# Weak Field Approximation and Tidal Forces

1. Mar 27, 2015

### nigelscott

The weak field approximation in the Newtonian limit shows that the coordinate acceleration along a geodesic is related to the gravitational force.

The geodesic deviation equation relates the coordinate acceleration between adjacent geodesics to tidal forces.

If I drop 2 balls together from the top of a building they fall towards the earth but on the way down there will be an attraction between the two. Is it correct to say that the free fall corresponds to acceleration, g, along the geodesic and the attraction between them is due to the deviation between their respective geodesics (not g) . But if that is true, isn't there also a tidal force associated with the vertical motion? How is this accounted for?

2. Mar 27, 2015

### wabbit

My understanding is that the tidal effect in vertical drop consists in an elongation in the direction of fall and a contraction in the transverse direction. The contraction is simply from the two trajectories getting closer as they converge towards the same point at the center of the earth, and the elongation from the fact the the lower part is subject to higher g than the higher part.( these effects are of course negligible for a drop from a building, but they account for "spaghettification" as one falls into a black hole).

https://en.wikipedia.org/wiki/Spaghettification

3. Mar 27, 2015

### Mentz114

Wabbit's description of tidal forces is accurate.
More accurately the geodesic deviation equation assigns the gravitational field to the Riemann tensor. The tidal part is a contraction with a 4-velocity and so depends on the state of the observer.

$T_{ab}=R_{cabd}U^d U^c$

(sorry if this is not relevant in the weak field theory)

Last edited: Mar 27, 2015
4. Apr 3, 2015

### nigelscott

OK thanks. I think my confusion lies between 'along' and 'between' geodesics. If I drop 2 bowling balls from the top of a building they will both travel along geodesics. There is a longitudinal gravitational force acting along the respective geodesics that pulls them towards the ground. There is also a gravitational force acting in the transverse direction due to the geodesic deviation that causes them to move closer to each other. I think I have mistakenly been treating this as 2 separate mechanisms and that the correct way of thinking about this is that solutions to the geodesic deviation equation results in 2 components of acceleration: longitudinal (radial) and transverse. Am I close to being correct?

5. Apr 3, 2015

### wabbit

Yes but I think this is clearer with the following distinction : considering each ball separately, each one has an acceleration, $\vec g_1\neq \vec g_2$ aligned with its own geodesic, and no other acceleration. Now if you consider the center of mass of the two balls, it has an acceleration $\vec g$ along its own geodesic, and then you can write $\vec g_1=\vec g+\vec h_1,\vec g_2=\vec g+\vec h_2$ and interpret $\vec g$ as their common longitudinal acceleration and $\vec h_1,\vec h_2$ as transverse tidal accelerations.

6. Apr 3, 2015

### nigelscott

Ok...makes more sense now. So the Christoffel symbols in the geodesic equation represent the longitudinal force only .

7. Apr 4, 2015

### wabbit

Yes, although I find it a little confusing to call it the longitudinal force in this context, since it is the only force there (tidal forces being an expression of how the gravitational force varies over space, not a separate force.)
The terms "longitudinal force" and "transverse force" seem more apt when used to refer to the components of the tidal force, i.e. to the differences in the acceleration of nearby test particles, those differences being parallel to the direction of their common motion (longitudinal tidal force) or transverse to that direction (tranverse tidal force).