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In the event horizon of the Schwarzschild-metric not only the time coordinate but also the radial space coordinate seems to change sign:

$$ds^2=\left(1-\frac{2M}{r}\right)\mathrm{d}t^2-\left(1-\frac{2M}{r}\right)^{-1}\mathrm{d}r^2-r^2\mathrm{d}\theta^2-r^2\sin^2\theta \, \mathrm{d}\phi^2$$

(in units wherein $c=G=1$). As we can see, for $0<r<2M$, $t$ be negative (just like the space coordinates) and $r$ will be positive (just like the time coordinate), although $r$ has a singular point at $r=2M$.

Does this sign change have some physical meaning?

Reacting comments

  1. Under "normal" circumstances (not stepping through the event horizon of a Schw BH), a such change like this don't happen, this is because I have the question, what about physical meaning could have a such change.
  2. As I understand the Kruskal-Szekeres metric, it has multiple sections. It effectively handles the problem, what happens if somebody steps through to event horizon. But stepping through the event horizon is impossible only for an external observer, the object falling in the Schw BH can watch its fall into the BH. What will it experience [as a theoretical possibility], as its radial spatial coordinate transforms to a temporal and vice versa?
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  • $\begingroup$ The singular point $r=2GM$ is only because we're using inappropriate coordinate systems. No curvature scalars are singular at $r=2GM$. See Kruskal coordinates for an elaboration on how to remove the singularity. $\endgroup$ – JamalS May 9 '14 at 9:17
  • $\begingroup$ There is no physical meaning. However, it is possible to interpret the coordinates as the union of two disjoint coordinate charts. So one can use them inside or outside, but not both, so it's not appropriate to use them in any situation that crosses the horizon. That $\partial r$ is timelike inside the horizon is just a matter of labels, not physics; one can easily rewrite $(1-2M/t)$ there, etc. $\endgroup$ – Stan Liou May 9 '14 at 9:51
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    $\begingroup$ The title is misleading. The signature doesn't change it is always, in your case, (+,-,-,-). $\endgroup$ – MBN May 9 '14 at 10:09
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    $\begingroup$ The change of sign of the components of the metric means that, in some sense, $t$ becomes a "spatial" coordinate and $r$ a "temporal" one: the "future" points towards decreasing $r$ instead of increasing $t$, you can see that looking at the light cones in Schwarzschild coordinates. See this figure: kierul.files.wordpress.com/2013/12/schwarzschilddiagram.jpg In addition, you cannot have a positive $ds^2$ without a non-null $dr^2$ due to the sign change, so inside a Schwarzschild black hole you have to move. $\endgroup$ – giordano May 9 '14 at 10:23
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    $\begingroup$ @giordano: that's just because the Kerr metric is non-diagonal. If you put the schwarzschild metric in Kerr-style coordinates, you'll see a similar effect. $\endgroup$ – Jerry Schirmer May 9 '14 at 13:50
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In Schwarzschild coordinates, the change of sign of the $g_{00}$ and $g_{11}$ components of the metric means that, in some sense, $t$ becomes a "spatial" coordinate and $r$ a "temporal" one: the "future" points towards decreasing $r$ instead of increasing $t$, you can see that looking at the light cones in Schwarzschild coordinates, see for example this figure

Schwarzschild light cones in Schwarzschild coordinates

Schwarzschild light cones in Schwarzschild coordinates (from MTW, page 848)

In addition, inside the surface $r = r_\text{S} = 2M$ you cannot have a positive $\mathrm{d}s^2$ without a non-null $\mathrm{d}r^2$ due to the sign change, so inside a Schwarzschild black hole you have to move. This, again, can be seen using the above light cones: the word line cannot keep constant $r$.

