Is Schwarzschild black hole solution geodesically complete?

Question

I try to answer the question whether spacetime of Schwarzschild vacuum solution is geodesically complete by analyzing its null geodesics. The infinitesimal length element is $$ds^2=(1-\frac{r_s}{r})~c^2 dt^2-(1-\frac{r_s}{r})^{-1} dr^2-r^2 d\Omega^{2}. \tag{1}$$ The solving of corresponding differential equation for $ds=0$ results in $$\pm~ c t=r+r_{s} \log{(r-r_s)}. \tag{2}$$

The "coordinate time" $t$ is coevally affine parameter which value for infalling ($+$) geodesic ranges from $-\infty$ to $+\infty$ for $r\in (+\infty,~ r_{s})$.

As the answers below have proved the statement in bold letters is not true! Coordinate time is a non-affine parameter. However, one can always re-parametrize $t$ to affine parameter $\tau(t)$ (see Appendix in Quasi-geodesics in relativistic gravity) which function is $$\pm~ c \tau=r+2 r_{s} \log{(r-r_s)}-\frac{r_{s}^2}{r-r_s}. \tag{3}$$

For ingoing geodesic the affine parameter $\tau \in (-\infty$,$+\infty$) generates geodesic $r(\tau)\in (+\infty,~ r_{s})$.

Thus, Schwarzschild vacuum solution represents a geodesically complete manifold with null geodesics that never cross the event horizon ($r=r_{s}$) which is not part of that manifold. Do I interpret it correctly (this time)?

Remark

The derived equation (3) is not correct as well because the used equation A.7 from the mentioned reference has a wrong sign (it should be plus instead of minus there). The correct formula for affine parameter is $$\tau\equiv A\int{e^{\nu+\lambda}} dr+B= A r +B \tag{4}.$$

My interpretation based on the at first derived equation (3) was therefore not correct.

Answer 1

The definition of geodesic completeness requires the $\textbf{affine parameter}$ which paramerizes geodesics to take values between $-\infty $ to $\infty$. Note also that a spacetime can also be geodesically incomplete if it is extendible. This is the case with Schwarschild spacetime. It is only valid for $r >r_s$. Therefore, it is trivially incomplete.

We, therefore, consider the Kruskal Spacetime which is the maximally extended version of Schwarschild spacetime and is inextendible. To see why it is inextendible, it suffices to look at geodesics inside the event horizon. A simple calculation shows that any $\textbf{causal curve}$ will eventually reach the black hole singularity in less than $\pi M$ affine parameter time. Therefore, the Kruskal Spacetime is not geodesically complete.

$\textbf{Edit}:$ Furthermore, the range for $t$ should be checked for one geodesic and not both of them at the same time. In you case, you should check that both geodesics (ingoing or outgoing) have t ranging from $-\infty $ to $\infty$. Now you may argue that light coming from past infinity and suddenly changing direction at $r = r_s$ and going outward will have t ranging from $-\infty $ to $\infty$ but this is not a geodesic.

Answer 2

No, to examine geodesic completeness you need to

1. Consider the geodesic equation (what you are solving is not the geodesic equation).
2. Consider all geodesics not just radial null rays.

Since you do neither of these things there is nothing you can say about the geodesic completeness of Schwarzschild spacetime based on your calculation.

(In particular, note that neither t nor r provide an affine parametrization of the curve defined by your equation (2).)

EDIT: The geodesic equations for a radial null geodesic in Schwarzschild spacetime are

\begin{align} \frac{d^2 t}{ds^2} &= -\frac{r_s}{r(r-r_s)}\frac{d t}{ds}\\ \frac{d^2 r}{ds^2} &= 0. \end{align}

The second equation implies that $r$ is in fact an affine parameter along the geodesic (so I take that back $r$ is an affine parameterization). Setting $s=r$ and solving the first equation gives your solution (2) (up to a typo).

This immediately shows that Schwarzschild exterior patch is geodesically incomplete, since we reach the $t=\infty$ edge of the patch at a finite value of the affine parameter $r=r_s$.