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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?

Pierre Paquette
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user6760
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4 Answers4

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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.

TimRias
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  • Could you elaborate a bit on the historical context? I mean, the argument made in @AgnibhoDutta's answer seems like something that should've been readily apparent, if the physicists knew that gravitational waves would stretch/contract objects as they passed through -- so I'm guessing that they didn't know that and that there was something else that needed to be worked out? – Nat Dec 13 '23 at 01:54
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    @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.) – TimRias Dec 13 '23 at 09:56
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    @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. – uhoh Dec 13 '23 at 10:04
  • Isn't the wave deforming space? Why is there friction? – Pablo H Dec 13 '23 at 19:01
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    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. – Ján Lalinský Dec 14 '23 at 01:31
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    @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). – Real Dec 14 '23 at 18:42
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    @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). – Ján Lalinský Dec 14 '23 at 19:51
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    @Real This phenomenon already happens in Frenkel's theory of point charged particles (and its siblings like the Tetrode, Fokker, Feynman-Wheeler theories), where a plane wave causes single charged particle to gain kinetic energy, but the energy gained may be much greater than what the wave transports (that may be even zero, if the wave is due to another single charged particle). – Ján Lalinský Dec 14 '23 at 19:53
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    @Real This is connected to the fact that in this class of theories, EM energy density is not everywhere non-negative, but can have both positive and negative value, depending on the position. So the accelerated charged particle can gain kinetic energy at the expense of decrease of the local EM energy near the particle below zero value. The wave of a single particle here just causes the transfer from local EM energy to local kinetic energy of another particle, it does not transport any additional positive EM energy. – Ján Lalinský Dec 14 '23 at 19:55
  • @JánLalinský Perhaps I can improve my argument. I think it's "something's gotta give". If generating gravitational waves took no energy from a system (hence we might be able to say gravitational waves carry no energy), and they consistently heated the beads, then you've got a kind of perpetual motion machine. If they do require energy to be created, then as someone mentioned there should be an associated reverse absorption process and hence they do carry energy (which is indeed the case). – Real Dec 14 '23 at 23:04
  • @Real Waves transporting energy and waves not doing so are both mathematically possible, they are just different waves (due to different sources). Both kinds can cause conversion of local invisible energy in the space into kinetic or internal energy of the beads and the rod. The perpetual motion machine producing unlimited amount of heat (or work) out of nothing is not implied, because the energy really comes from local energy surrounding the beads, and obeys the equation of local conservation of energy. – Ján Lalinský Dec 15 '23 at 00:24
  • @Real You might think the beads can oscillate under action of the primary wave for a long a time and evolve a lot of heat, and a lot of energy will be gotten from the local energy in the surrounding space. And that is mathematically possible for some value of "long" and "lot", but they are finite; eventually, the secondary waves created by the beads will reach the primary source, and make further zero-work generation of the primary wave impossible. Relativity postpones the need to do some work to make waves, but only for a finite time. – Ján Lalinský Dec 15 '23 at 00:36
  • @JánLalinský: Yes, strictly speaking, Feynman's argument only proves that either gravitational waves carry energy or conservation of energy is violated. Which isn't that surprising, because most novel arguments about how something must carry or store energy ultimately come down to something like "see, we'll need to tweak the definition of this useful conserved quantity that we call energy like this or it won't be conserved." For example, in exactly the same way, a falling apple accelerating proves that either energy is not conserved or the gravitational potential must store energy. – Ilmari Karonen Dec 15 '23 at 17:09
  • @IlmariKaronen I don't agree, Feynman's argument proves nothing. It assumes that kinetic/internal energy of the beads increases and then makes a logical jump that this must be due to energy carried over by the wave, otherwise "energy conservation" would not hold. But local energy conservation can hold even if the wave does not transport energy, because energy can flow in from everywhere around the beads, even without the wave bringing additional positive energy. See my comments in this thread, I explain why waves causing energy release in the beads and the rod need not transport energy. – Ján Lalinský Dec 15 '23 at 23:09
  • @JánLalinský Thanks for the answer. You're proposing the conservation of energy could be maintained by making the heating process bounded from the influence of the back-scaterred waves back into the source, eventually causing the source to stop emitting or imposing an energy cost to emission from those backscattered waves. 1/2 – Real Mar 11 '24 at 17:24
  • (2/2) First, this mechanism seems a bit fragile and implausible (that's intuition only, hopefully it makes sense). Consider, in reality this back scattering should (must?) undergo inverse-square losses, and be severely attenuated, which I believe makes it so the energy accounting couldn't work (for example, the further from the source, the less backscattering). It's also not clear how those waves could impart a new and permanent energy cost that wouldn't vanish after the backscattering waves themselves vanish. – Real Mar 11 '24 at 17:26
  • @Real there is a consistent mathematical theory of interacting point charged particles due to Frenkel, obeying the law of local conservation of energy, where regions of space where only single field due to single particle is present carry zero EM energy, and only in those regions of space where two or more particle fields overlap, there can be non-zero EM energy. Any heat evolution can be explained as conversion from EM energy into heat, and won't go on indefinitely unaffected because of the effect of the secondary field back on the source. – Ján Lalinský Mar 11 '24 at 18:06
  • @Real So there is no problem with energy accounting in this kind of theory; energy conservation holds in the sense of local conservation of energy, positive heat energy due to primary wave working on a particle is balanced by decrease of EM energy around the particle. For Frenkel's theory, see papers by Frenkel and Stabler, links at the end of this answer of mine: https://physics.stackexchange.com/questions/675178/intuition-for-energy-density-of-electromagnetic-fields/675241#675241 – Ján Lalinský Mar 11 '24 at 18:07
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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.

