Gravity vs Gravitational Waves

Question

I thought I had a reasonable understanding of relativity, the speed of light speed limit, and how this stuff related to gravity. Then I read through all the answers/comments for this question:

How does Zumberge's 1981 gravitational measurements relate to gravitational waves?

And now I'm more confused than ever. Here's the opening for the (currently) most upvoted answer:

"This represents a major misunderstanding of what a gravitational wave is. The effect presented is simply the semi-static gravitational field at earth due to the earth, moon and sun. It is predicted by Newtonian gravity. There is no 'wave' that propagated, it's the instant positions of the 3 bodies that change over 1 day (and over 1 year also). "

So... my understanding of relativity prohibits the existence of a "static gravitational field" based on the "instant positions" of some masses. If it were allowed, that would imply that information about gravity traveled at infinite velocity which violates relativity. All changes in gravitational fields must propagate at a maximum of the speed of light.

Fundamentally though, I guess my big confusion is that it seems like everyone in that thread keeps arguing about some difference between "measuring gravity" and "measuring gravitational waves". Is measuring gravity equivalent to measuring magnetism around a permanent magnet? Is measuring gravitational waves equivalent to a digital camera capturing photons?

edit diving a little deeper

There's more context in the linked question, but I guess to begin with, I always assumed that "regular gravity" was "transmitted" via virtual particles of some type of gauge boson. So I always assumed that "gravity waves" then must refer to "real" particles in the same way that photons are real particles. Furthermore, afaik, there's no possible way to detect a static magnetic field using a mechanism which could also detect a photon. That's the whole point of a virtual photon.

So in terms of the magnet vs camera analogy, that doesn't make any sense to me in the context of the linked question. Let me bring some of that background here.

"This is the SAME gravitational wave effect measured by the LIGO researches recently

This is not a gravitational wave, this is a measurement of tides caused by the movement of a single detector within the earth-moon dimple. If there had been a second detector, it would have been clear that these tides do not propagate across the earth at the speed of light, but at the speed of the movement of the moon."

"LIGO actually detects, then filters out, this local gravitational wave

This uses the term "gravitational wave" incorrectly. It is more correct to say, "LIGO actually detects, then filters out, this local gravitational tidal noise."

According to the magnet/camera analogy, LIGO shouldn't even be able to detect tidal effects. It's fundamentally the wrong type of particle. Also, if gravity waves are the "real" equivalent of a force carrying particle, then there would be no need for a second detector. You don't need two cameras to take one picture... That doesn't even make sense.

"Is measuring gravity equivalent to measuring magnetism around a permanent magnet? Is measuring gravitational waves equivalent to a digital camera capturing photons?" Yep! That's a good analogy. — , Apr 25 '16 at 18:51
@ACuriousMind "Changes in the EM field propagate at the speed of light. But just a bunch of charges moving around and creating a wave-like pattern in the field intensity does not necessarily constitute what one calls "electromagnetic wave"." There's more context in the linked question, but I guess to begin with, I always assumed that "regular gravity" was "transmitted" via gauge bossons, more specifically, by some kind of virtual particle — industry7, Apr 25 '16 at 19:28
Ugh, I accidently hit "enter" instead of "ctrl"+"enter", and then I didn't get my edit in during the 5 minute window. So sorry, ignore my last comment, I'm going to update my question instead. — industry7, Apr 25 '16 at 19:49
Check this link out (http://live.iop-pp01.agh.sleek.net/2015/08/20/catching-gravity-rolling-by/) a quick scan of your question highlighted a few things that this link should iron out. It's pretty easy reading. Do a control+F for "time" and just read the paragraphs that talk about deformation of space-time and such. That should make it clearer what the distinction is between an oscillation in gravity and a gravity wave — Alexander McFarlane, Apr 25 '16 at 21:28
@Alexander McFarlane - thanks for the link. Based on reading that, it seems that there are no gauge bosons involved in "transmitting" gravitational forces. The article also made it sound to me that LIGO is simply measuring changes in "normal" gravity (normal in the sense of how lay people think of gravity). That is, that "gravity waves" are just changes in the strength of gravity which manifest through propagation of space-time deformations. Would it be correct to say that fundamentally, LIGO is trying to prove that space-time deformations can't propagate faster than the light-speed limit? — industry7, May 17 '16 at 18:54
Haha I don't know the specifics of LIGO :) I would tag one of the previous commenters as gravitational fields are not really my thing I'm more on top of symmetries / QFT. As far as I was aware though I thought LIGO was just trying to detect them full stop. Perhaps one of the other commentators has more to add. — Alexander McFarlane, May 17 '16 at 19:22

Steve Harris · Answer 1 · 2017-10-24T03:20:09.833

In the matter of gravity vs. gravitational waves, I've always found it easier to think of (static) electric fields vs. EM waves.

