You are currently browsing the tag archive for the ‘CFC esféricas compactificadas’ tag.

La salida ahora para un tensor dos veces contravariante en la base ortonormal queda:

CovDerTen2SphCom1,

Para primera ecuación:

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

definimos como antes

V^i := \mathcal{D}_j \alpha \psi^{-6} \hat{A}^{ij},

de manera que la ecuación original la reescribimos como

\Delta \Theta_\beta = \frac{3}{2} \mathcal{D}_i V^i,

De esta manera, en nuestras coordenadas obtenemos:

\Delta \Theta_\beta = \frac{3}{2} \mathcal{D}_i V^i = \frac{3}{2} (\mathcal{D}_{\bar{r}} V^{\bar{r}} + \mathcal{D}_{\theta} V^{\theta} + \mathcal{D}_{\varphi} V^{\varphi}) =

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

donde

V^{\bar{r}} = \mathcal{D}_{\bar{r}} (\alpha \psi^{-6} \hat{A}^{\bar{r} \bar{r}}) + \mathcal{D}_{\theta} ( \alpha \psi^{-6} \hat{A}^{\bar{r} \theta} ) + \mathcal{D}_{\varphi} ( \alpha \psi^{-6} \hat{A}^{\bar{r} \varphi} ),

V^{\theta} = \mathcal{D}_{\bar{r}} (\alpha \psi^{-6} \hat{A}^{\theta \bar{r}}) + \mathcal{D}_{\theta} ( \alpha \psi^{-6} \hat{A}^{\theta \theta} ) + \mathcal{D}_{\varphi} ( \alpha \psi^{-6} \hat{A}^{\theta \varphi} ),

V^{\varphi} = \mathcal{D}_{\bar{r}} (\alpha \psi^{-6} \hat{A}^{\varphi \bar{r}}) + \mathcal{D}_{\theta} ( \alpha \psi^{-6} \hat{A}^{\varphi \theta} ) + \mathcal{D}_{\varphi} ( \alpha \psi^{-6} \hat{A}^{\varphi \varphi} ),

que desarrollando las covariantes quedan:

V^{\bar{r}} = \frac{(1 - \bar{r})^2}{a} \partial_{\bar{r}} (\alpha \psi^{-6} \hat{A}^{\bar{r} \bar{r}}) +

+ \frac{1 - \bar{r}}{a \bar{r}} [ \partial_{\theta} ( \alpha \psi^{-6} \hat{A}^{\bar{r} \theta}) + \alpha \psi^{-6} \hat{A}^{\bar{r} \bar{r}} - \alpha \psi^{-6} \hat{A}^{\theta \theta} ] +

+ \frac{1 - \bar{r}}{a \bar{r}} [ \csc \theta \partial_{\varphi} ( \alpha \psi^{-6} \hat{A}^{\bar{r} \varphi}) + \alpha \psi^{-6} \hat{A}^{\bar{r} \bar{r}} + \cot \theta \alpha \psi^{-6} \hat{A}^{\bar{r} \theta} - \alpha \psi^{-6} \hat{A}^{\varphi \varphi}] ),

V^{\theta} = \frac{(1 - \bar{r})^2}{a} \partial_{\bar{r}} (\alpha \psi^{-6} \hat{A}^{\theta \bar{r}}) +

+ \frac{1 - \bar{r}}{a \bar{r}} [ \partial_{\theta} ( \alpha \psi^{-6} \hat{A}^{\theta \theta}) + 2 \alpha \psi^{-6} \hat{A}^{\bar{r} \theta} ] +

+ \frac{1 - \bar{r}}{a \bar{r}} [ \csc \theta \partial_{\varphi} ( \alpha \psi^{-6} \hat{A}^{\theta \varphi}) + \alpha \psi^{-6} \hat{A}^{\bar{r} \theta} + \cot \theta \alpha \psi^{-6} \hat{A}^{\theta \theta} - \cot \theta \alpha \psi^{-6} \hat{A}^{\varphi \varphi} ] ),

V^{\varphi} = \frac{(1 - \bar{r})^2}{a} \partial_{\bar{r}} (\alpha \psi^{-6} \hat{A}^{\bar{r} \bar{r}}) +

+ \frac{1 - \bar{r}}{a \bar{r}} [ \partial_{\theta} ( \alpha \psi^{-6} \hat{A}^{\bar{r} \theta}) + \alpha \psi^{-6} \hat{A}^{\bar{r} \bar{r}} - \alpha \psi^{-6} \hat{A}^{\theta \theta} ] +

