Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

By 'Newton's Law of Gravity', I am referring to

The magnitude of the force of gravity is proportional to the product of the mass of the two objects and inversely proportional to their distance squared.

Does this law of attraction still hold under General Relativity's Tensor Equations?

I don't really know enough about mathematics to be able to solve any of Einstein's field equations, but does Newton's basic law of the magnitude of attraction still hold?

If they are only approximations, what causes them to differ?

share|improve this question
3  
If you're really interested in this stuff, check out Carroll's book "Spacetime and Geometry" for a pretty good intro to G.R. and the math behind it. –  j.c. Nov 3 '10 at 14:21
    
related: physics.stackexchange.com/q/68067/4552 –  Ben Crowell Jun 14 '13 at 21:48
    
See my answer here: physics.stackexchange.com/questions/7781/… where I derivedd Poisson's equation from EFE as an approximation. –  Dimensio1n0 Jul 1 '13 at 5:14
add comment

6 Answers

up vote 10 down vote accepted

Yes, in the appropriate limit. Roughly, the study of geodesic motion in the Schwarzschild solution (which is radially symmetric) reduces to Newtonian gravity at sufficiently large distances and slow speeds. To see how this works exactly, one must look more specifically at the equations.

share|improve this answer
1  
What exactly is geodesic motion, and the Schwarzschild solution? I'm sorry, I don't really have much of a background in physics past the nineteenth century. –  Justin L. Nov 3 '10 at 6:04
1  
The geodesic motion are the orbits. The Schwartzschild solution represents the gravitational field in free space where all the mass is concentrated in a spherical region. –  Sklivvz Nov 3 '10 at 9:45
    
In fact, if I recall correctly, you can derive from the Newton's law of gravity the Schwarzschild radius (radius of the event horizont) of the Schwarzschild solution for Einstein equations. –  user52 Nov 3 '10 at 14:48
1  
It's worth noting that "sufficiently large distance" is really pretty small. Experiments at the University of Washington have shown that gravity follows an inverse-square law for separations as small as about 50 microns (0.05 mm, on the thin side of the diameter of a human hair). –  Chad Orzel Nov 13 '10 at 13:16
1  
@Chad: the distance where Newton works depends on the mass, so the right measure is the ratio of the separation to the Schwartschild radius. The deviations from Newton's law fall as one over this ratio, so even when this ratio is relatively large, like Mercury's orbit, you can see deviations from Newton over the centuries. –  Ron Maimon Jan 17 '12 at 4:50
add comment

The main problem here is this: Newton gives us formulas for a force, or a field, if you like. Einstein gives us more generic equations from which to derive gravitational formulas. In this context, one must first find a solution to Einstein's equations. This is represented by a formula. This formula is what might, or may not, be approximately equal to Newton's laws.

This said, as answered elsewhere, there is one solution which is very similar to Newton's. It's a very important solution which describes the field in free space.

You can find more about this formula -- in lingo it's a metric, here: http://en.wikipedia.org/wiki/Schwarzschild_metric

The fact that they are approximations fundamentally arises from different factros: the fact that they are invariant laws under a number of transformations, but mostly special relativity concerns - in other words, no action at a distance - is a big one.

share|improve this answer
add comment

Newton's Law of Gravity is consistent with General Relativity at high speed too :)

Lets consider Newton equation of energy conservation for free fall from the infinity with initial speed of object equal to zero:

$\large {mc^2=E-\frac{GMm}{R}}$

or

$\large {mc^2=E-\frac{R_{g*}}{R}\;mc^2}$ where $\large {R_{g*}=GM/c^2}$

so

$\large {E=mc^2\left(1+\frac{R_{g*}}{R}\right)=mc^2\left(\frac{R+R_{g*}}{R}\right)}$

Now

$\large {mc^2=E\;\frac{R}{R+R_{g*}}=E\left(1-\frac{R_{g*}}{R+R_{g*}}\right)}$

and as the result

$\bf\large {mc^2=E-\frac{GM}{R+R_{g*}}\;\frac{E}{c^2}}$

Compare to

$\bf\large {mc^2=E-\frac{GMm}{R}}$

In the resulting equation energy ($E/c^2$) is attracted, not mass ($m$). That's why gravitational redshift is the same in Newton Gravity and in General Relativity (for $R>>R_g$).