Think of a static field "going out" from a charge. It isn't really going out. The field lines don't have "ends" that travel out at the speed of light. That's because they don't have ends at all. They all end at charges, and they stretch as far as needed. Charge is never created, it's always + - dipoles that are created (like electron-positron pair creation) so there's no problems with field lines that go out from an electron and go to the limits of the universe and end on some + charges out there. They've been stretched for that long, since the Big Bang. We don't think of how fast they move. They've been there since the beginning when the universe was small, and now they span it completely. A billion light years is nothing.

Similarly with static gravity. Mass-energy cannot just appear or disappear, so the field lines never have ends that have to move outward. They're always connected to mass and energy far away, and have had since the Big Bang to do so. There's no point in asking how fast static gravity moves. It's just "there" from here to the edge of the universe.

If you start to suddenly move, with respect to an already established static gravity or electric charge field-line, the field DIRECTION moves immediately with you, and so does the direction to the source. That's just Lorentzian relativity. The speed of light is not being violated. A source a billion light years away would suddenly start to look like it is moving, but that's because its field is already out to where you are, and the field where-you-are, tells you. It changes direction when you move. It responds immediately to relative uniform motion, via the mechanism of the field that is already extended to each.

But if the static or gravitational CHARGE moves (accelerates) then there is a "kink" or update that moves out from it at c. You don't see this at all from far away, until time d/c has passed. That's the EM wave or gravity wave. It's not a relative thing between source and viewer, because acceleration is not "relative" in relativity. You can't pretend that the observer accelerates and the source does not.

The Lienard-Weichert potentials for EM have two terms for this reason. One is the static one that depends only on relative velocity and points at the source (so long as relative velocity has been constant for long enough). The other one shows aberration (does not point at source), retardation, and is a disturbance in the field due to source acceleration (not observer acceleration).

The slow dance of Sun and Moon are a mix of both effects, in the near-field. The static effects point right at the sources, and are due to fields that already extend to infinity. They have no "speed." However, the second order effects due to small amounts of source acceleration (from orbital acceleration) are tiny, but they are genuine gravitational waves, and they move outward at speed c. They are retarded and would show aberration.

score 0 · Answer 2 · answered Mar 08 '19 at 22:15

You can not have gravitational waves without gravity, but you can have gravity without gravitational waves. Gravity is a fundamental force between masses, but gravitational waves are just a secondary effect related to the dynamics of a gravitational system in certain circumstances.

Whether gravity is transmitted instantaneously or not has been a controversial issue since Laplace. Even today scientists contradict themselves about this: for instance the ephemeris equations used by NASA to calculate the positions of the planets are all formulated instantaneously, not retarded (cf. Eq.(27) in https://naif.jpl.nasa.gov/pub/naif/generic_kernels/spk/planets/de430_and_de431.pdf ) but in other NASA resources you can read claims to the exact opposite ( https://einstein.stanford.edu/content/relativity/q2226.html ). Maybe it does not matter at all in these practical applications, as the positions are probably corrected (based on observations) often enough so that in general any errors due to mis-modelling of the physics can not accumulate (a strategy that is, in a different context, for instance also used to keep the GPS system accurate) .

In electrodynamics, the electrostatic potential for moving charges (the Lienard-Wiechert potential) always represents the potential of the charge at the position it had a time d/c ago, whether the charge is moving with constant velocity or not (this is indeed the assumption under which the Lienard-Wiechert potential is derived, see https://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential ) (so the remarks in Steve Harris's answer above are therefore not correct).

Gravity vs Gravitational Waves

2 Answers2

Linked