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I was reading about how a large amount of mass is lost as gravitational waves, X-ray radiation, and gamma radiation during a kilonova. I also read about the sticky bead analogy to better understand how energy is carried by gravitational waves. But then can this energy be transferred over to matter? The friction in the sticky bead analogy isn't really an electromagnetic wave, right, so it can't transfer anything to matter I guess?

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    $\begingroup$ Einstein's field equations are time reversal symmetric. If matter can transfer energy into a gravitational wave, a gravitational wave can transfer energy to matter $\endgroup$ Commented Dec 12, 2023 at 13:18
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    $\begingroup$ Doesn't the operation of LIGO require that gravitational waves do work? $\endgroup$
    – aroth
    Commented Dec 13, 2023 at 1:05
  • $\begingroup$ @aroth isn't that exactly what I ask in the 2nd question in the comment directly above yours? $\endgroup$
    – uhoh
    Commented Dec 13, 2023 at 1:07
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    $\begingroup$ @uhoh - Yes, I second the question / inference that because LIGO works gravitational waves can too. $\endgroup$
    – aroth
    Commented Dec 13, 2023 at 1:51

4 Answers 4

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For up to a long time this was a topic of controversy, involving formal mathematical arguments. It was settled by a simple physical argument credited to Feynman:

Later in the Chapel Hill conference, Richard Feynman used Pirani's description to point out that 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, contrary to the view expressed in 1955 by Rosen.

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    $\begingroup$ @Nat The focus of the debate at the time was whether gravitational waves could carry energy, fueled by the obvious observation that gravitational waves are solutions to the vacuum Einstein equations, and therefore ostensibly have a zero energy-momentum tensor. Feynman's argument approached the problem form the opposite end, exactly by asking the question can gravitational waves do work? (Cause clearly if they do, they must somehow transport energy.) $\endgroup$
    – TimRias
    Commented Dec 13, 2023 at 9:56
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    $\begingroup$ @TimRias there's no if about it - GWs carry huge amounts of energy! There would be no spirals-down and mergers if they didn't. $\endgroup$
    – uhoh
    Commented Dec 13, 2023 at 10:04
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    $\begingroup$ Feynman's argument is a logical jump, because the thought experiment demonstrates increase of internal energy of the stick and beads due to the wave, but not that this energy was carried over from distance by the wave. It is logically possible the wave just causes transformation of the pre-existing energy in the space near the stick and the beads, just as teacher saying "we're out of time, the lecture is over" makes most of the students to stand up and leave the classroom; while teacher's sound waves did cause the movement, they did not supply energy for it. $\endgroup$ Commented Dec 14, 2023 at 1:31
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    $\begingroup$ @JánLalinský But his argument works in vacuum (far from any other object) too right? So energy would need to have come from vacuum, which seems absurd. Moreover, such "unconditional transfer of energy from surroundings" might violate conservation of energy (in case there's no readily available energy), or perhaps the 2nd law of thermodynamics (both of which I understand may have caveats in GR). $\endgroup$
    – Real
    Commented Dec 14, 2023 at 18:42
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    $\begingroup$ @Real yes the argument and my objection both apply to bodies in vacuum (in the classical sense, no material medium or atoms present around the bodies). The energy to increase the internal energy of the bodies may come via influxf local gravitational/EM/any other kind of energy distributed in the space around the bodies, while the wave only causes the influx, not the presence of the energy itself. Energy so transformed per unit time (positive) may have greater magnitude than energy transported by the wave far from the heated body (the latter may even be zero, depending on the definition). $\endgroup$ Commented Dec 14, 2023 at 19:51
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I will proceed in two stages.

  • I will first establish that tidal effect can do work.
  • Then I will offer the consideration that a passing gravitational wave has a tidal effect.

Let a moon be orbiting a planet in an eccentric orbit.

For comparison I will first discuss the case of circular orbit, with the moon in tidal lock. In that motion there is tidal effect, but it has no consequences. The tidal effect causes an elongation of the moon, along the radial direction. With the moon in tidal lock the orientation of the tidal effect with respect to the interior of the moon remains constant, as a consequence there is no opportunity to do work.

