# The central point is singular or not?

Let take the metric \begin{eqnarray} \mathrm ds^2 = -f^2(r) dt^2 + g^2(r)dr^2 + r^2~\mathrm d\Omega^2 \end{eqnarray} with $f^2(r) = 1 -ar^3$ and $g^2(r) =\frac{b}{1-ar^3}$.

Is the central point $r=0$ regular or not? I am confused because I calculate The Ricci scalar as \begin{eqnarray} R= \frac{20ar^3+2b^2 -2}{b^2r^2} \end{eqnarray} which is singular at the origin.

• – Kyle Kanos Dec 20 '17 at 16:43

## 2 Answers

A scalar is a spacetime invariant, so if you find a scalar which diverges at some point, you know that your metric is not regular at that point.

If the expression you have obtained for the Ricci scalar is correct, then as $r \to 0$ it is clear that $R$ diverges unless $b=\pm 1$. So for values of $b$ other than those two, the metric is singular at $0$.

What about for those values of $b$? You can try to compute other scalars like $R_{\mu\nu}R^{\mu\nu}$ or $R_{\mu\nu\sigma\rho}R^{\mu\nu\sigma\rho}$ or any other which comes to your mind. If any of these diverges, you can say that your metric is singular. If they don't, then you know nothing. If you wanted to show that the metric is not singular for those values of $b$, it would be easiest to find a coordinate trasformation for which the metric is clearly regular.

It appears, on first glance, that Ricci scalar does tend to diverge as $r\to 0\,.$ I would calculate the Kretschmann scalar $${\cal K} = R_{abcd}\,R^{abcd}$$ in order to check. If it is well behaved at $r=0$ then I don’t think there would be a problem at $r=0$. But without calculating it I could not say for certain.

However, it would be appropriate if you could provide some background to your question, e.g. which space-time does the metric you mention above describe?

• I will check. What does it mean? The solution is regular or singular? – user55944 Dec 20 '17 at 14:31
• I am looking for interior solutions to star – user55944 Dec 20 '17 at 14:34
• The Kretchmann scalar is a space-time invariant. So if it doesn’t diverge then there is probably no issue. You should calculate it for yourself. If you managed to calculate the Ricci scalar, which is also an invariant of the space-time metric, I don’t think you will find it difficult to calculate the Kretchmann scalar. Moreover, if the expression for $R$ that you posted above is correct then the Kretschmann scalar will probably also be ill behaved at $r=0$. – Physics_Et_Al Dec 20 '17 at 14:38
• Perhaps you could also calculate the energy-momentum-stress tensor, $T_{ab}\,.$ This would yield physical information about the energy-matter distribution and pressures inside the star. It would also be worthwhile, once you have calculated $T_{ab}$, to determine which of the energy conditions are satisfied. – Physics_Et_Al Dec 20 '17 at 14:48