In this answer as well as on Wikipedia the radial coordinate of the Schwarzschild metric is described as follows:

...the r co-ordinate is the value you get by dividing the circumference of the circle by 2π.

This circumvents (no pun intended) the problem of talking about the path outwards from the singularity, but what about the circle itself: does it have the length we would measure when flying around it (however this is done) or is it a virtual circle assuming flat space, i.e. no mass at the center.


The circumference is the distance you would measure if you laid out a (very long!) tape measure along a circle centred on the black hole. Or if piloted your spaceship round the black hole it's the distance your navigation computer would record.

Extracting these distances is less work then you might think. Suppose we use polar coordinates so $r$ is the distance from the black hole, $\theta$ is the latitude and $\phi$ is the longitude. The metric is then:

$$ ds^2 = -\left(1-\frac{r_s}{r}\right)dt^2 + \frac{dr^2}{1-\frac{r_s}{r}}+r^2d\theta^2 + r^2\sin^2\theta d\phi^2 $$

Suppose we trace out a circle round the equator so $dt = dr = d\theta = 0$ and $\theta = \pi/2$, then the metric simplifies to:

$$ ds^2 = r^2d\phi^2 $$


$$ ds = rd\phi $$

As we go round the equator $\phi$ goes from zero to $2\pi$, so integrating to get the distance travelled gives:

$$ \Delta s = r\int_0^{2\pi} d\phi = 2\pi r $$

To show that the $r$ coordinate is not simply the radial distance let's use the same technique to measure the distance from $r = R$ to the singularity $r = 0$. We'll use a radial line at constant $t$, $\theta$ and $\phi$, so $dt = d\theta = d\phi = 0$, and substituting these into the metric gives:

$$ ds = \frac{dr}{\sqrt{1-\frac{r_s}{r}}} $$

so the proper distance from $r=R$ to the centre is:

$$ \Delta s = \int_0^R \frac{dr}{\sqrt{1-\frac{r_s}{r}}} $$

which is most definitely not just $R$. I won't evaluate the integral here because I've already done it in my answer to How much extra distance to an event horizon?. The result isn't particularly illuminating.


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.