The gravitational way has a direction (say, z) in which its propagating. Within the plane perpendicular to that direction (x-y), a circular ring of particles will at at one moment experience squeezing in one direction (x) and stretching the perpendicular direction (y). As the wave passes through and you move from the peak of the wave to the trough, the directions reverse, so the first direction (x) stretches and the other direction (y) squeezes. By "stretching" and "squeezing" I mean instantaneous additional (positive and negative) acceleration on top of the (much, much larger) acceleration from the background gravitational field provided by the Earth.
Just as a child can swing their legs at the resonant frequency of a swing to pump up their sinusoidal amplitude, a very weak gravitational wave can pump up a ring oscillator if it's oscillating at the ring's resonant frequency.
Exactly square gravitational waves are of course not possible, just as for electromagnetic waves. (They would have infinite energy at the corners.) But in principle you could get a close approximation. However, spacetime is incredibly stiff, and I think all the known real-world sources produce pretty smooth waves. I presume most violent events are mergers of existing black holes, and essentially always result from a smooth in-spiral rather, say, a sharp collision event. This is what the "chirp" signal looks like to the LIGO detector:
The effects of a square wave would be roughly as you would expect: instead of smoothly pumping up an oscillator, it would give it a sharp kick, just as with electromagnetism.
> Just as a child can swing their legs at the resonant frequency of a swing to pump up their sinusoidal amplitude, a very weak gravitational wave can pump up a ring oscillator if it's oscillating at the ring's resonant frequency.
So is it possible that a passing gravitational wave could initiate some natural process that otherwise might have not happened?
I was thinking more along the lines of a chemical or nuclear reaction (or some yet-to-be-conceptualized space-time reaction) at near an initiation point, but not quite there.
I assume the gravity wave could push the reaction to initiate by warping a subatomic element (like an electron orbital) into an otherwise impossible configuration on a scale of picometers for a split second.
The nuclear and electromagnetic forces in the vicinity are stupendously stronger than gravity at small scales. One could imagine that some electromagnetic or nuclear process might be heavily suppressed by some symmetry such that it could only be unlocked by gravitational forces (which would break that symmetry), but nothing like that comes to my mind. Basically, anything that gravity could do could be done by electromagnetic and nuclear forces vastly faster.
These gravity waves aren't strong enough to pull us off Earth. Everything on Earth is already subjected to gravity, and it doesn't cause nuclear reactions. Gravity waves won't cause impossible configurations to become possible.
Well it depends. How big a wave are we talking about? The ones so far have been very distant. But when to 100 solar mass black holes merge and form a 90 solar mass black hole they release 10 solar mass of energy (E=mc^2 strikes again). Being close to such a merge would be lethal.
However there's nothing that we know of that could happen close by, so the risk is near zero. Apparently the black hole at the center of the milky way is going to merge with another super massive blackhole in Andromeda in 4.5 billion years or so.
I think you are overestimating the amount of power in these gravitational waves by several dozen orders of magnitude.
To even detect these, we need to observe multiple pulsars over long periods of time in order to find minute effects only visible at galactic scales due to the nanohertz frequency of these waves. In other words, spacetime is being stretched and compressed at subatomic scales on a sinusoidal wave with a period of a month or so.
It’s a bit like asking if cosmic rays from the Pinwheel Galaxy are affecting cancer rates.
Makes sense. I am a bit of a drive by tourist on this; such that the only reason I asked this was the analogy of a kid pumping their feet stuck on me and messed up my assumed scale on effects.
I don't see it as impossible... Cosmic events frequently give rise to events that might not otherwise happen.
The real question is, is the change meaningful? If you have a tsunami but it doesn't change anything meaningful, is it an interesting observation outside of the event itself?
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?
> 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.
"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.
Agreed; it's very serendipitous that you can basically light up a room with the gravitational energy that the sun/earth pair radiate out into space.
Light up 1 room with an incandescent bulb, or your entire flat with LED bulbs nowadays. I would be very interested in seeing some napkin math, based on power efficiency progress and "rate of technological innovation", that attempted to project when we could feasibly run the equivalent of our present-day human civilization purely off of gravitational waves/radiation.
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 (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 :-)
I used a paper from 1968 ( https://doi.org/10.1098/rspa.1968.0004 ), and treating the centripetal acceleration due to orbital motion as uniform acceleration. It's good enough for the Earth-Sun system, but it'll fail in the case of something like close orbiting pulsars.
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.
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?)
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.)
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.
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.
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.
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.
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.
> 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?