Slight modification of Newton equation describes radial movement of an object at any speed with different initial conditions in the same way as General Relativity. Not only free fall from infinity with initial speed equal to zero.

$\bf\large {E_1\left(1-\frac{GM}{c^2(R_1+R_{gm}+R_{gM})}\right)=E_2\left(1-\frac{GM}{c^2(R_2+R_{gm}+R_{gM})}\right)}$

And it has no any singularity! So I like it :)

share|improve this answer
    
@voix LaTeX markup works here, just put code inside two dollar signs, like $E=mc^2$. –  mbq Nov 17 '10 at 11:34
1  
Einstein proved $E = mc^2$ (therefore making $m$ redundant and indeed, this symbol is used differently nowadays to mean invariant mass). I don't see any mathematical nor physical content in these equations (or more precisely one equation written out seven times). –  Marek Nov 17 '10 at 17:05
1  
@voix: then it makes even less sense because $E$ in your equation is just a sum of gravitational energy $\Phi$ and rest energy $mc^2$. Now, where did the kinetic energy go? You can't just make equations up like this. Besides $m$ in this sense is invariant (it doesn't depend on from where you look at it) but $E$ clearly isn't (you can see in your formula that $E$ depends on $v$). So your equation also doesn't obey the laws of relativity. –  Marek Nov 17 '10 at 17:48
1  
@voix: I still don't think the answer is great. But at least it's readable and quite interesting now. Anyway, thank you for trying to improve the answer; I appreciate that :-) –  Marek Nov 18 '10 at 19:41
1  
-1: Come on! The General relativistic energy conservation equation holds, but it has an additional potential contributions which you didn't consider. The answer above it giving wrong formulas, and I don't know why it is upvoted at all. –  Ron Maimon Jan 17 '12 at 4:52
show 14 more comments

Eric's answer is not really correct (or at least not complete). For instance, it doesn't tell you anything about the motion of two comparably heavy bodies (and indeed this problem is very hard in GR, in stark contrast to the Newtonian case). So let me make his statements a bit more precise.

The correct approach is to treat the Newtonian gravity as a perturbation of the flat Minkowski space-time. One writes $g = \eta + h$ for the metric of this space-time ($\eta$ being Minkowski metric and $h$ being the perturbation that encodes curvature of the space-time) and linearize the theory in $h$. By doing this one actually obtains a lot more than just Newtonian gravity, namely gravitomagnetism, in which one can also investigate dynamical properties of the space-time not included in the Newtonian picture. In particular the propagation of gravitational waves.

Now, to recover Newtonian gravity we have to make one more approximation. Just realize that Newtonian gravity is not relativistic, i.e. it violates finite speed of light. But if we assume that $h$ changes only slowly and make calculations we will find out that the perturbation metric $h$ encodes the Newtonian field potential $\Phi$ and that the space-time is curved in precisely the way to reproduce the Newtonian gravity. Or rather (from the modern perspective): Newtonian picture is indeed a correct low-speed, almost-flat description of GR.

share|improve this answer
add comment

All four answers agree in saying « no ». Newton's Law is not consistent with General Relativity. But all four answers point out that Newton's Law is sometimes a reasonable approximation and can be derived from Eintein's Equations by neglecting some terms and introducing some approximations.

share|improve this answer
add comment

May be the case that Gerber could not give an exact explanation for his formula, 18 years before GR, on the advance of Mercury's perihelium as we can see at mathpages. After reading the fine explanation on Lienard & Wiechert retarded potentials in the Hans de Vries online book I think that the treatment of the subject is not correct in the mathpages.

It appears to me that Walter Orlov, 2011 has a nice way to explain why Gerber's formula is correct to explain Mercury's orbit.

The answer is that they are mutually consistent because Gerber'gravity (post-Newtonian treatment with delayed potentials) is consistent with observations, the same as with GR's formulation.

Before I can ask 'Do I need GR to explain the observations?' I need to be sure that Orlov got it right.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.