A motivation to develop the Bohr model of the atom was according to Wikipedia:

The laws of classical mechanics (i.e. the Larmor formula), predict that the electron will release electromagnetic radiation while orbiting a nucleus. Because the electron would lose energy, it would rapidly spiral inwards, collapsing into the nucleus on a timescale of around 16 picoseconds. This atom model is disastrous, because it predicts that all atoms are unstable.

To overcome this difficulty Bohr had to postulate that electrons don't radiate in orbits where the angular momentum of the electron is a integer multiple of Plancks constant.

Now lets assume we had an atom of neutral particles holding together only by gravity. If a small light particle orbits a much more heavy one, I believe the light particle would follow a Schwarzschild geodesic. But neutral particles following a Schwarzschild geodesic don't seem to loose energy by radiating gravitational waves. So I think, if we had atoms holding together only by gravity, there would be no need to introduce Bohrs postulates and there would be no need to quantize the angular momentum of the orbits.

Is this really true? Would "gravitational atoms" be stable without quantization?

  • 1
    $\begingroup$ You are ignoring the gravitational field of the orbiting object here. You can only have a static Schwarzschild metric if the orbiting object's mass is negligible. $\endgroup$ – Count Iblis Aug 21 '16 at 20:21
  • $\begingroup$ Do gravitons, if they exist, not already effectively quantize gravity? I don't know enough except to ask the question, I don't have an opinion. $\endgroup$ – user108787 Aug 21 '16 at 20:22

Quantization in gravity is much harder to do in practice but should work like the electromagnetic counterpart. In fact, for weak (escape velocity << c) gravity (we think) it would behave identically to the electromagnetic case where charge is replaced by mass and the electric constant is replaced by the negative of the gravitational constant.

But neutral particles following a Schwarzschild geodesic don't seem to loose energy by radiating gravitational waves.

Yes they do, as long as the mass is non-zero. Accelerating mass emits gravitational waves just as accelerating charge emits electromagnetic waves.

The atoms wouldn't be stable w/o quantization, but since the gravity is very weak it would orbit a long time before decaying unless the mass is very large.

With quantization it can be stabilized, provided the "nucleus" can't trap the particle (analogous to electron capture). This is hard to do. If both are made out of normal matter, Van Der Waals forces, etc will trap the particle, as will a black hole nucleus. A lump of metal and an orbiting neutrino would work but it would be very hard to observe the neutrino's energy level; a self-gravitating neutrino nucleus and orbiting atom would be feasible to measure but very hard to build.

  • $\begingroup$ So there is a Larmor-like formula for gravitational radiation, too? $\endgroup$ – asmaier Aug 21 '16 at 21:11

As pointed out by Count Iblis in the comments, assuming the Schwarzschild metric is not correct in this situation. General relativity predicts that, without external gravitational fields, two bodies orbiting one another will emit gravitational radiation. Thus such a system is not stable in GR. A largescale example of this phenomenon could be seen in the recent observations of gravitational waves coming from merging black holes.

For some quantitative approximations of how fast the system deteriorates see, e.g. https://en.wikipedia.org/wiki/Two-body_problem_in_general_relativity.

  • $\begingroup$ Yes you can, it requires 50 rep and you already have it. $\endgroup$ – peterh Aug 21 '16 at 20:46
  • $\begingroup$ From this very answer apparently... $\endgroup$ – Adomas Baliuka Aug 21 '16 at 20:49
  • $\begingroup$ Yes, but I think your this question is a quite useful answer, so you don't need to convert it to a comment. $\endgroup$ – peterh Aug 21 '16 at 20:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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