# What is the exact gravitational force between two masses including relativistic effects?

I was wondering if there is a closed-form formula for the force between two masses $m_1$ and $m_2$ if relativistic effects are included. My understanding is that the classic formula $G \frac{m_1 m_2}{r^2}$ is just an approximation (which is good enough for probably even going to the Moon), but what would the "correct" formula be according to the general theory of relativity? Does a closed formula even exist? For example for an idealized situation of just two spherical masses with homogenous mass distribution?

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You could have a look at this wikipedia page. –  Kitchi Dec 28 '12 at 12:41
I am afraid that if that Wikipedia page answers my question, i am unable to even realize that it does answer my question :) I had been hoping for a simpler closed form function of $m_1$ and $m_2$ (and maybe other information necessary) to calculate force where the calculated force would slightly (for everyday use) differ from the result obtained by the classical Newtonian formula. –  Johsm Dec 28 '12 at 18:26
Here's a related question that might even be considered a duplicate: physics.stackexchange.com/questions/2684/… –  David Z Dec 29 '12 at 4:59
Related: Geodesic Equation. ! –  Dimensio1n0 Jul 18 '13 at 9:10

Worse than electrodynamics, general relativity is non-linear, in the sense that the field from multiple sources is not just the sum of fields from each isolated source. Even the simple case you are asking about, which is the two-body problem in general relativity, has not been solved exactly.

An even simpler case is the limit $m_2 \to 0$. In that case only $m_1$ affects the geometry of spacetime, and $m_2$ follows a geodesic in that spacetime. This has been solved exactly. The linked article gives details.

[Addendum] To directly answer the original question for the limit $m_2 \to 0$:

Of course if $m_2=0$, then the force between the two masses is $0$. But the thing about classical gravity is that the acceleration due to gravity is independent of mass. (This is the equivalence principle and is actually one of the starting points of the theory of general relativity.) So it still makes sense to ask what would be the acceleration of a negligible mass (aka, a test body) due to the gravity of another mass. The classical answer is $a_2 = \frac{Gm_1}{r^2}$.

In general relativity, the acceleration of a test body due to the gravity of a single spherical, homogenous, non-rotating mass is given exactly by the Schwarzschild solution, which link you can consult for details. The result of which is that

$$\ddot{r} = -\frac{Gm_1}{r^2} + r\dot{\theta}^2 - \frac{3Gm_1}{c^2}\dot{\theta}^2$$ $$\ddot{\theta} = -\frac{2}{r}\dot{r}\dot{\theta}$$ [CORRECTED AND SIMPLIFIED Jan 2]

where $r$ and $\theta$ are polar co-ordinates centred on the gravitating mass, the dots represent differentiation by the proper time of the test body.

So the first term is just the classical radial acceleration $-\frac{Gm_1}{r^2}$. The terms $r\dot{\theta}^2$ and $-\frac{2}{r}\dot{r}\dot{\theta}$ are the classical centrifugal and Coriolis acceleration for polar co-ordinates.

What's not classical is the extra term $\frac{3Gm_1}{c^2}\dot{\theta}^2$. Finally, there is the fact that differentiation is with respect to proper time of the test body. Different test bodies will experience time differently. They can be related by:

$$(1 - \frac{r_s}{r})\dot{t}^2 - \frac{\dot{r}^2}{(1 - \frac{r_s}{r}) c^2} - \frac{r^2\dot{\theta}^2}{c^2} = 1$$

The constant $r_s = \frac{2Gm_1}{c^2}$ is introduced for simplicity. It is called the Schwarzschild radius of the gravitating mass.

Here the co-ordinate $t$ is introduced as a reference time, so $\dot{t}$ is the rate of change of reference time with respect to proper time of the test body. For a distant ($r \to \infty$), stationary ($\dot{r}=0, \dot{\theta}=0$) test body, this becomes $\dot{t} = 1$, so the reference time can be interpreted as the time measured on a distant, stationary clock.

In the classical case of course every body experiences the same time, but it can also be compared to the special relativistic case, where the equation would be:

$$\dot{t}^2 - \frac{\dot{r}^2}{c^2} - \frac{r^2\dot{\theta}^2}{c^2} = 1$$

So what's new in general relativity is the factor $(1 - \frac{r_s}{r})$. To get some idea of scale, for the Earth, $\frac{r_s}{r}$ is about one and a half parts per billion on the surface of the Earth. (Note the Schwarzschild solution is only applicable outside the gravitating body.)

