da_willem said:
I looked it up for a Schwarzschild geometry and in this case it is indeed true that the curvature (as in the componnents of the Riemann tensor) gets less as you approach the center of mass (decreasing r).
About your second remark; I've heard before that the curvature makes you cover more space in equal units of time. and that's where the acceleration comes from. Can you (or somebody else) elaborate on this, maybe a little more mathematical, or provide me a link to a relevant webpage?
Where did you look this up?
For the Schwarzschild metric, which is a point mass, the components of the Riemann tensor increase as you go towards the center. In terms of the unit basis vectors of an observer the "r" components go as
\large R_{\hat t \hat r \hat t \hat r} = -2GM/r^3
MTW, pg 820, "The nonsingularity at the gravitational radius"
If you were moving towards the center of a distributed, non-pointlike body, I would expect different behavior, so I don't necessarily disagree with the remark you were responding to, I haven't worked that case out (but it sounds right). The Schwarzschild case, though, definitely has the Riemann components increasing as you decrease r.
as I will explain, this tensor component represents the tidal "stretching force" in the radial direction. There are other components of the tensor, but they behave in a similar manner.
The link between space-time curvature and acceleration is the geodesic deviation equation
Two nearby _freely moving_ obsevers separated by spacelike vector E
k will accelerate away from each other with an acceleration given by
\large \frac{D^2 E^k}{d \tau^2} = -R^{\hat j}{}_{\hat \tau \hat k \hat \tau} E^k
Here \large \tau is the proper time of the observer(s). Since they are intially co-moving and close together, their clocks run at the same rate.
The symmetries of the Riemann make this equivalent to
\large -R_{\hat \tau \hat j \hat \tau \hat k} E^k
When we translate this remark out of tensor language,when we look at an observer freely falling near a black hole, we say that two observers separated by a distance delta-r in the radial direction will accelerate away from each other. The magitude of the acceleration is -2GM/r^3 * delta-r, the value of the apprpriate component of the Riemann tensor * the separation, and it can be interpreted as the tidal stretching force / unit mass, or the tidal acceleration.
For a web reference, I can't come up with anything better than
John Baez's
http://math.ucr.edu/home/baez/gr/geodesic.deviation.html
I hope the tensor notation isn't too confusing, but I'm afraid it may be. A few more examples might help.
EDIT for readibility, I'm going to replace \tau with t from now on, \tau looks too much like r.
Our observer has coordinates
r, \theta, \phi, t and unit vectors of \hat r, \hat \theta, \hat \phi, \hat t
Suppose there were a "tidal torque" that caused the end of a rod pointed in the \hat r direction to accelerate in the \hat \theta direction? The associated component of the Riemann would be
\large R_{\hat t \hat \theta \hat t \hat r}.
There isn't any such component in actually, but there is a compresive tidal force, for a dispacement in the \hat \theta direction, that points in the same direction, which is given by
\large R_{\hat t \hat \theta \hat t \hat \theta} and has a value of GM/r^3.
You can basically think of the 16 componets R_{\hat t \hat i \hat t \hat j} as being a collection of tidal stretching/compressive forces, when i=j, and tidal torques, when i is not equal to j.
