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I already read this Phys.SE question but my question is different. I'm well aware that the effect of gravity in GR gets transmitted from point to point at the speed of light. But let's ignore GR for now. What about Newtonian gravity?

Imagine a universe in which Newtonian gravity(but neither GR nor SR ) holds. Imagine also that there's only one object(say, a planet or anything with mass) in this universe. Suddenly another planet(or any object with mass) pops out into existence(yes I know, but ignore the conservation of mass for now). How long will it take for the gravity to affect those two objects. Is there any way to calculate it from Newton's law of gravitation?

$$F= \frac{GMm}{r^2}$$

It seems to that this equation does not say anything about how fast gravity should get transmitted. Or maybe In this Newtonian universe it cannot be calculated from theory but rather be measured experimentally?

A lot of people say it gets transmitted instantaneously, but from where does that follow? How does one prove this only assuming Newton's laws of motion and his gravity law as our axioms? And no don't tell me that $c$ is the upper limit, I'm assuming a Newtonian universe here.

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Omar Nagib
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  • https://en.wikipedia.org/wiki/Action_at_a_distance#Newton – pfnuesel Jun 03 '15 at 01:42
  • Newton himself was troubled by the fact that he didn't know what transmitted gravity, and that it seemed to be instantaneous. He left the question unanswered. – Ernie Jun 03 '15 at 01:49
  • That it is instantaneous follows from the fact that the force only depends on the instantaneous positions of the mases –  Jun 03 '15 at 21:40

4 Answers4

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So the fact that you don't see anything about the speed of gravity in Newton's equation is a bit of a clue; Newtonian gravitational interactions propagate instantaneously (ie. with infinite speed). If there was some finite speed of propagation, we'd have to have a constant related to that speed in the force law, and we don't.

It's a little more obvious to see if you write Newtonian gravity in terms of Poisson's equation instead of Newton's force law;
$\nabla^2 \Phi(\textbf{r},t) = 4 \pi G \rho(\textbf{r},t)$
In case you've not seen Poisson's equation before, it's the field equation for gravity. That means that given a distribution of mass $\rho(\textbf{r},t)$, we can calculate the gravitational potential $\Phi(\textbf{r},t)$. From $\Phi$ we can calculate the force of gravity at each point, and if we place a point mass at the origin we recover Newton's force law.

The reason that it's obvious from Poisson's equation that Newtonian gravity is transmitted instantaneously is because it doesn't have any time derivatives ($\nabla^2$ contains only spatial derivatives). We solve the equation to calculate a spatial distribution of the gravitational potential at a given time, and then if the matter distribution in our spacetime changes, the gravitational potential changes instantaneously at all points in space.

Perhaps the prototypical field equation that does include time derivatives, and along with those a finite propagation speed, is the wave equation, where $\Theta$ is some general field in space and time;
$-\frac{1}{v^2}\frac{\partial^2}{\partial t^2}\Theta(\textbf{r},t) + \nabla^2 \Theta(\textbf{r},t) = 0$
Where we can now see that, rather than a set of functions $\Theta$ which we solve for spatially at each time time, we have to solve for the time and space parts of the equation together. Also note the appearance of $v$, the speed of propagation of the wave, which has no equivalent in Poisson's equation.

This problem of infinite propagation of Newtonian gravity was the main motivating reason for Einstien's search for a relativistic theory of gravity.

There's a lot more information in the wikipedia article called 'speed of gravity' if you're interested.

