Many experiments (e.g. changes in Mercury's orbit) show that Newton's Law of Gravity is not exact. According to the theory of relativity, what is the exact formula?

  • 5
    $\begingroup$ By "exact formula" do you mean the formula that is exactly the one used in general relativity, or do you mean "exactly correct"? There are many physicists who think that general relativity is not exactly correct; that's essentially the point of string theory. $\endgroup$ Aug 13, 2021 at 16:42
  • 4
    $\begingroup$ See en.wikipedia.org/wiki/Schwarzschild_geodesics for various comparisons between Newtonian gravity & GR around a spherically symmetric mass (the Schwarzschild solution). $\endgroup$
    – PM 2Ring
    Aug 13, 2021 at 16:43
  • 1
    $\begingroup$ All theories are approximations of reality. Newton is exact enough to land humans on the moon and get them back safely. It works well in many circumstances, and using the more complex (but arguably more precise) GR equations isn't always the best choice. Newton isn't "wrong" either, it just works better at different scales then GR. $\endgroup$
    – Polygnome
    Aug 16, 2021 at 11:04

2 Answers 2


Going from Newton's theory to Einstein's theory is not simple. It's not like you can just add a term to Newton's gravity, like $\textbf{F}=-{GmM \over r^3}\textbf{r} + \textbf{f}(\textbf{r})$ and obtain the right formula.

To say it with Feynman's words, you can't make imperfections on a perfect thing, you have to make another perfect thing.

And the new perfect thing, Einstein's gravity, is definitely more complicated than Newton's. To understand the extent of the difficulty, let me do a simple comparison between the theories.

Newton's theory of gravitation

The theory is concerned with two kinds of objects. The matter and the gravitational field. To determine the evolution of these two objects there are two equations, that are usually expressed in an inertial reference system.

The first equation shows how a particle of matter at position $x^i$ moves under the influence of the gravitational field $\Phi(\textbf{x},t)$

$${d^2x^i(t) \over dt^2} = -{\partial \over \partial x^i} \Phi(\textbf{x},t)$$

The second equation details how to compute the gravitational field $\Phi(\textbf{x}, t)$ at any point in space and time, if you know the density of matter $\rho(\textbf{x},t)$ at every point in space and time.

$$\nabla^2\Phi(\textbf{x},t) = -4\pi G\rho(\textbf{x},t)$$

It is important to notice that $\Phi$ and $\rho$ are both scalar functions: they are functions that map a real number to every point of space and time.

These two equations are everything that is needed to describe gravity. The first one shows how the matter moves and the second one describes how the gravitational field adjusts itself to the change of matter distribution. The modified gravitational field will then affect the motion of matter in a different way and so on.

Einstein's theory of gravity

Here the role of gravitational potential $\Phi$ is taken by the metric tensor $g_{\mu\nu}$ which is a collection of 16 real numbers, one for every combination of $\mu$ and $\nu$, that can take values 0,1,2 and 3. Actually not all 16 numbers are independent, because $g_{\mu\nu} = g_{\nu\mu}$, so there are only 10 independent numbers. We go from $\Phi$, that was only 1 real number for every point in spacetime, to $g_{\mu\nu}$ which carries 10 real numbers for every point in spacetime. This is already looking more complicated.

A similar argument can be made for $\rho$. It gets substituted with the stress-energy tensor $T_{\mu\nu}$, which again carries 10 independent real numbers.

The two equations above get also modified and become a lot more complicated. An important consideration is that Einstein's equations are usually written in a totally generic reference system, to just inertial ones. The first, the one that describes the motion of one particle under the influence of gravity becomes

$${d^2x^{\alpha} \over ds^2} = -\Gamma_{\mu\nu}^\alpha {dx^{\mu} \over ds}{dx^{\nu} \over ds}$$

where $\Gamma_{\mu\nu}^\alpha$ is a function of $g_{\mu\nu}$ and its first derivatives with respect to the coordinates $x^{\mu}$. Notice that in this equations the derivatives are not taken with respect to time ($x^0$), but to the parameter $s$ (that can be chosen to be equal to time, but this is often not the smartest choice).

The second equation, which describes how the distribution of matter $T_{\mu\nu}$ shapes the metric tensor $g_{\mu\nu}$ is

$$G_{\mu\nu} = {8\pi G \over c^4}T_{\mu\nu}$$

where $G_{\mu\nu}$ is a pretty complicated, non-linear function of $g_{\mu\nu}$, and its first and second derivatives. This last equation is actually a system of 10 coupled non-linear second order partial differential equations in 4 dimensions and it is a nightmare to solve.

Just compare it to the corresponding Newtonian equation, which is 1 linear second order partial differential equation in 3 dimensions.

These are the exact formulas according to the theory of general relativity.

  • 1
    $\begingroup$ Note/extra info: the JPL DE ephemerides for the Solar System bodies (e.g. Park et al. for DE440; Section 3.1) use a parameterized post-Newtonian (PPN) formulation for the acceleration of each body due to the others; the PPN $\beta,\gamma$ values are set at the 1.0 predicted by General Relativity. This formulation is described in more detail in Section 4.4 of Moyer 2000. There you can see the Newtonian force as the 1st term with the relativistic perturbations added $\endgroup$ Aug 13, 2021 at 22:31
  • 1
    $\begingroup$ @astrosnapper One should be careful in interpreting such equations. The resulting dynamics are accurate, but one should keep in mind that effective gravitational acceleration is a nebulous and ill-defined thing in GR. Equation 4-61 is (for $\beta = \gamma = 1$) simply the coordinate acceleration in Schwarzschild isotropic coordinates, but the analogous expression in static coordinates ends up looking appreciably different. Neither has a better claim to being the "true" acceleration, however: both yield correct position predictions, e.g. for gravitational lensing or orbital precession. $\endgroup$
    – jawheele
    Aug 14, 2021 at 4:53
  • $\begingroup$ Might be worth noting that it is equivalent to consider the connection as the "gravitational potential" $\endgroup$ Aug 16, 2021 at 11:47

In Einstein's formulation, there is an object called a tensor, that describes the curvature of four-dimensional spacetime. The values of this tensor depend on the distribution of stress, energy, mass and momentum within spacetime. Then there is a geodesic equation that describes the shape of the world line of a freely falling particle in spacetime.

Finding the effective "force" on the particle at given time is equivalent to solving these equations, and in general, no closed solution exists. It is possible to show that in the weak gravity/slow motion limit, Einstein's model simplifies to Newtonian gravity, so for the (relatively) weak gravity in the solar system, Newtonian gravity is an excellent approximation. It is possible to find approximate solutions to Einstein's equations that can explain the relativistic effects such as Mercury.

If you just want a slogan to print on a t-shirt, then the Einstein field equation is

$$R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} + \Lambda g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu}$$

Where $R_{\mu \nu}$ and $R g_{\mu \nu}$ are tensors that represent aspects of the curvature of space, $\Lambda g_{\mu \nu}$ represent the cosmological constant (possibly an aspect of dark energy) and $T_{\mu \nu}$ is the tensor for the distribution of stress, energy and mass in spacetime.

One big difference between Newton and Einstein's formulation is the treatment of time. Newton has particles in three dimensional space moving, so it is natural to talk about a particle's velocity. In Einstein, time is one of the dimensions, and instead of particles moving, there are static world-lines in four-dimensional space.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .