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Imagine a very long continuous stick with one end being pushed into the event horizon, all outside observers must agree that this end of the ruler come to a complete halt while I'm still pushing at the opposite end? Question is do I feel any opposite force from the stick while I push the entire stick into this event horizon? The stick is nigh indestructible just for this thought experiment.


marked as duplicate by Qmechanic Apr 14 at 10:13

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    $\begingroup$ The stick is nigh indestructible… A concept of indestructible (which I presume, could be formalized as “rigid”) already fails within special relativity, see. e.g., Ehrenfest paradox. $\endgroup$ – A.V.S. Apr 14 at 6:46
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/158195/2451 and links therein. $\endgroup$ – Qmechanic Apr 14 at 10:13

$\def\rs{r_{\rm s}}$ First of all, far from pushing the stick you'd to pull it to avoid a strong gravitational force makes it fall at great acceleration.

As for your question, the answer is no. There is no appreciable effect when something crosses the horizon. The only thing, that you certainly know, is that light emitted can't escape.

It's useful to recall the meaning of metric and coordinates. It's true that at the horizon the coefficient of $dr^2$ goes to infinity, but that has only to do with the meaning of the radial coordinate. A stick of proper length $dl$ will have its ends separated in $r$ by $$dr = dl\,\sqrt{1 - {\rs \over r}} \tag1$$ which goes to zero when $r\to\rs$.

In GR coordinates are just labels - they are devoid of physical meaning except the one given by metric. So the only meaning of $r$ should be read from eq. (1).

BTW, whereas (1) shows that for assigned $dr$ the proper length goes to infinity, the same isn't true for a finite length. By integrating (1) you can verify that a stick extending from $r=\rs$ to $r=r_1$ has proper length $$l = \sqrt{r_1 (r_1 - \rs)} + \rs\,\sinh^{-1}\!\sqrt{{r_1 \over \rs} - 1}.$$

The case of time dilation is different. When you read $$d\tau = dt\,\sqrt{1 - {r_{\rm s} \over r}}$$ and interpret it as showing that $dt$ increases, for a fixed $d\tau$ as $r\to\rs$, you've not yet proved time dilation, as $t$ is just a coordinate with no immediate physical meaning.

Another physical step is needed: you place a measuring instrument far away ($r_{\rm r}\gg\rs$) so that for it $d\tau\approx dt$. If two light signals are emitted near horizon spaced by $d\tau_{\rm e}$ they arrive to observer with the same $dt$ spacing (this is because metric doesn't depend on $t$) so that $$d\tau_{\rm r} = {d\tau_{\rm e} \over \sqrt{1 - \rs / r_{\rm e}}}$$ and we see that $d\tau_{\rm r} \to \infty$ as $r_{\rm e}\to\rs$.

  • $\begingroup$ I don't need this insult, so I've deleted my answer and comments. Please never contact me again or I will report you to the moderators for a repeated violation of the Code of Conduct. If you post an answer, answer the question, but not your misguided "analysis" of someone else's "understanding". $\endgroup$ – safesphere Apr 15 at 16:27

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