# Speed of Newtonian Gravity

1. Apr 10, 2010

### Nabeshin

I've been told for many years now that the speed of newtonian gravity is infinite. However, I've never received an explanation (or derivation) as to why this is that completely satisfies me. Even the explanations in textbooks (e.g Hartle) seem lackluster to me. So let's see...
The field equations for newtonian gravity are:
$$\nabla ^2 \Phi = 4 \pi G \mu(\vec{r})$$

Okay, great. How can I deduce from this that gravity propagates instantly from this? Of course, there is the analogous equation from electrostatics:
$$\nabla ^2 \Phi = \frac{\rho(\vec{r})}{\epsilon_0}$$
But this is only valid for electrostatics and we know from the full Maxwell's how we can derive a wave equation and extract a wave velocity.

What would be golden is if we could somehow get a "wave equation" for Newtonian gravity where one could explicitly see the velocity go to infinity. Unfortunately, there are no analogs to maxwell's equations that I can manipulate towards this end. Is it precisely because no dynamical equations exist that we sort of assume propagation is instantaneous? At any rate, someone try their hand at a convincing, hopefully rigorous, explanation.

2. Apr 11, 2010

### Cleonis

It is my understanding that Laplace (1749-1827) gave an argument based on the supposition that with a finite speed of gravity there would be an aberration effect.

If newtonian gravity would have a mediator that propagates at a finite speed then the gravitational force that the planets experience would not be directed towards the instantaneous position of the Sun. Instead it would be directed towards an earlier position of the Sun (as seen from the perspective of the planet.) That would result in a torque that would decrease the planetary orbits.

When Laplace did his computations the level of astronomical data was such that he could put a very sharp upper limit on how much the Moon's orbit could possibly be decaying. From that a lower limit for the propagation speed of newtonian gravity can be derived. This lower limit was many times the speed of light.

It is my understanding that for many years this was regarded as a definitive argument for gravity as an instantaneous effect.

Incidentally, the supposition of an aberration effect does not carry over to special relativity. The first to explore the possibility of a Lorentz invariant theory of gravity was PoincarÃ©. He pointed out that the argument given by Laplace did not apply for the Lorentz invariant theory.

3. Apr 11, 2010

### Tomsk

I think it's the other way round, we assume the propagation speed is infinite, therefore dynamical equations don't exist.

If gravitational effects did propagate at finite speed in Newtonian gravity, you would expect the potential in free space to be a solution to the wave equation
$$\nabla^2\phi - \frac{1}{v^2}\frac{\partial^2 \phi}{\partial t^2} = 0$$
So you could say the Poisson equation for the potential in free space IS the wave equation but with 1/v^2 = 0. The potential can have time dependence so the time derivatives aren't necessarily zero (in other words time derivatives don't enter the equations because gravity propagates at infinite speed), unlike in electrostatics where you assume the time derivatives are zero.

I think?

4. Apr 11, 2010

### Nabeshin

Ah, this line of reasoning is interesting! But why must we expect the potential in free space to be a solution to the wave equation? I'll give you that it certainly seems like the most likely prospect, and the first I would try given the problem, but must it be so? Once you do assume that it is of the form of the wave equation, then I like the argument that v -> infinity in order to recover the field equation we know to be true.

The shaky leg, for me, is assuming the form of the wave equation. In electromagnetism, we see the wave equation come out of the maxwell's equations. In GR, we see the wave equation come out of linearized small pertubations in the metric. I would really like to see the wave equation come out of something than simply assume its form because we see it so many other places.

5. Apr 11, 2010

### Nabeshin

Okay so historically this was how Laplace argued speed is very large, perhaps instantaneous. How did he arrive at many multiples of c, though? In GR, linearized gravity waves propagate exactly at c, so we see the same effect where the gravitational "force" (Forgive the terrible terminology, I think you know what I mean) is not directed exactly towards the sun. Why then can Laplace conclude many multiples of c are needed for stability while in GR we see obvious stability with v=c?

6. Apr 11, 2010

### D H

Staff Emeritus
The potential in free space with Newtonian gravity is

$$\phi(\mathbf x,t) = \sum_i \frac{G M_i}{||\mathbf x_i(t) - \mathbf x||}$$

where the sum is (conceptually) over all particles in the universe. In practice, there is no reason to go out that far because (a) the contribution of even the nearby stars is rather small, and (b) why bother? Newtonian gravity isn't quite the right model.

There is no time dependency in that equation. The equations of motion for a particle of miniscule mass are

$$\ddot{\boldsymbol x} +\\boldsymbol{nabla}_x \phi(\boldsymbol x,t)$$

The potential has an implicit time dependency because all of those other particles in the universe are also subject to gravitation. The gravitational equations of motion do not, however, have an explicit term like that 1/v2 term in the wave equation.

7. Apr 11, 2010

### Nabeshin

Hrm. So we see no 1/v2 term explicitly in the EOM, okay, this much is obvious..

Now that I stare at the equation of motion for a bit it does look to me like changes in potential must travel instantanously -- if we imagine our particles generating the potential to be whipping around in all kinds of horribly complex motion, our test particle knows instantly simply from the form of the potential, no?

Perhaps it would be constructive to wonder what this would look like in the event that there was a finite Newtonian speed to gravity and then compare.. It would seem the potential should have some "time lag" which is a function of each xi and some propagation velocity, no?
$$\phi(\bold{x},t)=\sum \phi_i = \sum \phi_i (\bold{x_i},t-\frac{x_i}{c} )$$

Or something like that?

8. Apr 11, 2010

### D H

Staff Emeritus
Pretty much. That, BTW, is a favorite trick amongst crackpots. They pretend that this is the only difference between general relativity and Newtonian gravity. General relativity introduces velocity-dependent effects as well as a finite speed of gravity. These velocity-dependent effects nearly (but not completely) cancel out the effects resulting from a finite speed. That they don't quite cancel is most noticeable at high speeds and close proximity to a central mass. In other words, Mercury.

9. Apr 12, 2010

### Nabeshin

Ah, neat. Cheers, and thanks!

10. Apr 13, 2010

### Cleonis

I was very impressed by Kevin Brown's discussion of that issue.
http://www.mathpages.com/home/kmath249/kmath249.htm
In that discussion, search the word "quadrupole", that is the part where the discussion turns to gravity.