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:

$$\mbox{d}\varepsilon^0 = \frac{}{}\varepsilon^1\wedge\varepsilon^0 = -\omega_{10}\wedge\varepsilon^1 - \omega_{20}\wedge\varepsilon^2 - \omega_{30}\wedge\varepsilon^3 [1]$$

$$\mbox{d}\varepsilon^1 = 0 = -\omega_{12}\wedge\varepsilon^2 - \omega_{13}\wedge\varepsilon^3 - \omega_{10}\wedge\varepsilon^0\quad [2]$$

$$\mbox{d}\varepsilon^2 = \frac{1}{rH_1(r)}\varepsilon^1\wedge\varepsilon^2 = \omega_{12}\wedge\varepsilon^1 - \omega_{23}\wedge\varepsilon^3 - \omega_{20}\wedge \varepsilon^0\quad [3]$$

$$\mbox{d}\varepsilon^3 = \frac{1}{rH_1(r)}\varepsilon^1 \wedge \varepsilon^3 + \frac{\cot\theta}{r}\varepsilon^2 \wedge \varepsilon^3 = \omega_{13}\wedge\varepsilon^1 + \omega_{23}\wedge\varepsilon^2 - \omega_{30}\wedge\varepsilon^0\quad [4]$$


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 $\omega_{30} = \Gamma_{300}\varepsilon^0$ and substituting this into [1] we have $\Gamma_{300} = 0$ since it is the sole coefficient of the 2-form basis element $\varepsilon^3 \wedge \varepsilon^0$" and "Similarly, from [1] we have $\omega_{20} = \Gamma_{202}\varepsilon^2$"

Unfortunately, this is not obvious to me :redface". And apparently we can put all this together to make

$$\omega_{10} = \frac{H_0(r)'}{H_1(r)}\varepsilon^0 + \Gamma_{101}\varepsilon^1$$

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

$$\mbox{d}\varepsilon^0 = \frac{H_0(r)'}{H_1(r)}\varepsilon^1\wedge\varepsilon^0 = ...$$

 

[Oxymoron]

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

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

$$\omega_{30} = \Gamma_{300}\varepsilon^0$$

Similarly from [3], we have

$$\omega_{20} = \Gamma_{200}\varepsilon^0$$

Substituting into [1] we have

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

$$\frac{H_0(r)'}{H_1(r)}\varepsilon^1 \wedge \varepsilon^0 = -\omega_{10} \wedge \varepsilon^1$$

$$\omega_{10} = \frac{H_0(r)'}{H_1(r)}\varepsilon^0$$

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

 

[Oxymoron]

Does anyone know how to calculate this:

$$\mbox{d}\left(-\frac{1}{r}\frac{1}{H_1(r)}\varepsilon^2\right)$$

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

$$\omega_{12} = \frac{1}{r}\frac{1}{H_1(r)}\varepsilon^2$$

Then from Cartan's second structural equation we have

$$R_{12} = \mbox{d}\omega_{12} + \omega_{13}\wedge\omega_{32} + \omega_{10}\wedge\omega_{02}$$

the last two terms vanish and we are left with

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

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

$$H_1(r) = \frac{1}{\exp\left(-\frac{1}{2}\lambda(r)\right)}$$

then

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

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

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

$$= \left(\frac{1}{r^2}\frac{1}{H_1(r)} + \frac{1}{2}\frac{1}{r}\frac{1}{H_1(r)}r'\right)\varepsilon^1 \wedge \varepsilon^2 - \frac{1}{r^2}\frac{1}{H_1(r)^2}\varepsilon^1 \wedge \varepsilon^2$$

since $\mbox{d}\varepsilon^2 = \frac{1}{rH_1(r)}\varepsilon^1\wedge\varepsilon^2$.

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

when I really want this bit to look like:

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

Notice the subtle difference? Because then

$$=\left(\frac{1}{r}\frac{1}{H_1(r)^2}r'\right)\varepsilon^1\wedge\varepsilon^2$$

Which is the correct answer.

Please help me!