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).

  • $\begingroup$ possible duplicate of I need help understanding a step in the derivation of the Schwarzschild solution $\endgroup$ Jan 21 '14 at 21:35
  • 2
    $\begingroup$ @JerrySchirmer That is totally different question, and it doesn't offer an answer to what I'm asking. $\endgroup$ 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. $\endgroup$ 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? $\endgroup$ Jan 22 '14 at 0:16
  • $\begingroup$ @JerrySchirmer By the way, if you write your comment as an answer I will accept it immediately. $\endgroup$ Jan 22 '14 at 0:17

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.

  • $\begingroup$ 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

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.