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


1 Answer 1


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.