The time and radial coördinate change are exchanged with one another behind the black hole horizon. So even stating that it's happening behind the horizon is misleading. There is, as a figure of speech, an ocean of time after the horizon. Yesterday, ten years ago, hundred years ago, they're all in there. From the outside, all stuff seems to freeze on the horizon, except the initial stuff that's in the hole, although that is cut off by the horizon as there is no inside anymore, after the formation. Seen from far away, that is.

But now I fall in. Can I can meet a particle that fell in hundred years ago, according to someone very far away? And will I be radiated away shortly after in the form of Hawking radiation (or better, information about my particles)?


1 Answer 1


Please do NOT confuse what is happening with coordinates in General Relativity, with what anyone measures.

When we refer to coordinates as "timelike" or "spacelike", this refers to whether an increment in those coordinates leads to an increase in the proper time (the time measured on a watch carried by an observer) or a decrease.

For example, in the Schwarzschild metric, expressed using "Schwarzschild" (or "Droste") coordinates, the interval of proper time can be written (in units where $c=1$) as $$ d\tau^2 = \left( 1 - \frac{r_s}{r}\right) dt^2 - \left(1 - \frac{r_s}{r}\right)^{-1} dr^2 - r^2 \sin^2\theta\ d\theta^2 - r^2 d\phi^2\ ,$$ where $r_s$ is the Schwarzschild radius, $d\tau$ is an increment of proper time, whilst $dr, d\theta, d\phi$ are increements in the spatial coordinates (loosely modelled on spherical polar coordinates, but note that $dr$ cannot be used as a physical radial separation and the convention $d\tau^2 = (d\tau)^2$ etc. is used.

Now when $r>r_s$, the first term on the right hand side, that multiplies $dt^2$ is positive. That means an increment in $t$ will lead to increasing $d\tau$ and this is a timelike change. On the other hand, the second term will be negative and an increment in $r$ would decrease $d\tau$ (note that $d\tau$ is always $\geq 0$) and would be a spacelike change.

When $r<r_s$ the behaviour of these two terms swaps over because the terms in brackets become negative when $r<r_s$ and thus the timelike and spacelike nature of the coordinates swaps over. The thing is, nobody measures this happening because Schwarzschild coordinates are only an appropriate measurement system for an observer far away from the black hole that never sees anything at $r<r_s$.

But now the meat of your question in your second paragraph. The answer is no (or at least no, for any sensible sized black holes in our universe). Something that fell into a black hole before you started falling will always be ahead of you and you will nver catch it up. Your story of the fall is quite different to that of an outside observer. Your own clock continues to tick forward, you cross the event horizon (providing the black hole is massive enough that tidal forces at the horizon are small) and then you continue to smaller and smaller $r$ as your $\tau$ increases until you are inevitably ripped apart by tidal forces as you approach the singularity. Any body that fell in before you has already met that fate.

Finally, it is meaningless to describe something happening at $r<r_s$ as far as a distant observer is concerned, where events are measured in a coordinate system that approximates to Schwarzschild coordinates. However, a falling observer can see things quite differently. In particular, they can observe things happening at $r<r_s$; but only after they too have passed below the event horizon. So in that regard it is not meaningless (in general) to refer to events happening at $r<r_s$.

Schwarzschild coordinates are a poor way of describing falling observers and events that occur at $r<r_s$ and lead to these confusions about time running backwards etc. Much better to use a coordinate system expressly designed to be continuous across the event horizon such as Gullstrand-Painleve coordinates.

  • $\begingroup$ Great answer, Professor! Thanx! $\endgroup$
    – Felicia
    Jun 13, 2022 at 19:24
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    $\begingroup$ @Felicia If this answer helped you, please click the checkmark to mark this post as complete. $\endgroup$ Jun 14, 2022 at 3:12
  • $\begingroup$ @ProfRob, if I understand it properly, the Schwarzschild metric, as written above, describes static spherically symmetric spacetime but only above event horizon ($r_{S}\le r \le \infty$). The spacetime behind the event horizon is no more static. I would expect there other metric. Do I mess something up? $\endgroup$
    – JanG
    Jul 17, 2022 at 14:53
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    $\begingroup$ @JanGogolin the metric has no time dependence either above or below the horizon. The metric as written in Schwarzschild coordinates works everywhere except at $r=r_s$ (and $r=0$). The metric works everywhere except $r=0$ in other coordinate systems - for example, Gullstrand-Painleve coordinates, as I said in my answer. $\endgroup$
    – ProfRob
    Jul 17, 2022 at 15:13
  • $\begingroup$ @ProfRob thanks for clarifying. $\endgroup$
    – JanG
    Jul 17, 2022 at 16:28

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