This is exact only for $m_2 \to 0$, but it remains a very good approximation as long as $m_2$ is much smaller than $m_1$, such as for a planet orbiting a star.

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Thank you - the term "two-body problem in general relativity" is useful for finding lots of resources and papers on the web that seem to be about what I was interested in. I would still be interested in some insight how one could simplify this problem by specializing it to a toy problem: For example, how complicated would it be to approximate the force between two homogenous spherical, non-rotating masses that do not move (because they are touching)? Would it be possible to give e.g. a series formula for this? Is it appropriate to ask this here or better in a separate question (or not at all)? –  Johsm Dec 30 '12 at 17:16
@Johsm: I think you're looking for the post-newtonian expansion. But the two-body probelm remains unsolved for two POINT masses, much less ones that have structure. –  Jerry Schirmer Dec 31 '12 at 19:57
Two spheres on top of each other: one case that would work is something like a boulder on the Earth. Another is if both of the masses have very weak gravity, like two asteroids on top of each other. The Schwarzschild solution should be adequate (if the classical answer isn't already adequate!). No other cases come to mind. For example if you tried to stand the moon on the Earth, they wouldn't sit still, they would collapse! –  Retarded Potential Dec 31 '12 at 22:19
My curiosity is essentially motivated because I tried to calculate how relativistic (and other) effects influence some of the results one would get for ordinary "school problems". E.g. adding velocities by using $\frac{v_1+v_2}{1 + \frac{v_1 v_2}{c^2}}$ instead of just $v_1 + v_2$. That is fun and can provide some insights to laymen like me. Plus: any schoolkid can calculate it. So I was wondering about essentially doing the same for gravitation and the answers I got here already provided lots of interesting insights! Unfortunately, there will no be many schoolkids who can calculate this ... –  Johsm Jan 2 '13 at 11:36
Well, if we use proper time of the body and Schwarzschild co-ordinates for distance and angle, as I've done, then the acceleration formula is almost the classical one, with a small correction term $(3GM/c^2)\dot{\theta}^2$ if there is non-radial motion. But proper time is different from "stationary distant observer" time. This is true in special relativity as well, but general relativity adds the correction factor $(1 - \frac{r_s}{r})$. I've updated my answer accordingly. Hope this provides some more insight! –  Retarded Potential Jan 2 '13 at 22:01

Yes the non-relativistic equation of motion

$$\frac{dp}{dt} = F = -\frac{G m_1 m_2}{r^2}$$

for a single particle interacting gravitationally with other is only valid for velocities much smaller than the speed of light and for small enough masses. Notice the minus sign in the expression for the non-relativistic force F --gravitation is attractive--.

First, general relativity is a (geo)metric theory. There is no gravitational forces in general relativity. In general relativity, bodies affected only by gravitation are moving freely, but in a curved spacetime

The general relativistic equation of motion for a single body is the geodesic equation

$$\frac{DP^\mu}{D\tau} = 0$$

Note the zero at the right, which is a consequence of the absence of gravitational forces in general relativity. The greek indices run over 0,1,2,3 --the spacetime coordinates-- and summation convention is being used. $P^\mu$ is the four-momentum, $\tau$ the proper time and $D$ denotes the covariant derivative, which includes the effects due to spacetime curvature

$$\frac{DP^\mu}{D\tau} = \frac{dP^\mu}{d\tau} + \Gamma_{\nu\lambda}^\mu U^\nu P^\lambda ,$$

where $\Gamma_{\nu\lambda}^\mu$ are the Christoffel symbols and $U^\nu$ the four-velocity.