When the moon is in an eccentric orbit the tidal effect has a significance that it doesn't have in circular orbit case. Closer to the planet the gradient in graviational potential energy is steeper. So: for a moon in eccentric orbit the magnitude of the tidal effect is not constant. During the journey towards closest approach the magnitude of the tidal effect increases. During the climb to furthest distance the magnitude of the tidal effect decreases.

The effect of the non-constant magnitude of tidal effect is that the moon is being "kneaded". The (periodic) change in magnitude of tidal effect has as result a (periodic) change in magnitude of physical elongation.

The connection with work done is as follows:
A celestial body tends to contract itself into spherical shape; the spherical state is a state of lowest possible potential energy. For a celestial body to shift to be in an elongated state is to be in a state of higher potential energy than the spherical lowest-possible-state.

The process of relaxing back to a lower elongation state is a process of releasing potential energy. That is the connection with work done.


Condition for dissipation of energy

If the moon would be in a state of superfluidity then the "kneading" due to the change in magnitude of tidal effect would not lead to dissipation of energy. (Superfluidity in the sense of the state of superfluidity that Helium goes to when cooled below 2.17 Kelvin.)

But of course the moon in this thought demonstration is not in a state of superfluidity. Just as dough increases in temperature as it is being kneaded the process of the moon being "kneaded" generates heat.



Gravitational wave and tidal effect

A gravitational wave has a tidal effect. In the absence of a gravitatioanal wave spacetime is uniform. A gravitational wave is a quadrupole wave with a corresponding tidal effect. Since the gravitational wve is propagating: the passage of a graviational wave means a passing oscillation in tidal effect.

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    $\begingroup$ some of the answers to Astronomy SE's Gravitational waves vs. "normal gravity" also touch on tidal forces and other physical effects of a passing GW. $\endgroup$
    – uhoh
    Commented Dec 13, 2023 at 0:43
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    $\begingroup$ Related: "Tidal heating", Wikipedia. $\endgroup$
    – Nat
    Commented Dec 13, 2023 at 16:04
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Yes, gravitational waves can do work. When they pass through an object, they cause stretching and contraction of that object which requires energy.

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    $\begingroup$ @TimRias no real-world material has zero internal friction so I think it is pretty clear (unless we're talking about a "transparent aluminum" universe here) $\endgroup$
    – uhoh
    Commented Dec 13, 2023 at 10:01
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    $\begingroup$ @TimRias when you stretch a spring work is done. when it regains its original form the potential energy stored in it is converted to kinetic energy. $\endgroup$
    – user355398
    Commented Dec 13, 2023 at 14:39
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    $\begingroup$ @AgnibhoDutta But when the gravitational wave is passing through the system, nothing is in a proper inertial system, so it is not immediately clear that such Newtonian logic can be applied. The moment you have energy dissipating in other modes of the system, it is clear that work has been done (which is the essence of the sticky bead argument). $\endgroup$
    – TimRias
    Commented Dec 13, 2023 at 14:48
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    $\begingroup$ It doesn't really cause stretching and contraction of the object, doesn't it? You will not be able to measure any stretching with a measuring rod, for example. It stretches space instead. $\endgroup$
    – fishinear
    Commented Dec 14, 2023 at 19:51
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    $\begingroup$ @AgnibhoDutta They don't cause the objects to get stretched in any measurable way, as far as I know. They just cause light to take a tiny bit more time to travel through the waves (and only in one direction) $\endgroup$
    – fishinear
    Commented Dec 15, 2023 at 9:45
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Very simple answer: Since they carry energy away when they are created, they can also deposit energy in the same way.

It's the principle of reversibility that applies throughout physics (with important exceptions). If you have a process, that same process can run in reverse. Because the equations that govern these processes are symmetric about time. Since gravitational waves are created by accelerated masses, they are also able to accelerate masses. So, if you have two bodies that rotate around each other at the same frequency that some merging black holes did, you will find that those bodies may get a tiny kick from the passing gravitational wave. Whether that kick is positive or negative will depend on the exact phase angle between the rotating bodies and the wave.

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