Compactificamos la primera coordenada mediante \boxed{r = \frac{a \bar{r}}{1 - \bar{r}}}.

El Laplaciano, con esta compactificación, queda:

\Delta = \frac{(a-\bar{r})^4}{a^4} \partial_{\bar{r} \bar{r}} + \frac{2}{\bar{r}} \frac{(a-\bar{r})^4}{a^4} \partial_{\bar{r}} + \frac{(a - \bar{r})^2}{a^2 \bar{r}^2} \partial_{\theta \theta} + \frac{(a-\bar{r})^2}{a^2 \bar{r}^2} \cot \theta \partial_{\theta} + \frac{(a - \bar{r})^2}{a^2 \bar{r}^2 } \csc \theta \partial_{\varphi \varphi}

y las fuentes:

\boxed{\Delta \Theta_{X} = 6 \pi \mathcal{D}^j S^*_j}

\Delta \Theta_X = 6 \pi f^{ji} \mathcal{D}_i S^*_j = 6 \pi ( \mathcal{D}_{\bar{r}} S^*_{\bar{r}} + \mathcal{D}_{\theta} S^*_{\theta} + \mathcal{D}_{\varphi} S^*_{\varphi} ) =

= 6 \pi \frac{1-\bar{r}}{a \bar{r}} ( (\bar{r}-\bar{r}^2) \partial_{\bar{r}} S^*_{\bar{r}} + 2 S^*_{\bar{r}} + \partial_{\theta} S^*_{\theta} + \cot \theta S^*_{\theta} + \csc \theta \partial_{\varphi} S^*_{\varphi} )

\boxed{\Delta X^{i} = 8 \pi f^{ij} S^*_j - \frac{1}{3} \mathcal{D}^i \Theta_X}

Pasando la derivada contravariante a covariante mediante la métrica, queda:

\Delta X^{i} = 8 \pi f^{ij} S^*_j - \frac{1}{3} f^{ik} \mathcal{D}_k \Theta_X.

Definimos ahora

S_X^i := 8 \pi f^{ij} S^*_j - \frac{1}{3} f^{ik} \mathcal{D}_k \Theta_X,

de manera que:

S_X^{\bar{r}} = 8 \pi f^{\bar{r} j} S^*_j - \frac{1}{3} f^{\bar{r} k}\mathcal{D}_{k} \Theta_X = 8 \pi S^*_{\bar{r}} - \frac{1}{3} \mathcal{D}_{\bar{r}} \Theta_X = 8 \pi S^*_{\bar{r}} - \frac{(1-\bar{r})^2}{a^2} \partial_{\bar{r}} \Theta_X,

S_X^{\theta} = 8 \pi f^{\theta j} S^*_j - \frac{1}{3} f^{\theta k} \mathcal{D}_{k} \Theta_X = 8 \pi S^*_{\theta} - \frac{1}{3} \mathcal{D}_{\theta} \Theta_X = 8 \pi S^*_{\theta} - \frac{1-\bar{r}}{a \bar{r}} \partial_{\theta} \Theta_X,

S_X^{\varphi} = 8 \pi f^{\varphi j} S^*_j - \frac{1}{3} f^{\varphi k} \mathcal{D}^{k} \Theta_X = 8 \pi S^*_{\varphi} - \frac{1}{3} \mathcal{D}_{\varphi} \Theta_X = 8 \pi S^*_{\varphi} - \frac{a-\bar{r}}{a \bar{r}} \csc \theta \partial_{\varphi} \Theta_X.

En este punto tenemos que el vector

(S_X^{\bar{r}}(\bar{r},\theta,\varphi),S_X^{\theta}(\bar{r},\theta,\varphi),S_X^{\varphi}(\bar{r},\partial_\theta,\partial_\varphi))

expresado en la base \{ \partial_{\bar{r}}, \theta, \varphi \}. Lo que hacemos ahora es expresar este vector en la nueva base \{ \partial_x, \partial_y, \partial_z \}, de manera que obtenemos

(S_X^{x}(\bar{r},\theta,\varphi),S_X^{y}(\bar{r},\theta,\varphi),S_X^{z}(\bar{r},\theta,\varphi)).

y como es esta base las ecuaciones están desacopladas y \Theta_X es un campo escalar, resolvemos independientemente:

\Delta X^{x} = S_X^{x},

\Delta X^{y} = S_X^{y},

\Delta X^{z} = S_X^{z}.

Finalmente, con el cambio de base inverso, calculamos a partir de (X^{x},X^{y},X^{z}) el vector (X^{\bar{r}},X^\theta,X^\varphi) .

