1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Einstein Equations of this metric

  1. Apr 12, 2015 #1
    1. The problem statement, all variables and given/known data

    (a) Find the christoffel symbols
    (b) Find the einstein equations
    (c) Find A and B
    (d) Comment on this metric

    2014_B5_Q2.png

    2. Relevant equations

    [tex]\Gamma_{\alpha\beta}^\mu \frac{1}{2} g^{\mu v} \left( \partial_\alpha g_{\beta v} + \partial_\beta g_{\alpha v} - \partial_\mu g_{\alpha \beta} \right) [/tex]

    [tex]R_{v \beta} = \partial_\mu \Gamma_{\beta v}^\mu - \partial_\beta \Gamma_{\mu v}^\mu + \Gamma_{\mu \epsilon}^\mu \Gamma_{v \beta}^\epsilon - \Gamma_{\epsilon \beta}^\mu \Gamma_{v \mu}^\epsilon [/tex]

    3. The attempt at a solution

    Part(a)

    After some math, I found the christoffel symbols to be:
    ##\Gamma_{11}^0 = \frac{A A^{'}}{c^2}##
    ##\Gamma_{22}^0 = \frac{B B^{'}}{c^2}##
    ##\Gamma_{33}^0 = \frac{B B^{'}}{c^2}##
    ##\Gamma_{01}^1 = \frac{A^{'}}{A}##
    ##\Gamma_{02}^2 = \frac{B^{'}}{B}##
    ##\Gamma_{03}^3 = \frac{B^{'}}{B}##

    Part (b)
    Now brace yourselves for the ricci tensors...
    [tex]R_{00} = -\partial_0 \left( \Gamma_{01}^1 + \Gamma_{02}^2 + \Gamma_{03}^3 \right) - \Gamma_{10}^1 \Gamma_{01}^1 - 2\Gamma_{20}^2 \Gamma_{02}^2 [/tex]
    [tex]R_{00} = -\frac{A^{''}}{A} - 2 \frac{B^{''}}{B}[/tex]

    By symmetry, ##R_{01} = R_{02} = R_{03} = R_{12} = R_{13} = R_{23} = 0##.

    Now to find the ##11## component:
    [tex]R_{11} = \partial_0 \Gamma_{11}^0 + \Gamma_{11}^0 \left( \Gamma_{10}^1 + \Gamma_{20}^2 + \Gamma_{30}^3 \right) - \Gamma_{11}^0 \Gamma_{10}^1 - \Gamma_{01}^1 \Gamma_{11}^0 [/tex]
    [tex] = \partial_0 \Gamma_{11}^0 + 2 \Gamma_{11}^0 \Gamma_{20}^2 - \Gamma_{11}^0 \Gamma_{10}^1 [/tex]
    [tex] R_{11} = \frac{A A^{''}}{c^2} + 2 \left( \frac{A}{B} \right) \frac{A^{'} B^{'}}{c^2} [/tex]

    By symmetry, to find ##22## and ##33## components, we swap ##A## with ##B##:
    [tex]R_{22} = R_{33} = \frac{B B^{''}}{c^2} + 2 \left( \frac{B}{A} \right) \frac{A^{'} B^{'}}{c^2}[/tex]


    The einstein field equations are given by:
    [tex]G^{\alpha \beta} = \frac{8 \pi G}{c^4} T^{\alpha \beta} - \Lambda g^{\alpha \beta} [/tex]

    Thus, the simultaneous equations we seek are:
    [tex] G^{00} = \frac{8 \pi G}{c^4} T^{00} [/tex]
    For ##\mu, v \neq 0## we have
    [tex] R_{\mu v} = 0[/tex]
    So we simply equate ##R_11 = 0##, ##R_22 = R_{33} = 0##.


    However, the equations don't match..
     
