I see tidal gravity mentioned in the literature sometimes, and sometimes people even say something like that is the “real gravity”. I am confused about the significance of tidal gravity, and what role it plays in curvature. Please correct me if I have any misconceptions, and I will try to explain my confusion.
Tidal gravity, by my understanding, is the difference in gravity between two points. As the strength of gravity falls as the inverse square of the distance from a spherical source, tidal in this sense indicates a strictly geometrical attribute. What I mean is that there is no difference between this effect and any other geometrical inverse square law, such as for the intensity of light as emitted by a point source. It is due to geometry. Individually, the radiated photons travel to infinity, but the overall intensity (flux) falls off as the inverse square of the distance, due to equal amounts of light passing through increasingly larger surface patch areas.
As for what curvature means, I like how it is presented in the Feynman lectures “curvature expressed in terms of the excess radius is proportional to the mass inside a sphere”. They also give the second law of gravitation as ” How objects move if there are only gravitational forces—namely, that objects move so that their proper time between two end conditions is a maximum.”
So it seems to me that tidal gravity is only kind of indirectly related to curvature. The spherical nature of the gravitational source of bodies such as planets and stars make the distribution of matter (energy) unevenly distributed. This (uneven density distribution of energy) gives rise to the curvature, or excess radius. The fact that the bodies are spheres (or point sources as viewed from a larger perspective) is also what gives rise to the inverse square law.
We could consider the situation for a gravitational body which is an infinite plane. In this case the gravitational strength would not fall as the inverse square of the distance. It seems to me that below the surface there would still be an excess radius, and therefore curvature would exist. But above the surface there are no tidal effects. Gravity is still attractive above the surface, but is space-time there curved?
I could be wrong, but from what I think I understand about tidal forces, there are three cases that I can think of. Please correct me if I am incorrect.
Gravitational tidal force 1: (The inverse square law) This is the decrease in gravitational strength as a body travels further from the center of an (idealized) spherical gravitational source.
Gravitational tidal force 2: (density distribution) These tidal forces are due to the uneven density distribution of mass/energy in the gravitational source. If the mass in the center of the earth is more dense than the mass closer to the surface, this will be a second kind of tidal force. Mountains, valleys, and in general any deviation from a perfectly spherical shape contribute to these tidal forces.
Gravitation tidal force 3: (non-linearity, or the gravity of gravity) I am a bit fuzzy on this, but it seems to me how this works is that above and beyond the direct gravitation due to mass/energy, there is additional gravitation due to the fact that that gravitation itself is a form of energy.
Errors in my assessment of tidal forces notwithstanding, my question is what is the relationship between gravitational tidal forces and curvature.
When I brought up the scenario of an infinite plane gravitational source, I did so because I thought (and still do think) that the relationship between tidal forces and curvature might be more easily understood if we could eliminate gravitational tidal force #1 above. With an infinite plane gravitational source the strength of gravity does not fall off as a function of distance from the surface.