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I was wondering how the finite speed of gravity waves influences the behaviour of galaxies, and came up with this thought experiment, that seems to give different results when looked at from different reference frames. Forgive me for using Newtonian terminology in the description.

Assume a circle of stars, all of equal mass, distributed equidistantly along the circle. We assume there is no mass in the centre for the moment. Assume the stars all have exactly the correct velocity to travel counterclockwise along the circle. See the 8 black dots representing the stars in this figure:

Circle of stars orbiting each other

As seen from the outside inertial frame, the stars are all orbiting around the common centre of mass, they experience a gravitational pull towards the centre of mass, and as a result, only experience acceleration directed towards the centre. Because the situation is fully symmetric, that centre of mass will not move, and the stars keep orbiting forever.

But if we view the situation from one of the stars (the bottom star in the figure above), the situation is different. The gravity from the other stars takes time to reach the star, so their gravitational pull should come from slightly behind their real positions. We assume that gravity travels with the speed of light, then it will appear to pull from a star’s visible location, the white dots in the figure above. As a result, the combined gravity from the other stars shouldn’t pull towards the centre of the circle, but slightly to the right, along the red arrow in the figure.

This should also imply that the star experiences, besides the radial acceleration, also acceleration tangent to the circle. That is, its speed along the circle should increase. Because this situation is fully symmetrical for all stars, all the stars should gain more and more speed over time, and eventually they should spiral away from each other.

Because this can obviously not be true, there must be a mistake in my way of thinking. What is it?

And how does the situation change if there is a black hole in the middle of the circle? Due to the stretching of space around a black hole, gravity should take even longer to reach the other side of the circle?

UPDATE: The answer to a similar question quoted by @benrg in the comments, links to the following explanation on the web: https://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html

The main explanation in this essay is:

In that case, one finds that the "force" in GR is not quite central—it does not point directly towards the source of the gravitational field—and that it depends on velocity as well as position. The net result is that the effect of propagation delay is almost exactly cancelled, and general relativity very nearly reproduces the newtonian result.

They also state that the same is true for electro-magnetism:

If a charged particle is moving at a constant velocity, it exerts a force that points toward its present position, not its retarded position, even though electromagnetic interactions certainly move at the speed of light. [...] a calculation shows that the force on A points not towards B's retarded position, but towards B's "linearly extrapolated" retarded position. [...] This is exactly what one finds when one solves the equations of motion in general relativity.

So, in short, the gravity will not appear to come from the actual retarded position, but from an "extrapolated" position.

Is it possible to explain that phenomenon from an intuitive physics point of view, avoiding faster-than-light communication and without going into tensor calculus?

fishinear
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  • Related: Given finite speed of gravity, why didn't Earth fell into the Sun already? – benrg Apr 20 '21 at 18:13
  • @benrg That may indeed be a duplicate of the question. The conclusion in the answer is less than satisfactory, though: "there must therefore be compensating terms that partially cancel the instability of the orbit caused by retardation." I am basically asking what these compensating terms are. – fishinear Apr 20 '21 at 19:01
  • See this question: https://physics.stackexchange.com/questions/263191/can-two-heavy-objects-circling-around-their-c-m-be-separated-because-of-the-spe – Deschele Schilder Apr 26 '21 at 20:01
  • @DescheleSchilder Thanks for that link, that is indeed an identical question with good answers as well. But all answers basically boil down to "do the math, and you'll see the tangential pull does not exist". I worked through Feynman Lectures in Physics II-21, and he shows the same for the electrostatic force. So I can see that is correct for the mathematics. But somehow that does not give a good feel for WHY that happens. – fishinear Apr 27 '21 at 10:11
  • The math is indeed not very intuitive. Why does each mass see the other mass exactly in the middle? You can indeed say that upon arrival on a star the gravity seems to originate from the middle because it's deflected but why should it be deflected in precisely the right amount? The point is (I think) that the position of a star is always in the direction of where you see the other star. Light is bent also, so the direction of seeing is the same as the direction of gravity source. – Deschele Schilder Apr 27 '21 at 10:38
  • Because of this the gravity doesn't seem to originate from somewhere behind the star (on the orbit of the star) but exactly from where the star seems to be. So you seem to accelerate forever toward the other star (assuming two stars). If it seems forever than it is forever. – Deschele Schilder Apr 27 '21 at 10:58
  • Isn't this also the case in your picture? In which case the photons and gravitons seem to emerge from a direction somewhere around the axis which connects the stars (if two are present, which makes things easier to visualize). I guess so... Then the only possible way to make both the photons and gravitons come from a point on the axis is indeed to make them curve in space. – Deschele Schilder Apr 27 '21 at 11:19
  • And that is indeed the non-Newtonian element. When spacetime is flat (Newtonian), as in your drawing, the gravity and light do indeed seem to come from behind the actual position of the star (from a point on the orbit that lies behind the connection line of the stars). So from each star, it looks as if the other star finds itself "off-center". But when the photons and changing gravity field are deflected the stars will be seen "at-center". It's still quite remarkable that they are deflected in just the right amount. – Deschele Schilder Apr 27 '21 at 11:40

