Given the Schwarzschild metric

$$ ds^{2} = -\Big(1-\frac{r_{s}}{r}\Big)c^{2}dt^{2} + \Big(1-\frac{r_{s}}{r}\Big)^{-1}dr^{2} + r^{2}(d\theta^{2} + \sin^{2}\!\theta \, d\phi^{2}), $$

we can apply the approximation $(1-x)^{-1}\approx 1+x$ and get

$$ ds^{2} = -\Big(1-\frac{r_{s}}{r}\Big)c^{2}dt^{2} + \Big(1+\frac{r_{s}}{r}\Big)dr^{2} + r^{2}(d\theta^{2} + \sin^{2}\!\theta \, d\phi^{2}). $$

Taking $r_{s} = 2GM/c^{2}$ and $\Phi(r) = -GM/r$ gives

\begin{align}\tag{1} ds^{2} = -\Big(1 + \frac{2\Phi(r)}{c^{2}}\Big)(c\, dt)^{2} + \Big(1 - \frac{2\Phi(r)}{c^{2}}\Big) dr^{2} + r^{2}(d\theta^{2} + \sin^{2}\!\theta \, d\phi^{2}). \end{align}

This is nice and all. However, page 126 in Gravity by Hartle and eqn (12) in page 2 in this link give us something a little different:

$$ ds^{2} = -\Big(1 + \frac{2\Phi(r)}{c^{2}}\Big)(c\, dt)^{2} + \Big(1 - \frac{2\Phi(r)}{c^{2}}\Big) (dx^{2} + dy^{2} + dz^{2}). $$

By changing to spherical coordinates, we get

\begin{align}\tag{2} ds^{2} = -\Big(1 + \frac{2\Phi(r)}{c^{2}}\Big)(c\, dt)^{2} + \Big(1 - \frac{2\Phi(r)}{c^{2}}\Big) dr^{2} + r^{2}\Big(1 - \frac{2\Phi(r)}{c^{2}}\Big)(d\theta^{2} + \sin^{2}\!\theta \, d\phi^{2}). \end{align}

Question: How do we get from (1) to (2)? In (2), why is there that factor in the angular part of the metric even though it is not there in (1)? Is there some substitution/rescaling I need to do (nothing obvious seems to work)? Or do we just say that (1) and (2) are merely approximately the same?

If this is a matter of approximation, then I am still confused, because I thought we can only ignore terms of order $O(1/c^{4})$. The difference between (1) and (2) is only of order $1/c^{2}$.


After looking around, I'm thinking that (2) was derived as an approximation to the metric in the so-called isotropic coordinates as described in this post. Indeed, when the appropriate approximation there is made, you get out (2).

It seems that the coordinates for (1) are not the same as the coordinates for (2).

  • 1
    $\begingroup$ They are related by a coordinate transformation (the link above also mentioned this). In non-isotropic coordinates (1), the closer you are to the spherical symmetric source, the less "Euclidean" the metric is: at every constant $t$ slice the radial part gets "distorted" more. But for isotropic case (2), every constant $t$ slice is Euclidean space up to rescaling. Newtonian limit is typically stated in the case of (2), since the usual 3-space is Euclidean in Newtonian physics. $\endgroup$ – Everiana Feb 16 '20 at 6:45

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