Cleonis
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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.

  • If the object returns to its former shape after the wave has passed, it is not at all clear that any work has been done. – TimRias Dec 13 '23 at 09:50
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    @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) – uhoh Dec 13 '23 at 10:01
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    @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. –  Dec 13 '23 at 14:39
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    @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). – TimRias Dec 13 '23 at 14:48
  • @TimRias "...is converted to kinetic energy and heat." With good spring material, internal friction is low and almost none of the work input is converted to heat. But for other metals it's much more heat, and for rubber bands (another kind of spring) they get quite hot. Type "Feynman rubber bands" into a search engine... – uhoh Dec 13 '23 at 19:06
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    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. – fishinear Dec 14 '23 at 19:51
  • @fishinear It is true that it stretches spacetime but that stretching in turn cause the objects to get stretched. Similar to how water waves can affect a rock on the surface. –  Dec 15 '23 at 06:29
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    @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) – fishinear Dec 15 '23 at 09:45
  • @fishinear, If you'd orbited in space a few radii away from a pair of merging black holes... It would suck to be you, really. As for tidal heating, Jupiter's moon Io is the best example. It's the most volcanic body in the Solar System, and by a huge margin! It's no different, tidal deformation is tidal deformation, be it in Newton's or GR. And if your premise that a body deforming with space doesn't feel anything, then what powers tidal disruption events? – kkm -still wary of SE promises Dec 15 '23 at 10:43
  • @kkm-stillwaryofSEpromises OK, that is fair enough. They cause disturbance of the EM bonds between molecules. Although I think there is a difference between close tidal influences, like the type you mention, that are changing direction and diverging, and gravitational waves from far away systems that are parallel. Need to think about it more. – fishinear Dec 16 '23 at 21:19
  • @fishinear, no they don't. GR is classical, it has no concept of a molecule or a bond. My point is, tidal effects of a GR wave are real, just like other GR and Newtonian tidal effects. Think of a remote GR source thusly: take a narrow cone, as if a searchlight from the source lit you, and look at its crossection. The circle will bounce into an ellipse: now it's taller, next it's wider. You're physically tied to this circle of space! You will feel how this wiggly circle pulls you alternating wide and tall. The area of the ellipse is constant, but shape isn't. (It's one wavemode of many.) – kkm -still wary of SE promises Dec 16 '23 at 21:36
  • @kkm-stillwaryofSEpromises Normal tidal effects occur because the gravitational curvature is diverging and moving, causing changing gravitational potentials. That does not happen with gravitational waves from far away sources. – fishinear Dec 18 '23 at 09:02
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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.

cmaster - reinstate monica
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