# Impact of electronic binding energies on mass

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.

$$M=M(k_i,...,k_n)$$

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 $$H_{\textrm{grav.}}=-\frac{GmM}{r} \implies -\frac{Gm\hat{M}}{r}= -\frac{Gm\hat{H}_{\textrm{EM}}/c^2}{ r}$$

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.
• 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 $GMm\over r^2$. – RogerJBarlow Feb 20 at 13:27