Take the hydrogen atom. It is easy to imagine that the gravitational pull it creates is smaller than the sum of those of the proton plus the electron, because a photon of 13.6 eV was created when the atom assembled, and has left the system. In other words, the binding energy between the proton and the electron has to be subtracted to the mass of the system.

But where does this appear in general relativity? How does it fit in the stress-energy tensor? I know how to define this tensor for one particle, or for an electromagnetic field, but I don't see how the binding energy fits.


1 Answer 1


Generally we just fudge the issue and ignore the binding energy. It helps that gravity is negligable on the subatomic scale$^1$ so this isn't a problem. We just take measured mass of the (in this case) hydrogen atom and bung it in $T_{00}$.

If you start worrying about binging energies you risk walking the path to madness. The majority of an atom's mass is in the nucleus, and 99% of the mass of a nucleon is the QCD binding energy. So if you start worrying about how to describe the electrostatic binding energy in an atom you have a far worse problem trying to describe the binding energy in a nucleon.

Of possible interest is that the kinetic energy of subatomic particles should also go into the stress-energy tensor because it will produce non-zero contributions to all the entries in the stress-energy tensor. I note someone has even put a paper on the Arxiv about this.

$^1$ I'm assuming here that the LHC hasn't produced any black holes.

  • $\begingroup$ I'm assuming here that the LHC hasn't produced any black holes Lubos says we couldn't even tell if it did. $\endgroup$
    – Kyle Kanos
    Commented Nov 13, 2015 at 18:21
  • $\begingroup$ @John, Does that mean we don't know how to include binding energies? Or simply that it is too complex to calculate? $\endgroup$
    – fffred
    Commented Nov 14, 2015 at 8:18
  • $\begingroup$ @fffred: the stress-energy tensor is (in principle at least) built up from an assemblage of point particles. The momentum flow and pressure elements have a simple interpretation as the momenta of the point particles. I suppose you could try and represent the binding energy using the virtual quanta that quantum field theory describes, though whether this is physically reasonable I'm not sure. In any case it would be prohibitively complicated ... $\endgroup$ Commented Nov 14, 2015 at 8:29
  • $\begingroup$ ... and in any case must reproduce the macroscopic mass deficit, so why bother. $\endgroup$ Commented Nov 14, 2015 at 8:29
  • $\begingroup$ @John, Does it mean these virtual quanta have negative mass? $\endgroup$
    – fffred
    Commented Nov 14, 2015 at 9:46

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