According to relativity, the inertial mass of an object that enters into the force equation to get acceleration is given by the total energy in the center of mass rest frame.

$$E=Mc^2=\sum_i (m_i c^2 + KE_i +U_i)$$

In other words, the inertial mass that is involved in gravity has a dependence on the energy/momentum of the constituents.


Outside of nuclear objects, usually only the first two terms are significant, so the binding energy can be neglected for practical purposes. But I am interested in its impact for precision measurements.

For example, Hydrogen has a binding energy of 13.6 eV, but a rest mass dominated by the proton of 932 MeV, so a few parts in $10^8$.

In principle, if a measurement was sensitive enough, then the mass could be measured by allowing the hydrogen to fall under the influence of gravity and measuring its acceleration. I am guessing $10^8$ sensitivity with a measurement like this is not possible currently, but would be glad to be wrong.

My question is the following: how do we understand the interplay between the excited states (electronic or otherwise) with forces like gravity that depend on mass? For example, what $M$ should be put in the gravitational interaction $-\frac{GmM}{r}$ classically or within the Dirac/Schrodinger equation? My guess would be that it would have to be promoted to an operator of some kind for everything to work out.

$$M\implies \hat{M}=\hat{M}(\hat{k}_i,...,\hat{k}_j)$$ or $$H_{\textrm{grav.}}=-\frac{GmM}{r} \implies -\frac{Gm\hat{M}}{r}= -\frac{Gm\hat{H}_{\textrm{EM}}/c^2}{ r}$$


1 Answer 1


The $m$ used in $G Mm\over r^2$ is absolutely the same as the $m$ in $E=mc^2$. The force on a hydrogen atom in its ground state is not the same as the force on a hydrogen atom in an excited state, and the mass on an uncombined proton+electron is different again. Only, as you observe, by a few parts in $10^8$, but in principle the difference is there - and ensures that they will all fall with the same acceleration.

When you have a system with interacting components (proton+electron, or something more complicated) then you can draw a conceptual line around it and say that whatever these constituents get up to in their interactions, exchanging energies and momenta, creating photons, even creating particle-antiparticle pairs, then provided everything stays within the system, the total mass and thus the total inertia and the force of gravity are not affected. The mass of the system may be made up from kinetic and potential energies and from rest masses but details of the division don't matter; only the total matters.

  • $\begingroup$ Thanks! Any comments on the quantum part of the question? It seems like the gravitational term depends on the full non-gravity hamiltonian, does that have any interesting consequences? $\endgroup$
    – KF Gauss
    Feb 20, 2019 at 12:54
  • $\begingroup$ If there are consequences they're not very interesting, as far as I can tell. The energy of the atom may well described by a simple classical formula, or better described by a fuller quantum formula, but to Newtonian gravity it's just a number in $GMm\over r^2$. $\endgroup$ Feb 20, 2019 at 13:27
  • $\begingroup$ I think you've convinced me that classically this effect ia boring. But quantum mechanically it seems like the M's become dynamical, and if both objects are quantum mechanical then the term is a product of their two Hamiltonians $\endgroup$
    – KF Gauss
    Feb 21, 2019 at 14:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.