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 spacetime distortion caused by the gold sphere be detectable by the change in the interference patterns of the LIGO detector, or would it 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.
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Would just putting it there cause much of a wave? – xxbbcc Apr 22 '19 at 20:46
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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. – Cristian Dumitrescu Apr 22 '19 at 20:52
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The question is whether the displacement caused by the gravitational field of the gold sphere is detectable. – Cristian Dumitrescu Apr 22 '19 at 21:06
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2Why should it matter the material the sphere is made of? – lurscher Apr 22 '19 at 21:11
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It doesn't, I chose gold due to its higher density. – Cristian Dumitrescu Apr 22 '19 at 21:16
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So you wouldn't need a big sphere for the same effect. – Cristian Dumitrescu Apr 22 '19 at 21:21
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@CristianDumitrescu: When providing clarification on a question, it's better to do it by editing the question rather than by replying in comments. – Apr 22 '19 at 22:42
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Static curvature from a small nearby source is probably more readily detected using other techniques: https://physics.stackexchange.com/questions/356166/has-the-curvature-of-spacetime-been-measured-at-the-human-scale/356250#356250 – Rococo Apr 23 '19 at 00:05
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Thank you for the references. – Cristian Dumitrescu Apr 23 '19 at 01:03
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Perhaps related: Why can't we produce gravitational waves in the lab? – Dhruv Saxena Apr 23 '19 at 12:15
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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.
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1This 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? – knzhou Apr 22 '19 at 22:55
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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. – Cristian Dumitrescu Apr 22 '19 at 23:18
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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.. – Cristian Dumitrescu Apr 22 '19 at 23:29
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@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. – Apr 23 '19 at 00:18
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A static sphere of gold does not produce gravitational waves. Since LIGO is set up to detect perturbations in the arm lengths at frequencies of 30-3000 Hz then the presence of just another mass near the beam is just another (very) small DC offset that would not be detected.
Still, it is of interest to ask whether introducing a mass close to the beam would cause a change that needed any adjustments. I suppose you could model the perturbation as the Shapiro delay close to a spherically symmetric mass.
The round-trip delay for light travelling close to (grazing) a sphere next to the beam path would be $$\Delta t \simeq \frac{4GM}{c^3} \left(\ln\left[\frac{x_1 + (x_1^2 + r^2)^{1/2}}{-x_2 + (x_2^2 + r^2)^{1/2}} \right] - \frac{1}{2}\left[\frac{x_1}{(x_1^2 + r^2)^{1/2}} + \frac{x_2}{(x_2^2 + r^2)^{1/2}}\right]\right)\ , $$ where $r$ is the radius of the sphere and $x_1+x_2$ is the LIGO arm length of 4 km. If the sphere is halfway between the arm mirrors, such that $x_1=x_2$ and assuming $r\ll 4$ km, then $$\Delta t \simeq \frac{4GM}{c^3} \left(\ln\left[\frac{2 + r^2/2x^2}{r^2/2x^2} \right] - \left[\frac{1}{(1 + r^2/2x^2)}\right]\right)\ , $$
$$\Delta t \simeq \frac{4GM}{c^3} \left(\ln\left[\frac{4x^2}{r^2} \right] - 1 \right)\ , $$ where $x=2$ km.
If we let $r=1$ m (a lot of gold, surely lead would do!) then $M \simeq 8\times 10^4$ kg and $\Delta t \simeq 3\times 10^{-30}$ s, or an equivalent length change of $10^{-21}$ m.
This is many orders of magnitude smaller than the $\sim 10^{-18}$-m length changes that LIGO is sensitive to when they happen at frequencies of $30-1000$ Hz. It is also many more magnitudes smaller than the imposed offset in path length that takes the interferometer slightly off the dark fringe in normal operating conditions. I would conclude therefore that this offset is unobservable even under ideal conditions. Lower frequency changes are in any case obliterated by seismic noise, gravity gradient noise and various noisy instrumental control loops by many orders of magnitude more.
ProfRob
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