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Considerando las dos bases, holonómica y ortonormal, para el espacio tangente introducidas en este post:

\{ e_i \} = \{ \partial_{\bar{r}}, \partial_{\theta}, \partial_{\varphi} \},

\{ \hat{e}_i \} = \{ \frac{(a-\bar{r})^2}{a^2}\partial_{\bar{r}}, \frac{a-\bar{r}}{a\bar{r}} \partial_{\theta}, \frac{a-\bar{r}}{a\bar{r}} \csc \theta \partial_{\varphi} \},

tenemos que:

\hat{e}_1 = \frac{(a-\bar{r})^2}{a^2}e_1, \hat{e}_2 = \frac{a-\bar{r}}{a\bar{r}} e_2 y \hat{e}_3 = \frac{a-\bar{r}}{a\bar{r}} \csc \theta e_3,

y, por tanto:

A_{\hat{i}}^{i} = \left(  \begin{array}{ccc} \frac{(a-\bar{r})^2}{a^2} & 0 & 0 \\  0 & \frac{a-\bar{r}}{a\bar{r}} & 0 \\  0 & 0 & \frac{a-\bar{r}}{a\bar{r}} \csc \theta \end{array}  \right) i A_{i}^{\hat{i}} = \left( \begin{array}{ccc}  \frac{a^2}{(a-\bar{r})^2} & 0 & 0 \\  0 & \frac{a\bar{r}}{a-\bar{r}} & 0 \\  0 & 0 & \frac{a\bar{r}}{a-\bar{r}} \sin \theta  \end{array}  \right).

Vamos a encontrar la expresión de las derivadas covariante en la base normalizada mediante cambios de base. Antes de empezar, dos consideraciones: en primer lugar, T^i y T^{\hat{i}} son tensores una vez contravariante, o lo que es lo mismo, son campos vectoriales. Por la linealidad de la conexión, podemos centrarnos en la derivada covariante de los elementos de la base. En segundo lugar, las derivadas parciales no afectan a los escalares, que están sobre la variedad y no en el espacio tangente.

Empezamos:

\mathcal{D}_{\bar{r}} T^{\bar{r}} = \mathcal{D}_{\hat{\bar{r}}} T^{\hat{\bar{r}}} A_{\bar{r}}^{\hat{\bar{r}}} A_{\hat{\bar{r}}}^{\bar{r}} = \mathcal{D}_{\hat{\bar{r}}} T^{\hat{\bar{r}}},

\partial_{\bar{r}} T^{\bar{r}} + \frac{2}{a-\bar{r}} T^{\bar{r}} = \partial_{\bar{r}} (T^{\hat{\bar{r}}} A_{\hat{\bar{r}}}^{\bar{r}}) + \frac{2}{a-\bar{r}} T^{\hat{\bar{r}}} A_{\hat{\bar{r}}}^{\bar{r}} = \partial_{\bar{r}}(\frac{(a-\bar{r})^2}{a^2} T^{\hat{\bar{r}}}) + \frac{2}{a-\bar{r}} \frac{(a-\bar{r})^2}{a^2} T^{\hat{\bar{r}}} =

= -\frac{2(a-\bar{r})}{a^2} T^{\hat{\bar{r}}} + \frac{(a-\bar{r})^2}{a^2} \partial_{\hat{\bar{r}}} T^{\hat{\bar{r}}} + \frac{2(a-\bar{r})}{a^2} T^{\hat{\bar{r}}} = \frac{(a-\bar{r})^2}{a^2} \partial_{\hat{\bar{r}}} T^{\hat{\bar{r}}},

con lo que:

\boxed{\mathcal{D}_{\hat{\bar{r}}} T^{\hat{\bar{r}}} = \frac{(a-\bar{r})^2}{a^2} \partial_{\hat{\bar{r}}} T^{\hat{\bar{r}}}}.

