Why Does D'Inverno Equate Lbarg in the Palatini Approach?

[maddy]

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!

 

[M98Ranger]

Firstly, it is important to note that the equations in this section are derived using the Palatini approach, which differs from the more commonly known Einstein-Hilbert approach. In this approach, the connection is treated as an independent variable, rather than being determined by the metric tensor as in the Einstein-Hilbert approach.

To understand why Lg and Q(a,a) are assumed to be zero, we must look at the fundamental principle of least action, which states that the true equations of motion are those which minimize the action. In this case, the action is given by the integral of the Lagrangian over spacetime. In order to obtain the correct field equations, the Lagrangian must be chosen in such a way that it is stationary, i.e. its variation with respect to the metric and connection vanishes.

In equation 11.39, we see that Lg and Q(a,a) are the only terms that contain both the metric tensor and connection. In order for the Lagrangian to be stationary, these terms must be set to zero. This leads to the equation Lg=0 and Q(a,a)=0, which are then substituted into the expression for Lbarg in equation 11.43.

In summary, the assumptions of Lg=0 and Q(a,a)=0 are necessary in order to obtain the correct field equations using the Palatini approach. Without these assumptions, the Lagrangian would not be stationary and the equations of motion would not be valid. I hope this helps to clarify your understanding of this section.