I Geodesic deviation in static spacetime

1. Oct 28, 2017

timmdeeg

Regarding Einstein's static universe John Baez explains in The Meaning of Einstein's Equation

To see this, consider a small ball of test particles, initially at rest relative to each other, that is moving with respect to the matter in the universe. In the 13 local rest frame of such a ball, the right-hand side of equation (2) is nonzero. For one thing, the pressure due to the matter no longer vanishes. Remember that pressure is the flux of momentum. In the frame of our moving sphere, matter is flowing by. Also, the energy density goes up, both because the matter has kinetic energy in this frame and because of Lorentz contraction. The end result, as the reader can verify, is that the right-hand side of equation (2) is negative for such a moving sphere. In short, although a stationary ball of test particles remains unchanged in the Einstein static universe, our moving ball shrinks.

So in this universe the ball of test particles shrinks perpendicular to the direction of motion, which I think means that geodesics converge accordingly. Now, in this spacetime the spatial geometry is spherical.
Can someone explain what happens to said ball and how works geodesic deviation in case of a static spacetime with flat spatial geometry instead, like the 3-torus?

2. Oct 28, 2017

Staff: Mentor

There is no "direction of motion". Baez is working in a frame in which the ball is initially at rest, and by conservation of momentum, the center of mass of the ball remains at rest in this frame. So the ball as a whole has no "direction of motion".

The shrinkage is of the volume of the ball in this frame. That means all of the particles in the ball start falling inward towards the center of the ball.

The geodesics of the particles in the ball do converge, yes--because all of them start falling inward towards the center of the ball.

No, it isn't. Baez is working in a local inertial frame, in which spacetime is flat. His thought experiment tells you nothing about the global geometry of the spacetime (or of "space" in any global coordinate chart).

3. Oct 29, 2017

timmdeeg

John Baez takes reference to Einstein's static universe and distinguishes two cases.
First, the center of the the ball is "at rest with respect to the local matter." In this case the second derivative of the ball's volume is zero.
Second, the ball moves with respect to the matter, then the second derivative of the ball's volume is negative. The motion of the ball is illustrated on page 14.

So what am I missing?

And also does it make sense to say the local curvature of spacetime depends on whether or not the ball is "at rest with respect to the local matter."? What is wrong to simply argue that the spacetime in Einstein's static universe is curved, because the energy density is not zero?

Thanks, that meets exactly what I was interested in. So the ball should shrink the same in the static 3-torus, because the cause of the shrinking is the same, "the matter is flowing by".

Last edited: Oct 29, 2017
4. Oct 29, 2017

Staff: Mentor

Ah, ok. Yes, in the second case the ball is moving relative to the "cosmological fluid", i.e. relative to the average of all the matter in the universe, and its behavior can be used to show the spatial curvature (if any) of the universe.

No, it doesn't. I misunderstood what you were referring to. As Baez explains, and as I noted above, the shrinkage of the moving ball shows spatial curvature. If there is no spatial curvature, as in a flat 3-torus, there is no shrinkage.

Note, however, that AFAIK there is no solution of the EFE that is both static and spatially flat. Invoking a 3-torus topology in order to make the universe spatially finite is not sufficient to allow it to be static. The positive spatial curvature of the Einstein static universe is necessary for it to be static (heuristically, because the positive curvature is necessary to balance the negative curvature due to the cosmological constant in order to allow the universe to be static).

5. Oct 29, 2017

Staff: Mentor

No. The curvature of spacetime does not depend on the motion of the ball. The effect of spacetime curvature on the ball can be affected by the motion of the ball, but not the spacetime curvature by itself.

6. Oct 29, 2017

timmdeeg

Great, thank you for this enlightening answer.