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Homework Help: Euler lagrange equation and einstein lagrangian

  1. Jan 17, 2010 #1
    Dear everyone

    can anyone help me with the euler lagrange equation which is stated in d'inverno chapter 11?
    in equation (11.26) it is said that when we use the hilbert-einstein lagrangian we can have:

    ∂L/(∂g_(ab,cd) )=(g^(-1/2) )[(1/2)(g^ac g^bd+g^ad g^bc )-g^ab g^cd ]

    haw can we derive this out of (11.24)?

    I tried to derive ∂L/(∂g_(ab,cd) ) out of (11.24), but I really don't know how to work with (∂g_(ab,cd) ?

    please help me!!!
  2. jcsd
  3. Jan 17, 2010 #2
    So let's think about it!

    In the equation (11.24), you have to forget about the terms of the last two lines because they are all made of the first derivatives of the metric tensor or the products of 'em. Keeping this in mind, make use of the relation between the determinant of the metric tensor and the christoffel symbols that is a simple equation involving first derivatives of gab. Remember that?!

  4. Jan 18, 2010 #3
    Thanks for helping me again!

    let me see
    here is what I do to get the equation:

    (∂L_G)/(∂g_(ab,cd) )=∂/(∂g_(ab,cd) ) {g^cd [1/2 g^ef {g_(cf,de)+g_(df,ce)-g_(cd,fe) }-1/2 g^ef {g_(cf,ed)+g_(ef,cd)-g_(ce,fd) } ] }

    now could you tell me what do you write for a term like this:
    ∂/(∂g_(ab,cd) )(g^cd g^ef g_(cf,de))

    maybe I do not know how to work with this.... :confused:
  5. Jan 18, 2010 #4
    A nice work up to here! So you've taken in what exactly I said about keeping those terms including a single metric tensor differentiated twice w.r.t the coordinates.

    I give you this sample and wish to come to me with the final result and calculation procedure:

    http://img513.imageshack.us/img513/907/calf.jpg [Broken]

    Last edited by a moderator: May 4, 2017
  6. Jan 18, 2010 #5
    Thanks again for helping. :smile:

    I did the exact thing but there is a minus sign which ruins everything!!
    here is what I have:

    (∂L_G)/(∂g_(ab,cd) )

    =1/2(g^ac g^db+g^ca g^db-g^ab g^dc-g^ad g^cb-g^cd g^ab+g^ad g^bc)

    =1/2 √(-g)(2g^ac g^db-2g^ab g^dc)

    which clearly is not (11.26)

    a problem is that -g^ad g^cb and +g^ad g^bc vanish together!! (the third term and the last one)
    and another one is that g^ac g^db+g^ca g^db=2g^ac g^db

    am I wrong?
  7. Jan 18, 2010 #6
    I want you to check your calculation again. Your result has a major error.

  8. Jan 18, 2010 #7
    I'm confused! This is exactly what I get from the deltas!
    would you please tke a look at this:

    (∂L_G)/(∂g_(ab,cd) )

    =1/2 g^cd g^ef [(δ_c^a δ_f^b δ_d^c δ_e^d+δ_d^a δ_f^b δ_c^c δ_e^d-δ_c^a δ_d^b δ_f^c δ_e^d )-(δ_c^a δ_f^b δ_e^c δ_d^d+δ_e^a δ_f^b δ_c^c δ_d^d-δ_c^a δ_e^b δ_f^c δ_d^d )]

    =1/2(g^ac g^db+g^ca g^db-g^ab g^dc-g^ad g^cb-g^cd g^ab+g^ad g^bc)

    and this is 1/2 √(-g)(2g^ac g^db-2g^ab g^dc)
    = √(-g)(g^ac g^db-g^ab g^dc)

  9. Jan 18, 2010 #8
    Everything is okay with your calculation now! The only thing to be remembered is that (∂L_G)/∂g_(ab,cd) is symmetric in the indices c and d. So you need to just put

    [tex]g^{ac} g^{db}[/tex] = [tex]1/2(g^{ac} g^{db}+g^{ad} g^{cb})[/tex]

    to get the equation (11.26) of D'inverno!

  10. Jan 19, 2010 #9
    oh my god!
    that's it!!!
    thank you very very much!
    (I don't know why, but I usually forget the symmetry! :biggrin:)

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