Resolving GR with two dropped objects

Question

I'm trying to understand GR and geodesics in relation to a specific example.

Two objects, suspended at different heights so that one is over the other while both of their velocities relative to the ground are zero. Both are released so that they enter free fall with a timing of release so both objects strike the ground at the same time.

In proxies for GR, such as two paths on a sphere which are pointing to the same location (walking to the north pole example), the rate which objects approach one another is dictated by the curvature. But in the example above, while both objects are approaching the same location, they approach it at different velocities.

For past questions I have asked about GR and geodesics I typically get an answer like "you are neglecting time and only looking at the curvature of space instead of the curvature of spacetime". I do understand that the curvature of time matters, and for past examples, like throwing objects into the air, I have been able to understand how time becomes relevant for them. (I think anyway, maybe this example shows I don't understand it)

I'm asking for help understanding how the curvature of time is changing the approach speed, not just that the answer lies in the curvature of time.

Answer 1

As you know already from @Dale's and my answers to your previous question (How do geodesics explain two identical balls thrown up at the different speeds?), all trajectories in 4-dimensional space-time with gravity $g=10\ \text{m/s}^2$ have the same very big radius of curvature $R=\frac{c^2}{g}$.

So let's draw two trajectories of two bodies with the same curvature radius $R$.

The bodies are released at different heights ($h_1$ and $h_2$) with zero velocity (i.e. with horizonzal slope). The one starting from higher level is released first, and the one starting from lower level is released at a later time. After some time they meet at the same point in space and time. They need different times of flight ($T_1$ and $T_2$) until they reach this meeting-point. And then they have different velocities (i.e. different slopes in $z$-$ct$ diagram).

You see, there is nothing mysterious about the fact that they reach the same point with different slope.

Edit (in response to comment):

You need to distinguish between slope and curvature.

Of course different trajectories can have different slopes at the same spot in space-time. Slope is a property of a trajectory, not a property of the spot in space-time. However, all trajectories passing through this spot in space-time share the same 4-dimensional curvature ($1/R$). Therefore it makes sense to say: curvature is a property of space-time itself, instead of a property of the trajectories.

Answer 2

In short, since you drop one after the other, at the moment the second object is dropped, the first object is passing by with some velocity. The two objects have different energy at the same time. In a 2-D analogy (like a North Pole example), it's like approaching the same point from many different angles, where each angle represents a different amount of energy at the same time. And although you can imagine an animation where each walker walks at the same speed in terms of interval per proper time, the physical velocity isn't the velocity you imagine anymore, but rather the angle in r-t space.

If you want to know some of the math for geodesics around a point mass, read this: https://en.wikipedia.org/wiki/Schwarzschild_geodesics

The Schwarzschild metric, ignoring the angular components (and only dropping things straight down), is:

$c^{2}d\tau^{2} = (1-\frac{r_{s}}{r})c^{2}dt^{2} - \frac{dr^{2}}{1-\frac{r_{s}}{r}}$

The bending of space and time appear in the $f(r)=(1-\frac{r_{s}}{r})$ terms.

Then using algebra you can see that

$\left(\frac{dr}{d\tau}\right)^{2} = f^{2}(r)c^{2}\left(\frac{dt}{d\tau}\right)^{2} - c^{2}f(r)$.

Since the metric does not vary with time, an energy-per-mass is conserved: $e = f(r)\frac{dt}{d\tau}c^{2}$ with $\frac{d}{d\tau}e=0$.

So $\left(\frac{dr}{d\tau}\right)^{2} = \frac{e^{2}}{c^{2}} - c^{2}f(r)$.

As a result, the velocity of an object changes with position and depends on the initial energy. Here's a somewhat more intuitive explanation. Zooming in on the spacetime close to the object, the object should be trying to move in a straight line: a constant amount of proper time, with a fixed amount in the $t$ direction and a fixed amount in the $r$ direction, as it would in flat space. However, from our perspective, due to the curvature of space and time, its fixed amount in the $t$ direction is less by $\sqrt{f(r)}$ (time dilation! clocks near black holes tick slowly) and its fixed amount in the $r$ direction is more by $\sqrt{\frac{1}{f(r)}}$. In less time, it travels more distance, so it is accelerating, and doing so according to how the metric is bending space and time measurements.