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Schwarzschild Solution using Cartan's Formalism

  1. Apr 29, 2006 #1
    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 :redface:

    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 Im 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!!!
    Last edited: Apr 29, 2006
  2. jcsd
  3. Apr 30, 2006 #2
    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]
  4. Apr 30, 2006 #3
    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 dont know why [itex]\omega_{30} = \Gamma_{300}\varepsilon^0[/itex] :redface:
  5. May 6, 2006 #4
    Does anyone know how to calculate this:


    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]


    [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


    Which is the correct answer.

    Please help me!
    Last edited: May 6, 2006
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