20
$\begingroup$

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?

$\endgroup$
3
  • 3
    $\begingroup$ 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. $\endgroup$
    – j.c.
    Nov 3, 2010 at 14:21
  • $\begingroup$ related: physics.stackexchange.com/q/68067/4552 $\endgroup$
    – user4552
    Jun 14, 2013 at 21:48
  • $\begingroup$ See my answer here: physics.stackexchange.com/q/7781 where I derived Poisson's equation from EFE as an approximation. $\endgroup$ Jul 1, 2013 at 5:14

7 Answers 7

17
$\begingroup$

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.

$\endgroup$
14
$\begingroup$

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.

$\endgroup$
5
  • 2
    $\begingroup$ 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. $\endgroup$
    – Justin L.
    Nov 3, 2010 at 6:04
  • 1
    $\begingroup$ 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. $\endgroup$
    – Sklivvz
    Nov 3, 2010 at 9:45
  • $\begingroup$ 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. $\endgroup$
    – user52
    Nov 3, 2010 at 14:48
  • 1
    $\begingroup$ 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). $\endgroup$
    – Chad Orzel
    Nov 13, 2010 at 13:16
  • 1
    $\begingroup$ @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. $\endgroup$
    – Ron Maimon
    Jan 17, 2012 at 4:50
5
$\begingroup$

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 factors: 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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 :)

$\endgroup$
14
  • 2
    $\begingroup$ 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). $\endgroup$
    – Marek
    Nov 17, 2010 at 17:05
  • $\begingroup$ @Marek, by $E$ I mean here $E=\frac{mc^2}{\sqrt{1-v^2/c^2}}$ $\endgroup$
    – voix
    Nov 17, 2010 at 17:36
  • 1
    $\begingroup$ @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. $\endgroup$
    – Marek
    Nov 17, 2010 at 17:48
  • $\begingroup$ @Marek, initial equation is the equation of energy conservation for free fall from infinity with initial speed of object equal to zero. Left side - energy of rest object at infinity, right side - full energy of object at $R$ plus potential energy. In case of classical Newton (low speed): $0=mv^2/2-\frac{GMm}{R}$ $\endgroup$
    – voix
    Nov 17, 2010 at 18:29
  • 1
    $\begingroup$ -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. $\endgroup$
    – Ron Maimon
    Jan 17, 2012 at 4:52
0
$\begingroup$

To be direct: the answer is no. There's no spatial curvature for Newtonian gravity, when it is rendered in geometric form as a curved space-time geometry. All the curvature is in time.

One of the most prominent ways this stands out is in the relation between radial distances versus circumferences. For Newtonian gravity, the circumferences $C_1$ and $C_2$ of two coplanar circular orbits will differ by $2π$ times their respective radii; or more precisely: the closest distance between the respective orbits will be $|r_1 - r_2|$, where $r_1 = C_1/(2π)$ and $r_2 = C_2/(2π)$.

For General Relativity, approximating the exterior of the gravitating body by the Schwarzschild metric, the closest distance between the orbits will be $$\left|\sqrt{r_1(r_1 - r_0)} - \sqrt{r_2(r_2 - r_0)} + r_0 \log{\frac{\sqrt{r_1} + \sqrt{r_1 - r_0}}{\sqrt{r_2} + \sqrt{r_2 - r_0}}}\right|,$$ where $r_0 = (2GM)/c^2$, with $G$ being Newton's constant and $M$ the mass of the gravitating body, where $r_1$ and $r_2$ are as above. By assumption, we're talking about the exterior of the body, so $r_1 > r_0$ and $r_2 > r_0$. In the limit as $c → ∞$, $r_0 → 0$, and the distance approaches the limit $$\left|\sqrt{{r_1}^2} - \sqrt{{r_2}^2}\right| = \left|r_1 - r_2\right|,$$ where $r_1 > 0$ and $r_2 > 0$.

The spatial curvature will also show up as a precessing of a fixed axis - such as that of a gyroscope - when taken around the gravitating body in an orbit. That test can be done ... and has been done, on the space shuttle.

The $r$ coordinate in the Schwarzschild metric, in General Relativity, is actually not radial distance at all, but is the "circumference radius" - that is: the circumference divided by $2π$. The two do not coincide in relativity, because of the spatial curvature.

$\endgroup$
3
  • $\begingroup$ "For Newtonian gravity, the circumferences C1 and C2 of two coplanar circular orbits will differ by 2π times their respective radii". And exactly the same is true in the Scharzschild metric if you substitute "radii" for "radial coordinates" - which you clearly know, but the way you've written suggests that it might be different. $\endgroup$
    – ProfRob
    Feb 8, 2023 at 18:17
  • $\begingroup$ That's false. In fact, the circumference isn't even an elementary function of the distance-radius. You'd have to invert the above function to get the circumference-radius in terms of the distance-radius. The Schwarzschild metric uses the circumference-radius, not the distance-radius. The equivalent metric for distance-radius, calling it (say) R, would have the line element dR² + r(R)² ((dθ)² + (sin θ dφ)²) - c² (1 - r(R)/r₀) dt², where r(R) is the circumference-radius, given as a (non-elementary) function of the distance-radius, R. The distance between the orbits is |r⁻¹(C₁/2π) - r⁻¹(C₂/2π)|. $\endgroup$
    – NinjaDarth
    Feb 8, 2023 at 18:34
  • $\begingroup$ Actually, you don't understand what you're saying. The expression you just gave is what I referred to as the "circumference radius". "Nobody doing GR talks about an orbital 'radius'" - is irrelevant since that has nothing to do with anything you're responding to. In fact, on the contrary: in GR, we do talk about the radial distance between two coplanar circular orbits - that is the length of the shortest geodesic between those two orbits, which is a radial geodesic. So, yes, you can talk about a "distance radius". $\endgroup$
    – NinjaDarth
    Feb 8, 2023 at 18:36
-1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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