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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?

AoZora
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1 Answers1

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

Michele Grosso
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