We know that kinetic energy is (or has) mass in special relativity.

Imagine the following scenario: Two point masses are measured to have rest masses $m_1$ and $m_2$ respectively when not in each other's influence. They are then placed at $r_0$ distance from each other. Then if we let them go, they accelerate towards each other increasing their kinetic energy. Since kinetic energy is mass, this means mass came out of the potential energy of the system. So, this means that potential energy is also mass.

Using this, I reached the following hypothesis:

Rest mass of a system of particles = Sum of rest masses of its constituents + Sum of kinetic energies of the constituents/c^2 + some function of the distances between pairs of constituent particles.

To me, this is crazy that mass should depend on distance between interacting particles. However, what is more crazy is that potential energy function, unlike mass, doesn't care whether we add a constant to it.

So, here are my questions:

  • Is the above blockquoted hypothesis correct?

  • If yes, then out of the infinite potential energy functions differing by constants, which one when divided by c^2 adds to the mass of the system?

  • $\begingroup$ "We know that a moving particle weighs heavier than a stationary one" See the answers under physics.stackexchange.com/q/133376 (and links therein). That is a valid interpretation of the theory, but one that is widely discouraged but regular practitioners. $\endgroup$ Commented Oct 17, 2017 at 4:05
  • $\begingroup$ Oddly enough, the way you have formed the blockquoted hypothesis is the way people would talk in the 'modern' parlance: with kinetic energy separated from the (invariant) mass of the particles. $\endgroup$ Commented Oct 17, 2017 at 4:07
  • $\begingroup$ Related: physics.stackexchange.com/q/66359 $\endgroup$ Commented Oct 17, 2017 at 4:34
  • $\begingroup$ Answered here: physics.stackexchange.com/q/356029 - Your hypothesis is incorrect. $\endgroup$
    – safesphere
    Commented Oct 17, 2017 at 4:53
  • $\begingroup$ @dmckee I think you are missing the point. I understand this confusion of invariant and variant mass in relativity. But that is, I think, only a confusion of convention. I think my question makes sense with either of the conventions. I am editing my question a bit to try to do just that. $\endgroup$ Commented Oct 17, 2017 at 8:17

1 Answer 1


This answer is basically a special relativity answer, and things get a bit more complicated when general relativity rears it head. You have to worry about whether you can even rely on energy conservation at all, for one (answer: in some space-times you can for carefully chosen frames of reference).

Taking as the definition of (invariant) mass for a particle in the usual manner as the modern parlance as the norm of the energy-momentum four vector divided by $c^2$: \begin{align} m &\equiv \frac{1}{c^2} ||\mathbf{p}|| \\ &= \frac{1}{c^2} \sqrt{ E^2 - (\vec{p}c)^2 } \;, \end{align} and noting that four-vectors simply add up across systems we define the mass of a system as the norm of the systems total energy-momentum four-vector divided by $c^2$ just as for a particle. \begin{align} M_{sys} &\equiv \frac{1}{c^2} ||\mathbf{P_{sys}}|| \\ &= \frac{1}{c^2} \sqrt{ E_{sys}^2 - (\vec{P}_{sys}c)^2 } \;. \end{align} This is obviously also a Lorentz scalar.

Now, as usual the energy of the system is found by adding up all the contributions from particles and from fields $$ \mathbf{P}_{sys} = \sum_i \mathbf{p}_i $$ (which accounts for the potential energies of the system both positive and negative) as well as any kinetic energy of the constituents relative the center-of-momentum frame of the system. (Motion of the CoM frame does not affect the mass of the system any more than it affect the mass of a single particle and for the same reason.) In particular this answers your question about the blockquote definition in the positive sense if "some function of the distances" is taken as a potential energy function.

A consequence of this is that the mass of a system is not the sum of the masses of it's parts because $||\mathbf{a}|| + ||\mathbf{b}|| \ne ||(\mathbf{a} + \mathbf{b})||$.

You question about the gauge of the potential is important because it is not true that absolute energy level doesn't matter if the mass of the system depends on it. But I've already shown you the way by discussing the energy of the field: we want the change in energy to go to zero if the parts don't interact, which leads to the Coulomb gauge for E&M. (Note that there is still field energy hanging about, it's just not separable from the basic mass associated with the charged particles in the first place.)

A (desirable!) consequence of this definition is that a system of particles orbiting one another keep a constant mass as they trade kinetic for potential energy and back again. Another consequence is that even though we are using the modern parlance and say emphatically that the atoms of a gas don't gain mass as you heat it, the gas taken as a system does gain mass proportional to the thermal energy added.

  • $\begingroup$ So the answer is that there are special constants to chose for the potential energy function (eg Coulomb gauge for electromagnetic potential energy), such that it describes the mass of the composite system? $\endgroup$ Commented Oct 17, 2017 at 9:46
  • $\begingroup$ What is the Coloumb gauge $\endgroup$ Commented Oct 17, 2017 at 9:47

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