Dumb metric
I saw on Stack Exchange the roughly following problem : Given a static, spherically symmetric spacetime where all quantities of the stress energy tensor are (locally) bounded ($T^{\mu\nu}T_{\mu\nu}$, $\operatorname{Tr}(T)$, etc), can a naked singularity appear?
The extremal Reissner-Nordström metric has a stress-energy tensor diverging upon the singularity, and the Kerr metric isn't spherically symmetric. Minkowski space with the origin removed would qualify, but regular boundary points are usually not considered singularities proper. In $2+1$-dimensions, it is possible to have such a spacetime by considering Minkowski space with a conical singularity (so-called quasi-regular singularities), but it's not clear if this can be done for $3+1$-dimensions, as the equivalent procedure would break the spherical symmetry.
But this reminded me of another way to build a singularity, the idea from Ellis and Schmidt's own Singular space-times of an oscillating singularity, so I attempted to work out such a spacetime. To simplify things a fair bit, consider a static, spherically symmetric spacetime of the form
\begin{equation} ds^2 = -dt^2 + e^{2f(r)} dr^2 + r^2(d\theta^2 + \sin^2\theta d\phi^2) \end{equation}The various components are fairly easy to work out, and can be found in most books on the topic of spherically symmetric metrics. The inverse metric is quite simply
\begin{equation} g^{\mu\nu} = \operatorname{diag}(-1, e^{-2f}, r^{-2}, r^{-2} \sin^{-2} \theta) \end{equation}The Christoffel symbols are
\begin{eqnarray} {\Gamma^r}_{rr} &=& f'\\ {\Gamma^\theta}_{r\theta} &=& r^{-1}\\ {\Gamma^r}_{\theta\theta} &=& -r e^{-2f}\\ {\Gamma^\phi}_{r\phi} &=& r^{-1}\\ {\Gamma^r}_{\phi\phi} &=& -r e^{-2f} \sin^2 \theta\\ {\Gamma^\theta}_{\phi\phi} &=& -\sin \theta \cos \theta\\ {\Gamma^\phi}_{\theta\phi} &=& \frac{\cos\theta}{\sin\theta}\\ \end{eqnarray}The Riemann tensor is
\begin{eqnarray} {R^r}_{\theta r \theta} &=& r e^{-2f} f'\\ {R^r}_{\phi r \phi} &=& r e^{-2f} \sin^2 \theta f'\\ {R^\theta}_{\phi \theta \phi} &=& (1 - e^{-2f}) \sin^2 \theta \end{eqnarray}The Ricci tensor is
\begin{eqnarray} R_{rr} &=& \frac{2}{r} f'\\ R_{\theta\theta} &=& e^{-2f} (r f' - 1) + 1\\ R_{\phi\phi} &=& \left[ e^{-2f} (r f' - 1) + 1 \right] \sin^2 \theta \end{eqnarray} And finally, the Ricci scalar is \begin{eqnarray} R &=& \frac{2}{r} \left( e^{-2f} f' + \frac{1}{r} \left[ e^{-2f} (r f' - 1) + 1 \right] \right) \end{eqnarray}Now what we want is for all these tensor components to remain bounded along any curve, but still fail to be continuous at some points. To do this, let's consider the classic bounded yet discontinuous function, and overall counterexample of many things,
\begin{equation} f(x) = \sin(x^{-1}) \end{equation}We have from the definition of limits that, given the limit
\begin{equation} \lim_{x \to c} f(x) = \ell \end{equation}This means that, for every sequence $(x_n)$ in $\mathbb{R} \setminus \{ c \}$ that converges to $c$, the sequence $(f(x_n))$ converges to $\ell$. Now given the sequence
\begin{equation} x_n = \frac{2}{(4n+1)\pi} \end{equation}The function sequence becomes
\begin{equation} f(x_n) = \sin(\frac{(4n+1)\pi}{2}) = 1 \end{equation}which isn't hard to show that it converges to $1$, but the sequence
\begin{equation} x_n = \frac{1}{2n\pi} \end{equation}give rise to the sequence
\begin{equation} f(x_n) = \sin(2n\pi) = 0 \end{equation}Appropriate sequences can converge from anything between $-1$ and $1$, and as $\mathbb{R}$ is a Hausdorff space, there can only be one limit, hence $f$ has no limit at $0$ and is therefore not continuous there. What we want is therefore to have a term of that form in $R$, without $R$ diverging due to other terms.