5 Answers5

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Your argument is inconsistent because the concept of the center of mass, as used by you, implies instantaneous interaction. If you want to use it with retarded interaction, then the center of mass has to refer to the retarded position of the star at the top, not the instantaneous one (after all, that's what the stars interact with). So you have to move the star positions of the 'source'-stars anti-clockwise so that the white dot at the top is opposite to the center of mass and the (irrelevant) black dot left of it.

However, there is a further inconsistency in your picture above: the retarded positions (white dots) as drawn by you refer to the black dot at the bottom, but in the inertial reference frame (fixed in the 'paper' plane), whereas they should refer to the moving star as the latter is the target of the gravity signal here. But for your example all the stars move with the same speed on the circle, so the distance between any two of them never changes. So effectively, all stars are at rest relatively to each other and retardation has therefore no visible effect.

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EDIT (detailing my answer in view of some of the other answers and comments)

If we restrict ourselves to just 2 masses orbiting each other, the argument by the OP (which goes actually back to Laplace more than 200 years ago) can be represented by the following diagram

Incorrect model for orbit retardation

According to this, mass $m_1$ is accelerated along its orbit as the force to the retarded position of mass $m_2$ is not radial anymore. Hence the orbit would be unstable. But this picture is incorrect. As mentioned already, it would apply to a stationary mass $m_1$ but not an orbiting one. As is obvious, the distance of mass $m_1$ to the retarded position $P_2'$ would be different here from that to the instantaneous position $P_2$. But for two masses in a circular orbit the distance between them must always be the same as $m_1$ moves the same distance within a given time as $m_2$. If we display the orbit of $m_2$ with regard to $m_1$ we get in fact the following picture

retarded orbit relative to m_1

Obviously, the retarded position of $m_2$ has the same distance from $m_1$ as the actual position. The situation is thus equivalent to one with instantaneous gravitational interaction and thus there is no effect on the dynamics of the orbit.

The gravitational interaction is something that happens between two masses. It would be incorrect to assume one mass sends out some 'gravitons' that then may (or may not) be absorbed by some other mass. If one wants to display the situation symmetrically in the center of mass reference frame, one should therefore use rather a picture like this

Correct representation in CM frame

The 'gravitons' are sent out by both masses when at the retarded positions and received when at the instantaneous positions. Assuming that the gravitons obey the invariance principle for the speed of light, the latter does not depend on the relative motion of the masses, so the gravitons are sent and received perfectly radially and thus no retardation effect occurs.

Effects would only occur for elliptical orbits, as there the distance between the masses is variable. For anyone interested, I have recently written a paper which calculates the effect of retardation on the orbits of all the planets; as it turns out, the only effect is a small (retrograde) precession of the orbits (note that this paper is as yet not accepted for publication in a journal, so use it at your own risk).

