Action variation in Brown-York formalism

Question

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?

Answer 1

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).