I've been told things such as the sum of the rest masses times c^2, or sum of the relativistic masses times c^2 (I forgot which), were actual equal to things such as the enthalpy of the system rather than the energy.

For a system of $n$ particles, having respective relativistic masses $m_0$ and the entire system having rest mass $M_0$, is it true $$\sum_{k=1}^n m_k = M_0 \\$$


Short answer: no.

The mass1 of a system depends on the masses of the constituents, their internal motions, and the interactions between them.

Further, is it also not true that the mass of the system is the sum of the so-called "relativistic masses" of the constituents $$ M \ne \sum_i \gamma_i m_i \;,$$ excepting cases where there is zero net potential energy in the system.

Explanation: The invariant mass of any object or system can be found by as the square of the energy-momentum four-vector (times appropriate factors of $c$) \begin{align} m^2 &= \frac{1}{c^4}\mathbf{p}^2\\ &= \frac{1}{c^4}\left(E^2 - (\vec{p}c)^2\right) \;, \end{align} which is a computation that comes out to the same value for all observer leading to the "invariant" appellation. The four-momentum of a system is equal to the sum of the four-momentum of the parts. $$ \mathbf{P} = \sum_i \mathrm{p}_i \;.$$ So expanding $M^2c^4 = \mathrm{P}^2$ we get \begin{align} M^2 &= \frac{1}{c^4} \left( \sum_i \mathbf{p}_i \right)^2\\ &= \frac{1}{c^4} \left[ \left( \sum_i E_i\right)^2 - \left( \sum_i \vec{p}_i c\right)^2 \right] \\ &= \frac{1}{c^4} \left[ \left( \sum_i E_i^2 \right) + 2\left( \sum_i \sum_{j>i} E_i E_j \right) - \left( \sum_i \vec{p}_i^2 c^2 \right) - 2\left( \sum_i \sum_{j>i} \vec{p}_i \cdot \vec{p}_j c^2\right) \right]\\ &= \frac{1}{c^4} \left[ \left( \sum_i E_i^2 - \vec{p}^2_i c^2 \right) \right] + \frac{1}{c^4}\left[ 2\left(\sum_i \sum_{j>i} E_i E_j - \vec{p}_i \cdot \vec{p}_j c^2\right)\right]\\ &= \left( \sum_i m_i^2 \right) + \frac{1}{c^4}\left[ 2\left(\sum_i \sum_{j>i} E_i E_j - \vec{p}_i \cdot \vec{p}_j c^2\right)\right] \end{align} Now that last line is equivalent to $( \sum_i m_i)^2$ only if the long term with the double sum is equal to $2 \sum_i \sum_{j>i} m_i m_2$, which is not generally the case.

Example cases: Because the factor $c^2$ is so absurdly large when expressed in any units suitable for day-to-day physics, the quantities of (kinetic or potential) energy that we encounter on a regular basis are two small to show up easily. But nuclear interaction involve large enough energies to make a measurable difference to the masses; if you have a good enough system for measuring mass (which we do in the form of mass spectrometers).

The capture of a neutron by a proton to make a deuteron is an example of a case where the resulting system is smaller than the sum of the parts (because about $2.2\,\mathrm{MeV}$ escapes as photons). \begin{align} m_p &= 1.6726 \times 10^{-27}\,\mathrm{kg}\\ m_n &= 1.6749 \times 10^{-27}\,\mathrm{kg}\\ m_\mathrm{D} &= 3.3436 \times 10^{-27}\,\mathrm{kg}\\ &< m_p + m_n \;. \end{align}

And the fission of large nuclei into two lighter nuclei and a small number of neutrons is a case where the system is larger than the mass of the bits. But I will leave running down the high-precision mass measurements to for some particular case to you.

1 This answer uses the language and notation of invariant masses. The thing called "rest mass" in the older language is simply called "mass" herein, and there is no equivalent name for "relativistic mass" at all.

  • $\begingroup$ It isn't $m^2 = (E^2-(pc)^2)/c^4$? The units don't seem to match otherwise $\endgroup$ – redsunx Feb 22 '18 at 19:49
  • $\begingroup$ @MCUrist Yes. Of course. Don't know what I was thinking. $\endgroup$ – dmckee --- ex-moderator kitten Feb 22 '18 at 21:06
  • $\begingroup$ OK. That is partially fixed the argument needs to be cleaned up and finalized still, but I have to go teach a class. $\endgroup$ – dmckee --- ex-moderator kitten Feb 22 '18 at 22:29

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