4

I'm trying to understand the difference between proper distance $d\sigma$ and coordinate distance $dr$ in Schwarzschild geometry. The bottom bit of the diagram represents flat space, the upper bit curved space. The inner circles represent Euclidean spheres of radius $r$, the outer circles radius $r+dr$.

Is the proper radius of these circles the same as $r$? I think I mean if I measured the radius of these circles with a real ruler would I get the coordinate distance $r$ of the Schwarzschild metric?

Schwarzschild radial distances

The more I think about this the more confusing I find it.

ProfRob
  • 141,325
Peter4075
  • 3,109

2 Answers2

5

In the Schwartzchild coordinates the r co-ordinate is the value you get by dividing the circumference of the circle by 2$\pi$. That is, it's the radius of the circles you've projected onto the base of your diagram.

As others have mentioned in the comments, deciding what you mean by the "real" distance from the circle to the singularity is a vexed issue. Far away from the singularity r agrees with what we think of as the radius, but of course that's only because space is (nearly) flat far away from the singularity. Once the curvature becomes significant r will not be the same value as you get by integrating the metric from the singularity out to your circle.

JR

John Rennie
  • 367,598
1

The proper distance (that you could measure with a ruler) is not $r_2 -r_1$ and you cannot measure the "radius" with a ruler because that would involve getting information out from within $r_s$, which isn't allowed.

What you can do is use your ruler to measure a circumference. If you divide that by $2\pi$ to get $r$.

If you do use a ruler to measure the proper distance in the radial direction (i.e. with no separation in coordinate time), you must integrate $$\int d\sigma = \int \frac{dr}{\sqrt{1 - r_s/r}}\ .$$ Since the denominator is $<1$ then the result is going to be bigger than the simple difference of the $r$ coordinate.

ProfRob
  • 141,325