Contrary to what is frequently claimed, there is thus no General Relativity needed to answer the OP's question, as the retarded force is a central force anyway in a circular orbit.

Thomas
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  • You mean that each star "sees" its private centre-of-mass, which is offset from the centre of the circle, and moves in a small circle around the centre? The star maintains the same distance from that moving centre of mass? But how can that result in the star moving in a circle from the outside point of view? The force is never towards the centre of the circle. – fishinear Apr 20 '21 at 18:35
  • @fishinear What I am saying is that you should replace all the black dots in your drawing with white dots, and put the black dots in advance of them (towards the left for the top). You always only interact with 'ghost' images if the signal speed is finite, so you have to treat the white spots as the 'real' positions here. Only they are dynamically relevant. – Thomas Apr 20 '21 at 18:47
  • "What I am saying is that you should replace all the black dots in your drawing with white dots, and put the black dots in advance of them" - yes, I understand that is the view from each individual star. What I don't understand is how that works when looking from outside the system. When looking at the system from outside, and taking the delayed gravity into account, the force is never towards the centre of the circle, so it should not be possible for the stars to move in circles. – fishinear Apr 20 '21 at 18:57
  • @fishinear If you do as I suggested (putting the white dots where the black dots are), your red arrow at the bottom will go through the center. – Thomas Apr 20 '21 at 19:01
  • "If you do as I suggested (putting the white dots where the black dots are), your red arrow at the bottom will go through the center." - yes, I understand that. But that is not how it looks like from outside the system; for somebody looking at the system from far away. – fishinear Apr 20 '21 at 19:07
  • @fishinear If you are asking how the system looks from an outside point, that's a different question, As it happens, I have recemtly published a paper about this which shows that a given moving particle distribution appears distorted. See https://www.physicsmyths.org.uk/retarded_positions.pdf – Thomas Apr 20 '21 at 19:24
  • "I have recently published a paper about this" - That is greatly appreciated, I will have to fully go through the paper fully, of course. But from a quick glance, you show the positions as viewed from a position inside the plane. I was referring to a position outside the plane, let's say from x0 = 0, y0 = 0, z0 = 100. From that point of view, the stars should appear to move in a circle, equidistant from each other. Yet they cannot, because the force on them does not point to the centre on the circle. – fishinear Apr 20 '21 at 19:32
  • Regarding the edit to your post: "So effectively, all stars are at rest relatively to each other." That would only be the case if you consider the whole system of stars to be a rotating reference frame. But in a rotating reference, wouldn't the gravity changes, just like the light, follow curved paths? That is, the gravity pull would still seem to come from the retarded position. – fishinear Apr 22 '21 at 16:35
  • @fishinear Your graphic does not show what an observer centrally above the circle sees. Such an observer would see the same displacement for all stars. It shows rather what a fixed (non-co-rotating) observer at the bottom of the circle would see (similar to the results shown in my paper linked above). An observer co-moving with the star would not see any effect at all because all stars are in the same gravitational free fall, so they are actually in an inertial frame, and according to the equivalence principle, this is indistinguishable from all stars being at rest (with no gravity present). – Thomas Apr 23 '21 at 21:02
  • I still fail to understand why the gravity would not appear to come from the retarded position. Simply said, how does the gravity "know" during travel to the other side of the circle that it needs to "change direction"? In your own paper, you also show that the star will "see" more mass on one side of the circle than the other side. Why does it not pull to that side then? – fishinear Apr 26 '21 at 18:56
  • @fishinear The asymmetry appears only for a stationary observer for whom the circle of masses is rotating/moving. Such an observer would indeed be pulled to one side as the receding half of the circle would have more mass than the approaching half But the stars in the ring itself would not see any distortion of the circle as the stars always have the same distance to each other. Hence there won't be any retardation effects within the star circle itself. – Thomas Apr 26 '21 at 21:17
  • "But the stars in the ring itself would not see any distortion of the circle" - I am indeed trying to understand why that is the case. If a stationary observer feels the pull to the side, why not a moving observer? The retardation should not be caused by the observer moving or not, but by the distance to the other star and their relative speed. Given that a moving observer has a higher relative speed with the other star, it would theoretically see a larger retardation than a stationary observer, not a smaller one. – fishinear Apr 27 '21 at 10:24
  • @fishinear It may seem strange that two observers in the same location see things differently, but it is actually not so unusual. Consider for example the Doppler effect: two observers will see the star in a different colour depending on their relative velocity. In this case the stationary observer will see the right side of the circle red-shifted and the left side blue-shifted. An observer moving with the star would not see any shift at all (if you ignore the small relativistic effect). The same applies to the retardation. Both effects require distance changes between source and observer, – Thomas Apr 29 '21 at 21:03
  • "It may seem strange that two observers in the same location see things differently" - the problem is not that they see it differently. The problem is that an observer on the moving star should see an even larger retardation than a stationary observer at the same point, because the relative speed of the other star is larger for the moving observer. – fishinear Apr 30 '21 at 18:58
