Do test particles initially comoving with a black hole accelerate away from it?

Question

This question feels ridiculous, but I really am confused. If you Google Image search "schwarzchild light cones" it shows how, relative to the frame comoving with the singularity, the speed of light seems to increase as you move outward from the event horizon. Thus, an outgoing null geodesic should accelerate away from the black hole. You can even see them doing that in the images.

I guess I'm okay with that; it's like, if the light is "above the escape velocity", then it will keep going forever, eventually righting itself to normal speed at infinity. But then, doesn't continuity imply that timelike geodesics would have to do the same thing, accelerate away from the singularity? That doesn't make sense if a black hole sucks things into it. But you can even see it from the connection; $\Gamma^r_{tt}$ works out to be $-\frac{r_s}{2r^2}(1-\frac{r_s}{r})$, which is negative outside the horizon, meaning the covariant derivative of $e_t$ itself is inward, meaning the parallel transport of $e_t$ actually picks up positive $r$-component, such that a $t$-geodesic should accelerate outward.

Answer 1

If you parallel transport $e_t$ in the $+t$ direction, it picks up a negative $r$ component, not positive. I think you just made a sign error in your calculation of $Γ^t_{rr}$. See, e.g., here.

Everywhere outside the horizon, worldlines of constant $(r,θ,\phi)$ accelerate outward, and therefore, everywhere outside the horizon, geodesics accelerate inward relative to those stationary references (though in another sense, of course, they don't accelerate).

The apparent decrease of the speed of light (narrowing of the light cones) close to the horizon in Schwarzschild coordinates is a coordinate artifact. In Gullstrand-Painlevé coordinates, the difference between the radial ingoing and outgoing speed of light is $2c$ everywhere, though you shouldn't assign any physical significance to that either.

Answer 2

I'm adding this as a new answer since it is completely different from the previous one, which missed the point of the question.

So, it is indeed true that both timelike and null outward-directed geodesics coming out of the horizon will have outward-directed coordinate acceleration $\mathrm{d}^2 r / \mathrm{d} t^2$, at least near the beginning of the trajectories.

I have made a sketch, with some hand-picked parameters, to describe how the coordinate velocity $\mathrm{d} r / \mathrm{d} t$ looks for these trajectories: the lightlike curve accelerates asymptotically to coordinate velocity 1, while the timelike curve accelerates only up to $r = 6GM$, where it starts decelerating and asymptotically approaches the Newtonian $v \propto \sqrt{1/r}$.

I'd advise against giving too much weight to this result, though: coordinate acceleration has little meaning, and even if these particles are "accelerating" in Schwarzschild coordinates they are still being redshifted throughout the trajectory.

The curves plotted here are $$ \frac{\mathrm{d}r}{\mathrm{d}t} = \frac{r - 2GM}{r} $$ for light and $$ \frac{\mathrm{d}r}{\mathrm{d}t} = \frac{r - 2GM}{r} \sqrt{\frac{2GM}{r}} $$ for matter. These should be, roughly, the coordinate velocities of radially outgoing trajectories in Schwarzschild --- I didn't re-derive them now, but one can do so straightforwardly by using Killing vectors together with imposing the square of the four-velocity of the particle to be constant, either 0 or -1, and getting a differential equation for $r(t)$.