Para:

\mathcal{D}_{\bar{r}} T^{\theta} = \mathcal{D}_{\hat{\bar{r}}} T^{\hat{\theta}} A_{\bar{r}}^{\hat{\bar{r}}} A_{\hat{\theta}}^{\theta} = \frac{a^2}{(a-\bar{r})^2} \frac{a-\bar{r}}{a\bar{r}} \mathcal{D}_{\hat{\bar{r}}} T^{\hat{\theta}} = \frac{a}{a\bar{r}-\bar{r}^2} \mathcal{D}_{\hat{\bar{r}}} T^{\hat{\theta}},

\partial_{\bar{r}} T^{\theta} + \frac{a}{a\bar{r}-\bar{r}^2} T^{\theta} = \partial_{\bar{r}} (T^{\hat{\theta}} A_{\hat{\theta}}^{\theta}) + \frac{a}{a\bar{r}-\bar{r}} T^{\hat{\theta}} A_{\hat{\theta}}^{\theta} = \partial_{\bar{r}} (\frac{a-\bar{r}}{a\bar{r}} T^{\hat{\theta}}) + \frac{a}{a\bar{r} -\bar{r}} \frac{a-\bar{r}}{a\bar{r}} T^{\hat{\theta}} =

= \partial_{\bar{r}} (\frac{a-\bar{r}}{a \bar{r}}) T^{\hat{\theta}} + \frac{a-\bar{r}}{a \bar{r}} \partial_{\hat{\bar{r}}} T^{\hat{\theta}} + \frac{1}{\bar{r}^2} T^{\hat{\theta}} = -\frac{1}{\bar{r}^2} T^{\hat{\theta}} + \frac{a-\bar{r}}{a \bar{r}} \partial_{\hat{\bar{r}}} T^{\hat{\theta}} + \frac{1}{\bar{r}^2} T^{\hat{\theta}} = \frac{a-\bar{r}}{a \bar{r}} \partial_{\hat{\bar{r}}} T^{\hat{\theta}},

tenemos:

\frac{a}{a\bar{r}-\bar{r}^2} \mathcal{D}_{\hat{\bar{r}}} T^{\hat{\theta}} = \frac{a-\bar{r}}{a \bar{r}} \partial_{\hat{\bar{r}}} T^{\hat{\theta}} \Leftrightarrow \boxed{\mathcal{D}_{\hat{\bar{r}}} T^{\hat{\theta}} = \frac{(a-\bar{r})^2}{a^2} \partial_{\hat{\bar{r}}} T^{\hat{\theta}}}

Procediendo de la misma manera, obtenemos:

\boxed{\mathcal{D}_{\hat{\bar{r}}} T^{\hat{\varphi}} = \frac{(a-\bar{r})^2}{a^2} \partial_{\hat{\bar{r}}} T^{\hat{\varphi}} },

\boxed{\mathcal{D}_{\hat{\theta}} T^{\hat{\bar{r}}} = \frac{a-\bar{r}}{a \bar{r}} [ \partial_{\hat{\theta}} T^{\hat{\bar{r}}} - T^{\hat{\theta}} ] },

\boxed{\mathcal{D}_{\hat{\theta}} T^{\hat{\theta}} = \frac{a-\bar{r}}{a \bar{r}} [ \partial_{\hat{\theta}} T^{\hat{\theta}} + T^{\hat{\bar{r}}} ] },

\boxed{\mathcal{D}_{\hat{\theta}} T^{\hat{\varphi}} = \frac{a-\bar{r}}{a \bar{r}} \partial_{\hat{\theta}} T^{\hat{\varphi}} },

\boxed{\mathcal{D}_{\hat{\varphi}} T^{\hat{\bar{r}}} = \frac{a-\bar{r}}{a \bar{r}} \csc \theta [ \partial_{\hat{\varphi}} T^{\hat{\bar{r}}} - \sin \theta T^{\hat{\varphi}}] },

\boxed{\mathcal{D}_{\hat{\varphi}} T^{\hat{\theta}} = \frac{a-\bar{r}}{a \bar{r}} \csc \theta [ \partial_{\hat{\varphi}} T^{\hat{\theta}} - \cos \theta T^{\hat{\varphi}}] },

\boxed{\mathcal{D}_{\hat{\varphi}} T^{\hat{\varphi}} = \frac{a-\bar{r}}{a \bar{r}} \csc \theta [ \partial_{\hat{\varphi}} T^{\hat{\varphi}} + \sin \theta T^{\hat{\bar{r}}} + \cos \theta T^{\hat{\theta}} }.

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septiembre 2017
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