Singularities and Horizons

Question

Well, the Morris and Thorne article on wormholes states that the function $\Phi(r)$ of metric $$ ds^2 = -e^{2\Phi(r)}c^2dt^2 + dr^2/(1 - b(r)/r)+r^2(d\theta^2 + sin^2\theta\hspace{0.1cm} d\phi^2) $$ must be finite for all values of $r \geq r_0$ to avoid singularities, but I did not understand why this avoids singularities ($g_{tt}$ being finite), while $b(r_0) = r_0$, which implies that $g_{rr}$ is not finite at this point, does not produce a singularity. What is the difference?

Answer 1

$b(r_{0}) = r_{0}$ is a coordinate singularity rather than a true physical one, in the sense that you could define a new set of coordinates such that the patch of physical spacetime you cover is different, and that singularity is avoided. A common simpler example of that is de Sitter space. Consider the global coordinates with closed slicing which cover the entirety of the manifold by embedding it in $\mathbb{R}^{1,4}$:

\begin{equation} \begin{split} x^{0} &= \ell \, \sinh{t} \\ x^{1} &= \ell \, \cosh{t} \cos{\chi} \\ x^{2} &= \ell \, \cosh{t} \sin{\chi} \cos{\theta} \\ x^{3} &= \ell \, \cosh{t} \sin{\chi} \sin{\theta} \cos{\varphi} \\ x^{4} &= \ell \, \cosh{t} \sin{\chi} \sin{\theta} \sin{\varphi} \\ \end{split} \end{equation}

The metric is then given as:

\begin{equation} ds^{2} = -\ell ^{2} \left ( -dt^{2} + \cosh ^{2} {t} d\Omega _{3} ^{2} \right ) \end{equation}

where $d\Omega _{3} ^{2}$ is the metric of a unit 3-sphere. As you can see, no singularity there. By contrast, if you choose the static coordinates:

\begin{equation} \begin{split} x^{0} &= \sqrt{ R^{2} - r^{2}} \, \sinh{\left (\frac{t}{R} \right )} \\ x^{1} &= \sqrt{ R^{2} - r^{2}} \, \cosh{\left (\frac{t}{R} \right )} \\ x^{2} &= r\sin{\theta} \cos{\varphi} \\\ x^{3} &= r \sin{\theta} \cos{\varphi} \\ x^{4} &= r \cos{\theta} \end{split} \end{equation}

then your metric acquires a coordinate singularity at $r=\pm R$:

\begin{equation} ds^{2} = -\left ( 1- \frac{r^{2}}{R^{2}} \right ) dt^{2} + \left ( 1- \frac{r^{2}}{R^{2}} \right )^{-1} dr^{2} + r^{2} d\Omega _{2} ^{2} \end{equation}

These coordinates cover only the region of dS space which is in causal contact with a static observer. The coordinate singularity in turn describes the two event horizons that appear (future and past) which is a feature of dS space. You can readily see this in the relevant Penrose diagram:

with the future and past horizons being $H^{+}$ and $H^{-}$ respectively, and the region covered by static coordinates is region I.

For your case, it might take some work to figure out exactly an appropriate coordinate system where the singularity is lifted, but from its form it seems that it is doable. On the other hand, $\Phi(r)$ can only be removed via a conformal/Weyl transformation, hence the stipulation that it is finite is necessary to keep the metric non-singular.

Answer 2

I am afraid I cannot explain it better as Francisco S. N. Lobo in his paper "Exotic solutions in General Relativity: Traversable wormholes and 'warp drive' spacetimes" therefore I quote him here verbatim:

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Consider the following spherically symmetric and static wormhole solution \begin{equation} ds^2=-e ^{2\Phi(r)} \,dt^2+\frac{dr^2}{1- b(r)/r}+r^2 \,(d\theta ^2+\sin ^2{\theta} \, d\phi ^2) \,, \label{metricwormhole} \tag{1} \end{equation} where $\Phi(r)$ and $b(r)$ are arbitrary functions of the radial coordinate $r$. $\Phi(r)$ is denoted the redshift function, for it is related to the gravitational redshift, and $b(r)$ is denoted the shape function, because as can be shown by embedding diagrams, it determines the shape of the wormhole (Morris and Thorne). The coordinate $r$ is non-monotonic in that it decreases from $+\infty$ to a minimum value $r_0$, representing the location of the throat of the wormhole, where $b(r_0)=r_0$, and then it increases from $r_0$ to $+\infty$. The proper circumference of a circle of fixed $r$ is given by $2\pi r$. Although the metric coefficient $g_{rr}$ becomes divergent at the throat, which is signalled by the coordinate singularity, the proper radial distance \begin{equation} l(r)=\pm\,\int_{r_0}^r{{dr}\over{(1-b(r)/r)^{1/2}}} \,, \label{eq:PD} \tag{2} \end{equation} is required to be finite everywhere. Note that as $0 \leq 1 -b(r)/r \leq 1$, the proper distance is greater than or equal to the coordinate distance, i.e., $|l(r)| \geq r - r_0$. The metric (1) may be written in terms of the proper radial distance as \begin{equation} ds^2=-e^{2\Phi(l)}dt^2 + dl^2 + {r^2}(l)({d\theta}^2+ {\rm sin}^2\theta\,{d\phi}^2)\,. \tag{3} \end{equation} The proper distance decreases from $l=+\infty$, in the upper universe, to $l=0$ at the throat, and then from zero to $-\infty$ in the lower universe. For the wormhole to be traversable it must have no horizons, which implies that $g_{tt}=-e^{2\Phi(r)}\neq 0$, so that $\Phi(r)$ must be finite everywhere.

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