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Calculating lie derivative

  1. May 24, 2009 #1

    trv

    User Avatar

    1. The problem statement, all variables and given/known data

    Calculate the lie derivative of the metric tensor, given the metric,

    [itex]
    g_{ab}=diag(-(1-\frac{2M}{r}),1-\frac{2M}{r},r^2,R^2sin^2\theta)
    [/itex]

    and coordinates (t,r,theta,phi)

    given the vector

    [itex]
    E^i=\delta^t_0
    [/itex]





    2. Relevant equations

    [itex]
    (L_Eg)ab=E^cd_cg_{ab}+g_{cb}d_aE^c+g_{ac}d_bE^c
    [/itex]


    3. The attempt at a solution

    [itex]
    (L_Eg)ab=E^cd_cg_{ab}+g_{cb}d_aE^c+g_{ac}d_bE^c
    [/itex]

    all derivatives above being partial

    Now the Last two terms go to zero, since E^i=Kronecker delta=constant and so its derivative is zero.

    So,
    [itex]
    (L_Eg)ab=E^cd_cg_{ab}
    [/itex]

    [itex]
    (L_Eg)ab=\delta^t_0 d_cg_{ab}
    [/itex]

    I'm unsure how to take it from here.

    Firstly, I'm unsure what
    [itex]
    \delta^t_0
    [/itex]

    means. Does it means we get the result 1 at t=0 and zero for all other times?

    How does it then affect the equation below.

    [itex]
    (L_Eg)ab=\delta^t_0 d_cg_{ab}
    [/itex]

    Please help.
     
    Last edited: May 24, 2009
  2. jcsd
  3. May 25, 2009 #2

    trv

    User Avatar

    anyone?
     
  4. May 25, 2009 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I think what they meant to write is [itex]E^i=\delta^i_0[/itex] i.e. the vector field with a constant value 1 in the time component. So the Lie derivative is pretty trivial.
     
  5. May 25, 2009 #4

    trv

    User Avatar

    Oh ok, so the 0 stands for the time component. That makes sense.
     
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