Karl Coryat said:
If I may put it in my own words, please correct me if I'm off:
You summarized it perfectly in my opinion but do take into account the things that A.T. added in his summary. You have to keep in mind that the most important flaw in the so called "rubber sheet analogy" is that it doesn't take into account space-
time structure (the spatial geometry embedding diagrams also fail to do this but they don't claim to describe space-time, they are meant to describe spatial geometries).
Some use of differential geometry actually makes it explicit why the "rubber sheet analogy" is flawed as far as depicting of space-time geometry goes. Take some arbitrary space-time with metric tensor ##g_{\mu\nu}## and imagine a very small spherical ball of massive test particles that are all
initially at rest with respect to one another (you may be asking yourself what it even means to have a spherical ball in curved space-time and this is a valid concern; we assume the particles are infinitesimally separated from one another in such a way so that they form an infinitesimal spherical ball in the usual vector space sense).
First note that each particle traces out some worldline in space-time so that the entire ball of particles trace out a family of worldlines that "knit together" to form a vector field ##\xi^{\mu}## which represents the 4-velocity
field of this ball of particles. Now we set up our system so that the particles making up the spherical ball are all initially at rest with respect to one another. We want to know how space-time curvature then causes these particles to start converging at that very instant once we let them go into free fall; put a different way, in what way does space-time curvature cause the worldlines of the particles (which are initially at rest relative to one another) to start converging towards one another? From here on out, we go to the instantaneous rest frame of anyone particle in the ball and calculate things from the perspective of an observer ##O## at rest in this instantaneous rest frame.
I won't derive the following relation here but if we assume the infinitesimal ball of particles has a volume ##V## then we can show that the divergence of ##\xi^{\mu}## is related to the rate of change of ##V## relative to ##O## by ##\nabla_{\mu}\xi^{\mu} = \frac{1}{V}\xi^{\mu}\nabla_{\mu}V = \frac{1}{V}\frac{d V}{d\tau}## where ##\tau## is the proper time read by ##O##'s clock. So if we want to see how space-time curvature causes the initially stationary ball of particles to start converging, which is equivalent to looking at the
second order rate of change of the volume of the ball (second order because space-time curvature is second order in ##\nabla_{\mu}##) then we want to focus on the rate of change of ##\theta\equiv \nabla_{\mu}\xi^{\mu}## (often called the expansion scalar) relative to ##O## i.e. we want to focus on ##\xi^{\mu}\nabla_{\mu}\theta = \frac{\mathrm{d} \theta}{\mathrm{d} \tau}##; keep in mind all of this is at the very instant that we let the initially stationary ball of particles go into free fall.
I also won't' derive the following expression (if you want to see where it come's from, read this:
http://en.wikipedia.org/wiki/Raychaudhuri_equation) but it can be shown that for a spherical ball of particles that are all initially at rest relative to one another and allowed to go into free fall, ##\frac{\mathrm{d} \theta}{\mathrm{d} \tau}## at that instant is given by ##\frac{\mathrm{d} \theta}{\mathrm{d} \tau} = -R_{\mu\nu}\xi^{\mu}\xi^{\nu}## where ##R_{\mu\nu}## is the Ricci curvature tensor. Now in the instantaneous rest frame that we are working in, ##\xi^{\mu} = (1,0,0,0) = e_{\hat{t}}## where ##e_{\hat{t}}## is the unit vector pointing along the time axis of this instantaneous rest frame so ##\frac{\mathrm{d} \theta}{\mathrm{d} \tau} = -R_{\hat{t}\hat{t}}##.
Now Einstein's equation ##R_{\mu\nu} = 8\pi(T_{\mu\nu} - \frac{1}{2}Tg_{\mu\nu})## combined with the strong energy condition (
http://en.wikipedia.org/wiki/Strong_energy_condition#Strong_energy_condition) implies that ##R_{\mu\nu}\xi^{\mu}\xi^{\nu} = R_{\hat{t}\hat{t}} > 0## if we are not in vacuum (which is the case we are considering here); this is basically the statement that observers always see
positive space-time curvature, as far as the Ricci curvature tensor goes. This implies that ##\frac{\mathrm{d} \theta}{\mathrm{d} \tau} = -R_{\hat{t}\hat{t}} < 0## i.e. the initially stationary particles composing the spherical ball start converging at the very instant we let them go into free fall.
The important point is that it was ##R_{\hat{t}\hat{t}}## (i.e. the time-time component of the Ricci curvature tensor in the instantaneous rest frame of ##O##) that caused the initially stationary particles to start converging towards one another. If we ignore the "time" part of space-time then we can't even explain why gravity as a manifestation of space-time geometry causes these particles to start converging! This is why the "rubber sheet analogy" is a completely incorrect representation of space-time geometry and the kinematics of freely falling particles in general relativity.