It takes time for a photon to move in your reference frame, but time within the photon's own reference frame is not advancing at all during that. Within the photon's reference frame, the photon exists instantaneously, simultaneously, at its emitter and absorber. Its whole existence "brings together" the spacetime it was emitted from and the spacetime it is absorbed into, at a single 4D pinch-point. It's like the whole universe is squished flat into two hyperplanes of "everything behind the photon at time of emission" and "everything ahead of the photon at time of absorption", and those two hyperplanes have no distance between them.
> Doesn't the atmosphere and passing through my glasses slow it down a wee bit?
When a photon is travelling through anything other than vacuum, it's not "slowed down." It's repeatedly being absorbed and re-emitted. (Or rather, it's being absorbed, and new photons that happen to be mostly equivalent are being emitted.) The refractive index of a material is effectively a measurement of the likelihood of absorption, times the average per-particle time-delay between absorption and re-emission.
No because a single photon doesn't *experience" the whole universe but only the points where it's emitted and absorbed and you could say all the points in between along the geodesic between the emission and absorption events
So if it is never absorbed, flying on thru the vast emptiness of space, does it experience the sum total of the existence of the universe ? (along that geodesic, of course)
Ah yeah, along that geodesic, which is far from the total of the universe. But I think I now understand that meant "total duration of the existence of the universe".
I think an answer to that question depends of what you mean by total duration of the universe.
If the photon never gets absorbed by anything, then it goes in forever. If it exists then the universe still exists forever since at least one photon exists forever.
No. From the photons perspective, there is no concept of time. Phase speed, group speed, shadows going faster than the speed of light, etc.. will all complicate using the concepts used to teach diffraction
Massless particles being required to travel at the speed of light is perhaps a lens to think about it.
There is a lot of confusion in this thread. One is the topic of photons vs proper time. tl;dr use affine time for things that move at c, and proper time for things that move at less than c, and remember that nobody's coordinate time is in any sense the time in relativity.
Some quotes from this thread:
> photons [have] no concept of time
and earlier
> Nothing that travels at light speed experiences time. For a photon, emission and absorption is a single event.
and other commenters in the same thread
> Photons ... "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
> time within the photon's own reference frame is not advancing at all
and even Don Lincoln in a linked video in this thread: "we have to be careful since the equations of relativity don't apply for travelling at the speed of light, but hopefully you see that this limit trick allows us to get arbitrarily close. So I think we can see that a photon experiences no time ..." Thus everyone above is in good company with these slogans. However, Don Lincoln almost certainly knows he needs to correct s/the/these/ (in the context of the lim v->c analysis in the video), and that his conclusion needs to be understood as "no proper time" in that context. But we also all know that it's a youtube pop sci outreach video, not a university lecture or crucial vital factual no-fake-news hackernews thread.
So, let me make the counter-propostion: photons evolve on their worldlines, so must experience some time.
Additionally, elastic Rayleigh scattering supports the idea that there may be one or more point-coincidences along the worldline of a photon. There can also be non-scattering point-coincidences where the photon's momentum energy is some fraction less than 1/1 of the energy-density (the stress-energy) at some point in spacetime along its worldline even if the photon does not interact non-gravitationally with the rest of the stress-energy there (e.g. at that point there could be one or more of a neutrino, free neutron, dark matter particle, or another photon). We should be able to describe such a point-concidence in coordinates adapted to our photon's worldline, just as we could adapt them to e.g. the free neutron's worldline.
Relativity gives us (for all practical purposes, fapp) total coordinate freedom. Point-coincidence physics are invariant under changes of coordinates.
So we can label any curve any way we like, without changing the physics of anything touching that curve.
Proper time \tau solves the timelike geodesic equation, which makes \tau handy for labelling points along a timelike geodesic, but \Delta\tau = 0 on null geodesics, so is not suitable for them.
There is a unique labelling of points along a null geodesic that does solve the geodesic equation, and that is the affine parameter. See https://physics.stackexchange.com/questions/17509/what-is-th... to save me a bunch of typing. Note that as the third answer says one can use the affine parameter to calculate and explain the gravitational or cosmological redshift as a consequence of the null geodesics picked out by the Einstein Field Equations.
That (and the equivalence principle) is also a satisfying way of understanding the relation E = hf (see the first equation at <https://en.wikipedia.org/wiki/Photon_energy#Formulas>) in a lab-scale patch of flat spacetime.
