6 ms·

Acoustic waves propagate through what are essentially elastic deformations of the material they travel through. Can gravitational waves be thought of as propaga

by derefr 3y ago

Acoustic waves propagate through what are essentially elastic deformations of the material they travel through. Can gravitational waves be thought of as propagating by elastically deforming spacetime? If this analogy holds, then can it be taken further? Acoustic waves dissipate their energy insofar as they trigger plastic deformation in a material. Could gravitational waves plastically deform... spacetime itself? Or would they just be deforming the material? Or is gravitational energy not dissipated into other forms of energy at all?

jessriedel 3y ago
Acoustic waves dissipate their energy into non-acoustic (molecular) degrees of freedom. In the absence of matter, I think there are no other degrees of freedom for gravitational waves to dissipate into, unless they are strong enough to form a black hole (which they can do under extreme conditions, I think). (This is waaaay outside my expertise though, so take my answer with a grain of salt. Everything I've said in this thread is basically based off the rudimentary understanding from taking a GR course in grad school. I've never done research on this topic.)
cyberax 3y ago
> Or is gravitational energy not dissipated into other forms of energy at all? Extremely weakly. The mechanism is similar to acoustic waves, but the coupling constant is so small, that the amount of dissipated power would be insignificantly small. In theory, you can use gravitational waves to extract energy. For example, you can wait for the "compression" part of the wave, and push a cart uphill during it. Then let it slide downhill when the compression peak passes. You'll be able to extract some useful energy, because the distance that you pushed the car uphill will be shorter than the "normal" distance. You can make more elaborate systems on this principle. E.g. a tuned resonator: two orbiting masses with a period selected to match the frequency of the gravitational waves.
- treeman79 3y ago
  The earth going around the sun produces gravitational waves, and some of the energy is lost. If all of that energy could be harnessed; it would be sufficient to power a small toaster oven.
  - staunton 3y ago
    > If all of that energy could be harnessed; it would be sufficient to power a small toaster oven. Do you have a source? Sounds like an interesting calculation.
    - diroussel 3y ago
      The earth going around the sun produces all the tides in all the oceans of our planet, excluding the lunar tides. The combined power of the tides is tremendous.
    - cyberax 3y ago
      The formula for the power of gravitational radiation for two orbiting objects is: W = 32 * G^4 / (5*c^5*r^5) * (M1*M2)^2 * (M1 + M2). "r" is the orbital radius, "G" is the Newton's constant, "M1" and "M2" are the masses of the orbiting objects. It's actually kinda amazing that once you substitute in all the values and do the math, all the scary large powers just somehow cancel out to leave a small macroscopic number (I remember getting around 60 watts). Edit: note, that both objects radiate. So it's 60 watts for both the Earth and the Sun, around 120W in total.
      - raattgift 3y ago
        Can I ask how you got there? G^4/c^5 makes sense, but the rest loses me. (Eventually after all the typing below I think that I traced the 32/5 and (M1*M2)^2 * (M1 + M2) to (eqn 16) in Peters & Matthews 1963 maybe? (e=0, a->r) https://doi.org/10.1103/PhysRev.131.435 https://doi.org/10.1103/PhysRev.131.435 (stick sci-hub.se in front of that if you need to). The authors take an approach comparable to the textbooks below.) Super-quick textbook review. Practically all of them start with the quadrupole moment and try to justify an energy which is quadratic in derivatives of that while still within a linear theory. Carroll and Mathtias Blau take slightly different-from-each-other paths through the transverse-traceless TT-gauge to P = dE/dt = -2/5 \frac{G^4 M^5}{r^5} (c=1) for a circular equal-mass soft binary. Wald uses the radiation gauge and so eqn 4.4.58 looks fairly different, and comes with the amusing Waldian line "A lengthy calculation (where many terms which integrate to zero are dropped) yields the final result,". Sigh. Blau's development looks a lot like MTW, but the latter gives us .... Exercise 36.6 ("Apply the full formalism ... to a binary star system with circular orbits. Calculate ... the total power radiated; the total angular momentum radiated ..."). Gee, thanks thick textbook. FWIW, 90 seconds of that (mostly trying to make sure both sides have the same dimension rather than extracting a power in watts because bad/lazy reasons and anyway I always think you had about the right order of magnitude) doesn't take me to anything like the form of your calculation. Rather than flip through other textbooks, let me rely on my maybe-shaky memory and say that most of them, at least the modern editions, follow the TT gauge approach and come up with an equation in a form similar to Carroll. What stands out here is that Earth-sun has a