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In these lecture notes the static isotropic metric is treated as follows (p. 71):

Take a spherically symmetric, bounded, static distribution of matter, then we will have a spherically symmetric metric which is asymptotically the Minkowski metric. It has the form (in spherical coordinates): $$ds^2=B(r)c^2dt^2-A(r)dr^2-C(r)r^2(d\theta^2+\sin^2\theta d\phi^2)$$ And then it goes on eliminating $C$ and expanding $A$ and $B$ in powers of $\frac{1}{r}$. No explanations are given on why we can assume that form for the metric. Could someone explain why, please?

Personally, I would rather assume the form (in cartesian coordinates): $$ds^2=f(r)dt^2-g(r)(dx^2+dy^2+dz^2)$$ which would certainly give a spherically symmetric metric, and then change to spherical coordinates, obtaining something looking like: $$ds^2=f(r)dt^2-g(r)(dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2)$$ which looks substantially different from the above. Is this approach wrong? Why?

By the way, don't be afraid of getting technical. I have a pretty good mathematical basis on the subject (a course of one year on differential geometry).

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  • $\begingroup$ possible duplicate of I need help understanding a step in the derivation of the Schwarzschild solution $\endgroup$ – Jerry Schirmer Jan 21 '14 at 21:35
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    $\begingroup$ @JerrySchirmer That is totally different question, and it doesn't offer an answer to what I'm asking. $\endgroup$ – Daniel Robert-Nicoud Jan 21 '14 at 21:37
  • $\begingroup$ Sorry. I probably shouldn't be on here in this state of tiredness. Close vote retracted. But I will note that you can always choose $R = r\sqrt{g(r)}$, do the coordinate transformation, and your metric will transform to the form you have above. $\endgroup$ – Jerry Schirmer Jan 21 '14 at 22:49
  • $\begingroup$ @JerrySchirmer Ah! I see. Thank you. So basically my method is correct, only the other Ansatz is equivalent and easier to work out? Also if I wanted, say, try to find out the metric for a rotating distribution of matter, would it be correct to start with something like $ds^2=f(r,z)dt^2-g(r,z)(dx^2+dy^2)-h(r,z)dz^2$ with $r^2=x^2+y^2$ and change to rotating cylindrical coordinates? Or is some other Ansatz better? $\endgroup$ – Daniel Robert-Nicoud Jan 22 '14 at 0:16
  • $\begingroup$ @JerrySchirmer By the way, if you write your comment as an answer I will accept it immediately. $\endgroup$ – Daniel Robert-Nicoud Jan 22 '14 at 0:17
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I am referencing B. Schutz, A First Course in General Relativity. 2009, p. 256-258.

Note first that the line element on a 2-sphere with radius of curvature $r$ is $dl^2=r^2(d\theta^2+\sin^2\theta d\phi^2)$. Since we want a metric that is spherically symmetric about a point in space, say $r'=0$, we must not be able to tell the difference (in terms of intrinsic geometry) between points located at the same distance $r'$ away from the centre of symmetry and at the same time $t$. Therefore every point in space-time is located on a spatial 2-sphere whose radius of curvature may depend on coordinates $r'$ and $t$, i.e. $r=r(r',t)$. The line element on such a 2-sphere is then $dl^2=r^2(d\theta^2+\sin^2\theta d\phi^2)$, where we have chosen to label each sphere by its radius of curvature $r$ instead of the actual radial distance $r'$ of its points away from the centre of symmetry.

Now consider two spheres corresponding to the same time $t$ but different distances $r'$ and $r'+dr'$. A priori, the coordinates on the two spheres might be defined with respect to different poles where $\theta=0$ and different half-planes where $\phi=0$. However, because the spheres are centred about the same point $r'=0$, we can always choose these references to be the same such that the curves $\theta=const.$, $\phi=const.$ (implicity, we also have $t=const.$) are orthogonal to all spheres at different distances $r'$. By definition, a tangent to these curves is the coordinate basis vector $\textbf{e}_r$. Therefore we have $g_{\theta r}=\textbf{e}_{\theta}\cdot\textbf{e}_r=0$ and $g_{\phi r}=\textbf{e}_{\phi}\cdot\textbf{e}_r=0$. We have thus reduced the most general metric in spherical coordinates, with all the possible cross terms, to the slightly simpler form: $$ds^2=g_{00}dt^2+2g_{0r}dtdr+2g_{0\theta}dtd\theta+2g_{0\phi}dtd\phi+g_{rr}dr^2+r^2d\Omega^2,$$ where I have defined $d\Omega^2\equiv d\theta^2+\sin^2\theta d\phi^2$.

Similarly, we argue that the curves $r=const.$, $\theta=const.$, $\phi=const.$, with tangent vector $\textbf{e}_t$ must be orthogonal to the two-spheres. Otherwise, $\textbf{e}_t$ would have components in the directions of $\textbf{e}_\theta$ and $\textbf{e}_\phi$. There would then be a preferred spatial direction on the 2-sphere, namely that direction parallel to the projection of $\textbf{e}_t$ on the sphere. But this is forbidden by spherical symmetry, therefore $g_{0\theta}=\textbf{e}_t\cdot\textbf{e}_\theta=0$ and $g_{0\phi}=\textbf{e}_t\cdot\textbf{e}_\phi=0$. We have now reduced the metric to: $$ds^2=g_{00}dt^2+2g_{0r}dtdr+g_{rr}dr^2+r^2d\Omega^2.$$ Lastly, a static spacetime is one where the metric is unchanged by a transformation $t \rightarrow -t$. This implies that $g_{0r}=-g_{0r}=0$. Therefore we get a line element that only contains $dt^2$, $dr^2$ and $r^2\Omega^2$, as required by the question.

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  • $\begingroup$ Note that \sin produces $\sin$ in MathJax and doubling the dollar (i.e., $$\mathrm{d}s^2=...$$) signs produces a centered equation $\endgroup$ – Kyle Kanos Jan 22 '17 at 20:06

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