3
$\begingroup$

In the original Brown-York paper on quasi-local charges, they start with this action

$$S = \frac{1}{16 \pi} \int_{D} \mathrm{d}^4x \sqrt{-g} R - \frac{1}{8 \pi} \int_{^3B} \mathrm{d}^3x \sqrt{-h} K + \frac{1}{8 \pi} \int_{\Sigma_{t_1}}^{\Sigma_{t_2}} \mathrm{d}^3x \sqrt{-\gamma} \Theta ,$$

And say that its variation leads to

$$ \delta S^1 = \text{(terms giving equations of motion)}$$ $$-\frac{1}{16\pi} \int_{^3B} \mathrm{d}^3 x \sqrt{-h} \left( K^{a b} - K h^{a b} \right)\delta h_{a b}$$

$$\int_{\Sigma_{t_2}} \frac{1}{16\pi} \sqrt{-\gamma} \left(\Theta^{a b} - \Theta \gamma^{a b} \right) \delta \gamma_{a b} \, \mathrm{d}^3 x $$ $$ - \int_{\Sigma_{t_1}} \frac{1}{16\pi} \sqrt{-\gamma} \left(\Theta^{a b} - \Theta \gamma^{a b} \right) \delta \gamma_{a b} \, \mathrm{d}^{3}x. $$

My question is how did the variation of the second integral yield $- \frac{1}{16\pi} \int_{^3B} \mathrm{d}^3 x \sqrt{-h} \left( K^{a b} - K h^{a b} \right)\delta h_{a b}$?

Here’s what I have tried

$$\delta (\sqrt{-h}K) = K \delta \sqrt{-h} + \sqrt{-h} \delta K.$$

The first term gives $K (\frac{1}{2} \sqrt{-h} h^{ab} \delta h_{ab})$, so that is fine. For the second term, the variation of $K$ is given by

$$\delta K = -h^{a b} \delta \Gamma^l_{a b} n_l$$ $$= -h^{a b} n_l \frac{1}{2} g^{l c} \left (\delta \partial_b g_{c a} + \delta \partial_a g_{c b} - \delta \partial_c g_{a b} \right )$$ $$\qquad=-{1 \over 2} h^{a b} \left( \delta \partial_b g_{d a} + \delta \partial_a g_{d b} - \delta \partial_d g_{a b} \right ) n^d $$ $$\qquad= \frac{1}{2} h^{a b} (\delta \partial_d g_{a b}) n^d.$$

This does not give $\frac{1}{2} K^{a b} \delta h_{ab}$ as I hoped for. Could somebody PLEASE help me?

$\endgroup$
1
  • $\begingroup$ This master thesis gives a detailed derivation in section 2.2. Title: The Boundary Terms of the Einstein-Hilbert Action Author: Simone S. Bavera $\endgroup$
    – 朔望-L
    Commented Feb 23, 2023 at 12:41

1 Answer 1

1
$\begingroup$

Consider the variation of the extrinsic curvature tensor

$$\delta K = \delta\left(\nabla_{\mu}n^{\mu}\right) = \nabla_{\mu}\delta n^{\mu} + \delta \Gamma^{\mu}_{\mu \nu} n^{\nu}$$

$$ \ \ \ \ \ \ \ \ \ \ = -\frac{1}{2}\nabla_{\mu}\left(n^{\mu}n_{\nu}n_{\alpha}\delta g^{\nu \alpha} \right) - \frac{1}{2}\nabla_{n}\delta g$$

$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -\frac{1}{2}K n_{\nu}n_{\alpha}\delta g^{\nu \alpha} - \frac{1}{2}\nabla_{n}\delta g - \frac{1}{2}n_{\nu}h^{\mu \alpha}\nabla_{\mu}\delta g^{\nu}_{\alpha} + \frac{1}{2}n_{\nu}\nabla_{\mu}\delta g^{\nu \mu}$$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}\left(n_{\nu}\nabla_{\mu}\delta g^{\mu\nu} - \nabla_{n} \delta g \right) - \frac{1}{2}K n_{\nu}n_{\alpha}\delta g^{\nu \alpha} - \frac{1}{2}\ \underbrace{ ^{(3)}\nabla_{\alpha}\left(n_{\nu}\delta g^{\nu \alpha} \right)}_{=0} + \frac{1}{2}K_{\mu\nu}\delta g^{\mu \nu}$

The covariant 3-derivative vanishes since the boundary of a boundary vanishes. Hence, we have,

$$\delta \left(K\sqrt{g} \right) = \frac{1}{2}\left(n_{\nu}\nabla_{\mu}\delta g^{\mu\nu} - \nabla_{n}\delta g \right) + \frac{1}{2}\left(K_{\mu\nu} - Kh_{\mu\nu} \right)\delta g^{\mu\nu}$$.

The second part of this equation is the desired output (note that the part with the normal and the derivative cancels out).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.