- #1
Mr-R
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Book: Landau Lifshitz, The Classical Theorey of Fields, chapter 11, section 95.
I have gone through the derivation of Einstein field equations but not without holes to fill and fix in my understanding. Let's start with the action for the grtavitational field ##S_g## which after some explanation they defined it as: $$\int R\sqrt{-g}d\Omega= \int G\sqrt{-g}d\Omega+\int \frac{\partial (\sqrt{-g}w^i)}{\partial x^i}d\Omega$$
Where as it is explained in the book, ##R## is the Ricci scalar. On the RHS, ##G## contains the metric tensor and its first derivative. The second term on the RHS has the form of the divergence of some function ##w^i## and by Gauss' theorem and the variational principle, it vanishes at the limits (Over all space-time coordinates) of the region of integration.
Amid varying we get :
$$\delta \int \sqrt{-g}Rd\Omega=\delta \int \sqrt{-g}Gd\Omega$$
Now they note that the quantity on the LHS is a scalar and therefore the RHS is also a scalar. But "G itslef is not a scalar". This confuses me because the last equality has the same form, which to me makes ##G## looks like a scalar when compared to the LHS (##R##). Does the very first (unvaried) equality show that ##G## is not a scalar? And ##G## does not have any indices. Doesn't that make it a scalar?
That's it for now. Thanks in advance.
I have gone through the derivation of Einstein field equations but not without holes to fill and fix in my understanding. Let's start with the action for the grtavitational field ##S_g## which after some explanation they defined it as: $$\int R\sqrt{-g}d\Omega= \int G\sqrt{-g}d\Omega+\int \frac{\partial (\sqrt{-g}w^i)}{\partial x^i}d\Omega$$
Where as it is explained in the book, ##R## is the Ricci scalar. On the RHS, ##G## contains the metric tensor and its first derivative. The second term on the RHS has the form of the divergence of some function ##w^i## and by Gauss' theorem and the variational principle, it vanishes at the limits (Over all space-time coordinates) of the region of integration.
Amid varying we get :
$$\delta \int \sqrt{-g}Rd\Omega=\delta \int \sqrt{-g}Gd\Omega$$
Now they note that the quantity on the LHS is a scalar and therefore the RHS is also a scalar. But "G itslef is not a scalar". This confuses me because the last equality has the same form, which to me makes ##G## looks like a scalar when compared to the LHS (##R##). Does the very first (unvaried) equality show that ##G## is not a scalar? And ##G## does not have any indices. Doesn't that make it a scalar?
That's it for now. Thanks in advance.