I Ricci Tensor in Vacuum Inside Earth - Ideas Appreciated

[AlanE]

Imagine a hole drilled through the Earth from which all air has been removed thus creating a vacuum. Let a cluster of test particles in the shape of a sphere be dropped into this hole. The volume of the balls should start to decrease. However, in his article "The Meaning of Einstein's Equation" page 18, John Baez and Emory Bunn indicate that the rate of decrease of the volume, per volume, should equal the t-t component of the Ricci tensor. But how can this be if the Ricci tensor is zero in a vacuum? Any ideas would be appreciated.

 

[pervect]

Imagine a hole drilled through the Earth from which all air has been removed thus creating a vacuum. Let a cluster of test particles in the shape of a sphere be dropped into this hole. The volume of the balls should start to decrease.

What makes you think the volume should decrease? After a little thought, I agree that the volume will decrease, by analogy with the case I mention below. Basically a disk of test particles at the same height will reach the center of the Earth at the same time, which winds up implying the volume at the center of the Earth is zero.

However, in his article "The Meaning of Einstein's Equation" page 18, John Baez and Emory Bunn indicate that the rate of decrease of the volume, per volume, should equal the t-t component of the Ricci tensor. But how can this be if the Ricci tensor is zero in a vacuum? Any ideas would be appreciated.

I've never thought much about a ball of test particles being dropped into the center of the Earth, but I have thought more about a ball of test particle orbiting the Earth in a non-rotating spaceship. I think the conclusions are similar.

What happens is that V(t) does decrease with time, but if you look at Einstein's equation, it only says that $\ddot{V}= 0$ at t=0, when all the particles in the ball are initially at rest relative to each other. So if we plot V as a function of t, initially V does not change with time, but this is only true at t=0, it is not necessarily true that $\ddot{V}= 0$ at any other time. There is no requirement that V be a constant of motion. The only requirement is that at t=0, the second time derivative of the volume is zero.

 

[AlanE]

Thanks for the reminder that merely having the second derivative equal to 0 doesn't imply that the volume won't decrease. I guess the equation for the change of volume could be something hypothetically like V(t) = V(1-kt^{3). or V(t) = V(1-kt**4) etc. I need to research this myself. Maybe someone knows the required equation even if only for Newtonian mechanics.}

 

[pervect]

Thanks for the reminder that merely having the second derivative equal to 0 doesn't imply that the volume won't decrease. I guess the equation for the change of volume could be something hypothetically like V(t) = V(1-kt^{3). or V(t) = V(1-kt**4) etc. I need to research this myself. Maybe someone knows the required equation even if only for Newtonian mechanics.}

I believe the evolution of the volume in the general case would be given by Raychauhurdi's equation, https://en.wikipedia.org/wiki/Raychaudhuri_equation. But the goal is to make things as simple as possible, which Baez does by choosing initial conditions appropriately - those initial conditions being that the ball of test particles start out with all the particles relatively at rest. Then a lot of complexities disappear, and the underlying simplicity emerges.

 

[PeterDonis]

a disk of test particles at the same height will reach the center of the Earth at the same time, which winds up implying the volume at the center of the Earth is zero.

Not quite. A disk is not a ball, and the version of Einstein equation under discussion only applies to a ball. A ball of test particles dropped from initial rest such that its center is at a given height will not all reach the center of the Earth at the same time (although the portion of the ball consisting of a disk transverse to the direction of fall will); the ball squeezes in the transverse direction but stretches in the radial direction.

I believe the evolution of the volume in the general case would be given by Raychauhurdi's equation

Strictly speaking, this won't be true if any of the test particles in the ball meet at the center of the Earth, because Raychaudhuri's equation only applies to a congruence of non-intersecting worldlines. However, we can use it up until that point.

Looking at the RHS of the equation, there is only one possible nonzero term: the shear term $- 2 \sigma^2$. The tidal tensor term is zero in vacuum (because it is the trace of the tidal tensor, which as noted is equal to $R_{mn} X^m X^n$, and $R_{mn} = 0$ in vacuum). So the key thing to compute is the shear scalar of the congruence in question. It starts out being zero (because, roughly speaking, the shear scalar is a sort of average shear in all directions, and the positive shear in the radial direction is exactly canceled by the negative shear in the transverse direction). However, I'm not sure whether it continues being zero once the fall is in progress or not.

 

[pervect]

Not quite. A disk is not a ball, and the version of Einstein equation under discussion only applies to a ball. A ball of test particles dropped from initial rest such that its center is at a given height will not all reach the center of the Earth at the same time (

although the portion of the ball consisting of a disk transverse to the direction of fall will); the ball squeezes in the transverse direction but stretches in the radial direction.

The later point is the one I was trying to make - if the disk, which is a sub-manifold of the ball, deforms to a point, we expect the volume of the deformed ball to wind up to be zero.

But this isn't inconsistent with Einstein's equation, for reasons that have already been discussed and which the OP seems happy with.