The fact that after having crossed the event horizon light cones point towards the $r = 0$ singularity is true also using other coordinates, such as Kruskal-Szekeres coordinates

Schwarzschild metric in Kruskal-Szekeres coordinates (see the full definition of the coordinates in the Wikipedia article): \begin{equation}\mathrm{d}s^{2} = \frac{4r_{\text{S}}^{3}}{r}\mathrm{e}^{-r/r_{\text{S}}}(\mathrm{d}v^{2} - \mathrm{d}u^{2}) - r^{2}\mathrm{d}\theta^{2} - r^{2}\sin^{2}\theta\mathrm{d}\phi^{2}\end{equation}

Schwarzschild light cones in Kruskal-Szekeres coordinates

Schwarzschild light cones in Kruskal-Szekeres coordinates. The $r = 0$ region is the one with the inward toothed border (from MTW, page 848)

and Eddington-Finkelstein coordinates

Schwarzschild metric in Eddington-Finkelstein coordinates (see the full definition of the coordinates in the Wikipedia article): \begin{equation}\mathrm{d}s^{2} = \biggl(1 - \frac{r_{\text{S}}}{r}\biggr)\mathrm{d}\tilde{v}^{2} - 2\mathrm{d}\tilde{v}\mathrm{d}r - r^{2}\mathrm{d}\theta^{2} - r^{2}\sin^{2}\theta\mathrm{d}\phi^{2}\end{equation} Schwarzschild light cones in Eddington-Finkelstein coordinates

Schwarzschild light cones in Eddington-Finkelstein coordinates (from MTW, page 849)

The sign change has a physical meaning in Schwarzschild coordinates because Schwarzschild $t$ and $r$ coordinates bear physical meanings ($t$ is the far-away time, $r$ the reduced circumference), while I'm not aware of any simple physical meaning of the Kruskal-Szekeres $u$ and $v$ coordinates or of the Eddington-Finkelstein $\tilde{v}$ coordinate. Note that $u$, $v$ and $\tilde{v}$ coordinates mix the original Schwarzschild $t$ and $r$ coordinates. Depending on the coordinates used, there isn't always a sign change in the metric components (in Kruskal-Szekeres coordinates there is no sign change at all), so don't take that change as a general rule.

The Kerr metric tensor, with Boyer-Lindquist coordinates (briefly described in this introduction to the Kerr spacetime by Matt Visser), reads

\begin{equation} g_{\mu\nu} = \begin{pmatrix} (\Delta - a^{2}\sin^{2}\theta)\Sigma^{-1} & 0 & 0 & a\Sigma^{-1}r_{\text{S}}r\sin^{2}\theta \\ 0 & -\Delta^{-1}\Sigma & 0 & 0 \\ 0 & 0 & -\Sigma & 0 \\ a\Sigma^{-1}r_{\text{S}}r\sin^{2}\theta & 0 & 0 & -\bigl((r^{2} + a^{2}) + a^{2}\Sigma^{-1}r_{\text{S}}r\sin^{2}\theta\bigr)\sin^{2}\theta \end{pmatrix} \end{equation} with \begin{align} \Delta &= r^{2} - r_{\text{S}}r + a^{2}, \\ \Sigma &= r^{2} + a^{2}\cos^{2}\theta. \end{align} The $g_{00}$ component changes its sign on the surfaces $$r_{\text{E}}^{\pm} = M \pm \sqrt{M^{2} - a^{2}\cos^{2}\theta}$$ Instead, $g_{11}$ changes its sign on the surfaces $$r_{\pm} = M \pm \sqrt{M^{2} - a^{2}}$$ which determine the outer (with $+$ sign) and the inner (with $-$ sign) event horizons. So they change their sign on two different surfaces. As Jerry Schirmer pointed out in a comment, this would occur also in the Schwarzschild geometry with Kerr-like non-diagonal coordinates (e.g., it occurs with Eddington-Finkelstein coordinates). This doesn't mean that the signature of the metric will change: there always will be a negative (positive) eigenvalue, and three other positive (negative) eigenvalues. In a non-diagonal metric tensor (like Schwarzschild one in Eddington-Finkelstein coordinates or the Kerr one in Boyer-Lindquist coordinates), the diagonal components of the metric tensor aren't necessarily the eigenvalues.

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