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Compactificamos mediante \boxed{x_i = a_i \, \mbox{arctanh} \, \bar{x}_i }.

En este caso, el operador Laplaciano de todas las ecuaciones queda:

\Delta = \frac{(\bar{x}^2 - 1)^2}{a^2} \partial_{\bar{x} \bar{x}} + \frac{2 \bar{x}(\bar{x}^2 - 1)}{a^2} \partial_{\bar{x}} +

+ \frac{(\bar{y}^2 - 1)^2}{b^2} \partial_{\bar{y} \bar{y}} + \frac{2 \bar{y}(\bar{y}^2 - 1)}{b^2} \partial_{\bar{y}} +

+ \frac{(\bar{z}^2 - 1)^2}{c^2} \partial_{\bar{z} \bar{z}} + \frac{2 \bar{z}(\bar{z}^2 - 1)}{c^2} \partial_{\bar{z}}

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{x}} S^*_{\bar{x}} + \mathcal{D}_{\bar{y}} S^*_{\bar{y}} + \mathcal{D}_{\bar{z}} S^*_{\bar{z}} ) =

= 6 \pi ( \frac{|\bar{x}^2-1|}{a} \partial_{\bar{x}} S^*_{\bar{x}} + \frac{|\bar{y}^2-1|}{b} \partial_{\bar{y}} S^*_{\bar{y}}+ \frac{|\bar{z}^2-1|}{c} \partial_{\bar{z}} S^*_{\bar{z}})

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

Pasamos la derivada contravariante a covariante mediante la métrica:

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

y como \Theta_X es un campo escalar, nos queda:

\Delta X^{\bar{x}} = 8 \pi f^{\bar{x} j} S^*_j - \frac{1}{3} f^{\bar{x} k}\mathcal{D}_{k} \Theta_X = 8 \pi S^*_{\bar{x}} - \frac{1}{3} \mathcal{D}_{\bar{x}} \Theta_X =

= 8 \pi S^*_{\bar{x}} - \frac{1}{3} \frac{|\bar{x}^2 - 1|}{a} \partial_{\bar{x}} \Theta_X,

\Delta X^{\bar{y}} = 8 \pi f^{\bar{y} j} S^*_j - \frac{1}{3} f^{\bar{y} k} \mathcal{D}_{k} \Theta_X = 8 \pi S^*_{\bar{y}} - \frac{1}{3} \mathcal{D}_{\bar{y}} \Theta_X =

= 8 \pi S^*_{\bar{y}} - \frac{1}{3} \frac{|\bar{y}^2 - 1|}{b} \partial_{\bar{y}} \Theta_X,

\Delta X^{\bar{z}} = 8 \pi f^{\bar{z} j} S^*_j - \frac{1}{3} f^{\bar{z} k} \mathcal{D}^{k} \Theta_X = 8 \pi S^*_{\bar{z}} - \frac{1}{3} \mathcal{D}_{\bar{z}} \Theta_X =

= 8 \pi S^*_{\bar{z}} - \frac{1}{3} \frac{|\bar{z}^2 - 1|}{c} \partial_{\bar{z}} \Theta_X.

\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{x} \bar{x}} = f^{\bar{x} m} \mathcal{D}_m X^{\bar{x}} + f^{\bar{x} n} \mathcal{D}_n X^{\bar{x}} - \frac{2}{3} \mathcal{D}_k X^{k} = \frac{2}{3}( 2 \mathcal{D}_{\bar{x}} X^{\bar{x}} - \mathcal{D}_{\bar{y}} X^{\bar{y}} - \mathcal{D}_{\bar{z}} X^{\bar{z}}) =

= \frac{2}{3}(\frac{2|\bar{x}^2-1|}{a} \partial_{\bar{x}} X^{\bar{x}} - \frac{|\bar{y}^2-1|}{b} \partial_{\bar{y}} X^{\bar{y}} - \frac{|\bar{z}^2-1|}{c} \partial_{\bar{z}} X^{\bar{z}}),

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

= \frac{|\bar{x}^2-1|}{a} \partial_{\bar{x}} X^{\bar{y}} + \frac{|\bar{y}^2 - 1|}{b} \partial_{\bar{y}} X^{\bar{x}},

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

= \frac{|\bar{x}^2-1|}{a} \partial_{\bar{x}} X^{\bar{z}} + \frac{|\bar{z}^2 - 1|}{c} \partial_{\bar{z}} X^{\bar{x}},

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

= \frac{2}{3}(-\frac{|\bar{x}^2-1|}{a} \partial_{\bar{x}} X^{\bar{x}} + \frac{2|\bar{y}^2-1|}{b} \partial_{\bar{y}} X^{\bar{y}} - \frac{|\bar{z}^2-1|}{c} \partial_{\bar{z}} X^{\bar{z}}),

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

= \frac{|\bar{y}^2-1|}{b} \partial_{\bar{y}} X^{\bar{z}} + \frac{|\bar{z}^2 - 1|}{c} \partial_{\bar{z}} X^{\bar{y}},

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

= \frac{2}{3}(-\frac{|\bar{x}^2-1|}{a} \partial_{\bar{x}} X^{\bar{x}} - \frac{|\bar{y}^2-1|}{b} \partial_{\bar{y}} X^{\bar{y}} + \frac{2|\bar{z}^2-1|}{c} \partial_{\bar{z}} X^{\bar{z}}).

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 las tratamos es este nuevo post.

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