+ \frac{1 - \bar{r}}{a \bar{r}} [ \csc \theta \partial_{\varphi} ( \alpha \psi^{-6} \hat{A}^{\bar{r} \varphi}) + \alpha \psi^{-6} \hat{A}^{\bar{r} \bar{r}} + \cot \theta \alpha \psi^{-6} \hat{A}^{\bar{r} \theta} - \alpha \psi^{-6} \hat{A}^{\varphi \varphi}] ),

que combinandolo con la anterior, queda (solo escribimos como empezaría debido a la longitud de la ecuación):

\Delta \Theta_\beta = \frac{3 - 3\bar{r} }{2a \bar{r}} \Big [ (\bar{r}-\bar{r}^2) \partial_{\bar{r}} [\frac{(1 - \bar{r})^2}{a} \partial_{\bar{r}} (\alpha \psi^{-6} \hat{A}^{\bar{r} \bar{r}}) + \ldots ] + \ldots \Big ]

Finalmente, las ecuaciones:

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

con las que procederemos de manera similar a como hemos hecho con las X^i, es decir, calculando las fuentes en una base, haciendo un cambio de base que las desacople (cartesianas), resolviendolas de manera independiente y volviendo a la base original:

S^i_\beta (\bar{r},\theta,\varphi) := 2\mathcal{D}_j ( \alpha \psi^{-6} \hat{A}^{ij} ) - \frac{1}{3} \mathcal{D}_i \Theta_{\beta},

que quedan:

S^{\bar{r}}_\beta= 2 \big [ \mathcal{D}_{\bar{r}} (\alpha \psi^{-6} \hat{A}^{\bar{r} \bar{r}}) + \mathcal{D}_{\theta} (\alpha \psi^{-6} \hat{A}^{\bar{r} \theta}) + \mathcal{D}_{\varphi} (\alpha \psi^{-6} \hat{A}^{\bar{r} \varphi}) \big ] - \frac{1}{3} \mathcal{D}_{\bar{r}} \Theta_\beta,

S^{\theta}_\beta = 2 \big [ \mathcal{D}_{\bar{r}} (\alpha \psi^{-6} \hat{A}^{\theta \bar{r}}) + \mathcal{D}_{\theta} (\alpha \psi^{-6} \hat{A}^{\theta \theta}) + \mathcal{D}_{\varphi} (\alpha \psi^{-6} \hat{A}^{\theta \varphi}) \big ] - \frac{1}{3} \mathcal{D}_{\theta} \Theta_\beta,

S^{\varphi}_\beta = 2 \big [ \mathcal{D}_{\bar{x}} (\alpha \psi^{-6} \hat{A}^{\bar{y} \bar{x}}) + \mathcal{D}_{\bar{y}} (\alpha \psi^{-6} \hat{A}^{\bar{y} \bar{y}}) + \mathcal{D}_{\bar{z}} (\alpha \psi^{-6} \hat{A}^{\bar{y} \bar{z}}) \big ] - \frac{1}{3} \mathcal{D}_{\bar{y}} \Theta_\beta,

donde las derivadas covariantes del tensor dos veces contravariante:

T^{ij}:=\alpha \psi^{-6} \hat{A}^{ij}

son como acabamos de hacer en la ecuación anterior y las del escalar \Theta_\beta es como ya hicimos con las X^i:

S^{\bar{r}} = 2 V^{\bar{r}} - \frac{(1-\bar{r})^2}{3a} \partial_{\bar{r}} \Theta_{\beta},

S^{\theta} = 2 V^{\theta} - \frac{1-\bar{r}}{3a\bar{r}} \partial_{\theta} \Theta_{\beta},

S^{\varphi} = 2 V^{\varphi} - \frac{1-\bar{r}}{3a\bar{r}} \csc \theta \partial_{\varphi} \Theta_{\beta}.

Hacemos a continuación el cambio:

[S^{\bar{r}}(\bar{r},\theta,\varphi),S^{\theta}(\bar{r},\theta,\varphi),S^{\varphi}(\bar{r},\theta,\varphi)] \rightarrow

\rightarrow [S^x(\bar{r},\theta,\varphi), S^y(\bar{r},\theta,\varphi), S^z(\bar{r},\theta,\varphi)],

y resolvemos:

\Delta \beta^{x} = S^{x}

\Delta \beta^{y} = S^{y}

\Delta \beta^{z} = S^{z},

deshaciendo el cambio:

[\beta^x(\bar{r},\theta,\varphi), \beta^y(\bar{r},\theta,\varphi), \beta^z(\bar{r},\theta,\varphi)] \rightarrow

\rightarrow [\beta^{\bar{r}}(\bar{r},\theta,\varphi),\beta^{\theta}(\bar{r},\theta,\varphi),\beta^{\varphi}(\bar{r},\theta,\varphi)]

para terminar.

Anuncios

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.

septiembre 2017
L M X J V S D
« Ago    
 123
45678910
11121314151617
18192021222324
252627282930