Consider a binary star system, as these stars go around one another they would emit gravitational waves. Since, the graviton is a spin 2 particle. Wouldn't the angular momentum of the stars decrease? And would this decrease be observable?


although the angular momentum change was predicted by GR, could there also potentially be an extra angular momentum effect caused by the quantum mechanical spin 2 effect of the gravitons (which could be looked for for a proof of quantum gravity

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    $\begingroup$ Dear Anant: After your edit, you clearly you had in mind a different, and highly specific, technical question from the one I answered. Also, I added the QG tag and suggest you think about a title like "Is all the angular momentum loss in Hulse-Taylor like star systems accounted for by GTR?" - this might help draw the right kind of expertise to your very interesting question (which, unfortunately, I can't answer but would love to see an answer to). $\endgroup$ – WetSavannaAnimal Apr 14 '14 at 23:09

Yes indeed to all your questions: mutually orbitting binaries do spin down, the system's orbital angular momentum thus decreases with time and the loss of energy and angular momentum is almost certainly owing to the emission of gravitational waves.

Look up the Hulse-Taylor binary system: its spin-down has been carefully observed and measured since its discovery in 1974 and the observed spin down carefully compared with the spindown foretold by General Relativity (one calculates, by GTR, the gravitational wave power emitted). So far, as you will quickly learn, the agreement between GTR gravitational wave power loss model and the observations has been astounding. This is considered by mainstream astronomy to be very strong evidence for the existence of gravitational waves and was the first experimental evidence for them. Early this year, direct observation of gravitational waves in the early cosmos is thought to have been made by the BICEP2 experiment when frozen ripples in the CBR.

Edit: as Warrick pointed out (thanks Warrick), a worthwhile piece of physics history to add is that Russell Hulse and Joseph Taylor were awarded the 1993 Nobel prize for physics for their analysis of the system, the wording of the prize was

"[for the discovery of] ... new type of pulsar, a discovery that has opened up new possibilities for the study of gravitation."

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    $\begingroup$ It's worth noting that Hulse and Taylor won the Nobel Prize in 1993 for the discovery and analysis of this system. $\endgroup$ – Warrick Apr 14 '14 at 19:18
  • $\begingroup$ The OP seems to have had a different question in mind than the one originally asked, but +1 for the answer to what was originally written. $\endgroup$ – user10851 Apr 14 '14 at 19:44
  • $\begingroup$ Also, note that the Hulse-Taylor curve is nearly perfect withouta accounting for any quantum effects: astro.cardiff.ac.uk/research/gravity/resources/hulse_taylor.jpg $\endgroup$ – Jerry Schirmer Apr 15 '14 at 16:43

In answer to the edit, any transitions due to single-graviton exchange will involve energies that are just impossibly small. To convince yourself of this, remember that the energy levels of the hydrogen atom are given by:

$$E = \frac{\mu k^{2}e^{4}}{2\hbar^{2}n^{2}} = \frac{13.6\,\,{\rm eV}}{n^{2}}$$

If you do the same for two solar mass neutron stars bound by gravity, you get (up to a fudge factor of order 1 to account for the reduced mass of the system):

$$E = \frac{ G^{2}M^{5}}{2\hbar^{2}n^{2}}$$

Now, this tells us that the energy for the first transition is:

$$\frac{G^{2}M^{5}}{2\hbar^{2}}\left(1 - \frac{1}{4}\right) = \frac{3G^{2}M^{5}}{8\hbar^{2}}$$

Now, let's dump this energy into kinetic energy in one of the stars:

$$\begin{align} \frac{1}{2}Mv^{2} &= \frac{3G^{2}M^{5}}{8\hbar^{2}}\\ v&=\frac{\sqrt{3}GM^{2}}{2\hbar} \end{align}$$

Comparing this to escape velocity from radius $r$, we have:

$$\begin{align} \frac{v}{v_{e}} &= \frac{\sqrt{3}GM^{2}}{2\hbar}\sqrt{\frac{r}{2GM}}\\ &=\frac{\sqrt{3GM^{3}r}}{2^{3/2}\hbar}\\ &=5.15\times 10^{79}\left(\frac{M}{M_{\bigodot}}\right)^{3/2}\left(\frac{r}{\rm 1\,\, AU}\right)^{1/2} \end{align}$$

So, it should be clear that you're not going to observe any quantum transitions in bulk astronomical orbits, which makes single-graviton effects nigh unobservable (and justifies me ignoring the reduced-mass effects).

  • $\begingroup$ Note that this is not definitive, as there are orbit stability effects from general relativity to care about, amongst other things, but this should be enough to show that these orders of magnitude combine badly. $\endgroup$ – Jerry Schirmer Apr 15 '14 at 1:24

You seem to be asking for a comparison of the angular momentum carried by gravitational waves in classical general relativity with the the angular momentum carried by quantum gravitons.

For the analogous problem in electromagnetism, you should read Beth's Mechanical Detection and Measurement of the Angular Momentum of Light, from early in the days of quantum theory. Beth used circular-polarized light to drive a torsion pendulum. The quantum model predicts that each photon exchanges angular momentum 2ℏ with a half-wave plate; classical electromagnetism predicts the same total angular momentum from the electric and magnetic fields; both are consistent with the size of the pendulum oscillations. The correspondence principle at work!

Similarly, a theory of quantum gravity should predict exactly the same angular momentum transfer from the binary star to the gravitational field as general relativity does --- especially for a macroscopic system.

If you go to do computations, you should remember that photons may also carry orbital angular momentum away from their emitters, and expect a similar phenomenon in graviton emission.


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