# Why special relativity is unsuitable to describe gravity

And more importantly, will the free faller detect the gravitational wave if the sun suddenly disappears? If so, how can he explain that wave?
The sun cannot simply disappear in GR. Every distribution of matter evolves from from a previous one, and evolves into a future distribution according to Einstein's equations.
In Newtonian gravity you can simply remove the sun, as time is not involved at all.

You're right that acceleration due to changes in motion can be described in special relativity's flat space and that gravity cannot. The Shapiro effect shows that light passing between two points is delayed if a gravity well is between the points. There are two ways to interpret this. One way is to think of the gravitational field as slowing the light passing through. The other way, the one consistent with Relativity, presumes the speed of light in vacuum is constant and that spacetime is dilated in a gravity well. That is, the path is longer.

The best model I've found for this is the rubber sheet model. If the sheet's stretching is directly proportional to the change in gravitational potential energy and the stretching is viewed from above (gravity dilates spacetime radially about the center of mass so is always viewed as stretching away from the observer).

I do not understand why propagation delay in mr"=-mM/r^2 with t replaced by t+r/c gives an unstable orbit since observation time and coordinates are delayed by the same amount. Please show the derivation and solution for the non-relativistic Newtonian case with propagating gravity.

Please show me how to derive and get the observation time dependent solution to the non-relativistic Newtonian 2-body problem (sun-earth) with gravity propagation delay = r/c. I want to understand the claimed instability!

PeterDonis
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I do not understand why propagation delay in mr"=-mM/r^2 with t replaced by t+r/c

Your statement makes no sense because "t" doesn't appear anywhere in the equation.

observation time and coordinates are delayed by the same amount

This doesn't make any sense either. Coordinates aren't "delayed". They're just labels for events.

For a quick layman's discussion of the issue, see here:

http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html

For more technical details see this paper by Carlip (referenced in the article linked to just now):

https://arxiv.org/abs/gr-qc/9909087

The introduction to this paper discusses the issue with propagation delayed Newtonian gravity and gives further references with more detailed analysis.

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PeterDonis
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I want to understand the claimed instability!

The simplest way to describe it is that, if we take Newtonian gravity and just put in a light-speed propagation delay with nothing else changed, then, for example, the force on the Earth due to the Sun no longer points towards the Sun's position "now"--it points towards the Sun's position one light-travel time ago (i.e., about 500 seconds in a frame in which the Sun is at rest). But this means the force is no longer pointing in the right direction to keep Earth's orbit a stable, closed ellipse--since the standard Newtonian direction is the only possible direction that does that.

Nugatory
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The simplest way to describe it is that, if we take Newtonian gravity and just put in a light-speed propagation delay with nothing else changed, then, for example, the force on the Earth due to the Sun no longer points towards the Sun's position "now"--it points towards the Sun's position one light-travel time ago (i.e., about 500 seconds in a frame in which the Sun is at rest). But this means the force is no longer pointing in the right direction to keep Earth's orbit a stable, closed ellipse--since the standard Newtonian direction is the only possible direction that does that.
Wouldn't that be mathematically identical to the problem of standard Newtonian orbit around the Sun's position 500 seconds ago? Both in direction and magnitude of force. I would think that the stability of the orbit would be identical in the two approaches to a solution.

PAllen
Wouldn't that be mathematically identical to the problem of standard Newtonian orbit around the Sun's position 500 seconds ago? Both in direction and magnitude of force. I would think that the stability of the orbit would be identical in the two approaches to a solution.
Think of it this way. What is the direction of Newtonian gravity? The vector from earth to sun now, one or 500 seconds ago? This gets rid the question do whose point of view. If you use a time retarded vector, you get highly unstable orbits. Newton was amazingly well aware of this issue, and reluctantly proposed the instantantaneous vector direction because that was necessary for orbital stability as had been observed.

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Think of it this way. What is the direction of Newtonian gravity? The vector from earth to sun now, one or 500 seconds ago? This gets rid the question do whose point of view. If you use a time retarded vector, you get highly unstable orbits. Newton was amazingly well aware of this issue, and reluctantly proposed the instantantaneous vector direction because that was necessary for orbital stability as had been observed.
At first I didn't see why the current relative position of the Sun mattered at all since the only interaction is with a "mythical" Sun positioned 500 seconds ago. I couldn't see why a planet wouldn't have just as good an orbit (although different) around the mythical Sun position as it would around the actual current Sun position. I suppose that there must be some immediate interaction with the current Sun position that is mixed in with the delayed gravitational force that would cause the situation to be essentially different.