  • But according to the mathematics, the moving observer does not "see" any retardation in the gravity at all. Or to be more precise: it "sees" the other star at its projected position, based on the retarded position, and the relative velocity it had at that time. I am trying to get a more fundamental understanding for why that is the case (see update to the question). – fishinear Apr 30 '21 at 18:58
  • @fishinear The explanation in terms of GR also considers, like you, the force on a stationary mass not an orbiting one, so it is based on a false premise. And the maths it uses is incorrect as well. It is similar to the false claim in electrodynamics that the retarded force would point to the instantaneous position. See my paper https://www.physicsmyths.org.uk/retarded_potentials.pdf for more (note that this is as yet not accepted for journal publication, so use it at your own risk). See also my edited answer. – Thomas May 03 '21 at 15:55
  • "It is similar to the false claim in electrodynamics that the retarded force would point to the instantaneous position" - I worked through the math of the electric field of a moving charge myself, based on Feynman Lecture in Physics II-21, and that claim is definitely not false. The force points to the projected position of the charge. I am trying to get a better understand of the existing physics, not invent new physics. – fishinear May 04 '21 at 16:14
  • Regarding your edit to your answer, many thanks for that, it is a thorough analysis. Unfortunately it still leaves me with the same question. General Relativity describes the situation as changes to the curvature of space time that travel with the speed of light. You seem to call those changes "gravitons" in your updated explanation. In your "view from the center of mass" picture, it is still not clear to me, why those "gravitons" that are emitted at P2' would arrive at P1 from the direction of the center of mass. In my view they should arrive from the direction of P2'. – fishinear May 04 '21 at 16:43
  • Or in other words, according to my view, at any time, space-time at the position of P1 should be curved according to the retarded position P2' of the other mass, and should therefore accelerate P1 in the direction of the retarded position P2'. I understand that the mathematics magically makes it pull towards the projected position instead, but that does not help my understanding of what is actually going on. It really just feels like some magical mathematical trick, without any real physical foundation. – fishinear May 04 '21 at 16:43
  • I also fail to see how the analysis from the point of view of the target mass helps. Your picture for that analysis clearly shows that the force on the target is sideways wrt. the instantaneous position of the source mass. That is, part of the force is orthogonal to the instantaneous direction to the source mass. As I see it, that orthogonal part should lead to an acceleration in that direction, and therefor an increase in the relative linear speed between the masses. That increase in relative speed should either lead to an increase in angular speed or in increased radius or both. – fishinear May 04 '21 at 17:51
  • @fishinear I also fail to see how the analysis from the point of view of the target mass helps. Your picture for that analysis clearly shows that the force on the target is sideways wrt. the instantaneous position of the source mass. Any point of the orbit of $m_2$ relative to $m_1$ has the distance $2r$ (if $m_1$=$m_2$) , so the connecting line to the retarded position of $m_2$ must go through the center of mass. It can not be sideways. So drawing it the way you did is wrong. – Thomas May 05 '21 at 07:19
  • @fishinear I worked through the math of the electric field of a moving charge myself, based on Feynman Lecture in Physics II-21, and that claim is definitely not false. Work through the paper I linked to in my comment above where I derived the correct form for the retarded potential and field of a moving charge. The retarded electric field vector points to the retarded position (as you would logically expect). In any case, as shown otherwise in my answer, it is irrelevant for the problem here. The orbit won't be affected by it (at least not if it is circular). – Thomas May 05 '21 at 07:23
  • "so the connecting line to the retarded position of 2 must go through the center of mass" - But, as you show in your picture, that line does not go through the instantaneous centre of mass, and will therefore impart a sideways acceleration on m1. That is the thing I have been trying to point out the whole time. And, in contrast to your picture, the mathematics actually says that it does go through the instantaneous centre of mass, because it points to the projected retarded position. I just want to understand why that is the case. – fishinear May 05 '21 at 13:58
  • @fishinear as you show in your picture, that line does not go through the instantaneous centre of mass, and will therefore impart a sideways acceleration on m1 As I said already earlier, using the instantaneous center of mass would imply instantaneous interaction. For retarded interaction you have to use the retarded center of mass. In any case, you can see from my figure that the force vector is always radial to the orbit of the other mass, so the orbiting mass does not change its orbital velocity due to the gravitational interaction (retarded or not). – Thomas May 06 '21 at 07:24
  • I think that I have a better understanding now on what is happening (see my own answer on the question). Thanks for your answer and the discussion, it helped shape my thoughts. – fishinear May 06 '21 at 14:46
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I think that a simpler example is two objects rotating in a circle, like the top and bottom stars in your diagram. This should be an allowable motion, but if we use a delayed gravity approach the motion cannot be maintained. That's why simply introducing a retarded gravity is not a solution to Newton's instantaneous action at a distance.

