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I have computed radial time-like geodesics for the Schwarzschild interior solution beyond the Buchdahl limit. While checking the result I noticed some strange behavior of the proper time. It seems to go backwards when the ratio $r_S/R$ exceeds $8/9$ (the Buchdahl limit). Here is my calculation:

For a static spherically symmetric spacetime described by the length element $$ ds^2=e^{2\nu}c^2dt^2-e^{2\lambda} dr^2-r^2 d\Omega^2 ,\tag{1}$$ the geodesics equation for radially ($d\Omega=0$) in-falling time-like geodesic is $${e}^{2\nu+2\lambda}~\frac{\dot{r}^2}{c^2}=k^2-{e}^{2\nu} \tag{2}.$$ In this equation, k is the constant of motion, $k={e}^{2\nu}\cdot\dot{t}$, and the dot is the derivative with respect to the proper time $\tau$. Using the dimensionless variable $\xi \equiv r/R$ and the parameter $\alpha\equiv r_{S}/R$, the Schwarzschild interior metric reads $$e^{-\lambda} =\sqrt{1-\alpha~\xi^2},~~~~~e^{\nu} = 3/2~\sqrt{1-\alpha}-1/2~\sqrt{1-\alpha~\xi^2}, \tag{3}$$ The geodesics equation ($2$) can be solved analytically by integration: $$\int_\xi^1 \frac{{\rm e}^{\nu+\lambda}~d\xi}{\sqrt{k^2-{\rm e}^{2\nu}}}=-\frac{1}{R}\int_{s}^{s_1} ds=\frac{c}{R}~(\tau-\tau_1).\tag{4}$$ If we set the integration constant $\tau_1$ to zero, the relation between proper time $\tau$ and dimensionless radius $\xi$ is given by $$\tau(\xi)=\frac{R}{c}\frac{2}{\sqrt{\alpha}}\frac{1}{9(1-\alpha)-(1-2k)^2}\bigg\{(3\sqrt{1-a}+1)\big[\rm{F}\big(\varphi(1)|m\big)-\rm{F}\big(\varphi(\xi)|m\big)\big]-2\big[\Pi\big(n,\varphi(1)|m\big)-\Pi\big(n,\varphi(\xi)|m\big)\big]\bigg\} \tag{5}$$ where $\rm{F(x|m)}$ and $\Pi(n;x|m)$ are elliptic integrals of the first and third kind, respectively, and the variable $\varphi$ and the constants $n$ and $m$ are defined as $$\varphi(\xi)\equiv\arcsin{\sqrt{\frac{(3\sqrt{1-\alpha}+1-2 k)\cdot(\sqrt{1-\alpha \xi^2}-1)}{(3\sqrt{1-\alpha}-1-2 k)\cdot(\sqrt{1-\alpha \xi^2}+1)}}},\tag{6}\\n\equiv\frac{3\sqrt{1-\alpha}-1-2 k}{3\sqrt{1-\alpha}+1-2 k},~~~m\equiv\frac{9(1-\alpha)-(1+2 k)^2}{9(1-\alpha)-(1-2 k)^2}.$$ It is the same result as in the paper A relativistic gravity train.

The graph below shows the relation between $r$ and $\tau$ in dimensionless units ($\tau R/c, r/R$) for three $\alpha$ values and $k=1$ (zero velocity at infinity).

enter image description here

We can see that the proper time to reach the center ($\xi=0$) from the boundary ($\xi=1$) is finite, as expected, but its behavior depends on $\alpha$: for $\alpha<8/9$ it goes forward, for $\alpha=8/9$ (Buchdahl bound) it stops, and for $\alpha>8/9$ it seems to go backward (dashed part of the line).

If proper time cannot go back, then one must conclude that the radial in-falling time-like geodesics in Schwarzschild interior spacetime end on the two-sphere with curvature radius defined by $e^{\nu}(\xi_0(\alpha))=0$, i.e. $r_0=R\sqrt{9-8/\alpha}$.

Would this conclusion be correct?

