The Riemannian Curvature in a solid sphere

Question

Is the Riemannian Curvature at the centre of a solid sphere zero?

Answer 1

If we assume the sphere has a constant density, and we can ignore contributions from the pressure, then the geometry inside a solid sphere is described by the Schwarzschild interior metric:

$$ ds^2 = -\left[\frac{3}{2}\sqrt{1-\frac{2M}{R}} - \frac{1}{2}\sqrt{1-\frac{2Mr^2}{R^3}}\right]^2dt^2 + \frac{dr^2}{\left(1-\frac{2Mr^2}{R^3}\right)} + r^2 d\Omega^2 $$

where $M$ is the mass of the sphere and $R$ is the radius of the sphere. As usual, the $r$ and $t$ coordinates are the radial and time coordinates for an observer far from the sphere.

At the centre of the sphere $r \rightarrow 0$ and the metric becomes:

$$ ds^2 = -\left[\frac{3}{2}\sqrt{1-\frac{2M}{R}} - \frac{1}{2}\right]^2dt^2 + dr^2 + r^2 d\Omega^2 $$

So far so good.

The trouble is that your question is ill defined because you don't state what you mean the Riemann curvature. The Riemann tensor is clearly not zero because the metric at $r = 0$ differs from the Minkowski metric. I haven't calculated the Christoffel symbols or the various curavture scalars because, well, life's too short, but it seems unlikely that they would be zero either. So to the extent that your question can be answered the answer is no.

Answer 2

$\newcommand{\d}[1]{\;\mathrm{d} #1} \newcommand{\c}[3]{\Gamma^#3_{#1 #2}}\newcommand{\riem}[4]{\mathcal R^{#4}_{\; #1 #2 #3}}$As it turns out setting $r=0$ and calculating the Riemann tensor as an attempt to cut some corners, which is what I did earlier¹, doesn't do the job. I'll start by taking the original metric

$$\d s^2 = -\left[\frac{3}{2}\sqrt{1-\frac{2M}{R}} - \frac{1}{2}\sqrt{1-\frac{2Mr^2}{R^3}}\right]^2 \d t^2 + \frac{dr^2}{\left(1-\frac{2Mr^2}{R^3}\right)} + r^2 \ d\Omega^2$$

and calculating the Riemann tensor therefrom.

Warning: I won't write the whole tensor as it takes too much space but rather write the components whilst assuming $r=0$ you can find all of the components below.

$$\riem t r t r = \riem t \theta t \theta = \riem t \phi t \phi = \frac{M-3\,M\,\sqrt{\frac{R−2\,M}{R}}}{2\,{R}^{3}} \\ \riem r r t t= \frac{3\,M\,\sqrt{\frac{R−2\,M}{R}}−M}{3\,{R}^{3}\,\sqrt{\frac{R−2\,M}{R}}−5\,{R}^{3}+9\,M\,{R}^{2}}\\ \riem r \theta r \theta = \riem r \phi r \phi =−(2M)/R^3$$

Leading to the conclusion that the curvature is not flat at the centre of a star.

As John Rennie indicated in the comments, first letting $r \to 0$ doesn't do you any good because then you are calculating the Christoffel Symbols for a particular point, for which there always exist a transformation such that the space time is flat.

All of the components of the Riemann tensor calculated by a computer algebra system can be found in pictures below, though I doubt that it can be useful to anybody. The notation used below is

$$ \text{riem}_{i,j,k,l} = \riem i j k l \quad \text{and} \quad [t,r,\theta,\phi] \to [1,2,3,4].$$