Why do you fall towards the center of a black hole?

I took a course on general relativity this spring, and a question came up in passing during our Q&A sessions with the TA that he couldn't answer right there, and then no one thought to bring it up with the lecturer. I only remembered it recently, I have no idea how to solve it, and it irks me.

Say we have a Schwartzschild black hole, and a small, massive particle at some spacetime point inside the event horizon. We have done several calculaions on this scenario during class. For instance that a free fall maximizes the eigentime until the particle hits the singularity, so I'm at least somewhat familiar with it. However, those calculations involved calculating $\left(\frac{dr}{d\tau}\right)^2$ and taking the square root. At no point did we show that $\frac{dr}{d\tau}$ is actually negative, i.e. that the particle will experience falling towards the center of the black hole.

Question: How can one show that $\frac{dr}{d\tau} < 0$ for the time-like world line of a massive object inside the event horizon?

If we consider the Schwarzschild metric and only radial movement ($c=1$ units), then $$-ds^2 = d\tau^2 = (1 -r_s/r)\ dt^2 - (1 - r_s/r)^{-1}\ dr^2 = g_{00}\ dt^2 + g_{11}\ dr^2,$$ where $r_s$ is the Schwarzschild radius.

If $r>r_s$ then $g_{00}>0$ and $g_{11}<0$. For a particle with mass, we know that $d\tau^2>0$, which means that $g_{00}\ dt^2 > g_{11}\ dr^2$. This allows $dr=0$ (a stationary particle) outside the event horizon.

Now if we let $dr<0$ such that the particle moves inside the event horizon, where $g_{00}<0$ and $g_{11}>0$. Now in order to minimise $dr^2$, we could set $dt^2=0$, but clearly then $dr^2$ must be $>0$ in order to ensure that $d\tau^2 >0$. This means that the particle cannot be stationary inside the event horizon, but it also means that $dr/dt$ cannot change sign because it $dr$ can't pass through zero.

Thus as $dr$ was negative to get inside the event horizon, it must continue to be negative as $t$ increases. You specifically asked about proper time $\tau$ rather than $t$, but as $d\tau$ must also be positive, then $dr/d\tau$ must also continue to be negative.

EDIT: Eddington-Finkelstein coordinates.

As Arthur points out, a problem with the above argument is that it is not possible to cross from $r>r_s$ to $r<r_s$ because of the coordinate singularity in Schwarzschild coordinates. Moving instead to Eddington-Finkelstein coordinates, we have $$d\tau^2 = (1-r_s/r)\ dt'^2 -(2r_s/r)\ dt'\ dr - (1 +r_s/r)\ dr^2 = g_{00}\ dt'^2 + g_{01}\ dt'\ dr +g_{11}\ dr^2$$

If $r>r_s$ then $g_{00}>0$ so there is no problem in having $dr=0$, $dt'>0$ to give $d\tau^2>0$ and a stationary particle is possible.

Now, there is no problem at $r=r_s$; so long as $dt'>0$ and $dr<0$ and $dr/dt' >-1$ then $d\tau^2>0$. i.e. The particle can only go inwards across the event horizon.

If $r<r_s$ then $g_{00}<0$, $g_{01}<0$ and $g_{11}<0$. Now the only way that $d\tau^2>0$ for $dt'>0$ is if $dr<0$, so the particle cannot be stationary and must continue to move inwards.

• But if my understanding of the Schwarzschild metric is right, you cannot, within the standard spherical coordinates, go from outside the event horizon to inside the event horizon. You can move to Eddington-Finkelstein and invoke continuity of the choice of future light-cone at each point, but I don't think your approach works in spherical coordinates. Aug 4 '16 at 16:07
• @Arthur - how does that seem? Aug 4 '16 at 18:26
• You're assuming that the particle fell in from outside the black hole, which the OP did not specify. If the black hole is actually a white hole, which has the same metric, then particles can originate inside the white hole and then exit it. Jan 1 '19 at 14:57
• @tparker Perhaps, but then the question asks about Schwarzschild black holes...?? Jan 1 '19 at 15:21
• True - I just think it's worth emphasizing that the Schwarzchild metric does not necessarily describe a black hole, so you can't understand why objects fall inward just from the metric - you need additional information as well. Jan 1 '19 at 15:57

You cannot show that the particle is falling towards the black hole because the equations of motion are time reversible. That is, they work equally well for a particle falling towards the black hole and one speeding away from the black hole.

There are various ways of calculating the trajectory, and typically you end up calculating $\left(dr/d\tau\right)^2$ as indeed you can recall doing. Then the negative square root give the velocity of the particle heading towards the black hole and the positive root give the velocity of the particle heading away from the black hole.

• "any trajectories that move to increasing $r$ are spacelike" Isn't that like saying that any trajectory that moves towards decreasing $t$ is spacelike outside the event horizon? There is a future light cone and a past light cone, and staying within the past light cone is still timelike, isn't it? Aug 4 '16 at 16:09
• @Arthur: Yes, fair point, and the time reversed trajectory is motion out of a white hole, which is a physically permissible solution though as far as we know not realised in nature. So ignore the last paragraph. You cannot show $dr/dt$ is always negative. Aug 4 '16 at 16:12
• It's true that you can't show that $dr/d\tau$ is negative, but I think you missed an important point, which is that you can show that $dr/d\tau$ has the opposite sign than $dt/d\tau$ does for the part of the trajectory outside the hole. So if you've designated a "forward flow of time" direction in the ordinary region outside the whole, and the part of the trajectory that lies in the hole is at later $\tau$ than the part that lies outside, then that designation does determine that $dr/d\tau < 0$. Feb 27 '17 at 0:45