PAllen
At first I didn't see why the current relative position of the Sun mattered at all since the only interaction is with a "mythical" Sun positioned 500 seconds ago. I couldn't see why a planet wouldn't have just as good an orbit (although different) around the mythical Sun position as it would around the actual current Sun position. I suppose that there must be some immediate interaction with the current Sun position that is mixed in with the delayed gravitational force that would cause the situation to be essentially different.
Just consider that for a circular orbit, the acceleration must in the direction of the center. The direction from prior position to the center as a current acceleration direction cannot possibly produce a circular orbit. A slight generalization establishes that an elliptical orbit is impossible. The deviations are large, not on the scale of perihelion advance. Physicists in the 1800s had even calculated that a propagation delay of ten billon times light speed would still be inconsistent with observations.

FactChecker
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Just consider that for a circular orbit, the acceleration must in the direction of the center. The direction from prior position to the center as a current acceleration direction cannot possibly produce a circular orbit. A slight generalization establishes that an elliptical orbit is impossible. The deviations are large, not on the scale of perihelion advance. Physicists in the 1800s had even calculated that a propagation delay of ten billon times light speed would still be inconsistent with observations.
Thanks. I'll have to think about it. I believe you. I certainly see that the delay would change the orbit. I also see that a delay in what is essentially a feedback system could change the stability of the orbit. Still, I have a hard time seeing how the actual orbit with the delay would be different from a theoretical orbit with no delay around a theoretical Sun at the position of the real Sun 500 seconds ago and why that theoretical orbit would have different stability properties.

PeterDonis
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Wouldn't that be mathematically identical to the problem of standard Newtonian orbit around the Sun's position 500 seconds ago?

No, because the Earth is not in the same position now as it was 500 seconds ago, and the force is acting on the Earth now.

FactChecker
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No, because the Earth is not in the same position now as it was 500 seconds ago, and the force is acting on the Earth now.
Oh! Thanks! I think I see. The velocities and behavior of the planet does not match the actual current distance from the Sun, it matches the gravity from the earlier distance. So the current velocities and distances do not indicate that the orbit would be stable. Only by considering the delay would a stable orbit with that behavior make sense.

PS. Sorry if I have hijacked this thread. I will bud out.

PeterDonis
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The velocities and behavior of the planet does not match the actual current distance from the Sun, it matches the gravity from the earlier distance.

Yes, you've got it.

How if instead of the sun, there is a magical structure pulls the earth with a long rope and rotates it. Still the pulling force has to propagate with some propagation speed due to the electromagnetic force at the rope's mollecular level. Will be any instability in the rotating earth orbit? In general will a rotating disc be unstable because the direction of the force acting on the edge is not on the same direction of the force a moment ago?

Buzz Bloom
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The physical situation may have stable orbits with or without delays in gravity. It's just that the actual stable orbit will not exactly match a calculated stable orbit unless the calculations include the correct delays. The observed orbit of a planet will not match the zero-delay Newtonian calculations of a stable orbit because the delays used in the calculations are not correct. But that does not mean that there would be no stable orbit if there were no delay in gravity -- just that the stable orbit would be different.

PS. I am probably using the term "stable" incorrectly. If the orbit is not an exact repeating ellipse, but rather a rotating rosette, I think of that as stable.

Staff Emeritus
This thread is really messy. If one is talking about Newtonian gravity - or for that matter, SR - one should not muddy the waters discussing spacetime curvature. The nit that the sun's mass is not exactly constant is singularly unhelpful. It's also negligible, as it is less than a part per trillion per decade.

Ibix answer is right - you cannot take Newtonian gravity, make it travel at c rather than instantaneously, and match observations. However, this is not (directly) due to "the position of the sun in the sky 8 minutes ago". First, that's dominated by the earth's rotation. Second, as pointed out, insofar as the sun is stationary, the gravitational field at point X is the same as it was 8 minutes ago, so the field the Earth traverses is the same as it was 8 minutes ago, so how can the orbit be different?

The answer to this apparent paradox is that the sun is not exactly stationary. Even in a one-planet solar system, the Earth and Sun revolve around their common center of mass. If you put in a propagation delay, the Earth sees its own gravity through the Sun's (small) motion with a 16 minute delay. This tends to perturb the orbit, and as described earlier, results in an instability over many millions of years. When you add the other 8 - sorry, 7 - planets, this instability just gets worse.

