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Schwarzschild in 5d

  1. Feb 13, 2017 #1
    1. The problem statement, all variables and given/known data
    The most general form:

    ##ds^2=e^{2A(r)}dt^2-e^{2B(r)}dr^2-r^2(d\theta_1^2 +sin^2\theta_1(d\theta^2_2+sin^2\theta_2d\phi^2))##

    Ricci tensors:

    ##R_{tt}=e^{2(A-B)}(\frac{3A'}{r}+A'^2-B'A'+A'')##
    ##R_{rr}=-A''+\frac{3B'}{r}+A'(B'-A')##
    ##R_{\theta_1 \theta_1}=2+e^{-2B}(-2+r(B'-A'))##
    ##R_{\theta_2 \theta_2} = sin^2 \theta_1 R_{\theta_1 \theta_1}##
    ##R_{\phi \phi} = sin^2 \theta_1 sin^2 \theta_2 R_{\theta_1 \theta_1}##

    To solve the Ricci tensors for ##A## and ##B##?



    2. Relevant equations


    3. The attempt at a solution

    I have tried preceeding as you do in the 4-d case. This is to do
    ##e^{-2(A-B)} R_{tt} + R_{rr} =0 ##
    which gives ##A=-B##

    and then to solve
    ##R_{\theta_1 \theta_1}=0##

    However doing this i get:

    ##1+e^{2A}(1-A') = 0##
    multiply though by ##e^{-3A}##
    to get
    ##d/dr(e^{-A}r)=-e^{-A}##
    ##e^{-A}r=\int -e^{-A} dr ##


    and I can't solve for ##A(r)## explicitly

    I can't see a better way to proceed as with the other components there are terms including ##A'' , A'^2## things.

    Many thanks
     
  2. jcsd
  3. Feb 20, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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