large mass ratio and noncircular orbit, and it's not really a binary system anyway, and so will defy these textbook schemes. Secondly all of these take P in the far field, because linearization. (Compare that with your edit). I also got way off into the weeds wanting to work with chirp mass (rather than q=m_1/m_2) which is what GW obs data analyses use because extracting the individual masses is hard. I also know a bit about EMRI BHBs (extreme mass-ratio inspiral) and in those dissipation is dealt with differently from textbooks even for soft binaries (e.g. soft->hard roughly PN & perturbative methods, EOB, GSF, numrel) absolutely none of which is of immediate practical use here. So that's some of what motivates my question about the origin of your calculation. ETA: I think most of what happened here is that my brain reads equations as words, and I simply forgot that I could actually rearrange teh ltteres! TGIF :-)
- derefr 3y ago
  Could there exist a material that resistively dissipates gravitational flux more strongly than normally, in the same sense that ferromagnetic materials resistively dissipate magnetic flux more strongly than normally? Something something "gravitational permeability"? (I guess such a material would interact with gravitational waves as if it had more mass than it does, without that affecting its inertial mass?)
nyrikki 3y ago
Hard or impossible to visualize, but as the temporal dimension is so much larger, most of the deformity is in the time dimension. But yes, spacetime itself is jiggly like Jell-O. As we can't visualize 4d spacetime, most analogies will be wrong. But as photons don't experience time themselves, thinking about the geodesic path is probably less error prone than thinking of it as squishing physical objects. Objects being pulled towards slower flows is the intuition that matches the math best for me.
- sunshinerag 3y ago
  what does "photos don't experience time themselves.." mean? why not?
  - alfiopuglisi 3y ago
    Photons travel at the speed of light, and at that speed, any "subjective" time is zero. In Einstein's theory of special relativity, the faster you go, the slower your proper time appears to an external observer. At the speed of light, this effect reaches infinity.
    - euroderf 3y ago
      Okay, so what does a photon observe when it partakes of Cherenkov radiation ?
    - criddell 3y ago
      > the faster you go, the slower your proper time appears to an external observer From my perspective, it takes about 8 minutes for a photon from the sun to hit my eye. From the perspective of the photon, a little time has passed, no? Doesn't the atmosphere and passing through my glasses slow it down a wee bit? Can the photon "know" that its position has changed between emission and absorption? From the photons point of view, I must be very, very close to the sun, right?
      - ben_w 3y ago
roywiggins 3y ago
They are predicted to leave permanent distortions in spacetime, though I think it's a more complicated mechanism. https://www.quantamagazine.org/gravitational-waves-should-permanently-distort-space-time-20211208/ https://www.quantamagazine.org/gravitational-waves-should-pe...
- piyh 3y ago
  Flashback to 3 body problem and harvesting energy from local c.
marsten 3y ago
In principle yes, gravitational waves dissipate energy into ordinary matter they pass through. However the coupling is extremely weak (that gravitational waves are so hard to detect is testament to this). In fact this weak coupling is what makes GWs so interesting for observational astronomy: They propagate from the source to our detectors virtually unchanged. (This is in contrast to EM radiation, which is very easy to scatter.) For example the farthest we can see back with EM radiation is about 200k years after the Big Bang, when the plasma of the early universe recombined into neutral hydrogen. By contrast gravitational waves can see back to the Big Bang itself, so it is a truly unique source of information as compared with light.
- HappMacDonald 3y ago
  I would rate the weak coupling as a far second or third point of interest behind linear signal fall-off vs inverse square for most other kinds of signal such as electromagnetic waves. Gamma ray burst is twice as far away? It's four times dimmer. A thousand times as far? A million times more dim. Gravitational wave signal from <event> is twice as far away? Makes it twice as hard to detect. A thousand times as far away? Only a thousand times as hard to detect.
  - marsten 3y ago
    To be clear, gravitational waves follow the same inverse square law as EM radiation, in terms of power. This is after all just a statement of conservation of energy. The linear drop-off you're referring to is when we look at it in terms of field strength (in this case the spacetime strain). Since power is proportional to field squared, this implies a linear drop-off in the field. It just so happens that for GWs it's easier to detect the field, whereas for (most) EM radiation it's easier to detect power. There are field-detection methods for EM radiation as well, which are useful for weak signals. Homodyne and heterodyne detection are good examples.

