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In a static, spherically symmetric space time we can choose the coordinates so that the metric takes the form: $$ds^2=-A(r)dt^2+\frac{dr^2}{B(r)}+C(r)\,[d\theta^2+\sin^2\theta\,d\varphi^2]$$ Sometimes we use a radial diffeomorphism $\;r\mapsto r'\;\;$ setting $\;\;C(r)=r'^2\;\;$.

The fact that makes this possible is the interdependence of the non trivial Einstein equations:

$$R^t_t=0\qquad R^r_r=0\qquad R^{\theta}_{\theta}=0$$

(for instance see this former question and the accepted answer).

For this reason it is clear that we can set one of the three functions $A\,,\,B\,,\,C\;$ to whatever value and still find solutions.

On the other hand it can be more difficult to understand which diffeomorphism brings one of the three functions to a chosen form. In particular in the case of the choice $$A(r)=B(r)$$ We would need a radial diffeomorphism $r \,\mapsto\,r'=u(r)$ so that $$A(u^{-1}(r'))=B(u^{-1}(r'))$$ Which does not look generally possible if the diffeomorphism involves only the radial coordinate.

In conclusion, which diffeomorphism realizes the choice?

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A diffeomorphism and a coordinate transformation are two different ways of doing precisely the same thing. A diffeomorphism is an active coordinate transformation, while a traditional coordinate transformation is passive. In the former you move the points on the manifold and then evaluate the coordinates of the new points; in the latter you keep the manifold fixed and change the coordinate map.

As for the derivation of the Schwarzschild metric, i.e. a spacetime shaped by a static and spherically symmetric massive object, the first step is a coordinate change in the radial coordinate $r$ that makes the factor multiplying the $r^2 d\Omega^2$ reduce to unity.

However the second step is not a coordinate transformation, but you have to apply the Einstein field equations in vacuum (the energy-momentum tensor is zero outside the massive body). That is, you have to calculate the Ricci tensor $R_{\mu \nu}$ expressed against the remaining two funtions, set its components to zero and thus get the formulation of the functions in terms of the radial coordinate.

Note: Implications of a static and spherically symmetric spacetime on the metric
Let us consider spherical coordinates $(t, r, \theta, \phi)$. The metric tensor $g_{\mu \nu}$ in presence of a static and spherically symmetric spacetime can be specialized as: 1) Static: $g_{t \nu} = 0$ for $\nu \ne t$; 2) Spherically symmetric: $g_{a b} = 0$ for $a \ne b$ with $a, b = r, \theta, \phi$. So, we remain with the three functions of the radial coordinate $A(r), B(r), C(r)$. The symmetry can not help more. As for $A(r)$ and $B(r)$ no symmetry can require they are equal. In fact $A(r)$ is related to the spacetime interval $ds$ along $dt$, that is the proper time, while $B(r)$ is related to the spacetime interval $ds$ along $dr$, that is the proper distance. It matters of two different physical measures!

Note: Coordinate transformation to have $A(r) = B(r)$
If you want to work on $A(r)$ and $B(r)$ you assume a coordinate transformation $r = r(r')$ such that $A(r(r')) = B(r(r')) / (dr(r')/dr')$. In this way in the new radial coordinate $r'$ you read A'(r') = B'(r'), where the latter includes the coordinate change of the differential. Of course the factor multiplying $d\Omega^2$ is $(r(r'))^2$.

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  • $\begingroup$ Thank you for the answer. I think this is clear. What I am not getting is which initial coordinate redefinition can give you A=B (in the notation of the post) to begin with. You can do it based on the fact that you have only two non trivial equations, sure; but this is due to the symmetries. Therefore I would expect to be able to exploit symmetriesat the level of changes of coordinates as well and start with coordinates for which A=B before solving the equations. If you are saying that this is not possible could you provide an argument? $\endgroup$ – AoZora Jul 14 at 8:46
  • $\begingroup$ @AoZora I added a note to my post: "Implications of a static and spherically symmetric spacetime on the metric". Please refer to it. $\endgroup$ – Michele Grosso Jul 14 at 19:40
  • $\begingroup$ Maybe the way I phrased my previous comment was misleading. I agree that nothing forces A to be equal to B. The fact is that you can choose them to be if you don't care to have $C=r^2$ . This choice is possible because you have two equations, and I am wondering if it can be interpreted in terms of a diffeomorphism that respects the symmetry of the system, like when we set $C=r^2$ before solving the equations . I hope this clarifies why I think your answer still does not address my point (maybe I am missing something?) . Thank you anyway for the effort! $\endgroup$ – AoZora Jul 15 at 7:41
  • $\begingroup$ @AoZora. I added a further note to my post: "Coordinate transformation to have $A(r) = B(r)$". Please refer to it. $\endgroup$ – Michele Grosso Jul 15 at 20:02
  • $\begingroup$ Thank you very much! It was very simple and it was all I was missing $\endgroup$ – AoZora Jul 16 at 7:25

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