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Stuck with Einstein Lagrangian

  1. Jan 14, 2004 #1
    Section 11.6 of Ray D'Inverno's book Introducing Einstein's Relativity

    This section shows Palatini's approach in using an equivalent Lagrangian to obtain the vacuum field equations of GR and the connection.

    From equation 11.39, we already have Lg=g(ab,b)T(c,ac)-g(ab,c)T(c,ab)-Lbarg+Q(a,a).(sorry, I tried to look up the instructions teaching how to type in equations here but couldn't find it)

    Lbarg is chosen in such a way that both it and Lg will give rise to the same field equations.
    The g here is the metric tensor density.
    Q(a)=g(bc)T(a,bc)-g(ab)T(c,bc) and Stokes integral around the boundary of delgamma for this will render it zero coz Q(a) vanishes on the boundary of delgamma.

    My problem is WHY the author equates
    Lbarg=g(ab,c)T(c,ab)-g(ab,b)T(c,ac) coz this seems use the assumptions that Lg=0 and Q(a,a)=0!

    If these assumptions were made, then only I could proceed to
    Lbarg=g(ab,c)T(c,ab)-g(ab,b)T(c,ac)
    =g(ab,c)T(c,ab)-1/2(g(ab,c)T(d,ad)+g(ab,a)T(d,bd))
    =g(ab,c)T(c,ab)-1/2(g(ab,c)T(d,ad)del(c,b)+g(ab,c)T(d,bd)del(c,a))
    =>delLbarg/delg(ab,c)=T(c,ab)-1/2.del(c,b)T(d,ad)-1/2.del(c,a)T(d,bd)
    which is equation 11.43 in the book.

    I just can't figure out WHY Lg=0 and Q(a,a)=0!
     
    Last edited: Apr 13, 2004
  2. jcsd
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