2
votes
0answers
46 views

Boundary term in Einstein-Hilbert action

Why is the boundary term in the Einstein-Hilbert action, the Gibbons-Hawking-York term, generally "missing" in General Relativity courses, IMPORTANT from the variational viewpoint, geometrical setting ...
8
votes
1answer
184 views

Evaluating the Einstein-Hilbert action

The Einstein-Hilbert action is given by, $$I = \frac{1}{16\pi G} \int_{M} \mathrm{d}^d x \, \sqrt{-g} \, R \, \, + \, \, \frac{1}{8\pi G}\int_{\partial M} \mathrm{d}^{d-1}x \, \sqrt{-h} \, K$$ ...
3
votes
1answer
77 views

Field equations in extended EH-GHY action. Is Schwarzschild a solution?

When taking the EH action, $$S_{EH} = \frac{1}{16\pi G}\int_M d^4x \sqrt{-g}R$$ and making a small variation in the metric while ignoring boundary terms, we obtain $$\delta S_{EH} = \frac{1}{16\pi ...
1
vote
0answers
126 views

Induced metric on the boundary of a manifold

The Gibbons-Hawking-York term which supplements the Einstein-Hilbert action is, $$S_{GH} = \frac{1}{8\pi G} \int_{\partial M} d^3 x\sqrt{-h} \, K$$ where $\partial M$ is the boundary of the manifold ...
0
votes
1answer
72 views

Why we can set variations for the metric and its derivatives to zero at infinity?

This question is the continuation of the following one. I still don't understand why $(1)$ may be set to zero. This refers to the zero value variations of metric and its derivatives on the infinitely ...
1
vote
1answer
216 views

Einstein action and the second derivatives

I have naive question about Einstein action for field-free case: $$ S = -\frac{1}{16 \pi G}\int \sqrt{-g} d^{4}x g^{\mu \nu}R_{\mu \nu}. $$ It contains the second derivatives of metric. When we want ...
2
votes
1answer
375 views

How do I calculate the induced metric in the Gibbons–Hawking–York boundary term?

This question concerns the expression for the induced metric in the explicit variation of the GHY boundary term. Just how is that expression derived in detail from the definition of the induced metric ...