I have a question on gravitational wave (GW) detection.

I would assume that gravitational waves do not impact rigid bodies. I mean, GWs should be extremely weak with respect to electromagnetic forces that binds together atoms and molecules, thus rigid bodies should be completely "untouched" by GWs, a part for very extreme cases (i.e., black hole collision much nearer to the earth).

This kind of reasoning is the very same that one finds when talking about expansion of the universe: the expansion of the universe does not cause the earth (or any rigid body) to expand, because the "force" of the expansion is ridiculous low with respect to other forces (electromagnetic bonds or even the gravitational pull of the sun).

If this is true, then I don't understand why, when talking about LIGO GW detection, people always talk about "stretching" of the arms: the earth should not be affected by GW passing. In my understanding, the stretching would be of the space, but not of physical objects, thus it would e.g. affect the mirrors if they are somehow "detached" by the earth and free of moving in the space, but I don't know if this is the case.

If this is not true, and GWs do stretch the earth and rigid objects, I would expect that the amount of the stretching depended on the type of matter that the object is made of: I mean, GW impacting on wood sticks would have a different impact than on steel sticks. Thus the amount of stretching detected in one of the LIGO detectors would be different than the one in the other (because the soils have different compositions): Is this the case?

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    $\begingroup$ Gravitational waves can have an effect on “rigid bodies” (as in, “solid”; as you probably know, perfectly rigid bodies are impossible). Keep in mind that the first attempts at detecting gravitational waves (all controversies aside), were solid aluminum cylinders. They didn’t work as detectors not because the basic theory was flawed but because of insufficient sensitivity. $\endgroup$ Nov 1, 2021 at 0:16
  • $\begingroup$ Another reason for the failure of the Weber bars is probably that there is no GW source for frequencies around 1500 Hz. All that's left is noise. $\endgroup$
    – 9herbert9
    Nov 18, 2023 at 11:44

3 Answers 3


The mirrors at the ends of LIGO's arms have been mounted in such away that their motion along the direction of arms is (almost) completely decoupled from their physical surroundings. The motion of mirrors in LIGO in these directions can thus be considered in perfect free fall, and is thus susceptible to effects of passing gravitational waves.

When speaking about the stretching of LIGO's arms, people are not referring to the any actual stretching of the vacuum tubes, but to the stretching of the "empty" space between the mirrors. It is also worth keeping in mind that the actual change in path length between the mirrors due to a passing gravitational wave is less than the width of a proton!

  • $\begingroup$ The physical space which the material of the vacuum tubes occupy will also be distorted by any gravitational wave that passes through it $\endgroup$
    – Jojo
    Oct 31, 2021 at 21:26
  • $\begingroup$ A gravitational wave is a motion in spacetime, so wouldn't there also be a brief dilation in time? $\endgroup$ Oct 31, 2021 at 22:45
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    $\begingroup$ @foolishmuse Gravitational waves are customarily written in gauge (the traceless transverse , or TT gauge) in which their effect is purely spatial. $\endgroup$
    – TimRias
    Nov 1, 2021 at 8:03
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    $\begingroup$ @Joe, sure it will, but whatever effect that has on the vacuum tubes is (almost) completely isolated from the motion of the mirrors, and therefore irrelevant to the measurements of LIGO. $\endgroup$
    – TimRias
    Nov 1, 2021 at 8:05
  • $\begingroup$ @mmeent I was not familiar with the term traceless transverse so I looked it up. Does it just mean that the time dilation impact is so small that it can be ignored? $\endgroup$ Nov 1, 2021 at 15:16

The mirrors in a gravitational wave interferometer are a good approximation to inertial test masses. Their "connection" to the Earth, via the points from which they are suspended, is almost non-existent, thanks to $13+$ orders of magnitude of damping at the frequencies of the gravitational waves.


It depends on the GW frequency. If the GW frequency is well above the resonant frequency of the solid object, the stress will not accelerate the object's parts rapidly enough for the object to respond. So, the object will experience a time-varying internal stress because it is getting longer and shorter as the GW passes.

If the GW frequency is low compared to the resonant frequency, the object will simply respond by staying the same size. But accelerometers on different parts of the object will record different accelerations, as this does not correspond to geodesic trajectories for all parts of the object.

And, in resonance, the wave may excite a substantial oscillation of the body. This is the principle behind the Weber Bar.

  • $\begingroup$ Re "the resonant frequency": Isn't it more like a low-pass filter ("corner" frequency)? $\endgroup$ Nov 1, 2021 at 11:11
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    $\begingroup$ @PeterMortensen The low-pass filter without peaking requires pretty heavy damping of the resonance. Weber Bars were designed for minimum damping. $\endgroup$
    – John Doty
    Nov 1, 2021 at 11:14
  • $\begingroup$ The very high quality $Q=10^6$ of the Weber bars ensures an extremely low bandwidth. A bar only responds in the frequency range $1499.9985 Hz<f<1500.0015$ Hz. It is very unlikely that a GW source exists anywhere that produces a long-lasting signal in this narrow frequency range. Short signals transport too little energy to generate significant amplitudes. $\endgroup$
    – 9herbert9
    Nov 18, 2023 at 12:06

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