There are few questions in Phys.SE concerning the speed of gravity, and the answers are traditionally that the speed of gravity equals to the speed of light. But in that case I have three more specific questions which I didn’t find in those discussions, so let me ask them here separately.

  1. All cosmological and astrophysical N-body simulation codes (e.g., the popular Gadget-2 code) calculate gravitational force acting on a given particle from all other particles taking their positions at the same moment of time. Well, that’s perhaps adequate if one simulates a small stellar cluster, but how can that be adequate when one simulates large-scale structure of the Universe? But most weird thing is that I failed to find any justification for that either in manuals or in Internet discussions – as if it is just obvious that gravitational interaction acts instantaneously irrelevant of the separation.

  2. Astronomers calculate orbits and predict positions of celestial objects without taking into account the finite speed of gravity, and yet they get accurate results. They do make correction for the time that will take light to travel from the object to the Earth to observe it in its predicted position, but if similar correction is introduced for gravity itself, the result becomes in fact incorrect.

  3. If the speed of gravity does not exceed the speed of light, how can black holes produce gravitational fields? The mass of the black hole is under the event horizon, and yet it manages to update its external gravitational field, whereas in fact the black hole should not reveal itself even gravitationally in this case.

The last two questions are actually spurred by the paper of Tom Van Flandern The speed of gravity – what the experiments say (Phys. Lett. A 250, 1998) that I have come across recently. In fact, there are more interesting points raised in the paper, e.g. concerned with aberration - gravity has no observed aberration though it should have it if it propagates at the speed of light, and the ultimate conclusion is that the speed of gravity is at least ten orders of magnitude higher than the speed of light.

There is some reaction to the paper. For example, Steven Carlip’s Aberration and the speed of gravity, Phys. Lett. A 267, 2000. But Carlip focuses on aberration and concludes that in general relativity aberration might in fact be cancelled by velocity-dependent interactions. But what about the three questions outlined above? For me the most striking fact is that in predicting positions of celestial objects the speed of gravity is taken to be infinite and predictions turn out to be correct. If that is the case, I don’t understand at all why we are still taught that nothing can travel faster than light. Can anyone clear up the situation in more or less simple terms?

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    $\begingroup$ Comment to the post (v3): Subquestion 3 is a duplicate of this Phys.SE question. $\endgroup$
    – Qmechanic
    Mar 4 '14 at 19:26

I don't think I can rigorously prove that simulation engines don't need to worry about the (possibly? I don't know if there's a reliable measurement) finite speed of gravity, but I can offer some lines of thought that point in that direction.

I'll start with your question 3. Suppose that gravity does have finite speed equal to $c$. Your question seems to be asked from a somewhat newtonian point of view, but looking at it with GR in mind, it looks a lot better. In GR, mass causes a deformation of spacetime. If gravity has speed $c$, what we're really saying is that changes to the deformation propagate at speed $c$. In terms of the curvature of spacetime, there is nothing special about the event horizon of a black hole, everything is well-behaved until you get to the singularity. The event horizon only becomes relevant when you start talking about how other particles propagate in the curved spacetime. Try replacing "gravitational field" with "curved spacetime" in your thinking and see if you're happier. Incidentally, I don't know if/how this reconciles with the hypothesized graviton particle. Maybe this is part of the difficulty in formulating a quantum theory of gravity?

Now on to your question 1. I think the answer is some combination of the homogeneous property of the Universe on large scales, and the shell theorem. Let's look at a "test particle" beside an extended, spherically symmetric self-gravitating mass distribution. Suppose that both have been in place for at least time $D/c$, where $D$ is the distance to the most distant part of the mass distribution (it turns out that "horizons" in the sense of the sphere of influence of a particle end up not being problematic, more on this later). So the test particle "knows" about the mass distribution. Now let's go through the exercise of calculating the acceleration of the particle with finite-speed gravity and infinite-speed gravity.

Infinite-speed gravity:

The mass distribution is spherically symmetric, so by the shell theorem any particle outside it feels a force equivalent to that exerted by a point mass with the same total mass placed at the centre of the distribution. Well, that was easy.

Finite-speed gravity:

Well, as long as the total mass of the distribution doesn't change, and it remains spherically symmetric, and its centre doesn't move, and it doesn't expand so much that the test particle ends up inside it, it just keeps behaving as a point mass at its centre at all times. So it doesn't matter how fast the signal propagates, the acceleration computed for any time will be valid for all times. But there are quite a few assumptions that need to be valid before this can be used. Let's have a look at each one.

