In the text "Exploring Black Holes" by Taylor ,Wheeler I came across the following thought experiment

"We consider two imaginary concentric spherical shells and suppose we measure the "reduced circumference " $r$ of smaller sphere to be $1 km$ less than that of bigger sphere (based on tape measure distance around the two spheres). Now we lower a plumb ball from the outer sphere and first time truly measure the radial distance between two spheres."

It was concluded that we wouldn't find the distance to be $1km$ (we would have if the space were flat). I'm still unable to figure out how is it possible?

  • $\begingroup$ That is the essence (or even one possible definition) of (intrinsic) spatial curvature. Like angles of a triangle not adding up to 180 degrees, or parallel lines meeting or diverging. $\endgroup$
    – m4r35n357
    Feb 3, 2021 at 13:14

1 Answer 1


The same thing is true on Earth. Let's say you plant a flag on the North pole, and then walk some distance $r$ away. The set of all points which are at a distance $r$ from the flag is a circle. What is the circumference of this circle?

The answer isn't $2\pi r$. This can be seen by letting $r = \pi R/2$, which means that you're standing on the equator. The circumference of this circle is then given by $4r$!

More generally, the metric on the 2-sphere is $$g = \pmatrix{R^2 & 0 \\ 0 & R^2 \sin^2(\theta)}$$

in angular coordinates $(\theta,\phi)$, where $R$ is the radius of the Earth. The infinitesimal distance element is therefore $ds^2 = R^2d\theta^2 + R^2\sin^2(\theta) d\phi^2$. If we walk from the North pole for some distance $r$, we will be moving along a line of constant $\phi$ so the line element is $ds = Rd\theta$, which means that $r=R\theta \iff \theta = r/R$ where $\theta$ is the polar angle corresponding to your chosen circle. The circumference of that circle is obtained by walking along a line of constant $\theta$ from $\phi=0$ to $2\pi$, so $ds = R\sin(\theta) d\phi$ and so the circumference is given by $C = 2\pi R\sin\left(\frac{r}{R}\right)$.

So to summarize, on the surface of the Earth, the circumference of a circle is related to its radius $r$ via $$C(r) = 2\pi R\sin\left(\frac{r}{R}\right)$$ which is different from the flat space relationship $C_{\mathrm{flat}}(r) = 2\pi r$. If $r\ll R$ then these expressions are approximately the same, which reflects the local apparent flatness of the Earth, but they are ultimately different.

Exactly the same thing is true in the Schwarzschild coordinates around a black hole. The curvature of space is such that the circumference of a circle is no longer simply proportional to the radial distance from its central point. The relationship is more complex than the one presented here, but the concepts are more or less identical - with the exception of some technicalities which would arise if you try to cross the event horizon in these coordinates.

  • $\begingroup$ Nice answer. In the conventional Schwarzschild coordinates $r$ isn’t the actual radius but the angular part of the metric is so chosen that the circumference works out to be $2\pi r$. Could you tell me the form of the angular part metric when the $r$ coordinate is so chosen that it is the actual radius but then the angular part gets changed so that the circumference is not $2 \pi r$. I was wondering g what would be the form of the metric in that case. $\endgroup$
    – Shashaank
    Feb 3, 2021 at 15:10
  • $\begingroup$ @Shashaank If you define the coordinate $\rho(r) = \int_{r_s}^r \frac{dr'}{\sqrt{1-\frac{r_s}{r}}} = r_s\ln\left(\sqrt{\frac{r}{r_s}} + \sqrt{\frac{r}{r_s}-1}\right)+\sqrt{r(r-r_s)}$ and express the metric in terms of $\rho$, you'll get your answer (as long as you stay outside of the event horizon). The metric will take the form $ds^2 = -\sqrt{1-\frac{r_s}{r}} dt^2 + d\rho^2 + r(\rho)^2(d\theta^2 + \sin^2(\theta) d\phi^2)$. Of course, $r(\rho)$ does not look like it can be obtained in a pleasant closed form. $\endgroup$
    – J. Murray
    Feb 3, 2021 at 15:43
  • $\begingroup$ ahha. Yes I see. I should have thought of that obvious coordinate transformation. Yes,I don’t think it would be good to invert to get $ r$ in terms of $\rho$. But by any chance have you seen the formula that circumference would take place in these quite complex coordinates? $\endgroup$
    – Shashaank
    Feb 3, 2021 at 15:50
  • $\begingroup$ @Shashaank Sure - on a curve of constant $\rho$ and $\theta$, the circumference would be $2\pi r(\rho)\sin(\theta)$, which is not simply $2\pi\rho\sin(\theta)$ as it would be in flat space. $\endgroup$
    – J. Murray
    Feb 3, 2021 at 17:11
  • $\begingroup$ oh yeah. Thanks. I was just trying to invert r for rho.. .. But yeah this is right.. $\endgroup$
    – Shashaank
    Feb 3, 2021 at 17:53

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