    Last edited: Apr 12, 2015
  2. jcsd
  3. Apr 16, 2015 #2
  4. Apr 17, 2015 #3
    bumpp
     
  5. Apr 18, 2015 #4
  6. Apr 19, 2015 #5
    bumpp
     
  7. Apr 20, 2015 #6
  8. Apr 22, 2015 #7
  9. Apr 23, 2015 #8
  10. Apr 24, 2015 #9
    bumpp on part (b)/(c)
     
  11. Apr 25, 2015 #10
    Would appreciate help with my "ricci-nightmare"
     
  12. Apr 26, 2015 #11
    Anyone managed to get a different result for the ricci tensors yet?
     
  13. Apr 30, 2015 #12
    anyone else had a go with the ricci tensors?
     
  14. May 4, 2015 #13
    tried again, still didn't get the required ricci tensors.
     
  15. May 4, 2015 #14

    thierrykauf

    User Avatar
    Gold Member

    Here is what I find for one term for Ricci
     
  16. May 7, 2015 #15
    I think the term is not appearing, do you mind posting it again?
     
  17. May 10, 2015 #16
    bumpp ricci
     
  18. May 10, 2015 #17

    thierrykauf

    User Avatar
    Gold Member

    Hold on. Here is what I find
     
  19. May 10, 2015 #18

    thierrykauf

    User Avatar
    Gold Member

    Here is what I find [tex]R_{00}=\Gamma^x_{x0}\Gamma^x_{x0} + \Gamma^x_{x0}\Gamma^y_{y0} + \Gamma^z_{z0}\Gamma^y_{y0}[/tex]
     
  20. May 14, 2015 #19
    Thanks alot for replying. I'll give it a go later today and post my updated work.
     
  21. May 14, 2015 #20

    thierrykauf

    User Avatar
    Gold Member

    Please do. And let's see what you have.
     
  22. May 15, 2015 #21
    My method is to go symbol by symbol and evaluate all possibilities for each symbol. Let's try to find ##R_{tt}##. For the first symbol, since all christoffel symbols are functions of ##t##, the first term ##\partial_t \Gamma^t_{tt}=0##. Third term is zero as there is a ##\Gamma^{\epsilon}_{tt}## term.Then we have

    [tex]R_{\nu \beta} = \partial_\mu \Gamma^\mu_{\beta \nu} - \partial_\beta \Gamma^\mu_{\mu \nu} + \Gamma^\mu_{\mu \epsilon} \Gamma^\epsilon_{\nu \beta} - \Gamma^\mu_{\epsilon \beta} \Gamma^\epsilon_{\nu \mu} [/tex]

    [tex]R_{tt} = -\partial_t \left( \Gamma^x_{xt} + \Gamma^y_{yt} + \Gamma^z_{zt} \right) - \Gamma^\mu_{\epsilon t} \Gamma^\epsilon_{t\mu}[/tex]

    Last term is non-zero only if ##\mu = \epsilon = x,y,z##.

    [tex] = -\partial_t \left( \Gamma^x_{xt} + \Gamma^y_{yt} + \Gamma^z_{zt} \right) - \left[ (\Gamma^x_{tx})^2 + (\Gamma^y_{ty})^2 + (\Gamma^z_{tz})^2 \right] [/tex]
     
    Last edited: May 15, 2015
  23. May 16, 2015 #22
  24. May 18, 2015 #23
    bumpp on
    [tex]R_{tt} = -\partial_t \left( \Gamma^x_{xt} + \Gamma^y_{yt} + \Gamma^z_{zt} \right) - \left[ (\Gamma^x_{tx})^2 + (\Gamma^y_{ty})^2 + (\Gamma^z_{tz})^2 \right] [/tex]
     
  25. May 22, 2015 #24
    bumpp
     
  26. May 23, 2015 #25
    [tex]R_{tt} = -\partial_t \left( \Gamma^x_{xt} + \Gamma^y_{yt} + \Gamma^z_{zt} \right) - \left[ (\Gamma^x_{tx})^2 + (\Gamma^y_{ty})^2 + (\Gamma^z_{tz})^2 \right] [/tex]

    bumpp
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted