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Homework Help: Schwarzschild Solution using Cartan's Equations

  1. Apr 26, 2006 #1
    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 dont know how to simplify all the above equations. Any guidance from here would be very helpful.
     
    Last edited: Apr 27, 2006
  2. jcsd
  3. Apr 27, 2006 #2
    I tried computing the derivatives of the dual basis directly, but Im not sure if that helps me:

    [tex]\mbox{d}\varepsilon^1 = \mbox{dd}\varepsilon^1 = 0[/tex]

    since [itex]\mbox{d}\circ\mbox{d}=0[/itex].

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

    Similarly,

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

    [tex]\mbox{d}\varepsilon^0 = \mbox{d}(H_0(r)\mbox{d}t)[/tex]
    [tex] = \mbox{d}H_0(r)\mbox{d}t \wedge H_0(r) \mbox{d}r\mbox{d} [/tex]
    [tex] = \frac{H_0(r)'}{H_0(r)}\varepsilon^0 \wedge \frac{H_0(r)}{H_1(r)}\varepsilon^1[/tex]
    [tex] = \frac{H_0(r)'}{H_1(r)}\varepsilon^1 \wedge \varepsilon^0[/tex]

    Note: The prime refers to differentiation with respect to r
     
    Last edited: Apr 27, 2006
  4. Apr 27, 2006 #3
    To summarize:

    [tex]\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][/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]

    Do these look right to anyone?
     
    Last edited: Apr 27, 2006
  5. Apr 28, 2006 #4
    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

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

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

    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.
     
  6. Apr 28, 2006 #5

    nrqed

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    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
     
  7. Apr 29, 2006 #6
    Ok, I have posted a similar thread in the Diff. Geometry and Tensor board.
     
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