(One of examples that potential energy contributes to mass.)

Does hot object weigh more than cold one?

(One of examples that kinetic energy contributes to mass.)

If these are true and justified by mass-energy equivalence principle, could the underlying principle be justified?

I've learned and derived $E=mc^2$ but only for the case that $m$ is interpreted as rest one. Could I generalize this argument to include kinetic and potential energy? :

$$mc^2=E =∑m_0c^2+∑K+V=∑\gamma m_0c^2+V$$

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    $\begingroup$ I'm sure this has been asked before, though searching the site for it is proving frustrating. I have found the question Has $E=mc^2$ been experimentally verified for macroscopic objects with potential energy? which includes a question on spring mass change due to compression. $\endgroup$ – John Rennie Apr 26 '16 at 17:49
  • $\begingroup$ Could I justify the last equation through any argument? I couldn't find the answer for my main question. $\endgroup$ – Sophomore_Jinx Apr 26 '16 at 17:57
  • $\begingroup$ As soon as you talk about how much something weighs you're using gravity, and to incorporate relativistic effects you need general relativity, not just special relativity. $\endgroup$ – tparker Apr 26 '16 at 18:22
  • $\begingroup$ Related: physics.stackexchange.com/q/159734/2451 $\endgroup$ – Qmechanic Apr 26 '16 at 18:24
  • $\begingroup$ I'll repeat a comment I made on 159688: most of the mass of 'ordinary' stuff around you is binding energy. It's not that you have to try hard to work this out, it's that you've never weighed anything without this effect dominating. $\endgroup$ – dmckee --- ex-moderator kitten Apr 26 '16 at 21:09

In relativity we only use the rest mass, also known as the invariant mass, of an object. In days past the concept of a relativistic mass was used, but this is now strongly deprecated as it has caused endless confusion. For example an obvious question is whether the increase of relativistic mass with speed can cause an object to become a black hole (tl;dr it cannot).

The rest mass does not include contributions from kinetic energy or potential energy, but it does include contributions from changes in internal energy. In the case of your spring suppose it start off with some mass $m_0$ corresponding to an energy $E=m_0c^2$. If we compress the spring we are doing work on it - the work done is $W = \tfrac{1}{2}kx^2$ where $k$ is the spring constant and $x$ is the distance we've compressed the spring. The internal energy of the spring has increased by $\tfrac{1}{2}kx^2$ and that means its mass has increased by $W = \tfrac{1}{2}kx^2/c^2$.

Anything that changes the internal energy of an object, e.g. heating and cooling it as well as doing work on it, will change the rest mass, though usually by too small an amount to be detectable.

The rest mass is useful because it is a Lorentz scalar, that is all observers will agree on its value. However kinetic energy is not an invariant because obviously it depends on the relative speed of the observer and the object. Potential energy is ill defined because it has a global gauge symmetry i.e. we can choose any value we want to use as the zero for our potential energy.


Yes it will do. You can take the potential energy to be zero when the spring is neither compressed or stretched. In special relativity the total invariant mass of the system would then include a contribution from the potential energy / $c^2$. The concept of mass in general relativity is quite subtle, but for weak gravitational fields, the Newtonian limit for nearly flat spacetimes can be used and so you also just add the spring potential energy / $c^2$: https://en.wikipedia.org/wiki/Mass_in_general_relativity#The_Newtonian_limit_for_nearly_flat_space-times


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