  • The total mass of the distribution doesn't change. Well, assuming no mass sources or sinks seems reasonable. There's another way the mass can "change", but it's covered by one of the later points, so I'll get to it in a minute.
  • The distribution remains spherically symmetric. This is a lot to ask for, especially since asymmetry is amplified in gravitational collapse. But on large enough scales, the Universe is thought to be homogeneous, and homogeneity implies spherical symmetry (for any arbitrary spherical sub-volume of the homogeneous distribution). So we might need to worry about small scales, but on large scales we're ok. More on this later.
  • The centre doesn't move. Again, homogeneity saves us on large scales. The mass distribution is uniform (and remains that way), so by definition if one parcel of mass moves away, another moves in to take its place.
  • The test particle doesn't end up inside the mass distribution. Clearly a test particle in the Universe is inside the mass distribution of the Universe. But if we ignore the immediate vicinity of the particle (so, still thinking about large-scale interactions) then this is safe.

Ok, so it looks like everything works out for large scales. So what do we need to do in a simulation to handle this? As John Rennie points out in his answer, we use the FLRW metric. Applying this means we just need to know a few parameters (the density, or energy-density, in different components, e.g. matter, radiation, $\Lambda$). This gives the global curvature of the spacetime. Since the curvature is the same everywhere at a given time, it doesn't matter how fast it propagates. The curvature is time-dependent, but its time evolution can be calculated independently of the details of the mass distribution, provided homogeneity on large scales is preserved (as you can tell, if the Universe is inhomogeneous on large scales, simulators are in BIG trouble). In a simulation engine like Gadget-2, a lot of the complications of computing in an expanding metric are removed by transforming to "comoving coordinates". This doesn't affect the solution, just makes it easier to compute. The physical solution can be recovered by making the inverse transformation back to "proper coordinates".

Ok, what about locally, where things get inhomogeneous? It turns out we still don't need to worry about the speed of gravity, based on a bit of error analysis. Consider a pair of simulated particles. The ratio of the acceleration one of the particles feels if gravity is infinite speed to the acceleration it would feel if there was finite speed is (considering just magnitude): $$\frac{a_{inf}}{a_{fin}} = \frac{r(t-\Delta t)^2}{r(t)^2}$$ where: $$\Delta t = \frac{r(t-\Delta t)}{c}$$ and $c$ is the speed of gravity. In other words, the acceleration due to the other particle where it is now compared to the acceleration due to the other particle where it was when its "gravitational signal" was emitted. To first order in $\Delta t$, the position where the particle was and where it is are related by: $$r(t-\Delta t) = r(t)-v(t-\Delta t)\Delta t$$ Putting this all together we get: $$\frac{a_{inf}}{a_{fin}} = \left(1+\frac{v(t-\Delta t)}{c}\right)^{-2}$$ So as long as the particles are not moving relativistically (with respect to the speed of gravity, which could be different from the speed of light), the error incurred is quite small (that is, quite small compared to some of the other errors that come up in the calculation). In fact, things are even better than this. When considering an entire simulation, what you'd have to worry about are not so much individual relativistic particles, but rather relativistic changes to matter distributions. And it turns out to be difficult to come up with mechanisms that could drive such rapid changes on scales meaningful to cosmology. Since basically the only mechanism to work with is gravity, you would need relativistically changing mass distributions to drive relativistic changes in other mass distributions. Since there aren't any in typical initial conditions, they don't tend to arise.

Finally, a word on horizons. The times (scale factors/redshifts) probed by typical cosmological simulations are late enough that the horizon scale is large enough for homogeneity to hold inside it. This means that horizons are of little consequence since, to a good approximation, any additional force from some mass entering an observers horizon at one point will be compensated by the same amount of mass entering diametrically across the horizon. There are no sudden changes to the potential, which is what we'd be worried about if the speed of gravity were finite.

