Schwarzschild Solution using Cartan's Formalism

[Oxymoron]

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!

 

[Oxymoron]

EDIT! Equation 1 is meant to read

[tex]\mbox{d}\varepsilon^0 = \frac{H_0(r)'}{H_1(r)}\varepsilon^1\wedge\varepsilon^0 = ...[/tex]

 

[Oxymoron]

Ok I think I am part of the way to understanding it:

From [4], for some unknown reason, we have

[tex]\omega_{30} = \Gamma_{300}\varepsilon^0[/tex]

Similarly from [3], we have

[tex]\omega_{20} = \Gamma_{200}\varepsilon^0[/tex]

Substituting into [1] we have

[tex]\frac{H_0(r)'}{H_1(r)}\varepsilon^1 \wedge\varepsilon^0 = -\omega_{10}\wedge\varepsilon^1 - \Gamma_{200}\varepsilon^0\wedge\varepsilon^2 - \Gamma_{300}\varepsilon^0\wedge\varepsilon^3[/tex]

[tex]\frac{H_0(r)'}{H_1(r)}\varepsilon^1 \wedge \varepsilon^0 = -\omega_{10} \wedge \varepsilon^1[/tex]

[tex]\omega_{10} = \frac{H_0(r)'}{H_1(r)}\varepsilon^0[/tex]

How does this look? Besides for the fact that I don't know why [itex]\omega_{30} = \Gamma_{300}\varepsilon^0[/itex]

 

[Oxymoron]

Does anyone know how to calculate this:

[tex]\mbox{d}\left(-\frac{1}{r}\frac{1}{H_1(r)}\varepsilon^2\right)[/tex]

I am currently trying to work out the Connection two-forms. I have

[tex]\omega_{12} = \frac{1}{r}\frac{1}{H_1(r)}\varepsilon^2[/tex]

Then from Cartan's second structural equation we have

[tex]R_{12} = \mbox{d}\omega_{12} + \omega_{13}\wedge\omega_{32} + \omega_{10}\wedge\omega_{02}[/tex]

the last two terms vanish and we are left with

[tex]R_{12} = \mbox{d}\left(-\frac{1}{r}\frac{1}{H_1(r)}\varepsilon^2\right)[/tex]

I used Maple 10 and tried differentiating this. The peculiar thing is this: If I made

[tex]H_1(r) = \frac{1}{\exp\left(-\frac{1}{2}\lambda(r)\right)}[/tex]

then

[tex]R_{12} = \mbox{d}\left(-\frac{1}{r}\frac{1}{H_1(r)}\varepsilon^2\right) = \frac{1}{r}\frac{1}{H_1(r)^2}r'\varepsilon^1 \wedge \varepsilon^2[/tex]

which is the answer I want! BUT if I differentiate manually I keep getting:

[tex]\mbox{d}\left(-\frac{1}{r}\frac{1}{H_1(r)}\right) = \left(\frac{1}{r^2}\frac{1}{H_1(r)} + \frac{1}{2}\frac{1}{r}\frac{1}{H_1(r)}r'\right)\varepsilon^1 \wedge \varepsilon^2 + \left(-\frac{1}{r}\frac{1}{H_1(r)}\right)\mbox{d}\varepsilon^2[/tex]

[tex]= \left(\frac{1}{r^2}\frac{1}{H_1(r)} + \frac{1}{2}\frac{1}{r}\frac{1}{H_1(r)}r'\right)\varepsilon^1 \wedge \varepsilon^2 - \frac{1}{r^2}\frac{1}{H_1(r)^2}\varepsilon^1 \wedge \varepsilon^2[/tex]

since [itex]\mbox{d}\varepsilon^2 = \frac{1}{rH_1(r)}\varepsilon^1\wedge\varepsilon^2[/itex].

[tex]=\left(\frac{1}{r}\frac{1}{H_1(r)^2}\left[\frac{1}{r} + \frac{1}{2}r' - \frac{1}{r}\frac{1}{H_1(r)}\right]\right)\varepsilon^1\wedge\varepsilon^2[/tex]

when I really want this bit to look like:

[tex]=\left(\frac{1}{r}\frac{1}{H_1(r)^2}\left[\frac{1}{r} + r' - \frac{1}{r}\right]\right)\varepsilon^1\wedge\varepsilon^2[/tex]

Notice the subtle difference? Because then

[tex]=\left(\frac{1}{r}\frac{1}{H_1(r)^2}r'\right)\varepsilon^1\wedge\varepsilon^2[/tex]

Which is the correct answer.

Please help me!