# If gravity is due to curvature, how does gravity work in situations with no curvature?

The strength of the gravitational field falls off as the inverse square of the distance from a spherical source. It only falls off as the inverse of the distance from an extended cylindrical or line source, and it does not fall off at any distance from an infinite plane.

As a gravitational source, instead of a massive spherical planet or star other some other body, I would like to consider a massive body in the shape of an infinite plane. Since this may be viewed as an unphysical situation, let us just say that the body has finite size but very large. Instead let us consider an approximation such as a wafer shaped mass which extends in a plane for millions of miles (let’s say the X,Z plane) and having a depth of 100 miles in the Y direction.

In this situation, if a test mass were above the surface, and sufficiently far from the edges (let’s say above the direct center of the wafer/planet), then the strength of gravity in that region would not decrease as a function of distance from the surface.

In a region above the center of this wafer planet, I presume that the Weyl tensor vanishes, and therefore there is no curvature. If that is incorrect, I appreciate some clarification. If I am basically correct so far, then my question is how does gravity work in this case, i.e. how does gravity work if there is no curvature?

"Gravity is the result of intrinsic (no fifth dimension needed) curvature >in four-dimensional spacetime, not three-dimensional space."

and

“If you consider space and time as a four dimensional geometry, gravity IS >the curvature of those dimensions.”

Also, many other sources indicate the same idea, that curvature is the cause of gravity. For example from Wikipedia

“Gravity is most accurately described by the general theory of relativity >(proposed by Albert Einstein in 1915) which describes gravity, not as a >force, but as a consequence of the curvature of spacetime caused by the >uneven distribution of mass/energy; and resulting in time dilation, where >time lapses more slowly in strong gravitation.”

By the way, in the scenario I propose, I believe that there is still curvature below the surface described by the Ricci tensor. Could the Ricci tensor curvature below the surface be the source of gravity as seen above the surface?

[Edit/update] After some research regarding infinite planes as suggested, as well as other material, I am beginning to feel it might be correct to say that curvature does always exists where there is gravitation. To put it another way, a body in a homogeneous gravitational field would not gravitate in any direction, because by the definition of being homogeneous, every direction would look exactly the same.

• Have a read through The general relativistic infinite plane. There is no solution for just an infinite 2D mass distribution. To get a solution you need to add pressure terms to the stress-energy tensor and a cosmological constant. – John Rennie Jan 15 '16 at 12:21
• The full tensor is the Riemann tensor. In some situations, the Ricci tensor may disappear. Such solutions are called vacuum solutions, since the Stress Tensor and thereby the Ricci tensor vanishes. The Riemann tensor however, will not vanish unless the space is indeed empty and flat. The non-vanishing of the Riemann tensor is what characterises curvature and therefore gravity. Also what John Rennie said above. – Horus Jan 15 '16 at 12:24
• the model is useful in cosmology, ie to study thin gaseous disks with embedded planets – user46925 Feb 3 '16 at 1:45