<p>Please do NOT confuse what is happening with <em>coordinates</em> in General Relativity, with what anyone measures.</p>
<p>When we refer to coordinates as &quot;timelike&quot; or &quot;spacelike&quot;, 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.</p>
<p>For example, in the Schwarzschild metric, expressed using &quot;Schwarzschild&quot; (or &quot;Droste&quot;) coordinates, the interval of proper time can be written (in units where <span class="math-container">$c=1$</span>) as
<span class="math-container">$$ d\tau^2 = \left( 1 - \frac{r_s}{r}\right) dt^2 - \left(1 - \frac{r_s}{r}\right)^{-1} dr^2 - r^2 \sin^2\theta\ d\theta^2 - r^2 d\phi^2\ ,$$</span>
where <span class="math-container">$r_s$</span> is the Schwarzschild radius, <span class="math-container">$d\tau$</span> is an increment of proper time, whilst <span class="math-container">$dr, d\theta, d\phi$</span> are increements in the spatial coordinates (loosely modelled on spherical polar coordinates, but note that <span class="math-container">$dr$</span> cannot be used as a physical radial separation and the convention <span class="math-container">$d\tau^2 = (d\tau)^2$</span> etc. is used.</p>
<p>Now when <span class="math-container">$r&gt;r_s$</span>, the first term on the right hand side, that multiplies <span class="math-container">$dt^2$</span> is positive. That means an increment in <span class="math-container">$t$</span> will lead to increasing <span class="math-container">$d\tau$</span> and this is a timelike change. On the other hand, the second term will be negative and an increment in <span class="math-container">$r$</span> would decrease <span class="math-container">$d\tau$</span> (note that <span class="math-container">$d\tau$</span> is always <span class="math-container">$\geq 0$</span>) and would be a spacelike change.</p>
<p>When <span class="math-container">$r&lt;r_s$</span> the behaviour of these two terms swaps over because the terms in brackets become negative when <span class="math-container">$r&lt;r_s$</span> 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 <em>never</em> sees anything at <span class="math-container">$r&lt;r_s$</span>.</p>
<p>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 <span class="math-container">$r$</span> as your <span class="math-container">$\tau$</span> 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.</p>
<p>Finally, it is meaningless to describe something happening at <span class="math-container">$r&lt;r_s$</span> as far as a <em>distant</em> 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 <em>can</em> observe things happening at <span class="math-container">$r&lt;r_s$</span>; but only after they too have passed below the event horizon. So in that regard it is <em>not</em> meaningless (in general) to refer to events happening at <span class="math-container">$r&lt;r_s$</span>.</p>
<p>Schwarzschild coordinates are a poor way of describing falling observers and events that occur at <span class="math-container">$r&lt;r_s$</span> 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 <a href="https://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates" rel="noreferrer">Gullstrand-Painleve coordinates</a>.</p>

CC BY-SA 4.0
2022-06-13T19:24:48.083Z
Great answer, Professor! Thanx!
Felicia
CC BY-SA 4.0
2022-06-14T03:12:52.377Z
@Felicia  If this answer helped you, please click the checkmark to mark this post as complete.
Nilay Ghosh
CC BY-SA 4.0
2022-07-17T14:53:40.537Z
@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?
JanG
CC BY-SA 4.0
2022-07-17T15:13:42.487Z
@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.
ProfRob
CC BY-SA 4.0
2022-07-17T16:28:14.163Z
@ProfRob thanks for clarifying.
JanG
CC BY-SA 4.0
2022-06-13T18:05:32.647Z
2022-06-13T18:05:32.647Z
ProfRob
<p>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.</p>
<p>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)?</p>

<black-hole><hawking-radiation>
What does it mean that the radial coordinate becomes the time coordinate behind the horizon?
Felicia
# What does it mean that the radial coordinate becomes the time coordinate behind the horizon?

(Asked by: username_0 on 2022-06-13)

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)?

**Tags:** black-hole, hawking-radiation

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## 1 answers

### Answer by username_1 on 2022-06-13. Score: 5

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 the behaviour of these two terms swaps over because the terms in brackets become negative when $r 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.

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 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; 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.

Schwarzschild coordinates are a poor way of describing falling observers and events that occur at $r 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](https://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates).

> **Comments:**
> **username_0**: Great answer, Professor! Thanx!  
> **username_2**: @Felicia  If this answer helped you, please click the checkmark to mark this post as complete.  
> **username_3**: @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?  
> **username_1**: @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.  
> **username_3**: @ProfRob thanks for clarifying.  

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