One meter rod at long distance is thrown to the Schwarzschild black hole. How its length near the black hole seems to distant observer?
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$\begingroup$ The relationship between the coordinate interval $dr$ and the proper distance $ds$ is $$ds=\frac{dr}{\sqrt{1-r/r_g}}$$. $ds$ is bigger than $dr$. Does not it mean radial proper distance seems bigger to the distant observer? $\endgroup$– ConstantinCommented Apr 23 at 16:39
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1$\begingroup$ A rod of any length (even light years long in the radial direction) shrinks to a zero length at the horizon in any remote coordinates. Both ends of the rod cross the horizon at the same local time. Simon’s answer below is correct. $\endgroup$– safesphereCommented Apr 24 at 3:56
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$\begingroup$ $ds=\frac{dr}{\sqrt{1-r/r_g}} $ what is physical meaning of this? As I understood ds is bigger than dr. ds proper distance dr coordinate difference. How the length is contracted if ds is than dr? $\endgroup$– ConstantinCommented Apr 24 at 8:02
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1$\begingroup$ The correct formula is $ds=\frac{dr}{\sqrt{1-r_s/r}}$ where $r_s$ is the Schwarzschild radius (the reduced circumference of the horizon). Consider $r$ is about $1\text{%}$ larger than $r_s$ or $r=100/99 r_s$, then $dr=0.1 ds$. Consider $ds=1$ meter where the rod is locally. Then remotely we see this distance occupied by the rod (the length of the rod) as $dr=0.1 ds$ or $0.1$ meter. So the rod appears $10$ times shorter in the remote coordinates. $\endgroup$– safesphereCommented Apr 24 at 17:13
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1 Answer
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If your question is about length contraction, that depends on the local velocity $\rm v$ since the observed length $\rm \Delta r=r_2-r_1$ to proper length $\rm \Delta R$ relation is
$$\rm \Delta R=\frac{\Delta r}{\int_{r_1}^{r_2}\sqrt{1-r_s/r} \ dr \cdot\sqrt{1-v^2/c^2}}$$
if you define the observed length to be the rod's radial coordinate length from $\rm r_1$ to $\rm r_2$ and if we assume purely radial motion and alignment of the rod.
If the $\rm \Delta r$ is neglible compared to $\rm r_s$, meaning the rod is much smaller than the black hole, you can drop the integral and simply multiply the gravitational and the kinetic components.
If your question is about gravitational lensing, the answer not only depends on the position $\rm r$ and velocity $\rm v$ , but also on the angle $\theta$ from which you observe. In that case there is no simple equation and you'll have to raytrace the scenario.
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$\begingroup$ I am analysing the video (youtu.be/q6RufF4a6LM?si=JDkS6jSPvkNOyNNB). At 9:34 he says "from the perspective of distant observer ruler closer to central mass will appear contracted" $\endgroup$ Commented Apr 23 at 16:09
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$\begingroup$ The relationship between the coordinate interval $dr$ and the proper distance $ds$ is $$ds=\frac{dr}{\sqrt{1-r/r_g}}$$ $\endgroup$ Commented Apr 23 at 16:25
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1$\begingroup$ @Constantin - the equation from your video goes d'accord with mine if you set v=0, the video shows the differential form while my answer shows the integral from. $\endgroup$– YukterezCommented Apr 23 at 17:34
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1$\begingroup$ @Constantin - the proper distance ds or dR is larger than the observed coordinate distance dr in your own comment above and also in your first comment to your question, since you divide the dr by a term smaller than 1 the ds or dR must be larger than 1, see here. You seem to confuse the observed length with the proper length though. The observed length in that terms is represented by the radial grid if you view the paraboloid from the top. $\endgroup$– YukterezCommented Apr 24 at 14:39
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1$\begingroup$ @Constantin “why the length contracted near the black hole?” - Just in case you are asking about the fundamental reason, it is because the vacuum spacetime is an incompressible fluid. Its volume does not change no matter how strong the gravity is. If $dx$ and $dy$ are the tangential coordinates, then the infinitesimal spacetime volume $dt\cdot dr\cdot dx\cdot dy$ is conserved. Its change is defined by the Ricci tensor which is always zero in vacuum. If time is dilated (stretched) 10 times, then length must be contracted also 10 times for this volume to remain unchanged. $\endgroup$ Commented Apr 25 at 3:18
