- #1
Oxymoron
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Im having some trouble coming up with my six independent connection 1-forms.
I have been given a metric:
[tex]g = -H_0(r)^2dt\otimes dt + H_1(r)^2 dr\otimes dr + r^2 d\theta\otimes d\theta + r^2\sin^2\theta d\phi \otimes d\phi[/tex].
I need to find [itex]H_0(r)[/itex] and [itex]H_1(r)[/itex], which are functions of r and not t, so the solutions are static. I must calculate everything using Cartan's formalism.
So the first thing I did was choose my orthonormal basis:
[tex]e_0 = \frac{1}{H_0(r)}\partial_t \quad e_1 = \frac{1}{H_1(r)}\partial_r \quad e_2 = \frac{1}{r}\partial_{\theta} \quad e_3 = \frac{1}{r\sin\theta}\partial_{\phi}[/tex]
so that my dual basis is:
[tex]\varepsilon^0 = H_0(r)\mbox{d}t \quad \varepsilon^1 = H_1(r)\mbox{d}r \quad \varepsilon^2 = r\mbox{d}\theta \quad \varepsilon^3 = r\sin\theta\mbox{d}\phi[/tex]
Now, using Cartan's structural relations I calculated:
[tex]\mbox{d}\varepsilon^0 = -\omega_{10}\wedge\varepsilon^1 - \omega_{20}\wedge\varepsilon^2 - \omega_{30}\wedge \varepsilon^3\quad [1][/tex]
[tex]\mbox{d}\varepsilon^1 = -\omega_{12}\wedge\varepsilon^2 - \omega_{13}\wedge\varepsilon^3 - \omega_{10}\wedge\varepsilon^0\quad [2][/tex]
[tex]\mbox{d}\varepsilon^2 = \omega_{12}\wedge\varepsilon^1 - \omega_{23}\wedge\varepsilon^3 - \omega_{20}\wedge \varepsilon^0\quad [3][/tex]
[tex]\mbox{d}\varepsilon^3 = \omega_{13}\wedge\varepsilon^1 + \omega_{23}\wedge\varepsilon^2 - \omega_{30}\wedge\varepsilon^0\quad [4][/tex]
But now I am stuck. I should be able to find 6 independent connection 1-forms but I don't know how to simplify all the above equations. Any guidance from here would be very helpful.
I have been given a metric:
[tex]g = -H_0(r)^2dt\otimes dt + H_1(r)^2 dr\otimes dr + r^2 d\theta\otimes d\theta + r^2\sin^2\theta d\phi \otimes d\phi[/tex].
I need to find [itex]H_0(r)[/itex] and [itex]H_1(r)[/itex], which are functions of r and not t, so the solutions are static. I must calculate everything using Cartan's formalism.
So the first thing I did was choose my orthonormal basis:
[tex]e_0 = \frac{1}{H_0(r)}\partial_t \quad e_1 = \frac{1}{H_1(r)}\partial_r \quad e_2 = \frac{1}{r}\partial_{\theta} \quad e_3 = \frac{1}{r\sin\theta}\partial_{\phi}[/tex]
so that my dual basis is:
[tex]\varepsilon^0 = H_0(r)\mbox{d}t \quad \varepsilon^1 = H_1(r)\mbox{d}r \quad \varepsilon^2 = r\mbox{d}\theta \quad \varepsilon^3 = r\sin\theta\mbox{d}\phi[/tex]
Now, using Cartan's structural relations I calculated:
[tex]\mbox{d}\varepsilon^0 = -\omega_{10}\wedge\varepsilon^1 - \omega_{20}\wedge\varepsilon^2 - \omega_{30}\wedge \varepsilon^3\quad [1][/tex]
[tex]\mbox{d}\varepsilon^1 = -\omega_{12}\wedge\varepsilon^2 - \omega_{13}\wedge\varepsilon^3 - \omega_{10}\wedge\varepsilon^0\quad [2][/tex]
[tex]\mbox{d}\varepsilon^2 = \omega_{12}\wedge\varepsilon^1 - \omega_{23}\wedge\varepsilon^3 - \omega_{20}\wedge \varepsilon^0\quad [3][/tex]
[tex]\mbox{d}\varepsilon^3 = \omega_{13}\wedge\varepsilon^1 + \omega_{23}\wedge\varepsilon^2 - \omega_{30}\wedge\varepsilon^0\quad [4][/tex]
But now I am stuck. I should be able to find 6 independent connection 1-forms but I don't know how to simplify all the above equations. Any guidance from here would be very helpful.
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