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    $\begingroup$ Have you checked scalar curvature invariants (e.g. Kretschmann scalar) for the “beyond Buchdahl limit” metric? If there is a singularity then here is your problem. $\endgroup$ Commented Apr 27 at 19:16
  • $\begingroup$ Not yet. I will do it. $\endgroup$ Commented Apr 28 at 7:13
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    $\begingroup$ Forget the Kretschmann scalar, the Ricci scalar itself is singular at $r_0$. $\endgroup$ Commented Apr 28 at 14:40
  • $\begingroup$ You are both rigth. The Kretschmann scalar K, as well as Ricci, are indeed divergent at the points defined by $e^{\nu}=0 $, my turning points, so there is a gravitational singularity there. What is interesting, however, is that this point moves to the surface of the star as $r_S/R$ goes to 1. It seems that the initial central singularity moves from center to the event horizon, something Leonard Sussking had postulated. $\endgroup$ Commented Apr 28 at 15:48
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    $\begingroup$ The interior solution is unphysical if the body's radius is below 9rs/8 since it assumes constant density and time independence, which do not fit together below this limit since that requires negative pressure, see the p in the Tᵤᵛ. $\endgroup$ Commented Apr 28 at 19:37

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The author's conclusion is not correct. Proper time does not stop at the event horizon. Something else happens. I hypothesize that when the radial component of the four velocity, $dr/ds$, becomes infinite, the infalling geodesic instantly switches to its antipodal counterpart, the outgoing geodesics, like a worldline of transcendent tachyon would do.

As the parameter $\alpha~(\equiv r_{S}/R)$ approaches $1$ (the Schwarzschild event horizon), the result is the same as that described by Gerard t’Hooft in his paper What Happens in a Black Hole When a Particle Meets Its Antipode: "(…) a curve that crosses the horizon from region I into region II does not enter the 'interior' of the black hole but instead goes directly to the points outside the antipode (...).", page $5$.

The picture below shows the instantaneous switch to antipodal points (vertical dotted lines) and geodesics that started at rest in infinity (solid lines). These geodesics result in a velocity equal to the $\sqrt{\alpha}$ upon crossing the star's surface ($r=R$). enter image description here

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    $\begingroup$ I have not studied your solution, but please see if my thoughts are relevant. During the collapse, the horizon forms at the center and expands to the Schwarzschild circumference. All matter always remains outside the horizon for several good reasons. At this point the original star matter just outside the horizon is in a free fall at near the speed of light, so it becomes practically a null dust (a set of relativistic bullets) whose stress-energy tensor is zero. The star becomes a vacuum solution with a thin layer of matter just outside the horizon and a zero stress-energy tensor. $\endgroup$ Commented Jun 26 at 22:02
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    $\begingroup$ The key point here is that the stress-energy tensor of the matter of the collapsed star vanishes as soon as the horizon reaches the Schwarzschild circumference. At this time the behavior of the proper time of a falling particle becomes Schwarzschild. Specifically, the particle reaches the horizon in a finite proper time in the infinite cosmological future. What happens “after” that is a matter of speculation. One option is that the proper time ends at the horizon. The other is that the inner singular spacetime exists “after” the cosmological eternity (?) and the particle hits the singularity. $\endgroup$ Commented Jun 26 at 22:12
  • $\begingroup$ The quantum approach suggests the former, essentially that the horizon is a special type of a physical singularity with nothing inside. Please let me know if any part is unclear. +1 $\endgroup$ Commented Jun 26 at 22:18
  • $\begingroup$ Please note that the $r_s/R$ ratio is not trivial, because the space is curved and the definition of what radius is can vary. The spacelike radius of the Schwarzschild horizon is always zero, so the value of $r_s/R$ depends on the definition of “radius”. Also, once the horizon forms at the center and starts “expanding”, its circumference grows, but its spacelike radius remains zero. This is one of the reasons the original star matter stays outside - the “inside” is not closer to the center (the radius is zero), the inside is the same place, just in the future, so there is nothing there ever. $\endgroup$ Commented Jun 27 at 0:53
  • $\begingroup$ @safesphere Your thoughts are relevant and very welcome. "The inside is the same place." This is exactly what the dotted lines on my diagram symbolize. If it takes no time to move from point A to point B, then they are the same point; there is no space between them. Topologically speaking, this means that the horizon behaves like a projective two-sphere. $\endgroup$ Commented Jun 27 at 11:09

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