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Throughout much of the 20th century there was debate about whether Gravitational Waves were real, and whether or not they carrier energy and could be detected. It is often presented that Feynman's 'sticky bead argument' was the strongest convincer:

a passing gravitational wave should in principle cause a bead on a stick (oriented transversely to the direction of propagation of the wave) to slide back and forth, thus heating the bead and the stick by friction. This heating, said Feynman, showed that the wave did indeed impart energy to the bead and stick system, so it must indeed transport energy

My question is: why doesn't the stick itself move in exactly the same way the bead does? i.e. at the location of the bead, why is there ever a relative velocity between the bead and the stick?

The same point seems to be brought up in this (presumably flawed) paper.

I'm kind of assuming the answer is that the electromagnetic force (which keeps the stick 'rigid'), acts with coordinate/comoving distance instead of proper distance? If so, how do we know that that is true?

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    $\begingroup$ I don't know about you, but I have been following the issue during the last quarter of the 20th century and there was no memorable debate that I can remember. Gravitational waves were pretty standard mainstream physics. If you browse something like part VIII of MTW, it's pretty much all there. $\endgroup$
    – CuriousOne
    Commented Apr 12, 2016 at 1:19
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    $\begingroup$ It does move, but a rigid stick of finite size will move differently compared to a pointlike bead. So, there will be a relative motion due to the tidal effect of the gravitational waves. $\endgroup$
    – Count Iblis
    Commented Apr 12, 2016 at 1:22
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    $\begingroup$ @CuriousOne What is your point? Are you disagreeing that this was a debate? As you could have found out from the link, the issue was mostly resolved by about 1960, though many counter papers still exist through the 80s. The sticky bead argument is also in MTW, without a detailed explanation. $\endgroup$
    – DilithiumMatrix
    Commented Apr 12, 2016 at 1:32
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    $\begingroup$ @CountIblis my question is basically, 'why does the stick move differently?' Thanks for your response, but I feel like it begs the question. What are the tidal effects from a plane GW wave? $\endgroup$
    – DilithiumMatrix
    Commented Apr 12, 2016 at 1:33
  • $\begingroup$ I am disagreeing that there was a debate "throughout much of the 20th century". At most there could have been a debate for about half of it, and I am pretty sure that even that wasn't the case if you look at the literature closely. If anything, one has to ask why there would have been a debate. Curvature is curvature, whether from a static or from a dynamic solution. I don't think one can deny that the static curvature is utterly detectable... it will twist your ankle when you are making a mis-step on the stairs, for sure. $\endgroup$
    – CuriousOne
    Commented Apr 12, 2016 at 1:38

3 Answers 3

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I would guess that the argument applies to the case where the frequency of the gravitational wave is low compared to the natural frequency of the stick, or put another way, the rate of change of length due to the gravitational wave is small compared to the speed of sound in the stick.

If you take some line normal to the GW and of length $\ell_0$ then the GW will produce an oscillating strain something like:

$$ \gamma(t) = \gamma_0 \sin (\omega t) $$

and the length of the line will be:

$$ \ell(t) = \ell_0 \left(1 + \gamma \right) = \ell_0 \left(1 + \gamma_0 \sin (\omega t) \right) $$

So the rate of change of the length of the line will be:

$$ \frac{d\ell}{dt} = \ell_0 \omega \gamma_0 \cos (\omega t) $$

In other words, a point at one end of the line will move relative to a point at the other end of the line with a relative velocity $v_r = \ell_0 \omega \gamma_0 \cos (\omega t)$.

Now lay down our stick along the line $\ell$. If $v_r$ is much greater than the speed of sound in the stick the intermolecular forces acting in the stick cannot act fast enough to stop the stick stretching and shrinking with the gravitational wave. That means the stick and bead will move together and the bead won't slide.

Conversely, if $v_r$ is much less than the speed of sound in the stick then the intermolecular forces will resist the gravitational wave and stay the same length. In that case the stick will move relative to the bead and we'll see the relative motion and associated frictional losses that Feynmann describes.

Since the frequency and intensity of the gravitational wave are probably outside our control the only variable we can change is the length of the stick. If we make the stick long enough it will oscillate with the GW and if we make it short enough it will stay the same length.

