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According to the current interpretation of the Schwarzschild vacuum solution inside the event horizon, the Schwarzschild coordinate $r$, while becoming "the time coordinate (...) also retains its geometrical significance" (Rindler, "Relativity: special, general and cosmological", p.261, Rindler's emphasis).

Now, with $r$ in this double role, what is $t$ supposed to represent? It cannot represent time, because time is already represented by $r$; and it cannot represent any spatial degree of freedom, because these are exhausted by $r$, and by the two angular Schwarzschild coordinates.

See Box 7 in Chapter 7 of Taylor, Wheeler, Bertschinger, "Exploring Black Holes", 2nd edition, and don't forget to notice the Figure inside that Box. The authors really believe that inside the event horizon the r coordinate keeps measuring some sort of "distance to the origin".

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  • $\begingroup$ Can you write the metric in the coordinates you're talking about? $\endgroup$ Jun 3, 2021 at 17:02

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$r$ is just a time coordinate inside the event horizon. $t,θ,\phi$ are the spatial coordinates.

I assume what Rindler means by "retains its geometrical significance" is that the 3D surface at fixed $r$ is geometrically a cylinder of infinite length, whose cross sections are spheres of radius $r$, both inside and outside the horizon. But outside the horizon the metric signature on the cylinder is $-{+}+$ (it's the world-sheet of a sphere), while inside the horizon the signature is $+{+}+$ (it's a Euclidean cylinder, the cross-sectional radius of which decreases at later times (that is, smaller $r$), and goes to $0$ at the singularity).

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    $\begingroup$ @FernandoPassos Topology of space alone is not well defined, because what is space and what is time may be different in different coordinate systems. If you are in a moving train, are you really moving in space or only in time? It depends on whom you ask. The Schwarzschild solution is a solution of the equation $R=0$ with the Ricci tensor, not Einstein tensor. The Ricci tensor describes vacuum with no matter. There is no point particle in the Schwarzschild solution, only vacuum. The mass of a black hole is not the mass of some particle at the origin, but the energy of the gravitational field. $\endgroup$
    – safesphere
    Jun 11, 2021 at 4:50
  • $\begingroup$ When we derive the Schwarzschild soln, we also have the fact that $dr$ is the radial distance as measured by the far away observer. If $r$ is to become timelike suddenly at one point then how are you to reconcile the fact that $dr$ should measure a spacelike distance as measured by the far away observer. Alternatively if $r$ represents the reduced area coordinate ( such that area of 3 spheres is $4*\pi*r^2$), then inside the horizon the area of 3 spheres is given by $4\pi r^2$ but now $r$ is timelike then what does it say about the area of spheres. Does it mean that at constant $t$ (cont) $\endgroup$
    – Shashaank
    Jun 13, 2021 at 17:07
  • $\begingroup$ Which is spacelike now the area of each sphere decreases $\endgroup$
    – Shashaank
    Jun 13, 2021 at 17:09
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inside the event horizon, the Schwarzschild coordinate 𝑟, while becoming "the time coordinate (...) also retains its geometrical significance"

This is correct. But, …

the r coordinate keeps measuring some sort of "distance to the origin".

This is not.

The geometrical significance of the Schearzschild $r$ coordinate never was as “some sort of distance to the origin”. The geometrical significance is that for constant $t$ and $r$ the surface defined by $\theta$ and $\phi$ is a 2 sphere with area $4\pi r^2$. This geometrical significance is retained inside the horizon, and allows us to easily write the $\theta$ and $\phi$ terms of the metric, even inside the horizon.

what is 𝑡 supposed to represent? It cannot represent time, because time is already represented by 𝑟; and it cannot represent any spatial degree of freedom, because these are exhausted by 𝑟, and by the two angular Schwarzschild coordinates.

$t$ is indeed a spatial degree of freedom. Locally, inside the horizon you can form a tetrad with a clock and three orthogonal rulers. The tetrad can be aligned and scaled so that locally one of the rulers indicates the Schwarzschild $t$ and the clock represents the Schwarzschild $r$. Forces can be applied in the $t$ direction and in all local respects it is indeed a spatial direction.

Again, the geometrical significance that $r$ retains is not as a distance to the origin nor as a spatial degree of freedom. It is simply the relationship to the surface area of the $\theta$, $\phi$ spheres.

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  • $\begingroup$ When we derive the Schwarzschild soln, we also have the fact that $dr$ is the radial distance as measured by the far away observer. If $r$ is to become timelike suddenly at one point then how are you to reconcile the fact that $dr$ should measure a spacelike distance as measured by the far away observer. Alternatively if $r$ represents the reduced area coordinate ( such that area of 3 spheres is $4*\pi*r^2$), then inside the horizon the area of 3 spheres is given by $4\pi r^2$ but now $r$ is timelike then what does it say about the area of spheres. Does it mean that at constant $t$ (cont) $\endgroup$
    – Shashaank
    Jun 13, 2021 at 17:14
  • $\begingroup$ Which is spacelike now the area of each sphere decreases. That is you a sphere centred at some distance from the black hole inside the horizon. And as time progresses forward (which means as r goes to 0 and r is timelike) the area of each sphere decreases while $t$ which is spacelike determines the position of the center of the sphere. Thus the area of each sphere inside the horizon decreases with time. Does the $r$ coordinate means that inside the horizon. $\endgroup$
    – Shashaank
    Jun 13, 2021 at 17:17
  • $\begingroup$ @Shashaank said “When we derive the Schwarzschild soln, we also have the fact that 𝑑𝑟 is the radial distance as measured by the far away observer.” I have never seen any derivation that does that. I would be quite skeptical of any such derivation. For a good derivation see this arxiv.org/abs/gr-qc/9712019 $\endgroup$
    – Dale
    Jun 13, 2021 at 17:20
  • $\begingroup$ Ok but what about the interpretation of $r$ as depicting the area of $2$ sphere inside the horizon as I have continued the comments ( after alternatively) $\endgroup$
    – Shashaank
    Jun 13, 2021 at 17:25
  • $\begingroup$ Does the radius of the 2 sphere decrease inside the horizon with time at all $\endgroup$
    – Shashaank
    Jun 13, 2021 at 19:56

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