by derefr 3y ago

jessriedel 3y ago

Acoustic waves dissipate their energy into non-acoustic (molecular) degrees of freedom. In the absence of matter, I think there are no other degrees of freedom for gravitational waves to dissipate into, unless they are strong enough to form a black hole (which they can do under extreme conditions, I think). (This is waaaay outside my expertise though, so take my answer with a grain of salt. Everything I've said in this thread is basically based off the rudimentary understanding from taking a GR course in grad school. I've never done research on this topic.)

cyberax 3y ago

> Or is gravitational energy not dissipated into other forms of energy at all? Extremely weakly. The mechanism is similar to acoustic waves, but the coupling constant is so small, that the amount of dissipated power would be insignificantly small. In theory, you can use gravitational waves to extract energy. For example, you can wait for the "compression" part of the wave, and push a cart uphill during it. Then let it slide downhill when the compression peak passes. You'll be able to extract some useful energy, because the distance that you pushed the car uphill will be shorter than the "normal" distance. You can make more elaborate systems on this principle. E.g. a tuned resonator: two orbiting masses with a period selected to match the frequency of the gravitational waves.

treeman79 3y ago
The earth going around the sun produces gravitational waves, and some of the energy is lost. If all of that energy could be harnessed; it would be sufficient to power a small toaster oven.
- staunton 3y ago
  > If all of that energy could be harnessed; it would be sufficient to power a small toaster oven. Do you have a source? Sounds like an interesting calculation.
  - diroussel 3y ago
    The earth going around the sun produces all the tides in all the oceans of our planet, excluding the lunar tides. The combined power of the tides is tremendous.
  - cyberax 3y ago
    The formula for the power of gravitational radiation for two orbiting objects is: W = 32 * G^4 / (5*c^5*r^5) * (M1*M2)^2 * (M1 + M2). "r" is the orbital radius, "G" is the Newton's constant, "M1" and "M2" are the masses of the orbiting objects. It's actually kinda amazing that once you substitute in all the values and do the math, all the scary large powers just somehow cancel out to leave a small macroscopic number (I remember getting around 60 watts). Edit: note, that both objects radiate. So it's 60 watts for both the Earth and the Sun, around 120W in total.
    - raattgift 3y ago
      Can I ask how you got there? G^4/c^5 makes sense, but the rest loses me. (Eventually after all the typing below I think that I traced the 32/5 and (M1*M2)^2 * (M1 + M2) to (eqn 16) in Peters & Matthews 1963 maybe? (e=0, a->r) https://doi.org/10.1103/PhysRev.131.435 https://doi.org/10.1103/PhysRev.131.435 (stick sci-hub.se in front of that if you need to). The authors take an approach comparable to the textbooks below.) Super-quick textbook review. Practically all of them start with the quadrupole moment and try to justify an energy which is quadratic in derivatives of that while still within a linear theory. Carroll and Mathtias Blau take slightly different-from-each-other paths through the transverse-traceless TT-gauge to P = dE/dt = -2/5 \frac{G^4 M^5}{r^5} (c=1) for a circular equal-mass soft binary. Wald uses the radiation gauge and so eqn 4.4.58 looks fairly different, and comes with the amusing Waldian line "A lengthy calculation (where many terms which integrate to zero are dropped) yields the final result,". Sigh. Blau's development looks a lot like MTW, but the latter gives us .... Exercise 36.6 ("Apply the full formalism ... to a binary star system with circular orbits. Calculate ... the total power radiated; the total angular momentum radiated ..."). Gee, thanks thick textbook. FWIW, 90 seconds of that (mostly trying to make sure both sides have the same dimension rather than extracting a power in watts because bad/lazy reasons and anyway I always think you had about the right order of magnitude) doesn't take me to anything like the form of your calculation. Rather than flip through other textbooks, let me rely on my maybe-shaky memory and say that most of them, at least the modern editions, follow the TT gauge approach and come up with an equation in a form similar to Carroll. What stands out here is that Earth-sun has a large mass ratio and noncircular orbit, and it's not really a binary system anyway, and so will defy these textbook schemes. Secondly all of these take P in the far field, because linearization. (Compare that with your edit). I also got way off into the weeds wanting to work with chirp mass (rather than q=m_1/m_2) which is what GW obs data analyses use because extracting the individual masses is hard. I also know a bit about EMRI BHBs (extreme mass-ratio inspiral) and in those dissipation is dealt with differently from textbooks even for soft binaries (e.g. soft->hard roughly PN & perturbative methods, EOB, GSF, numrel) absolutely none of which is of immediate practical use here. So that's some of what motivates my question about the origin of your calculation. ETA: I think most of what happened here is that my brain reads equations as words, and I simply forgot that I could actually rearrange teh ltteres! TGIF :-)
derefr 3y ago
Could there exist a material that resistively dissipates gravitational flux more strongly than normally, in the same sense that ferromagnetic materials resistively dissipate magnetic flux more strongly than normally? Something something "gravitational permeability"? (I guess such a material would interact with gravitational waves as if it had more mass than it does, without that affecting its inertial mass?)

