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Just a question about calculating Killing vectors

  1. Dec 10, 2011 #1
    1. The problem statement, all variables and given/known data
    I am new to the whole GR concept so I just would like to know if what I did is correct or utterly ******** (and if so, why).
    I have to find all Killing Vectors to a given metric (1) using equation (2). (2) leads to 3 partial differential equations which I used to calculate them.


    2. Relevant equations
    (1) [tex] ds^2=-dt^2 +dx^2 [/tex]

    (2) [tex]L_{\xi}g_{\mu \nu}=\xi^{\alpha}\partial_{\alpha}g_{\mu \nu}+g_{\mu \alpha}\partial_{\nu}\xi^{\alpha}+g_{\alpha \nu}\partial_{\mu}\xi^{\alpha}=0[/tex]

    3. The attempt at a solution

    I used equation to solve the three equations [tex]L _\xi g_{tt}=0[/tex], [tex]L _\xi g_{xx}=0[/tex] & [tex]L _\xi g_{tx}=0[/tex].
    That leads to

    [tex]\partial_t \xi^t=0[/tex]
    [tex]\partial_x \xi^x=0[/tex]
    [tex]\partial_t \xi^x=\partial_x \xi^t[/tex]

    the first 2 lead to

    [tex]\xi^t=f(x)[/tex]
    [tex]\xi^x=g(t)[/tex]

    and inserting that into the last equation shows that this one has to be equal to a constant since you can completely seperate the x and t depention. So with integration I get:

    [tex]\xi^x=g(t)=At+B[/tex]
    [tex]\xi^t=f(x)=Ax+C[/tex]

    So I have 3 constants and the solution:

    [tex]\xi=A \binom{t}{x} + B \binom{0}{1}+C\binom{1}{0}[/tex]

    and therefore those 3 vectors are the 3 linear independant Killing vectors.
    Is that correct?
     
    Last edited: Dec 10, 2011
  2. jcsd
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