would be converted to other forms of energy inside the box – ultimately the thermal energy.

High temperature means that particles are moving at higher speeds and they also have a higher mass, due to the relativistic dependence of mass on the speed. This increase of the mass due to the higher velocity/temperature of the particles would exactly compensate the total decrease of the rest mass.

Inside the box, the gravitational field would be different because in general relativity, it is not only the mass/energy density but also the pressure – all the components of the stress-energy tensor – that influence the local gravitational field. However, assuming that the box would be perfectly incompressible etc., the gravitational field outside the box would be fully dictated by the total mass of the box, and not the details about the pressure inside the cavity, so it wouldn't change. Because of the equivalence principle, the mass measured by a scale and/or the resistance towards acceleration would remain unchanged, too.

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    $\begingroup$ Does this mean that the radiation inside the box, despite its "form" being either mass or energy, contributes to the gravitational field which then influences the scale? So all the radiation that is left in the box, and all the radiation that goes into processes like heating the interior walls of the box etc. will all contribute the same amount of "gravity", as before the explosion? $\endgroup$ Commented Jul 2, 2013 at 8:46
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    $\begingroup$ Yes, as I hopefully wrote, radiation in the box does contribute the same to the gravity. Radiation has nonzero stress-energy tensor (momentum and energy density and flux) and the nonzero stress-energy tensor is what causes gravity in GR. $\endgroup$ Commented Jul 7, 2013 at 5:50

I agree to many points mentioned in the previous post but the answer is: "Yes, the weight(mass) would change."

My reasons are simple. You state that an unbreakable container is being used, so that the products of the explosion will be contained inside. However unbreakable this container may be it will not be able to contain all of the radiation produced in the blast. Alpha and beta radiation are likely to be contained but not so the gamma rays and perhaps other form of radiation. If the container gets hot, it will emit a broad spectrum of electromagnetic waves, infra red, and perhaps visible light (it might be glowing red hot). As all these kinds of radiation carry energy away, the mass (and weight) will change according to the mentioned mass-energy equivalence E=m*c^2.

Another loss would be neutrinos produced during the radioactive decay. Neutrinos are small elementary particles that only weekly with other matter. No container can contain them. Every second huge numbers of neutrinos produced by the sun flow through the earth, flow through every person on the planet, and only a tiny percentage of them interacts with the cores of atoms in their path.

Those neutrinos have a tiny mass and would change the weight.

Your container would not only have to be unbreakable, but also impenetrable to radiation and neutrinos and on top of that, it should not conduct any heat (which is practically impossible). In this improbable case the "weight" would not change.

Good luck with it. ;-)

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    $\begingroup$ I think it's pretty clear from his question that this is a thought experiment alone, designed to understand the relationship between energy and gravity. I'm not sure why everyone's focusing on the irrelevant "engineering" problem of designing a box. $\endgroup$
    – Nick
    Commented Aug 14, 2014 at 17:45

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