Suppose we are inside the Schwarzschild radius of a black hole and throw a ball radially outward. It is said that the ball has no possibility to increase its radial coordinate. It must continuously decrease its radial coordinate and reach finally to the center.

I don't understand why this is so. The metric inside the Schwarzschild radius for radial motion is:


I do understand that the ball must follow timelike world-line, i.e., we must have $ds^2>0$, whatever way the ball moves. And for this to happen it is necessary that $dr\ne0$. But the metric does not appear to put any restriction on whether $dr$ shall be positive or negative, because $dr$ appears there as a squared term.

So why the ball can't move with positive $dr$, i.e, radially outward?

  • $\begingroup$ Can you elaborate a bit why increasing $r$ is going back in time. I have seen in null congruence diagrams that even decreasing ‘r’ can be going back in time. $\endgroup$ – Hrishi Jul 29 '19 at 14:09
  • $\begingroup$ There are really two conceptual issues here. The first one is in the first sentence, the second in the remainder of the question. The answers haven't really addressed the first issue: Suppose we are inside the Schwarzschild radius of a black hole and throw a ball radially outward. If you're inside the horizon, you can throw in the $+t$ or $-t$ direction, because $t$ is spacelike. You can't throw in the $\pm r$ directions, because $r$ is now the timelike coordinate. $\endgroup$ – Ben Crowell Jul 29 '19 at 14:34
  • $\begingroup$ The very recent paper "Inside astronomically realistic black holes", at arxiv.org/pdf/1907.05292.pdf, may be of interest to you, and includes some video links. $\endgroup$ – Edouard Jul 29 '19 at 17:42

You are quite correct that the metric alone cannot tell you which way the ball is moving. That's because the metric describes both the black hole and the white hole. But what we can work out from the metric is that once inside the event horizon the velocity $dr/dt$ can never change sign. That means if you cross the horizon headed inwards into a black hole, i.e. with a negative value of $dr/dt$, the radial velocity can never increase to zero and become positive so you can never head outwards again. The same argument applies to white holes. In this case a ball inside the horizon and headed outwards can never come to a halt and fall inwards again.

To show this rigorously is involved, but if you'll forgive a bit of arm waving we can make a reasonable non-rigorous argument. The metric for a radial trajectory $(d\theta = d\phi = 0)$ is:

$$ c^2d\tau^2 = c^2\left(1 - \frac{r_s}{r}\right)dt^2 - \frac{dr^2}{1 - r_s/r} $$

For the direction of the velocity to change we require that the object be momentarily stationary, i.e. $dr=0$, but this gives us:

$$ c^2d\tau^2 = c^2\left(1 - \frac{r_s}{r}\right)dt^2 $$

The problem is that once inside the horizon $1 - r_s/r$ is negative and this will give us a negative value for $d\tau^2$. Since this is impossible the conclusion is that $dr$ can never be zero i.e. the ball can never change the direction of the radial velocity.

  • $\begingroup$ Thanks for the nice explanation. True that particles entering the horizon from outside can’t go back as $dr$ is not allowed to become stationary. But what about particles already there inside it? If one of them has positive velocity will it come out? In this case ‘dr’ will not require to change sign as ‘r’ is continually increasing. $\endgroup$ – Hrishi Jul 29 '19 at 14:02
  • $\begingroup$ To show this rigorously requires requires the Hawking Penrose singularity theorems, This doesn't have anything to do with the Penrose singularity theorem (or Hawking, which is about big bang singularities). $\endgroup$ – Ben Crowell Jul 29 '19 at 14:29
  • $\begingroup$ @John Rennie Is reversal of motion (from inward to outward) within a BH not possible in Einstein-Cartan theory (which allows spatial extent for individual fermions)? Poplawski (who uses that theory, which he describes as a "version" of GR) implies, in several papers written between 2009 and 2019, that it is. My familiarity with physics notation (which seems to vary a lot by context) leaves his reasoning imperfectly clear to me, but the basic idea seems to be that the trajectory of fermions materialized by the BH's gravity is reversed upon contact with the spin of the larger stellar ones. $\endgroup$ – Edouard Jul 29 '19 at 15:46
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    $\begingroup$ I guess you you could ask about particles trapped inside the horizon as the black hole forms, but this turns out to be surprisingly complicated because models of black hole formation are complicated. In the simplest model, the Oppenheimer-Snyder metric, the horizon starts at the centre of the collapsing object and grows outwards. That means everything inside the horizon must have a negative radial velocity and therefore can only fall inwards. $\endgroup$ – John Rennie Jul 29 '19 at 16:14
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    $\begingroup$ @Edouard I'm not familiar with Einstein Cartan theory so I can't comment. You could try asking a question on this subject. $\endgroup$ – John Rennie Jul 29 '19 at 16:16

