Section 11.6 of Ray D'Inverno's book Introducing Einstein's Relativity(adsbygoogle = window.adsbygoogle || []).push({});

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!

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# Stuck with Einstein Lagrangian

Can you offer guidance or do you also need help?

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