GR doesn't actually have to fix this up. There's no reason to demand that it give you stable orbits (although if you want to consider it as a viable description of nature you do, but the theory itself doesn't have to). It does and it doesn't - orbits in GR are not ellipses but rosettes, but these rosettes are stable over long time scales. It is a success of GR that it matches the data, both the long-term stability and the rosette nature of the orbits.

FactChecker
This thread is really messy. If one is talking about Newtonian gravity - or for that matter, SR - one should not muddy the waters discussing spacetime curvature. The nit that the sun's mass is not exactly constant is singularly unhelpful. It's also negligible, as it is less than a part per trillion per decade.

Ibix answer is right - you cannot take Newtonian gravity, make it travel at c rather than instantaneously, and match observations. However, this is not (directly) due to "the position of the sun in the sky 8 minutes ago". First, that's dominated by the earth's rotation. Second, as pointed out, insofar as the sun is stationary, the gravitational field at point X is the same as it was 8 minutes ago, so the field the Earth traverses is the same as it was 8 minutes ago, so how can the orbit be different?

The answer to this apparent paradox is that the sun is not exactly stationary. Even in a one-planet solar system, the Earth and Sun revolve around their common center of mass. If you put in a propagation delay, the Earth sees its own gravity through the Sun's (small) motion with a 16 minute delay. This tends to perturb the orbit, and as described earlier, results in an instability over many millions of years. When you add the other 8 - sorry, 7 - planets, this instability just gets worse.

GR doesn't actually have to fix this up. There's no reason to demand that it give you stable orbits (although if you want to consider it as a viable description of nature you do, but the theory itself doesn't have to). It does and it doesn't - orbits in GR are not ellipses but rosettes, but these rosettes are stable over long time scales. It is a success of GR that it matches the data, both the long-term stability and the rosette nature of the orbits.
I agree with FactChecker. To see this in a simple way, imagine we are dealing with 1-D force acting on an object moving along x-axis back and forth around the center. If the force is instantenous, the object will continue moving to the right side of the axis to some point and then back in the opposite direction to the negative side of x-axis and so on. If there is a propagation delay, again the particle will continue moving to the positve direction until the force catches it and pulls it back but it may take it further to the right this time compared with the first case. In both cases, the conservation of the energy holds the particle in stable motion.

Staff Emeritus
You should do the math, then. There's a famous paper by I.J. Good in the 70's that works this all out.

The problem shows up in the calculation very quickly. If A and B are in orbit around a common center of mass, and the force between A and B is not aligned with the present positions of A and B, there is a torque on the system. That torque is destabilizing.

FactChecker
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You should do the math, then. There's a famous paper by I.J. Good in the 70's that works this all out.

The problem shows up in the calculation very quickly. If A and B are in orbit around a common center of mass, and the force between A and B is not aligned with the present positions of A and B, there is a torque on the system. That torque is destabilizing.
In this context, does the term "unstable" mean that the radius of the orbit changes or does it also include a constant radius rosette pattern? I can easily understand a torque causing the rosette pattern. Does it also imply an unstable radius?

Staff Emeritus
There is a torque through an angle, and that means work is done. This work expands the radius of the orbit and separates the bodies.

Nik_2213 and FactChecker
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There is a torque through an angle, and that means work is done. This work expands the radius of the orbit and separates the bodies.
Ha! That is a very straightforward and, I think, intuitively convincing argument. Thanks.

Formally the problem becomes an iterative functional equation :

$$\ddot{\vec{r}}(t+\frac{2r(t)}{c})=-\frac{Gm\vec{e_r}(t)}{(2r(t))^2}$$

This has memory properties so that for example the initial condition is given by $$\{\vec{r}(t)|t\in [-\infty;0]\}$$

I didn't find a documentation about how to solve this exactly or even with a first order term which gives a component of the force along the jerk.

We could write the special relativistic dynamics in the same way but general relativity solves this time delay in a completely different way getting rid of higher derivatives and memory.

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Hi. Fake gravitation such as motions of constant acceleration or rotation can be fully understood by applying SR. Real gravity seems to have similar behaviour with fake ones thus introduced GR. Best.
There is no such thing as "fake" gravity, is there?

sweet springs
Ibix