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    $\begingroup$ You can stop after saying that "the GW will produce an oscillating strain " because that's already going to capture energy from the GW waves, so there is no need for the sliding bead anymore. $\endgroup$
    – Count Iblis
    Commented Apr 12, 2016 at 16:39
  • $\begingroup$ Thanks John, that's very helpful. But there's still an the assumption here that intermolecular forces will act to counteract the strain. In other words, if I stretch the stick slightly, the EM forces will tend to pull it back together because I'm trying to increase the intermolecular separation, i.e. increase the $r$ in $q_1 q_1/r^2$. But how do we know that $r$ is the same as $\mathcal{l}$ in your equations, and not the coordinate distance in the metric? $\endgroup$
    – DilithiumMatrix
    Commented Apr 12, 2016 at 17:24
  • $\begingroup$ The distance between two atoms/molecules/whatever is the proper length of the straight line joining them calculated using the metric. If the metric changes faster than the atoms can move in response then this proper length changes. As an aside, this is exactly what happens in the expanding universe. So the strain created by the GW is a real strain that changes the distance and therefore the potential energy of the atom/molecules. $\endgroup$
    – John Rennie
    Commented Apr 12, 2016 at 17:30
  • $\begingroup$ Of course... the effective distance has to be the invariant proper length, same as for all other 'observations' of GR effects. So if you believe GR produces e.g. redshift, you have to believe the beads and stick are physically moved, which takes energy, which has to then be carried by the GW. Got it. $\endgroup$
    – DilithiumMatrix
    Commented Apr 12, 2016 at 17:54
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One way to think about it in more intuitive terms than comoving/proper is to drop down to the micro level and consider how molecules hold their shape in the first place. They are bound together by stable electron configurations; which in turn obtain their shapes from quantum-electro-dynamical processes; an interfering sum of photons bouncing back and forth between the charged components of the system.

So the resting bond distance between (wlog) two hydrogen atoms is the result of some interference pattern between virtual photons moving back and forth around those protons and electrons. The propagation of those photons from one location to the next is metric-dependent. One can think of a gravitational wave passing as locally slowing down or speeding up light; or 'inserting' or 'removing' units of length locally. So such a modification of the metric should shift the interference pattern of virtual photons that keeps the molecule together; in such a way as to push the atoms back in their 'neutral' position; which is the position that keeps the integral of the metric between the atoms at whatever fixed value characteristic of H2.

Another way of putting it: if the metric expands, a photon going from atom one to atom two will arrive 'late', with a bigger phase shift, and the net effect of this is really indistinguishable from stretching the atomic bond by more conventional means.

So indeed if the wave is slow relative to the natural frequency of two masses constrained by a rod, the masses will move relative to two masses not so constrained; and if the wave is fast relative to the natural frequency, a plain old strain gauge should register a signal.

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DilithiumMatrix asked: "why doesn't the stick itself move in exactly the same way the bead does?"

Because the rigid stick is held together by its internal forces while the bead is free floating. The left end of the stick needs to hold a constant proper distance to the right end, while the bead on the left doesn't care about the bead on the right. Therefore the stick and the beads experience acceleration relative to each other.

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  • $\begingroup$ Thanks. However, there is still something missing here: why does the stick care about the proper distance as opposed to any other distance. Intuitively it's obvious in the sense that coordinate distance doesn't mean anything physically, and proper distance is in a way 'unique'... but I still wish there was some clear way of showing that it has to be the proper distance that the stick (i.e. EM forces) care about. $\endgroup$
    – DilithiumMatrix
    Commented Sep 26 at 17:03
  • $\begingroup$ @DilithiumMatrix - The stick only cares about the distances in his own rest frame, so take a slice of constant t (which is easy since the gtt=1 with gravitational waves) and integrate the √gxx, √gyy or √gzz between the two ends of the stick, then make sure that stays constant at all t. If the stick cared about the coordinate distances instead he would have a hard time since you can choose different coordinates and projections thereof. $\endgroup$
    – Yukterez
    Commented Sep 26 at 23:04

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