Not_Einstein
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  • So, what is the solution that General Relativity provides for this case? – fishinear Apr 20 '21 at 18:37
  • That's a rather big question that others are more qualified than me to answer. But put simply, in GR gravity is not considered a force. Masses curve spacetime and objects move through this curved spacetime. – Not_Einstein Apr 21 '21 at 00:31
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There's no contradiction here. You would observe a tangential acceleration, which would mean that the radius of the circle would increase, and the configuration would fly apart. If nothing else, using the retarded time makes the particles slightly farther apart than they would be using instantaneous time, which would make the gravitational force slightly less than what is needed to maintain circular motion.

All you've found is that this is an unstable configuration if you use the retarded time.

Zo the Relativist
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  • This extra tangential acceleration obviously does not occur in reality, otherwise galaxies could not exist. And when you work through the mathematical equations, then it does not occur either. The stars are attracted to the "projected retarded position", that is, to almost exactly the instantaneous position. I am trying to develop an intuitive understanding for why this happens. – fishinear Apr 27 '21 at 10:02
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You treated the problem as if spacetime is Newtonian (non-curved). In reality, spacetime isn't Newtonian. If you approach the problem in a general relativistic way, you'll find that the orbits will be perfectly circular. The general relativistic calculation is complicated but you can intuitively state that the extra speed gained by the retardation (in the Newtonian spacetime) is canceled by the curvature of spacetime.
As you already stated in your comments, the only way for the stars to see the other stars exactly in the middle and let the gravity come from the middle too (so all the separate CM's will fall together) is to make the photons and the changing gravity field (which can be described by photons, but this is not necessary) curve in the curved spacetime between the stars. This has to be the case because we know that two stars will always stay in a circular orbit (if no gravitational waves are emitted, which isn't so, but that's another story). So somehow, spacetime is curved in such a way that photons and gravitons seem to have their origin in the center of the stars. Quite remarkable! But general relativity shows that it is so.