Otherwise, how do you explain any \Delta f if photons "have no concept of time"? You and others in this thread appear to have been arguing that in Special Relativity the standard inertial frame for massive particles is inappropriate for showing the time-evolution of massless particles. That's true. But the point of relativity is that we can deploy (fapp) any system of coordinates and if we are doing covariant physics (i.e. using tensors; one might start with chapter 11 of J.D. Jackson's textbook which is freely available online (and 2nd ed is on the Internet Archive) and is very widely used in teaching) then it almost doesn't matter what system of coordinates we use.
Almost: we can choose practically useless coordinates, like labelling a curve in a non-monotonic way, or labelling points non-uniquely. In fact, any f(\tau) does both of those on a null geodesic: every point gets labelled with a 0. That's not the photon's fault, that's the fault of trying to use an inappropriate system of coordinates. To be fair, such coordinates seem like obvious choices by a person familiar with their use in inertial frames for massive objects, but who then may be misled into thinking the inappropriateness of the coordinates for objects on null geodesics determines the physics of those objects.
Unfortunately, this mistake is very common, and has led to poor slogans which have been repeated many times by several people in this discussion.
If one wants to sloganize, "proper time is inappropriate for photons because they are massless" (cf. §1.2 on the inverse square law and photon mass in Jackson) "but just as nobody's proper time is preferred in relativity, neither is any proper time; and for photons affine time is a useful substitute".
> Massless particles being required to travel at the speed of light is perhaps a lens to think about it.
Indeed, and I just did that for you, although Jackson and I would flip that around to say that c is the speed of massless particles and experimentally (and for theoretical reasons) photons are massless.
FWIW (I'm probably the only one ever likely to read this), in Ch. 7.0 Lightman, Press, Price, and Teukolsky's Problem Book in Relativity and Gravitation puts the previous comment in reverse (which I love and will steal). The entire brief section quoted below is glorious in its economy of English, too.
Forgive the lazy \latex anyone who actually sees this, including future me.
"If u is the tangent vector to a curve, a tensor Q is said to be parallel propagated along the curve if \nabla_u Q = 0. If the tangent vector is itself parallel propagated, \nabla_u u = 0 (tangent vector "covariantly constant") the curve is a geodesic, the generalization of a straight line in flat space. If x^\alpha(\lambda) is the geodesic (with u^\alpha = dx^\alpha / d\lambda) then the components of the geodesic equation are
"Here \lambda must be an affine parameter along the curve; for non-null curves this means \lambda must be proportional to the proper length.
"If a curve is timelike, u is its tangent vector, and a := \nabla_u u = Du/d\tau, then a vector V is said to be Fermi-Walker transported along u if \nabla_u V = (u \otimes a - a \otimes u) \cdot V."
See also problem 7.11 and its solution.
Also of interest is Matthias Blau et al. 2006, "Fermi coordinates and Penrose limits". doi:10.1088/0264-9381/23/11/020 hep-th/0603109 which adapts Fermi coordinates to null geodesics. (abs. "(Fermi coordinates are direct measures of geodesic distance in space-time)... We describe in some detail the construction of Fermi coordinates" §4, "We now come to the general construction of Fermi coordinates associated to a null geodesic \gamma in a space-time with Lorentzian metric g_munu. Along \gamma we introduce a parallel transported pseudo-orthonormal frame ... Fermi coordinate are uniquely determined by a choice of pseudo-orthonormal frame along the null geodesic \gamma" "For many (in particular more advanced) purposes it is useful to rephrase the above construction of Fermi coordinates in terms of the Synge world function").
This isn't quite true. At a macro level, light slows down. At a micro level, photons travel at the speed of light, get absorbed and re-emitted, and change directions. These effects average out to looking like photons traveling slower.
At the micro level there are infinitely many paths the photon follows and most of them involve it traveling considerably faster than the speed of light. While they cancel out, if you omit them then the calculations don’t give correct results, so they’re happening inasmuch as anything can be said to happen scientifically. If you want to object that that contradicts GR well there is a Nobel prize or ten waiting for whoever squares that circle.
I did, and enjoyed it as a good pop-sci outreach video, and do not fault it given that's what it is. The lim v->c analysis is standard and well-presented, but his conclusion about photons' time is liable to confuse people (including other physicists who aren't relativists). I refer you to my comment elsewhere in this thread <https://news.ycombinator.com/item?id=36537015> which has a paragraph about this video (your link having partially motivated my comment), and which explains that it is common for relativists to use affine time for massless particles.
"However, spacetime is incredibly stiff, and I think all the known real-world sources produce pretty smooth waves."
