- #1
Oxymoron
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I posted a thread in the Homework section on my attempt to find the Schwarzschild solution using Cartan's method instead of the orthodox Christoffel symbol method. Unfortunately I wasn't getting any help
Then I was asked to move the thread to this section because I may get more attention, but this might infringe upon the rule that no homework be posted here. So I will keep it brief:
In that thread (which you are welcome to check out - for my working and stuff) I found 4 equations:
[tex]\mbox{d}\varepsilon^0 = \frac{}{}\varepsilon^1\wedge\varepsilon^0 = -\omega_{10}\wedge\varepsilon^1 - \omega_{20}\wedge\varepsilon^2 - \omega_{30}\wedge\varepsilon^3 [1][/tex]
[tex]\mbox{d}\varepsilon^1 = 0 = -\omega_{12}\wedge\varepsilon^2 - \omega_{13}\wedge\varepsilon^3 - \omega_{10}\wedge\varepsilon^0\quad [2][/tex]
[tex]\mbox{d}\varepsilon^2 = \frac{1}{rH_1(r)}\varepsilon^1\wedge\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 = \frac{1}{rH_1(r)}\varepsilon^1 \wedge \varepsilon^3 + \frac{\cot\theta}{r}\varepsilon^2 \wedge \varepsilon^3 = \omega_{13}\wedge\varepsilon^1 + \omega_{23}\wedge\varepsilon^2 - \omega_{30}\wedge\varepsilon^0\quad [4][/tex]
The problem I am having is that from this information I should be able to find 6 independent connection 1-forms. The advice I have been given is this: "It should be obvious from [4] that [itex]\omega_{30} = \Gamma_{300}\varepsilon^0[/itex] and substituting this into [1] we have [itex]\Gamma_{300} = 0[/itex] since it is the sole coefficient of the 2-form basis element [itex]\varepsilon^3 \wedge \varepsilon^0[/itex]" and "Similarly, from [1] we have [itex]\omega_{20} = \Gamma_{202}\varepsilon^2[/itex]"
Unfortunately, this is not obvious to me :redface". And apparently we can put all this together to make
[tex]\omega_{10} = \frac{H_0(r)'}{H_1(r)}\varepsilon^0 + \Gamma_{101}\varepsilon^1[/tex]
If it is possible for anyone to explain to me where they got this connection 1-form from I would be sooo happy and relieved!
Then I was asked to move the thread to this section because I may get more attention, but this might infringe upon the rule that no homework be posted here. So I will keep it brief:
In that thread (which you are welcome to check out - for my working and stuff) I found 4 equations:
[tex]\mbox{d}\varepsilon^0 = \frac{}{}\varepsilon^1\wedge\varepsilon^0 = -\omega_{10}\wedge\varepsilon^1 - \omega_{20}\wedge\varepsilon^2 - \omega_{30}\wedge\varepsilon^3 [1][/tex]
[tex]\mbox{d}\varepsilon^1 = 0 = -\omega_{12}\wedge\varepsilon^2 - \omega_{13}\wedge\varepsilon^3 - \omega_{10}\wedge\varepsilon^0\quad [2][/tex]
[tex]\mbox{d}\varepsilon^2 = \frac{1}{rH_1(r)}\varepsilon^1\wedge\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 = \frac{1}{rH_1(r)}\varepsilon^1 \wedge \varepsilon^3 + \frac{\cot\theta}{r}\varepsilon^2 \wedge \varepsilon^3 = \omega_{13}\wedge\varepsilon^1 + \omega_{23}\wedge\varepsilon^2 - \omega_{30}\wedge\varepsilon^0\quad [4][/tex]
The problem I am having is that from this information I should be able to find 6 independent connection 1-forms. The advice I have been given is this: "It should be obvious from [4] that [itex]\omega_{30} = \Gamma_{300}\varepsilon^0[/itex] and substituting this into [1] we have [itex]\Gamma_{300} = 0[/itex] since it is the sole coefficient of the 2-form basis element [itex]\varepsilon^3 \wedge \varepsilon^0[/itex]" and "Similarly, from [1] we have [itex]\omega_{20} = \Gamma_{202}\varepsilon^2[/itex]"
Unfortunately, this is not obvious to me :redface". And apparently we can put all this together to make
[tex]\omega_{10} = \frac{H_0(r)'}{H_1(r)}\varepsilon^0 + \Gamma_{101}\varepsilon^1[/tex]
If it is possible for anyone to explain to me where they got this connection 1-form from I would be sooo happy and relieved!
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