  • $\begingroup$ John Rennie beat me to the punch, I was part way through writing this when his nice answer came up, then was delayed in completing my answer. But I have a few complementary things to add, so I'll put this up anyway. $\endgroup$
    – Kyle Oman
    Mar 4 '14 at 21:20
  • $\begingroup$ Thanks, I have to digest it. For now I would like to understand how you derived the first formula - why the ratio of accelerations is proportional to the ratio of squared distances. P.S. By the way, the Universe does seem to be inhomogeneous on large scales as suggested by the recent discovery of Large Quasar Group and Hercules-Corona Borealis Great Wall. $\endgroup$
    – ThisGuy
    Mar 6 '14 at 4:35
  • $\begingroup$ @ThisGuy For the derivation, it's just the ratio of a=-GM/r^2 for the two cases. r is either evaluated at the present time (instantaneous transmission), or in the past equal to the time it took for the signal to arrive (my def'n of delta-t). The particle sees the other particle "where it was" when the signal was emitted. And my tongue-in-cheek response to your comment on inhomogeneity is that... well... you need to look on larger scales! ;) $\endgroup$
    – Kyle Oman
    Mar 6 '14 at 6:34

It's tempting to think of gravity as some kind of interaction between the two bodies involved - maybe some form of signal (gravity wave?) sent between the two bodies. If this were the case then you would indeed have to allow for a propagation delay as the signals were sent between the two bodies. However this is not how gravity works.

A massive object curves spacetime in its vicinity. For example the Earth produces a curvature in spacetime (approximately) described by the Schwarzschild metric. Another object moving near the Earth isn't interacting with the Earth - it's interacting with the spacetime curvature caused by the Earth. Furthermore this interaction is local, that is our test object interacts with the spacetime curvature at its location and therefore with zero delay.

Changes in the curvature do indeed propagate at the speed of light, but once the curvature has been established there is no delay for objects interacting with that curvature.

This immediately answers your questions 2 and 3 (I'll come back to 1). Planets, asteroids or whatever moving around in the Solar System are moving in a straight line in the curved spacetime at their position. The spacetime curvature deforms the straight line into the elliptical orbits we see the planets following. The planets don't need to know the Sun is there - if some hypothetical alternative means could be found of creating the same spacetime curvature at the planets location it would move in the same way. We don't need to worry about the delays of the planets interacting with the Sun because they aren't interacting with the Sun. They are just interacting with their local spacetime curvature.

This also addresses your concern about how gravity escapes from a black hole. Your question says:

The mass of the black hole is under the event horizon, and yet it manages to update its external gravitational field

but it doesn't update the external gravitational field. The Sun curves spacetime in our vicinity and (assuming spherical symmetry - which is a good assumption in the Solar System) this curvature doesn't depend on how big the Sun is or how dense it is, just it's mass. If the Sun suddenly collapsed into a black hole the spacetime curvature in our vicinity would not be affected and the Earth would continue to orbit just as it currently does. You don't have to worry about how gravity would escape from the black hole because it doesn't have to. The Earth is just moving according to its local spacetime curvature and it neither knows nor cares what's going on inside the event horizon.

Finally back to your first question. If your system is very large then you need a full General Relativistic treatment to describe it. For example to describe the whole universe you solve Einstein's equations with a few assumptions about the distribution of matter and get the FLRW metric. This then tells you things such as the expansion rate.

However, for smaller systems the changes in the overall spacetime curvature are relatively slow and predictable, and you can do a fair job of describing the motion without needing to worry about the speed of propagation of those changes. For example we often assume a high level of symmetry (as in the FLRW metric) so the same changes are happening everywhere at the same rate and there are no propagation delays.

  • $\begingroup$ Well, I now had time to think over your answer thoroughly :) I think I am still not satisfied with my question #2 on predicting the orbits of celestial bodies, even when I think of gravity in terms of spacetime curvature. Certainly, if Earth was stationary (whatever it means :) ), then the curvature that it produces would also be stationary. However, the Earth moves, and so does the curvature that it produces. So if I want to predict the position of a body orbiting the Earth (i.e., moving in straightly in spacetime curved by the Earth), I must take into account that this curvature changes. $\endgroup$
    – ThisGuy
    Jun 11 '14 at 6:40
  • $\begingroup$ (... continue). And it changes with the speed of light (presumably). But, as far as I know, astronomers do not take that into account, and they get correct answer. $\endgroup$
    – ThisGuy
    Jun 11 '14 at 6:43
  • $\begingroup$ @John Rennie One issue with your explanation here - it treats gravity different than any other static field, when it doesn't have to. Static electric fields behave exactly the same way - you don't use the retarded positions of particles to determine where the source is (as if there were propagation delay) for any static field produced by a body at constant velocity, only for waves produced by accelerated motion. $\endgroup$
    – JPattarini
    Dec 16 '16 at 14:25

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