Suppose  you put  a gold sphere inside the LIGO interferometer, not in the path of the laser beams, but sufficiently close to  (in the vicinity of) one of the laser beams. Would the space  - time distorsion caused by the gold sphere be detectable by the change in the interference patterns of the LIGO detector , or it would be too weak for detection? If the effect is detectable, this would be an experimental way to study the gravitational field of small objects. Some ballpark estimation of the magnitude of the effect would be appreciated.

  • $\begingroup$ Would just putting it there cause much of a wave? $\endgroup$
    – xxbbcc
    Apr 22 '19 at 20:46
  • $\begingroup$ LIGO detects tiny spatial displacements caused by gravitational waves, but the gravitational field of the gold sphere might also cause some displacement that could be seen in the interference patterns of the laser beams. $\endgroup$ Apr 22 '19 at 20:52
  • $\begingroup$ The question is whether the displacement caused by the gravitational field of the gold sphere is detectable. $\endgroup$ Apr 22 '19 at 21:06
  • 2
    $\begingroup$ Why should it matter the material the sphere is made of? $\endgroup$
    – lurscher
    Apr 22 '19 at 21:11
  • $\begingroup$ It doesn't, I chose gold due to its higher density. $\endgroup$ Apr 22 '19 at 21:16

LIGO doesn't detect gravitational fields, it detects curvature. Specifically, it detects the type of spacetime curvature associated with tidal forces. (This is why each detector has two arms.) Also, LIGO doesn't detect DC effects, it only detects AC signals within a certain range of frequencies. The most difficult thing about the LIGO experiment is that it also serves as the world's most sensitive vibration sensor. This is why they have two (now 3, IIRC?) facilities far apart. They look for correlated signals between the different facilities.

So if you move a dense object close to one of the detectors, as in the Cavendish experiment, what will happen is that that detector will be momentarily disabled by the vibration, just as it would if a truck drove by and stopped suddenly. Once the detector recovered from the insult, there would be no AC signal, probably no signal that would mimic the kind of tidal distortions it's looking for, and no correlation with the other site(s). So you would get no signal.

  • 1
    $\begingroup$ This is a good answer to the OP's question, but just wondering, would a set of nearby wiggling (at ~100 Hz) heavy spheres be detectable? $\endgroup$
    – knzhou
    Apr 22 '19 at 22:55
  • $\begingroup$ Thank you @BenCrowell So in the idle mode, the orthogonal arms of the interferometer are calibrated so that no photons reach the detector, due to destructive interference. When the arms change length periodically (due to a passing gravitational wave), that's when the AC signal is generated by the detector, because there will be photons hitting it intermittently. And only certain AC frequencies pass the filters, the ones most likely caused by gravitational waves, according to simulations. $\endgroup$ Apr 22 '19 at 23:18
  • $\begingroup$ OK, so the exact LIGO design currently in use cannot be used in order to detect gravitational fields due to small objects. But you could, in principle, change some aspects of the LIGO design so you could detect and study gravitational fields of small objects. Any small displacement would make the destructive interference of the laser beams less perfect, so more photons would hit the detectors, when small gravitational fields are present in the path of the beams.. $\endgroup$ Apr 22 '19 at 23:29
  • $\begingroup$ @knzhou: just wondering, would a set of nearby wiggling (at ~100 Hz) heavy spheres be detectable? The vibration through the ground would cause the detector to shut down, and the signal would also be rejected because it wasn't detected at the other site(s). My example of the detector shutting down because of a truck braking was not made up. It was an actual example I read in a newspaper article. $\endgroup$
    – user4552
    Apr 23 '19 at 0:18

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.