In English, why does stiffness correlate to smooth waves? What does stiff spacetime mean? I'd have thought a square wave would be "stiff" as it's quite the opposite of smooth.
In order for a square wave to travel through a material, it would need to allow for infinitely high frequencies (see the animation on wikipedia[0] for a visual demonstation). The lower the maximum frequency possible, the more every wave will resemble a sine wave because it's the most basic shape: any non-smooth wave has higher-frequency components to give it its shape. Filter out those higher frequencies, and you're left with the basic smooth sine wave again.
A stiff material tends to dampen high frequencies, simply because it cannot deform fast enough to follow the wave's shape. In a way, the medium acts as a low-pass filter; compare, fow example, how fast you can clap your hands in air vs in water: the stiffness of water slows down your movements so you cannot reach high clapping frequencies.
> A stiff material tends to dampen high frequencies
All materials roll off their frequency response. But this is backwards - stiffer materials have higher resonant frequencies given equal density. It’s basically a word to describe a high spring constant.
I think what the GP was stating here were two independent reasons why we would find it hard to know what would happen if a gravitational square-ish wave interacted with spacetime: 1. spacetime is "stiff", and 2. there are no gravitational square-ish wave generators around to observe.
The two things are kind of related, though. One "natural" way to create a square wave in nature, is to "interrupt" a material transmitting a sinusoid wave, at the peak of its transmittance. And one way to do that, is to break through the modulus of elasticity of the material transmitting the wave, such that it switches from the elastic-deformation domain (transmitting the wave) to the plastic-deformation domain (ripping apart.)
Imagine a speaker cone tearing at the peak of a high-amplitude drum beat. The cone pushes "out" — and then doesn't push back "in", because instead the air behind it rips forward through it. The air created by the speaker cone wants to rush back "in", but now there's no longer a speaker cone acting as a waveguide for the inward flow, so the natural turbulence cancels out much of the "falling" energy of the wave, making it look much more like a square-wave drop.
I believe that the GP is saying that, because spacetime has such a high effective "modulus of elasticity", we haven't yet observed any practical way to perturb so as to create the conditions that would generate a gravitational square wave.
Imagine using a flick of your wrist to create a traveling transverse square wave in a string vs. in a long metal dowel. It's actually not so hard to create a smooth sinusoidal wave in a metal dowel. (It will be long wavelength, but it will be easily visible if you put your eye near one end and look down it.) But it would be impossible for you to make a square wave.
Thank you for that! As an EE and radio enthusiast, I feel like I have a fairly good grasp of how gravitational waves behave from that description. The way in which the mechanics of these different energies all sort of share characteristics is rather beautiful.
Could you set up massive pendulums to convert this gravitational wave energy into mechanical energy? Would such a device diminish the strength of the wave downstream?
Yes! In principle at least, and only miniscule/undetectable amounts of energy. The first gravitational wave detectors build were these "Weber bars" [1] - big block of aluminum that were supposed to resonant mechanically (like a bell) when a gravitation wave of the right frequency passed through them.
Looks like these things haven't detected a real gravitational wave, but if a strong enough one at the right frequency came through, they might start ringing like (very quiet) bells!
The gravitational way has a direction (say, z) in which its propagating. Within the plane perpendicular to that direction (x-y), a circular ring of particles will at at one moment experience squeezing in one direction (x) and stretching the perpendicular direction (y). As the wave passes through and you move from the peak of the wave to the trough, the directions reverse, so the first direction (x) stretches and the other direction (y) squeezes. By "stretching" and "squeezing" I mean instantaneous additional (positive and negative) acceleration on top of the (much, much larger) acceleration from the background gravitational field provided by the Earth.
Here's a visualization:
https://www.researchgate.net/publication/313828462/figure/fi...
Just as a child can swing their legs at the resonant frequency of a swing to pump up their sinusoidal amplitude, a very weak gravitational wave can pump up a ring oscillator if it's oscillating at the ring's resonant frequency.
Exactly square gravitational waves are of course not possible, just as for electromagnetic waves. (They would have infinite energy at the corners.) But in principle you could get a close approximation. However, spacetime is incredibly stiff, and I think all the known real-world sources produce pretty smooth waves. I presume most violent events are mergers of existing black holes, and essentially always result from a smooth in-spiral rather, say, a sharp collision event. This is what the "chirp" signal looks like to the LIGO detector:
https://www.youtube.com/watch?v=TWqhUANNFXw
The effects of a square wave would be roughly as you would expect: instead of smoothly pumping up an oscillator, it would give it a sharp kick, just as with electromagnetism.