\underline{\hat{A}^{ij} = \mathcal{D}^i X^j + \mathcal{D}^j X^i - \frac{2}{3} \mathcal{D}_k X^k f^{ij}}

volvemos a pasar las derivadas contravariantes a covariantes:

\hat{A}^{ij} = f^{im} \mathcal{D}_m X^j + f^{jn} \mathcal{D}_n X^i - \frac{2}{3} f^{ij} \mathcal{D}_k X^{k}

y obtenemos:

\hat{A}^{\bar{r} \bar{r}} = f^{\bar{r} m} \mathcal{D}_m X^{\bar{r}} + f^{\bar{r} n} \mathcal{D}_n X^{\bar{r}} - \frac{2}{3} \mathcal{D}_k X^{k} = \frac{2}{3}( 2 \mathcal{D}_{\bar{r}} X^{\bar{r}} - \mathcal{D}_{\theta} X^{\theta} - \mathcal{D}_{\varphi} X^{\varphi}) =

= \frac{2(1-\bar{r})}{3a\bar{r}} [ 2(\bar{r}-\bar{r}^2)\partial_{\bar{r}} X^{\bar{r}} - 2 X^{\bar{r}} - \partial_{\theta} X^{\theta} - \cot \theta X^{\theta} - \csc \theta \partial_{\varphi} X^{\varphi} ]

\hat{A}^{\bar{r} \theta} = f^{\bar{r} m} \mathcal{D}_m X^{\theta} + f^{\theta n} \mathcal{D}_n X^{\bar{r}} = \mathcal{D}_{\bar{r}} X^{\theta} + \mathcal{D}_{\theta} X^{\bar{r}} =

= \frac{1-\bar{r}}{a\bar{r}} [ (\bar{r}-\bar{r}^2) \partial_{\bar{r}} X^{\theta} - X^{\theta} + \partial_{\theta} X^{\bar{r}} ],

\hat{A}^{\bar{r} \varphi} = f^{\bar{r} m} \mathcal{D}_m X^{\varphi} + f^{\varphi n} \mathcal{D}_n X^{\bar{r}} = \mathcal{D}_{\bar{r}} X^{\varphi} + \mathcal{D}_{\varphi} X^{\bar{r}} =

= \frac{1-\bar{r}}{a\bar{r}} [ (\bar{r}-\bar{r}^2) \partial_{\bar{r}} X^{\varphi} - X^{\varphi} + \csc \theta \partial_{\varphi} X^{\bar{r}} ],

\hat{A}^{\theta \theta} = f^{\theta m} \mathcal{D}_m X^{\theta} + f^{\theta n} \mathcal{D}_n X^{\theta} - \frac{2}{3} \mathcal{D}_k X^{k} = \frac{2}{3}( - \mathcal{D}_{\bar{r}} X^{\bar{r}} + 2 \mathcal{D}_{\theta} X^{\theta} - \mathcal{D}_{\varphi} X^{\varphi}) =

= \frac{2(1-\bar{r})}{3 a \bar{r}} [ -(\bar{r}-\bar{r}^2) \partial_{\bar{r}} X^{\bar{r}} + X^{\bar{r}} + 2 \partial_{\theta} X^{\theta} - \cot \theta X^{\theta} - \csc \partial_{\varphi} X^{\varphi} ]

\hat{A}^{\theta \varphi} = f^{\theta m} \mathcal{D}_m X^{\varphi} + f^{\varphi n} \mathcal{D}_n X^{\theta} = \mathcal{D}_{\theta} X^{\varphi} + \mathcal{D}_{\varphi} X^{\theta} =

= \frac{1-\bar{r}}{a\bar{r}} [ \partial_{\theta} X^{\varphi} - \cot \theta X^{\varphi} + \csc \theta \partial_{\varphi} X^{\theta} ],

\hat{A}^{\varphi \varphi} = f^{\varphi m} \mathcal{D}_m X^{\varphi} + f^{\varphi n} \mathcal{D}_n X^{\varphi} - \frac{2}{3} \mathcal{D}_k X^{k} = \frac{2}{3}( - \mathcal{D}_{\bar{r}} X^{\bar{r}} - \mathcal{D}_{\theta} X^{\theta} +2 \mathcal{D}_{\varphi} X^{\varphi}) =

= \frac{2(1-\bar{r})}{3 a \bar{r}} [ -(\bar{r} - \bar{r}^2) \partial_{\bar{r}} X^{\bar{r}} + X^{\bar{r}} - \partial_{\theta} X^{\theta} + 2 \cot \theta X^{\theta} + 2 \csc \theta \partial_{\varphi} T^{\varphi} ]

Las dos ecuaciones no lineales correspondientes al factor conforme \psi y al lapse \alpha, como no contienen derivadas covariantes, quedan como las teniamos:

\boxed{\Delta \psi = -2 \pi E^* \psi^{-1} - \frac{1}{8}(f_{il} f_{jm} \hat{A}^{lm} \hat{A}^{ij}) \psi^{-7} }

\boxed{\Delta (\alpha \psi) = [ 2 \pi (E^* + 2 S^*) \psi^{-7} + \frac{1}{8}(f_{il} f_{jm} \hat{A}^{lm} \hat{A}^{ij}) \psi^{-8} ] (\alpha \psi) }

Finalmente, para el shift \beta y su ecuación auxiliar tenemos:

\boxed{\Delta \Theta_{\beta} = \frac{3}{4} \mathcal{D}_i \mathcal{D}_j (2 \alpha \psi^{-6} \hat{A}^{ij} )}

\boxed{\Delta \beta^i = \mathcal{D}_j ( 2 \alpha \psi^{-6} \hat{A}^{ij} ) - \frac{1}{3} \mathcal{D}^i \Theta_{\beta} }

que lo tratamos en este post.

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