Fall in Space-time

1. Jul 1, 2015

Noctisdark

Hi there, the most basic formula from newtonian gravitation is, for a free falling object, x = gt^2/2, where x is the displacement in the vectical direction, in the case of general relativity, this just happens to be for weak gravitational fields, for the sake of simplicity (trying to express my thoughts) let's first talk about a weak one, gt^2/2 look like a parabola, so there is some extreme point which is the minimum, you cannot go further, if space time is a curved manifold too, then there must be some extreme points such the manifold is minimum (like the tip of U), one can say that if spacetime is like that, then we can fall onto it, until we reach that minimum point when we cannot fall anymore, is that point special, what does it represent, how is the time flow relative to other observers away from the curvature, how is space there?

2. Jul 1, 2015

A.T.

You are confusing the extrinsic curvature of a trajectory with the intrinsic curvature of space time.

At the center of the Earth you have a locally minimal clock rate.

3. Jul 1, 2015

Noctisdark

That local minimum clock rate exist at the center of mass, so at that point of space, time is running so slow relative to some observers, away from the center time is running somewhat "right" to them, so can i conclude that spacetime is a measure of how time flows at a point of space relative to an observer in a flat time ?, for that observer can he say that t = t(x,y,z)?,

4. Jul 1, 2015

5. Jul 1, 2015

1977ub

Inside a hollow sphere there are no experienced gravitational accelerations. I take it though that a clock at the center ticks more slowly than one at the edge?

6. Jul 1, 2015

Staff: Mentor

Correct--more precisely, spacetime is flat.

No. Clocks everywhere inside the hollow sphere tick at the same rate. But they all tick more slowly than clocks outside the sphere (and clocks outside the sphere tick more slowly the closer they are to the sphere's surface).