Is $\theta=0$ a coordinate singularity in the Schwarzschild metric?

Question

We often hear this example of coordinate singularity. We can describe the 3D Euclidean space using rectangular or spherical coordinates. But in the spherical coordinate, the north pole $(r,\theta=0, \phi)$ is a coordinate singularity because one point here corresponds to infinitely many values of $\phi$.

In the Schwarzschild metric, $$ ds^2=-(1-\frac{2M}{r})dt^2+\frac{dr^2}{(1-\frac{2M}{r})}+r^2d\Omega^2, $$

we know that $r=2M$ is a coordinate singularity.

I have two questions.

(1).The metric determinant is $g=r^4\sin^2\theta$, which is not zero or divergent at $r=2M$. Can we ascertain a coordinate singularity only on the basis of the divergence of some metric component?

(2). According to the metric determinant, $g=0$ at $\theta=0$. Why is $\theta=0$ not a coordinate singularity as is shown in the above example?

Answer 1

You are correct. The pole, in your case $\theta=0$ is indeed a coordinate singularity.

That is one reason that it seems silly to obsess over the coordinate effects that occur in Schwarzschild coordinates at the event horizon. Believing that the coordinate effects at the horizon are somehow physical is akin to believing that the coordinate effects at the poles are physical. There is no magical force pushing you away from the pole that becomes infinite as you get near the pole, but if you take the coordinates as physical fact there appears to be such a force.

Spherical coordinates also have one other issue besides the poles. One of the defining requirements of a coordinate chart is that they must be invertible: each event covered by the chart has one coordinate and each coordinate in the chart has one event. That is a problem because if you allow $\theta \in \mathcal R$ then there will be multiple coordinates that point to any given event. So instead you can require $0 \le \theta < \pi$. But this is problematic because coordinate charts are defined on open regions and the $0=\theta$ boundary is not open. So we must instead require $0<\theta<\pi$. This gives a mathematically valid coordinate chart, but excludes all of the points on the boundary. So the chart does not cover the entire manifold.

Answer 2

1- you cannot use metric or other tensor components to test for singularities. You must use scalar quantities, because they are the same in any coordinate system. For example with Schwarzschild coordinates: take the Kretschmann invariant $K=R_{abcd} R^{abcd}=48M^2/r^6$ and see that it diverges when $r\rightarrow 0 $. This shows that r=0 is a physical singularity.

Check Gravitational singularity

When $r\rightarrow 2M$, the components of Riemann tensor in the orthonormal system of a falling observer are finite too, so no infinite tidal forces. No physical curvature at the horizon.

2- the metric determinant for a regular coordinate point must be finite (of course) and non-zero. This is because it is proportional to a area/volume/hypervolume spanned by the coordinate axis. If it is $0$, it means that these "volumes" have collapsed, ie. bad coordinates at that point.