# Describing the shape of a singularity

Hawking and Ellis write about the difficulty of describing the shape of a singularity when presented with a manifold that has curves of finite length that don't reach a point in the manifold.

[Investigating/defining what is] meant by the size, shape, location, and so on of a singularity [...] would be fairly easy if the singular points were included in the manifold. However it would be impossible to determine the manifold structure at such points by physical measurements. In fact there would be many manifold structures which agreed for the non-singular regions but which differed for the singular points. For example, the manifold at the $t=0$ singularity in the Robertson-Walker solutions could be that described by the coordinates $$\{t, r\cos \theta, r\sin\theta\cos \phi, r\sin\theta\sin \phi\}$$ or that described by $$\{t, Sr\cos \theta, S r\sin\theta\cos \phi, Sr\sin\theta\sin \phi\}.$$ In the first case the singularity would be a three-surface, in the second case a single point.

Of course, as a physicist I'm sensitive to anything that couldn't be resolved by physical measurements. But I don't actually see the example.

Can anyone explain how two different coordinates yield a differently shaped singularity?

• I am not sure what your question is. When t=0, the first set is a 3 surface, the second is just (0,0,0,0) i.e. one point. – MBN Nov 10 '15 at 19:40
• @MBN Feel free to assume that I can't remember what the authors mean by the S or even what the coordinates are. The first one looks like Minkowski space where you use spherical coordinates for the spatial part so you get a full 3d euclidean space as your t=0 surface. The second looks the same except you call your spatial radial coordinate Sr instead of calling it r, but names don't matter. – Timaeus Nov 10 '15 at 19:45
• @MBN I gave an example that is close to answering the literal question without showing an ounce of insight into the example Hawking and Ellis give. – Timaeus Nov 10 '15 at 20:12
• MBN's comment is correct, the $S$ is the FLRW scale factor, more commonly written $a(t)$, which has $S(t = 0) = 0$. Does this dissolve the question? (If no, please reformulate it because it is currently unclear what it is.) – ACuriousMind Nov 11 '15 at 0:30
• Possible duplicate of Is black hole singularity a single point? – user4552 Aug 12 '17 at 23:07

While I can't speak to the specific example listed, or the particular meaning of the $S$ term, there are examples of spacetimes that agree up until the singularity.

First consider a spacetime that is topologically $\mathbb R^4$ with time being the radial coordinate and then for each time you get a three sphere where you then adjust the scale factor in the metric so the sphere is however large the universe is at that time. For instance if at time $t$ the universe has a volume $V(t)$ then you can have 4d spherical coordinates $(t, \alpha, \beta, \gamma)$ that relate to $(w,x,y,z)$ by $(w,x,y,z)$=$(t\cos\alpha, t\sin\alpha\cos\beta, t\sin\alpha\sin\beta\cos\gamma, t\sin\alpha\sin\beta\sin\gamma).$

The normal metric for 3d spherical coordinates is $dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2)$ and similarly if your 4d space was euclidean you would have a metric of $dt^2+t^2(d\alpha^2+\sin^2\alpha(d\beta^2+\sin^2\beta d\gamma^2))$ and now we just need to make it Lorentzian and give it the right size at each time.

$-dt^2+\left(\frac{V(t)}{2\pi^2t^3}\right)^{2/3}t^2 \left(d\alpha^2+(\sin^2\alpha)(d\beta^2+(\sin^2\beta) d\gamma^2)\right)$

Now this metric can satisfy a cosmological metric provided $V(t)$ has the correct dynamics. And the coordinate $t=0$ corresponds to a single event. So the singularity is a single point.

Next consider a spacetime that consists of all the $(w,x,y,z)$ where $w^2+x^2+y^2+z^2\geq 1.$ This time we can call the radial coordinate in 4d $r$ and let $t=r-1$ and we can have the metric

$-dt^2+\left(\frac{V(t)}{2\pi^2(t+1)^3}\right)^{2/3}(t+1)^2 \left(d\alpha^2+(\sin^2\alpha)(d\beta^2+(\sin^2\beta)d\gamma^2)\right).$

At each time $t$ it is a surface in $\mathbb R^4$ that is a three sphere exactly one unit larger in radius than the last example. But as a spacetime it has the exact same volume at time $t$ as the previous one (it still takes the volume you want and divides it by the volume of the actual surface and scales the metric by that to give the right total volume).

But now the $t=0$ surface is the unit sphere topologically. But the metric doesn't actually look any different (the terms with $t+1$ actually all cancel out).

Can anyone explain how two different coordinates yield a differently shaped singularity?

Take a simple example over $$\mathbb R^2$$ that can easily be generalized to 4 dimensions in many ways:

Consider polar coordinates $$(r,\phi)$$ and remove the point $$r=0$$. If this was our singularity, then it is a single point. Now apply the coordinate transformation

\begin{align} \mathbb R^2\!\setminus\!\{0\} &\to \mathbb R^2\!\setminus\!\{r\leqslant1\} \\ (r,\phi) &\mapsto(r+1,\phi) \end{align}

which is smooth, invertible and can be extended to $$r=0$$. It maps the point $$r=0$$ to the unit circle $$r=1$$, i.e. if $$r=0$$ is a singularity, then just by picking other coordinates the singularity is changed to a circle. We could even transform the singularity to a line: Regard the plane as complex numbers and apply the mapping $$z\mapsto z+1/z.$$ This mapping is smooth and invertible for $$|z|>1$$ and maps the circle $$|z|=1$$ to the line $$\mathbb R \cap\{|z|\leqslant 1\}$$.

This means that the topology and the dimension of the singularity have no physical meaning because they depend on choice of coordinates.