# Diffeomorphism in static spherically symmetric space-time

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?

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 $$(r(r'))^2$$.

• 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? – AoZora Jul 14 '19 at 8:46
• @AoZora I added a note to my post: "Implications of a static and spherically symmetric spacetime on the metric". Please refer to it. – Michele Grosso Jul 14 '19 at 19:40
• 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! – AoZora Jul 15 '19 at 7:41
• @AoZora. I added a further note to my post: "Coordinate transformation to have $A(r) = B(r)$". Please refer to it. – Michele Grosso Jul 15 '19 at 20:02
• Thank you very much! It was very simple and it was all I was missing – AoZora Jul 16 '19 at 7:25