# Schwarzschild Solution Convention

In looking at the components of the Schwarzschild Metric, one finds $g_{00} = (1 - \frac{r_s}{r})c^2$. Wikipedia states that $r$ is measured as the circumference, divided by $2π$, of a sphere centered around the massive body. So if I'm to calculate the the time dilation experienced at one point relative to another ( let's say with $dr = 4\times 10^4 m$ ), am I plugging in $4\times 10^4$ or $\frac{4 \times 10^4}{\pi}$? I just feel like if we're just varying the radial coordinate, and we know precisely by what distance, we don't need a factor of inverse $\pi$.

• I think we need to see your time dilation equation, but I'd wager you don't need the $\pi$. – Ryan Unger Feb 1 '15 at 2:33
• $d\tau^2 = (1 - \frac{r_s}{r})dt^2$ and we hold the scenario fixed in space, allowing the space coordinates to vanish. – Doryan Miller Feb 1 '15 at 2:47
• That was the formula I had in mind. No $\pi$. Note that in the Schwarzschild metric, if you let $M\rightarrow 0$, you recover spherical coordinates. $r$ is, in a sense, just the radial coordinate. You don't need $\pi$. I'm not sure why you thought you would need $\pi$... – Ryan Unger Feb 1 '15 at 2:51
• Wikipedia has a very misleading definition. I should've put it quotes, but, "$r$ is the radial coordinate (measured as the circumference, divided by $2π$, of a sphere centered around the massive body.)" – Doryan Miller Feb 1 '15 at 2:53

You don't need $\pi$ for that formula.
Let me provide an alternative definition of $r$ for you. Let $M$ be Schwarzschild spacetime. An orbit of $M$ is a two-dimensional manifold $\Sigma_r$ generated by $SO(3)$. We construct $M$ by foliating with these 2-spheres. Each $\Sigma_r$ has area $A=4\pi r^2$. This radius $r$ defines the Schwarzschild coordinate $r$.
(Note that one should actually say that orbits are generated by $SO(3)$ acting on the spatial part of $M$, i.e. the part that has $SO(3)$ as an isometry group. This detail is very nit picky though.)