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Curveture math problem

  1. Mar 19, 2005 #1
    Hello
    I have a small pb:
    Let [tex]\displaystyle{M=\mathbb{R^{2}\{(x,y);x=0\quad ou \quad y=-1}[/tex]
    Let D such that Christoffel symboles different from 0 are\\\\
    [tex]\displaystyle\Gamma_{11}^{1}(x,y)=-\frac{1}{x}\\\\\\\\
    \Gamma_{22}^{2}(x,y)=\frac{1}{1+y}[/tex]\\\\\\
    How calculate curvurture?????????\\
     
  2. jcsd
  3. Mar 19, 2005 #2
    The Riemann-curvature is given by:

    [tex]R^a_{b c d} = \frac{\partial \Gamma^a_{b d}}{\partial x^c} - \frac{\partial \Gamma^a_{b c}}{\partial x^d} + \Gamma^a_{c e} \Gamma^e_{b d} - \Gamma^a_{d e} \Gamma^e_{b c}[/tex]
     
  4. Mar 20, 2005 #3
    excuse me [tex]\displaystyle\Gamma_{11}^{1}(x,y)=-\frac{1}{x}\quad and \quad Gamma_{22}^{2}(x,y)=\frac{1}{1+y}[/tex]
    And
    [tex]\displaystyle{M={R^{2}\{(x,y);x=0\quad or \quad y=-1}}[/tex]
    So How we use this formula in this case
    and how we obtain this formula???
     
  5. Mar 21, 2005 #4
    a, b, c, d and e are indices, so in this case they can take the values 1 or 2: [tex]x = x^1[/tex] and [tex]y = x^2[/tex]. Repeates indices are summed over (Einstein summation convention).

    So for example:

    [tex]R^1_{1 2 2} = \frac{\partial \Gamma^1_{1 2}}{\partial x^2} - \frac{\partial \Gamma^1_{1 2}}{\partial x^2} + \sum_{e=1}^2 (\Gamma^1_{2 e} \Gamma^e_{1 2} - \Gamma^1_{2 e} \Gamma^e_{1 2})[/tex]

    which in this case is 0 because only [tex]\Gamma^1_{1 1}[/tex] and [tex]\Gamma^2_{2 2}[/tex] are different from 0.

    If you want to see a derivation of this formula, you could have a look at (chapter 3 of) Sean M. Carroll's lecture notes: http://pancake.uchicago.edu/~carroll/notes/
     
  6. Apr 5, 2005 #5
    But in my case
    R=0?
     
  7. Apr 5, 2005 #6

    dextercioby

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    Are u looking for the tensor,or for its contractions (Ricci tensor,Ricci scalar)...?

    Daniel.
     
  8. Apr 6, 2005 #7
    What are those hieroglyphic looking symbols in the first post? I clicked on it and the latex code doesn't look anything like it. :confused:
     
  9. Apr 6, 2005 #8

    dextercioby

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    U didn't close \mathbb function right after R...:tongue:

    Daniel.
     
  10. Apr 6, 2005 #9
    [tex]\displaystyle{M=\mathbb{R}^{2}\{(x,y);x=0\quad ou \quad y=-1}[/tex]

    Ah, there we go. I still don't know what that means, but it looks more recognizable. I was hoping those were tensor diagrams like Penrose uses. I don't get how they work, but they sure look nifty.
     
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