nyrikki 3y ago

Hard or impossible to visualize, but as the temporal dimension is so much larger, most of the deformity is in the time dimension. But yes, spacetime itself is jiggly like Jell-O. As we can't visualize 4d spacetime, most analogies will be wrong. But as photons don't experience time themselves, thinking about the geodesic path is probably less error prone than thinking of it as squishing physical objects. Objects being pulled towards slower flows is the intuition that matches the math best for me.

sunshinerag 3y ago
what does "photos don't experience time themselves.." mean? why not?
- alfiopuglisi 3y ago
  Photons travel at the speed of light, and at that speed, any "subjective" time is zero. In Einstein's theory of special relativity, the faster you go, the slower your proper time appears to an external observer. At the speed of light, this effect reaches infinity.
  - euroderf 3y ago
    Okay, so what does a photon observe when it partakes of Cherenkov radiation ?
  - criddell 3y ago
    > the faster you go, the slower your proper time appears to an external observer From my perspective, it takes about 8 minutes for a photon from the sun to hit my eye. From the perspective of the photon, a little time has passed, no? Doesn't the atmosphere and passing through my glasses slow it down a wee bit? Can the photon "know" that its position has changed between emission and absorption? From the photons point of view, I must be very, very close to the sun, right?
    - ben_w 3y ago

roywiggins 3y ago

They are predicted to leave permanent distortions in spacetime, though I think it's a more complicated mechanism. https://www.quantamagazine.org/gravitational-waves-should-permanently-distort-space-time-20211208/ https://www.quantamagazine.org/gravitational-waves-should-pe...

piyh 3y ago
Flashback to 3 body problem and harvesting energy from local c.

marsten 3y ago

In principle yes, gravitational waves dissipate energy into ordinary matter they pass through. However the coupling is extremely weak (that gravitational waves are so hard to detect is testament to this). In fact this weak coupling is what makes GWs so interesting for observational astronomy: They propagate from the source to our detectors virtually unchanged. (This is in contrast to EM radiation, which is very easy to scatter.) For example the farthest we can see back with EM radiation is about 200k years after the Big Bang, when the plasma of the early universe recombined into neutral hydrogen. By contrast gravitational waves can see back to the Big Bang itself, so it is a truly unique source of information as compared with light.

HappMacDonald 3y ago
I would rate the weak coupling as a far second or third point of interest behind linear signal fall-off vs inverse square for most other kinds of signal such as electromagnetic waves. Gamma ray burst is twice as far away? It's four times dimmer. A thousand times as far? A million times more dim. Gravitational wave signal from <event> is twice as far away? Makes it twice as hard to detect. A thousand times as far away? Only a thousand times as hard to detect.
- marsten 3y ago
  To be clear, gravitational waves follow the same inverse square law as EM radiation, in terms of power. This is after all just a statement of conservation of energy. The linear drop-off you're referring to is when we look at it in terms of field strength (in this case the spacetime strain). Since power is proportional to field squared, this implies a linear drop-off in the field. It just so happens that for GWs it's easier to detect the field, whereas for (most) EM radiation it's easier to detect power. There are field-detection methods for EM radiation as well, which are useful for weak signals. Homodyne and heterodyne detection are good examples.