Reference: Moore, Thomas A., *A General Relativity Workbook*, University Science Books (2013) – Chapter 21; Problem 21.1.

The Einstein equation is

where we have yet to determine the constant . To do this, we need to show that the Einstein equation reduces to Newton’s law of gravity for weak gravitational fields. Actually, there are three conditions that should hold in the Newtonian limit. First, as we’ve said, the gravitational field is weak, meaning that spacetime is nearly flat. Second, objects should travel with a speed much less than the speed of light (the spatial four-velocity components for ). The second condition implies that the only non-negligible component of the stress-energy tensor is . For example, for a perfect fluid, we can assume that it’s effectively at rest, so the off-diagonal elements are all zero. For the diagonal spatial compoments, we have (for ; the other 2 components have the analogous formulas):

This equation is for a cubic volume of side length containing particles of momentum . Since and , (the requirement of a weak gravitational field means that can’t be very large, since we can’t have that much mass). As the spatial diagonal elements , the pressure and , the energy density, this condition translates to .

With these assumptions, we can try to show that the relativistic equation of geodesic deviation reduces to the Newtonian version. That is, we want to show that

where is the separation of two infinitesimally close geodesics (this is the tidal force.) and is the Newtonian gravitational potential.

Starting with 3, we can eliminate terms where or (because for ) to get

where we’re now considering only the spatial components: . Note that summing over is the same as summing it over since due to the anti-symmetry of the Riemann tensor under interchange of its last two indices, . Also, because we’re in the non-relativistic limit, the proper time and coordinate time are essentially the same thing: , so the time derivative is with respect to .

Comparing this to 4, we have (renaming to in the last equation):

Newton’s law of gravity in differential form is

(Again, the contraction over index in the second to last line can be taken over 3 or 4 coordinates, since .) We can raise both indices on the Ricci tensor in the usual way:

so we can write 10 in upper index form as

Now looking back at 1 and using the condition that is the only significant entry in the stress-energy tensor, we have

so

Comparing the second and fourth lines, we see that and

and the Einstein equation becomes

Actually, we can’t take ; all we can say is that for gravitational systems on the scale of the solar system (where Newtonian theory works well, except in the case of the orbit of Mercury) . To get a feel for how small needs to be, suppose we have a spherical gravitational potential in empty space (). Then in spherical coordinates

This is the radial component (the only non-zero component) of the gradient of the potential, and the negative gradient of the gravitational potential is the gravitational field, which is the acceleration of gravity. In the solar system, Newton’s theory says that the acceleration due to the sun is

so if but its effect is not felt in Newton’s theory, we must have

in order for Newton’s theory to be valid in the solar system. Distances in the solar system are around , and in general relativistic units is so this means