Atom's do not seem to emit gravitational waves. But they do contain changing mass quadrupoles, though very small ones. Obviously, the probability for emission of such waves is very small, as the gravitational coupling is very small. But why is there no emission at all?

We can split the discussion into two cases:

(1) The ground state. Atoms in the lowest state do not emit electromagnetic waves because there is no lower state to fall into. Is this also the reason for the lack of gravitational waves? Or is it that the ground state has no changing quadrupole? Or both?

(2) Atoms in excited electromagnetic states. Is the lack of gravitational wave emission (instead or in parallel to electromagnetic emission) just a question of low probabilities? Or is there some other reason?

  • $\begingroup$ Does an atom have a changing quadrupole moment? Surely the eigenfunctions of the Hamiltonian are time independant? $\endgroup$ Aug 23, 2014 at 6:43
  • $\begingroup$ How do you know that atoms do not emit gravitational waves? Did you measure it? $\endgroup$
    – CuriousOne
    Aug 23, 2014 at 14:12

1 Answer 1


The gravitational wave (graviton) emission by atoms is completely analogous to the electromagnetic case except that the relevant observables of the atoms are not dipoles but quadrupoles etc.

The ground state doesn't emit gravitons because of energy conservation: there is no lower-energy state that the atom could fall into after it emits a positive-energy graviton.

Other states can emit gravitons but the probability is tiny – zero for all practical purposes – because the gravitational force between the elementary particles is so much weaker (by 40+ orders of magnitude) than the electromagnetic one. Correspondingly, the lifetimes are longer by this factor, too.

"Changing quadrupole" is a straw man. When we neglect the interaction with the radiation field, all energy eigenstates have time-independent quadrupoles as well as dipoles and all other observables! The dipole and quadrupole (whether it's photon or graviton radiation) are only changing at the moment of the emission – the emission is a "quantized" or "concentrated" process of the change of the dipole or the quadrupole. But before this process takes place, nothing is changing about the multipoles of the atom.

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    $\begingroup$ +1 Nice and clear. By the way congratz for passing the 100k reputation limit ;) oh are you the first one ever? $\endgroup$
    – Ellie
    Aug 23, 2014 at 20:23
  • $\begingroup$ Thanks - only on Physics Stack Exchange, I guess. There are some higher scores at maths and elsewhere. $\endgroup$ Aug 24, 2014 at 4:54

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