# Schwarzschild Solution using Cartan's Equations

1. Apr 26, 2006

### Oxymoron

Im having some trouble coming up with my six independent connection 1-forms.

I have been given a metric:

$$g = -H_0(r)^2dt\otimes dt + H_1(r)^2 dr\otimes dr + r^2 d\theta\otimes d\theta + r^2\sin^2\theta d\phi \otimes d\phi$$.

I need to find $H_0(r)$ and $H_1(r)$, 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:
$$e_0 = \frac{1}{H_0(r)}\partial_t \quad e_1 = \frac{1}{H_1(r)}\partial_r \quad e_2 = \frac{1}{r}\partial_{\theta} \quad e_3 = \frac{1}{r\sin\theta}\partial_{\phi}$$

so that my dual basis is:
$$\varepsilon^0 = H_0(r)\mbox{d}t \quad \varepsilon^1 = H_1(r)\mbox{d}r \quad \varepsilon^2 = r\mbox{d}\theta \quad \varepsilon^3 = r\sin\theta\mbox{d}\phi$$

Now, using Cartan's structural relations I calculated:

$$\mbox{d}\varepsilon^0 = -\omega_{10}\wedge\varepsilon^1 - \omega_{20}\wedge\varepsilon^2 - \omega_{30}\wedge \varepsilon^3\quad [1]$$
$$\mbox{d}\varepsilon^1 = -\omega_{12}\wedge\varepsilon^2 - \omega_{13}\wedge\varepsilon^3 - \omega_{10}\wedge\varepsilon^0\quad [2]$$
$$\mbox{d}\varepsilon^2 = \omega_{12}\wedge\varepsilon^1 - \omega_{23}\wedge\varepsilon^3 - \omega_{20}\wedge \varepsilon^0\quad [3]$$
$$\mbox{d}\varepsilon^3 = \omega_{13}\wedge\varepsilon^1 + \omega_{23}\wedge\varepsilon^2 - \omega_{30}\wedge\varepsilon^0\quad [4]$$

But now I am stuck. I should be able to find 6 independent connection 1-forms but I dont know how to simplify all the above equations. Any guidance from here would be very helpful.

Last edited: Apr 27, 2006
2. Apr 27, 2006

### Oxymoron

I tried computing the derivatives of the dual basis directly, but Im not sure if that helps me:

$$\mbox{d}\varepsilon^1 = \mbox{dd}\varepsilon^1 = 0$$

since $\mbox{d}\circ\mbox{d}=0$.

$$\mbox{d}\varepsilon^2 = \mbox{d}r\wedge\mbox{d}\theta + r\wedge \mbox{dd}\theta$$
$$= \mbox{d}r \wedge\mbox{d}\theta$$
$$= \frac{1}{H_1(r)}\varepsilon^1 \wedge \frac{1}{r}\varepsilon^2$$
$$= \frac{1}{rH_1(r)}\varepsilon^1\wedge\varepsilon^2$$

Similarly,

$$\mbox{d}\varepsilon^3 = \frac{1}{rH_1(r)}\varepsilon^1 \wedge \varepsilon^3 + \frac{\cot\theta}{r}\varepsilon^2 \wedge \varepsilon^3$$

$$\mbox{d}\varepsilon^0 = \mbox{d}(H_0(r)\mbox{d}t)$$
$$= \mbox{d}H_0(r)\mbox{d}t \wedge H_0(r) \mbox{d}r\mbox{d}$$
$$= \frac{H_0(r)'}{H_0(r)}\varepsilon^0 \wedge \frac{H_0(r)}{H_1(r)}\varepsilon^1$$
$$= \frac{H_0(r)'}{H_1(r)}\varepsilon^1 \wedge \varepsilon^0$$

Note: The prime refers to differentiation with respect to r

Last edited: Apr 27, 2006
3. Apr 27, 2006

### Oxymoron

To summarize:

$$\mbox{d}\varepsilon^0 = \frac{H_0(r)'}{H_1(r)}\varepsilon^1 \wedge \varepsilon^0 = -\omega_{10}\wedge\varepsilon^1 - \omega_{20}\wedge\varepsilon^2 - \omega_{30}\wedge \varepsilon^3 \quad [1]$$

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

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

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

Do these look right to anyone?

Last edited: Apr 27, 2006
4. Apr 28, 2006

### Oxymoron

Perhaps I could help this process along with a further hint which I really like some explanation of:

Apparently in the next step of my calculations I should be able to recognize that from [4] we have

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

and substituting into [1] we see that $\Gamma_{300} = 0$ since it is the sole coefficient of the 2-form basis element $\varepsilon^3 \wedge \varepsilon^4$.

All of this, I dont get :( If anyone is willing to explain how this all works, then I may be able to do a lot of simplification and come up with, hopefully, 6 independent connection 1-forms.

5. Apr 28, 2006

### nrqed

Just a suggestion, Oxymoron....You may want to post this on the Differential Geometry boards. There are very knowledgeable people there that will surely help you out.

Patrick

6. Apr 29, 2006

### Oxymoron

Ok, I have posted a similar thread in the Diff. Geometry and Tensor board.