Second, the field theory of gravity provides a non-geometrical description of gravity. There are gravitational forces in the field theory. In this theory, bodies affected only by gravitation are moving in a flat spacetime --sometimes this is named the flat spacetime approach to gravitation-- but feel a gravitational force associated to gravitons

The field-theoretic equation of motion for a single body in a gravitational field is the Kalman equation

$$A_\mu^\nu \frac{dP^\mu}{d\tau} = F^\nu = - B_{\mu\lambda}^\nu U^\mu P^\lambda$$

where $F^\nu$ is the exact gravitational force with

$$A_\mu^\nu = \left( 1 - \frac{1}{c^2} \psi_{\lambda\gamma} U^\lambda U^\gamma \right) \eta_\mu^\nu - \frac{2}{c^2} \psi_{\mu\gamma} U^\gamma U^\nu + \frac{2}{c^2} \psi_\mu^\nu$$

and

$$B_{\mu\lambda}^\nu = \frac{2}{c^2} \psi_{\mu,\lambda}^\nu - \frac{1}{c^2} \psi_{\mu\lambda}^{,\nu} - \frac{1}{c^2} \psi_{\mu\lambda,\gamma} U^\gamma U^\nu$$

Here $\psi_{\alpha\beta}$ is the gravitational field potential and the comma denotes the ordinary flat spacetime partial derivative.

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@Johsm General relativity is a geometric theory where the effect of gravitation is not interpreted as a real force but as spacetime curvature. I have added a picture. Only non-geometrical approaches such as Newtonian gravity or the field theory introduce gravitational forces. An excellent discussion of the Newtonian vs GR picture is given in Wald textbook on gravitation, but it is not for layman!In fact –  juanrga Dec 31 '12 at 17:51
You would try to understand that forces are never directly measured, but inferred from apparatus raw data: e.g. accelerations of a test particle, elongation of a hook... –  juanrga Dec 31 '12 at 18:10
Treating gravity as a force is okay when relativistic effects can be neglected. However the idea of gravity as a force breaks down when you include (special) relativity. People tried to make such theories for years but they all failed. The fact that gravity couples universally to all objects in direct proportion to their inertia makes it very different from any other force. Indeed, once you include the action of gravity on light then flat spacetime can no longer work. You need curvature. Read up on the equivalence principle to learn more. Unfortunately I don't know a good lay account off hand. –  Michael Brown Jan 1 '13 at 3:00
@MichaelBrown There is no problem with relativistic gravitational forces. The equivalence principle is at the foundation of the geometric picture of GR, but this principle can be derived from the field approach when higher-order graviton correction are ignored. The field approach predicts the observable light bending. As stated by Feynman in his lectures on gravitation: "It is one of the peculiar aspects of the theory of gravitation, that is has both a field interpretation and a geometrical interpretation. [...] The geometrical interpretation is not really necessary or essential to physics". –  juanrga Jan 1 '13 at 13:58
@juanrga Isn't the (quantum) field theory approach you mention a perturbation theory about a flat metric? When you sum up all the one-point graviton functions with any non-trivial matter (such as a planet), you'll get a curved metric - the Schwarschild metric for example. My understanding is that the geometrical picture naturally springs out of the qft of a massles spin-2 boson. Which viewpoint you take to be more fundamental or natural depends on whether you think a fixed flat space background is sacred or not. –  Michael Brown Jan 1 '13 at 14:08

As in Electrodynamics, the forces become retarded, the corresponding equations become complicated and include radiation too.

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Good! That makes it even more interesting for a curious layman like me! Radiation? Cool!| Gimme the equations or rather, the solutions for the equations! Please? –  Johsm Dec 28 '12 at 13:34
I can't answer your question, but it might help others who can if you told us your physics background. –  Shivam Sarodia Dec 28 '12 at 14:56
My physics background is zero, I am afraid. Just what is left from school a few decades ago and what I read out of curiosity in the time since. The motivation for my question is: I found it surprising how impossible it was (for me at least) on the whole big internet to find a formula that gives the exact force $f$ between two masses $m_1$ and $m_2$ according to current physical knowledge, instead of just the classical Newtonian approximation. –  Johsm Dec 28 '12 at 18:20
@Johsm: There are no exact solutions, and the equations are too complicated to see the force. Physically the force acting on a mass depends on the retarded distance $R(t')$ and other retarded variables. It has not only a "radial" direction, but also a "perpendicular" one with respect to the line connecting two bodies. –  Vladimir Kalitvianski Dec 28 '12 at 18:29

I think if we consider a much bigger mass then the force becomes relativistically: $$F=-\frac{GMm}{r^2}+\frac{4G^2M^2m}{r^3c^2}$$

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your notation is a mess! –  Симон Тыран May 1 at 21:16

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