You are correct that although the metric can tell you whether or not a given line is timelike, it cannot on its own determine which is the future-pointing direction and which the past-pointing direction. This is determined by other considerations, such as the initial conditions. We can tell which is the future-pointing direction within the horizon by continuity from outside the horizon. Once it is determined for one worldline, then it is also determined for adjacent worldlines and hence for the whole spacetime, under ordinary conditions. However I don't know whether, from a mathematical point of view, there might be exceptions to this in special cases, such as spacetime configurations with closed timelike lines or naked singularities, but there is good reason to believe those don't occur in the physical world.

  • $\begingroup$ Regarding initial conditions, an infalling body will have $dr/d\tau < 0$ as it crosses the event horizon. It can be shown that inside the event horizon $|dr/d\tau| \ge \sqrt{\frac{2GM}{r} - 1}$, so $dr/d\tau$ remains negative. $\endgroup$ – Puk Jul 29 '19 at 9:14
  • $\begingroup$ Thanks. That’s true for particles coming from outside the horizon. But what about those already inside it. Can they have positive $dr/d\tau$ and emerge out of the horizon? $\endgroup$ – Hrishi Jul 29 '19 at 14:29
  • $\begingroup$ On the direction of time, I think of it as simply chosen or provided when one provides a spacetime. To affirm the continuity argument above, I would draw a Penrose diagram, and draw upward pointing arrows (a timelike vector field) consistently. $\endgroup$ – Colin MacLaurin Jul 31 '19 at 9:53
  • $\begingroup$ Hrishi, yes, if you consider the maximally extended spacetime. Then particles inside the white hole must have $dr/d\tau > 0$. They will emerge either into "our universe", the parallel exterior region, or in a very special case pass straight into the black hole interior. $\endgroup$ – Colin MacLaurin Jul 31 '19 at 9:56

The metric is in fact $$ds^2=-c^2(r_s/r-1)dt^2+(r_s/r-1)^{-1}dr^2$$ Or writing in terms of proper time $$d \tau^2 = -(r_s/r-1)dt^2+(r_s/r-1)^{-1}dr^2$$

When $r<r_s$, then since $d\tau > 0$ for an object with mass and $dt^2 \geq 0$ then $dr$ cannot equal zero.

This means that $dr/dt$ cannot change sign once $r<r_s$. Thus once moving into a black hole ($dr <0$), then the radial coordinate cannot increase.

  • $\begingroup$ Your answer seems to be relying on the idea that $dr$ is continuous, and so by the IVT, if it is negative for some value of $\tau$ and positive for another value, then there is some value of $\tau$ for which $dr$ is zero. I think the answer would be improved if this argumentation were explicit. $\endgroup$ – Acccumulation Jul 29 '19 at 17:12
  • $\begingroup$ I don't follow this argument. When you write "$dt^2\ge 0$", do you mean $(dt/d\tau')^2\ge 0$, where $\tau'$ is the proper time of the ball? Also incidentally, it is possible to have worldlines with $dt/d\tau' < 0$ inside the horizon $\endgroup$ – Colin MacLaurin Jul 31 '19 at 9:49
  • $\begingroup$ @ColinMacLaurin how can $dt^2$ be <0 ? $\endgroup$ – Rob Jeffries Jul 31 '19 at 12:57
  • $\begingroup$ @RobJeffries there is no squared bit in my expression, just $dt/d\tau'$. That line was incidental, and may not relate to your comment $\endgroup$ – Colin MacLaurin Aug 1 '19 at 21:42
  • $\begingroup$ @colinmaclaurin You said you didn't follow the argument that $dt^2 >0$. I don't follow yours. My answer is a simply a more succinct version of John Rennie's subsequent answer. $\endgroup$ – Rob Jeffries Aug 2 '19 at 7:25

To illustrate the points of the previous answers, you can think about flat minkowskian metric and just equivalently ask why an object can't move backwards in time, because the line element $ds^2$ only depends on $dt$ squared! the answer is that the equations of motion themselves are symmetric in time, and the direction we call 'future' or 'past' in time depends on other considerations, such as the direction in which entropy increases.

so similarly, you can draw a timelike curve with one end inside the event horizon and the other end outside it. But in order for it to be consistent with our usual definition of time direction, you will have to conclude that it describes an object falling into the black hole, because we can only cross the event horizon going inwards and not vice versa.

(what you can't do is draw a timelike curve in which $r$ both increases and decreases inside the event horizon. that would not be timelike)


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