What will be the lengthdifference in the paths through spacetime of a simultaneously shot bullet and dropped mass when the mass hits the floor? A bullet is shot down vertically from the same spacetime point from which we drop a mass from rest. The bullet arrives firstly on the ground and when the mass hits at the same place the clocks of both clocks stop ticking.
It's obvious that both of them have travelled the same length through space, but what about the lengths of the paths through spacetime? The falling mass travels on a geodesic and should be shorter than the path the bullet has travelled on. Which seems to imply that the clock on the bullet shows more elapsed time than the clock on the mass.
I'm not sure if it's important to specify the reference frame from which we look at the situation. After all, the clocks will show a difference independently of the observers frame and the height is the same for both. But say we look at it from the ground.
How can we calculate the difference in the two pathlengths? Say they start at $t=0$ and $r=h$ and the bullet's initial velocity is $v_0$ (it's clock starts ticking St the moment it has this velocity). Do we have to integrate small pieces of $ds$ on which $t$ is constant, making use of the metric? The situation for the falling mass, traveling on a geodesic and along $r$ only, seems straightforward, though I'm not to familiar with the geodesic equation (from Wiki):
On an n-dimensional Riemannian manifold $M$, the geodesic equation written in a coordinate chart with coordinates $x^{a}$ is:
$$\frac {d^{2}x^{a}}{ds^{2}}+\Gamma _{bc}^{a}{\frac {dx^{b}}{ds}}{\frac {dx^{c}}{ds}}=0$$
where the coordinates $x^a(s)$ are regarded as the coordinates of a curve $\gamma (s)$ in $M$ and $\Gamma _{bc}^{a}$ are the Christoffel symbols. The Christoffel symbols are functions of the metric and are given by:
$$\Gamma _{bc}^{a}={\frac {1}{2}}g^{ad}\left(g_{cd,b}+g_{bd,c}-g_{bc,d}\right)$$
where the comma indicates a partial derivative with respect to the coordinates:
$$g_{ab,c}=\frac {\partial {g_{ab}}}{\partial {x^{c}}}$$
The metric is the Schwarzschild metric of which we need only the $r$ part:
$$g=-c^{2}{d\tau }^{2}=-(1-\frac {r_{s} }{r})c^{2}\,dt^{2}+(1-\frac {r_{s} }{r})^{-1}dr^2$$
Is this enough information to solve the problem and is so, how to proceed? Is it even possible to calculate the length of the bullet's path?
 A: The difference will be tiny. The paths through spacetime are very close to straight lines.
The path through space is curved and very short - maybe a $100$ m or so for the bullet. On the other hand, the path through spacetime is much, much longer. It includes the distance in the timeward direction. Since $s^2 = -c^2 d\tau^2 +$ (small number), the length grows $300,000$ km longer each second.
Without gravity, an object at rest would follow a straight line. A freely falling object deviates a few meters from this in $1$ sec. You can show the radius of curvature of this path is about $1$ light year. Gravity on Earth is extremely weak.
It doesn't much matter if one bullet is freely falling, or the other is accelerated upward at $1$ g by the floor. The paths are so near straight that the difference in length is hard to see. The two paths do lead to slightly different spatial positions before one bumps the floor and is deflected back toward the destination of the other. The bullet does land on the dropped mass.
There will be a tiny difference in the ages of the bullets when they collide. Time runs slower in a gravitational well. Since one bullet spent some time at a higher altitude, its clock ticks a little faster. It arrives at the endpoint having traveled through a little less time.
The Harvard Tower Experiment showed that time runs about $1.000000000000005$ times faster at an altitude $22$ m higher. That gives you an idea of the difference in path lengths.