Now unlike $\sin(x^{-1})$, $x^n \sin(x^{-1})$ does have a limit at $0$ for $n > 0$, as
\begin{eqnarray} |x^n \sin(x^{-1})| &=& |x^n| |\sin(x^{-1})| &\leq & |x^n| \end{eqnarray}and as $x^n$ converges to $0$ for $x \to 0$, then so does this function. What we want is therefore some function of this form such that $R$ will still fail to converge at some points but will nethertheless remain locally bounded. Consider the function
\begin{equation} f(r) = r^n \sin(r^{-1}) \end{equation}with derivative
\begin{equation} f'(r) = r^{n-2} \left[ nr \sin(r^{-1}) - \cos(r^{-1})\right] \end{equation}Then the Ricci scalar is
\begin{eqnarray} R &=& 4 e^{-2r^n \sin(r^{-1})} r^{n-3} \left[ nr \sin(r^{-1}) - \cos(r^{-1})\right] - \frac{2}{r^2} (e^{-2r^n \sin(r^{-1})} + 1) \end{eqnarray}Now for $n > 0$, the second term does converge to $0$ (outside of $0$ it is entirely composed of continuous functions so no issues there), as can be seen by
\begin{eqnarray} |\frac{2}{r^2} (e^{-2r^n \sin(r^{-1})} + 1)| &=& 2|\frac{1}{r^2}| |e^{-2r^n \sin(r^{-1})} + 1|\\ &\leq& 2|\frac{1}{r^2}| (|e^{-2r^n \sin(r^{-1})}| + 1 \end{eqnarray}...
Now to show that this is a naked singularity, let's report to the definition of such. There is no strong standard definition for them, but we can just go with one of the most common one :
in other words, there is a singular point in spacetime that $p$ could observe an object or signal disappear into. Due to the appearance of an exponential and that sine inside, we're going to need some way to counterbalance it with the timelike component if we hope to keep everything timelike. Let's consider the curve
\begin{equation} x^\mu(\tau) = (e^{(-\lambda)^n}, -\frac{\lambda}{e}, 0, 0) \end{equation}This is hopefully an infalling, radial, future-pointing causal curve. As it has tangent $(-nx^{n-1}e^{(-\lambda)^n}, -e^{-1}, 0, 0)$, its norm is
\begin{eqnarray} |\gamma'| &=& -e^{2(-\lambda)^n} + e^{2(-\lambda)^n \sin((-\lambda)^{-1})} e^{-2}\\ &=& e^{2(-\lambda)^n} \left[ -1 + e^{-2} e^{\sin((-\lambda)^{-1})} \right] \end{eqnarray}For $n$ even, this is future-pointing. The second term is always negative as the upper bound of $\sin$ will simply be $1$ so that this term will be at most $-1 + e^{-2}$. We're quite lucky as no term of the Riemann tensor or any derived term depends on $t$ so that any transport along this curve will not get too much extra terms. In our case, for the Ricci tensor, this will be
\begin{equation} R(\gamma(\lambda)) = 4 e^{-2(-\lambda)^n \sin((-\lambda)^{-1})} (-\lambda)^{n-3} \left[ nr \sin((-\lambda)^{-1}) - \cos((-\lambda)^{-1})\right] - \frac{2}{(-\lambda)^2} (e^{-2(-\lambda)^n \sin((-\lambda)^{-1})} + 1) \end{equation}From the same arguments as before, we have that this function fails to be $C^0$ at $\lambda = 0$, so that if we wish for the curvature to be well-defined everywhere, we need to remove $r = 0$ (as can be checked by simply shifting the time of our curve, it needs to be removed as every spacelike hypersurface). All we need now is to show that there is a point to the future of this curve, so that we simply need to find a future-oriented curve from points of this curve to some point $p$.
From the properties of causal relations, we have that $p \ll q$ and $q \ll r$ implies $p \ll r$, so that we will only need to show that a small section of the curve $(-\varepsilon, 0)$ has a future point to it (we could of course do it at the actual point $\gamma(0)$, but this point is of course not in the manifold). For any point $\gamma(-\varepsilon)$, we are at the coordinates
\begin{equation} x^\mu(-\varepsilon) = (e^{\varepsilon^n}, \frac{\varepsilon}{e}, 0, 0) \end{equation}We will simply try to find a point sufficiently further in the future to accommodate all points on the curve. Take a point at a distance of $1$ from the singularity and at a future time $T$, we will simply need a curve such that
\begin{eqnarray} x^\mu(0) &=& (e^{\varepsilon^n}, \frac{\varepsilon}{e}, 0, 0)\\ x^\mu(\tau_1) &=& (T, 1, 0, 0) \end{eqnarray}As we still need it to be timelike, it will need to be of a similar form to before, although this time outgoing (let's assume $\varepsilon < 1$ since it can be arbitrarily small).
\begin{equation} x(\tau) = (e^{\varepsilon^n + \tau} + \tau (\frac{T - e^{\varepsilon^n + \tau_1}}{\tau_1}), \frac{1 - \frac{\varepsilon}{e}}{\tau_1} \tau + \frac{\varepsilon}{e}, 0, 0) \end{equation}with tangent
\begin{equation} \dot{x}(\tau) = (e^{\tau} + (\frac{T - e^{\varepsilon^n + \tau_1}}{\tau_1}), \frac{1 - \frac{\varepsilon}{e}}{\tau_1}, 0, 0) \end{equation}Last updated : 2019-06-21 16:46:18