Deschele Schilder
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  • Would if be fair to say that, the gravity variations due to the moving star travel a curved path through curved space-time? And would it be true that, due to this curved path, they would be arrive approximately from the direction of the center of the circle? If true, is there a way to develop an intuitive picture of this? I would expect space-time to be flat in the middle, and to curve outwards around that, rather than the other way around. – fishinear Apr 27 '21 at 09:58
  • Good point! You would indeed expect that for the stars to see the source of the gravity in the middle the influence must be bend in such a way as to arrive from the middle. It's mainly the time curvature part of the spacetime curvature though that causes the stars to rotate around each other. So I'm not that sure. As said, the calculation is complicated (and maybe the calculation for what you describe too). – Deschele Schilder Apr 27 '21 at 10:20
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Feynan describes a similar situation for the electrostatic charge in his Lectures on Physics II-21. After studying his mathematics in detail, I think I have a better understanding now how this works. The key is to realise that the other mass has a geometric size. I’ll try to describe our rotating stars situation, concentrating on how the gravitational field of the opposite star is shaped.

The gravitational potential of a mass

Imagine a single stationary mass (like a star), and consider its gravitational potential. I like to imagine that as a steel ball making an indentation on a rubber sheet, although real connoisseurs of General Relativity will likely cringe at that analogy:

Gravitational potential of a stationary mass

The gravitational potential is the same at the same distance from a stationary mass. If we place another mass anywhere on that, then it is the gradient (the slope of the rubber sheet) of the potential at that point which determines the direction of the acceleration it will obtain. For a stationary mass, that direction is always towards the centre of mass.

Now consider a moving mass instead, as viewed from a stationary point in its path in front of it. The gravitational potential at that point will decrease until the mass passes and will then increase again. In Newtonian physics, with infinite gravity speed, the gravitational potential at that point will always be identical to the stationary potential at the instantaneous distance. In Relativistic physics, where gravity travels at limited speed, you might expect (I did) that the gravitational potential is the same as the stationary potential, but centered on its retarded position. With the retarded position, we mean the position its centre-of-mass had when the gravity left it.

The effect of the geometric size of the mass

But the mass has a geometric size, and the front of the mass is closer to our stationary point than the back of the mass. Because the gravitational potential changes travel with finite speed, the change caused by the front (which lowers the potential), reaches the point before the change caused by the back (which raises the potential). As a result of that difference, the potential in front of the mass is lower than the stationary potential of the retarded position.

For the same reasons, for a stationary point behind the traveling mass, the potential is higher than the stationary potential would be. With the higher potential behind the mass, and the lower potential in front of the mass, there is effectively an extra gradient in the direction of travel of the mass.

Now, in our rotating stars situation, when we view a moving star from an opposite star, we are not at a point in front of or behind the retarded position, but at a point to the side. In that case, the gravitational potential itself is the same as the stationary potential of the retarded position. But the extra gradient is still there as well. The stationary gradient is directly towards the retarded position, but the extra gradient turns that direction more towards the direction of travel of the other mass. The combined gradient causes acceleration towards the projected retarded position. That is, towards the instantaneous position that the mass will have assuming it continues to travel with the same speed and direction. With low accelerations, that position is approximately identical to the real instantaneous position.

The resulting gravitational potential

If we take a snap-shot (at a stationary time) of the gravitational potential of the traveling star, then we get a picture similar to the following:

Gravitational potential of moving mass

The star is traveling along the red line, with its retarded position at 0, and its instantaneous position at point 3. Imagine the other star is located at the green -6 mark. Note the (small) gradient at that point towards the positive red direction.

PS 1: Note that the gravitational field is compressed in the direction of travel; it is shorter in the red direction than in the green direction. This is due to the Lorentzian length contraction.

PS 2: in the above I stressed that the extra gradient is due to the geometric size of the mass. Although that is true, Feynman shows that, if the size is small enough, that the magnitude of the effect is just dependent on the speed, not on the geometric size of the mass. That is, the effect is caused because the mass has a size, but its magnitude is independent of that size. That means that even a moving point mass will cause a similar extra gradient.

fishinear
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