A Sophisticated Response II
[EDIT: I should have started by saying: dx should probably ignore the following, but PF readers who have more familiarity with elementary gtr might benefit from this post--- sorry, dx, if I had been paying attention I would have noticed how unreasonable it was for me to write as if you could be expected to follow what I said here! So don't worry if you don't follow a word of it.]
My first response was probably too sketchy even for those PF readers who have taken more gtr than dx has, so let me try to elaborate a bit for the possible benefit of such readers.
In gtr the tidal tensor appears as one of the tensors in the Bel decomposition of the Riemann tensor wrt a timelike congruence (possibly having nonzero vorticity) and it can be written
<br />
E [\vec{X} ]_{ab} = R_{ambn} \, X^m \, X^n<br />
Then the Jacobi geodesic formula (see for example Eq. (8.43) in MTW) becomes:
<br />
\frac{D^2 \eta^a}{ds^2} = - {{E[\vec{X}]}^a}_b \, \eta^b <br />
Here, \vec{\eta} is a spacelike separation vector between two nearby timelike geodesics, and where \vec{X} is the tangent vector to the world line of our spaceship (railway car).
Assuming the form of the tidal tensor which I gave in my Post #2 (and which takes the same form in either gtr---- using a Schwarzschild vacuum model--- or in Newtonian theory; deriving the latter claim is easy, in fact I have taught it to first year calculus students), let us consider a thought experiment conducted over a short time period inside our spaceship when it is at distance r=r_0 from an isolated massive source of mass m, in which a cloud of test particles dropped from rest at time zero. We have the simple initial value problem (in either Newtonian gravitation or gtr):
<br />
\begin{array}{ccc}<br />
\ddot{x} = 2 \, m/r_0 \, x, & \ddot{y} = -m/r_0 \, y, & \ddot{z} = -m/r_0 \, z, \\ <br />
x(0) = x_0, & y(0) = y_0, & z(0) = z_0, \\<br />
\dot{x}(0) = 0, & \dot{y}(0) = 0 & \dot{z}(0) = 0<br />
\end{array}<br />
where we regard r_0 as well as m as a constant over the duration of the experiment. This has the solution
<br />
\begin{array}{rcl}<br />
x(s) & = & x_0 \, \cosh \left( \sqrt{2m/r_0} \, s \right) \\ <br />
y(s) & = & y_0 \, \cos \left( \sqrt{m/r_0} \, s \right) \\<br />
z(s) & = & z_0 \, \cos \left( \sqrt{m/r_0} \, s \right)<br />
\end{array}<br />
which to second order in the elapsed time is
<br />
\begin{array}{rcl}<br />
x(s) & = & x_0 \, \left(1 + \frac{2 \, m}{r_0} \cdot \frac{s^2}{2} \right) \\<br />
y(s) & = & y_0 \, \left(1- \frac{m}{r_0} \cdot \frac{s^2}{2} \right) \\<br />
z(s) & = & z_0 \, \left(1-\frac{m}{r_0} \cdot \frac{s^2}{2} \right)<br />
\end{array}<br />
Thus, the particles diverge in the x direction and converge orthogonally; a thin spherical shell of test particles is deformed into a prolate ellipsoidal shell with the long axis oriented parallel to the x direction. The fact that the tidal tensor is
traceless (which in both Newtonian gravitation and gtr is equivalent to choosing a
vacuum solution) ensures that the volume inside such a shell does not change over a short duration experiment, although its shape is deformed.
Notice that, in our short duration experiment, the behavior of the test particles is symmetrical wrt +/- x direction. Furthermore, our model is symmetrical under time reversal. Notice also that I wrote the second order approximation to show that there is a huge conceptual difference between the Newtonian view (according to which tidal forces are acting as given by the tidal tensor) and the gtr view (according to which the world lines of the test particles are timelike geodesics and thus are subjected to no physical force; in gtr all gravitational phenomena are modeled using curvature, although whenever gtr and Newtonian theory give the same results, to some approximation, one can interpret geodesic deviation in quasi-Newtonian fashion).
(Pedantic caveat: to save the trouble of worrying, we can assume that the test particle exhibit nonrelativistic motion during the duration of our thought experiment, so we don't need to worry about "proper time" versus "time".)
The problem as dx stated it is perhaps not perfectly clear, but as I interpret it, the idea is to get the student to analyze this experiment and to draw these conclusions:
- without looking out the window of the spaceship, using one of our short duration experiments you can orient a rod inside the spaceship such that one of its ends points at the massive object--- but you can't say which end, at this order of approximation!,
- the behavior of the cloud of test particles is symmetric wrt the time of dropping the cloud of test particles, at this order of approximation,
- if over a much longer period of time than the brief duration of our thought experiment, the spaceship rises radially and then falls back, and if we repeatedly perform short duration experiments of the kind modeled above, then there is no way to distinguish between the event "r=r0 and increasing" versus the event "r=r0 and decreasing", at this order of approximation, except by examining the changing tidal forces inferred over a long term period from numerous short duration experiments as per above.
IOW, if you imagine repeatedly performing short duration experiments as above, you can indeed in principle track the evolution over time of the tidal tensor, and you can infer in which direction the source must lie, but you can't tell "up" from "down" or "going up" from "going down" from a
single such experiment.
and I hope it will help.