Mithoron
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Jojo
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    I'd be glad if you present me with an example of an equation which entails time derivatives and as a consequence of that fact has a finite speed of propagation of its effect, Can you please add that in your answer? – Omar Nagib Jun 03 '15 at 11:44
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    (1) Analogously, if We were to write coulomb's law in the form of a Poisson's equation we will have : $\nabla^2 V(\textbf{r},t) = $ $\rho(\textbf{r},t)\over \epsilon$ where $V$ is the electric potential, This is identical to the gravitational equation. does this mean that the electrostatic force gets transmitted at an infinite speed? (2) It's not obvious to me Why the existence or non-existence of time derivatives dictate the speed of propagation. can you give me more details? – Omar Nagib Jun 03 '15 at 14:46
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    The wave equation is an example of such an equation, I have added it to my answer – Jojo Jun 03 '15 at 15:50
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    So let's talk electromagnetism a little. You bring up a good point - Poisson's equation does appear identical for the two cases. The problem is the labeling of the functions. Your $V$ and $\rho$ in the electrostatic Poisson's shouldn't be functions of time.
    What we find is that Poisson's equation for gravity is the only field equation for Newtonian gravity, whereas the full set of field equations for electromagnetism is really Maxwell's equations, which indeed include time derivatives. You have taken the electrostatic approximation in producing your version of Poisson's equation     – Jojo Jun 03 '15 at 15:54
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    To do that we assume $\dot{\mathbf{E}} = \dot{\mathbf{B}} = 0$, which gives us that $\nabla \times \mathbf{E} = 0$ by Maxwell's equations, and hence that we can write $\mathbf{E}(\mathbf{r}) = - \nabla V(\mathbf{r})$, for the electro static potential $V$. We can then substitute this into another Maxwell's equation to give the form of Poisson's equation that you stated. As there is no time dependence in electrostatics, we aren't making any statement about the speed of propagation of the electromagnetic force in writing Poisson's equation for electrostatics – Jojo Jun 03 '15 at 15:59
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    If instead of taking the electrostatic solution to Maxwell's equations, we can take the (time-dependent) vacuum solution, which sets $\rho = J = 0$. Then we can solve for the electric and magnetic fields using vector calculus identities, and we find that we get $-\epsilon_0\mu_0\frac{\partial^2}{\partial t^2} E(\textbf{r},t) + \nabla^2 E(\textbf{r},t) = 0$ (and the same equation for the magnetic field). Which we recognise as the wave equation, with a speed of propagation equal to $v = \frac{1}{\sqrt{\epsilon_0\mu_0}}$ – Jojo Jun 03 '15 at 16:07
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    Not-so-coincidentally, we find that $\frac{1}{\sqrt{\epsilon_0\mu_0}} = c$, the speed of light. This constancy of the speed of light predicted by Maxwell's equations was upsetting physicists at the end of the 19th century, because it disagreed with the idea of absolute space - paving the way (along with agreement from experiment) for Einstein to postulate special relativity to allow $c$ to remain constant in all frames of reference – Jojo Jun 03 '15 at 16:09
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    Extrapolating from your comments, the reason the existence of time derivatives matters is owing to this following fact: If our field equations contain time derivatives(like that of Maxwell's), then with some algebraic manipulation we can make this field equation(say Maxwell's) take the form of the $wave$ $equation$ and we know the wave equation by definition contain a finite speed of propagation. Is this correct summary of your words? – Omar Nagib Jun 03 '15 at 16:09
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    I think you should move all your excellent comments in your answer. – Omar Nagib Jun 03 '15 at 16:16
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    It's a little over-simplified to say that we can always reduce a given field equation to the wave-equation but essentially you have the right idea. In (special) relativistic physics we expect our field equations to be Lorentz-invariant, which implies we always have the speed of light as our speed limit. All (special) relativistic field equations (the I can think of now) have essentially the same form as the wave equation - including source terms. There are plenty of non-relativistic equations that can't be manipulated to wave equations, eg. the diffusion equation – Jojo Jun 03 '15 at 16:21
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In the Newtonian theory, gravity is instantaneous. This is because the force law says that whenever you have two objects a distance $r$ apart, there will be a force between them inversely proportional to $r^2$. The force at any given instant depends on the distance at that same instant, which means that if you move one of the bodies a little bit, the other one will know about it immediately.

Of course, this is not what actually happens. It's just what you can deduce from Newtonian mechanics.

PM 2Ring
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Javier
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Newtonian gravity must necessarily be instantaneous, otherwise planetary orbits wouldn't obey Kepler's laws.

In fact, Laplace computed that if propagation of gravity was simultaneously according to Newton's laws and at the speed of light Earth would fall into the Sun within a few centuries.

This would happen because Earth would get attracted by the Sun not toward the current position of the Sun, but toward the position where Sun used to be about 8 min ago (because it takes about 8 min for light to propagate from Sun to Earth). Thus the direction of the gravitational force would be at a slight angle away from the direction toward the center of the Sun, and that misdirection would be enough to destabilize Earth orbit and make it fall into the Sun within centuries.

Michael
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    This is very interesting. Can you provide me with links or sources to delve deeper into this matter? Can you show me where I can find those proofs of Laplace?(with modern notation if possible). – Omar Nagib Jun 03 '15 at 22:26
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    Here is a short source. Here is one that assumes Newtonian laws and estimates that speed of gravity must be at least 20000000000 faster than speed of light. – Michael Jun 03 '15 at 22:47
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    With respect to the paper you referenced, does this mean We can have a consistent theory of Newtonian gravitation in which the speed of gravity is finite? In other words, does this mean that Newton's law of gravitation do not necessarily imply infinite speed of propagation? – Omar Nagib Jun 03 '15 at 23:15
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    @Michael : you must read again your link. Some , arguing about gravitational aberration lacks, say that speed is infinite. The author demonstrates why gravitational aberration effects are canceled in GR and almost all causal gravity theories. –  Jun 04 '15 at 04:49
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To add to Joe's very good answer, another way of seeing this is to consider both the SR and GR theories as you take the limit of letting the speed of light ($c$) approach $\infty$.

In that limit SR Minkowski space (with Lorentz transformations) obviously reduces down to a three dimensional space (with Galilean transformations) plus an absolute time that is the same for all observers - in other words, Newtonian mechanics.

Also (but not as obviously) in that limit, GR reduces to Newtonian gravitational theory. Therefore in that limit the speed of light and the speed of transmitting gravity would both be infinite ($\infty$). QED.

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FrankH
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    Indeed, assuming $c=\infty$ would recover Galilean transformations in SR(I'm not sure whether Newton's law of gravity would come out of GR in this case). However, I cannot see the logical connection between your assumption (i.e: $c=\infty$) and your conclusion(i.e:speed of Newtonian gravity is infinite). So can you explain more please? – Omar Nagib Jun 03 '15 at 22:53