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Following (1) and (2), the Schwarzschild metric can be written as: $$ ds^2 = a(r)\cdot c^2 dt^2 - b(r)\cdot dr^2 - r^2\cdot(d\theta^2 + sin^2(\theta)\cdot d\varphi^2) $$ where: $$ a(r) = \begin{cases} \frac14\left(3\sqrt{1-\frac{r_s}{r_g}} - \sqrt{1-\frac{r^2 r_s}{r_g^3}}\right)^2 \qquad \text{if } 0\leq r \leq r_g\\ 1-\frac{r_s}{r}\qquad \text{if } r_g\leq r \end{cases} $$ $$ b(r) = \begin{cases} \left(1-\frac{r^2 r_s}{r_g^3}\right)^{-1} \qquad \text{if } 0\leq r \leq r_g\\ \left(1-\frac{r_s}{r}\right)^{-1} \qquad \text{if } r_g\leq r \end{cases} $$ Here :

  • $r_s=2GM/c^2$ is the Schwarzschild radius of a spherical body of mass $M$
  • $r_g$ is the value of $r$ at the surface of the spherical body

The part $r\in [0,r_g]$ is the interior Schwarzschild solution while the part $r\geq r_g$ is the exterior Schwarzschild solution.

Assume $r_s < r_g$.

Now, I graph $a(r)$ and $b(r)$. I see that $a(r)$ and $b(r)$ are continuous functions. I see that $a(r)$ looks fairly smooth. However, at the transition $r=r_g$, the function $b(r)$ does not look smooth. Is it normal? Why would $b(r)$ be not differentiable at the transition from $r<r_g$ to $r>r_g$?

Here is a screenshot for $r_s=1$ and $r_g=2$.

enter image description here


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1 Answer 1

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The distribution of mass has a discontinuity at $r=r_g$ so it entirely reasonable that the metric has a feature that reflects this.

The main point is that while the $a(r)$ function does not have a continuous derivative (slope), it is continuous, so there is no discontinuity in the metric itself. That is, the metric does not have to be smooth, only continuous.

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  • $\begingroup$ Yes, I was thinking about the non-continuous distribution of mass. However, in classical mechanics the potential energy is differentiable when going from inside $r^2/2 + const.$ to the outside $-1/r$. So I was thinking maybe this should be reflected in the above metric. But yes, if the mass is not continuous, so is $T_{\mu\nu}$, so is $G_{\mu\nu}$, so should some first derivatives of $g_{\mu\nu}$. $\endgroup$
    – Noé AC
    Jun 2, 2019 at 20:21
  • $\begingroup$ The usual matching in cases like follows Israel-Darmois junction conditions (see e.g. arXiv:1604.08930 [gr-qc]) and in the present case (see e.g. arXiv: 1412.7083 [gr-qc]) $a(r)$, $b(r)$, and $a'(r)$ are continuous while $b'(r)$ is discontinuous. The discontinuity in $b(r)$ is proportional to the difference in energy density at $R$ (see eq. (45) of arXiv: 1412.7083 [gr-qc]) which is very large for the interior Schwarschild solution as the density drops from its large constant value to zero at the stellar surface. $\endgroup$
    – N0va
    Jun 11, 2022 at 1:26

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