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In this paper section 3.1 we are given the action $$ \mathcal{I}= -\frac{1}{16\pi G_D}\left[\int_{\mathcal{M_1}}d^Dx\sqrt{g_1}(R_1-2\Lambda_1)+\int_{\mathcal{M_2}}d^Dx\sqrt{g_2}(R_2-2\Lambda_2)+2\int_Sd^{D-1}y\sqrt{h}(K_1-K_2)-2(D-2)\int_Sd^{D-1}y\sqrt{h}\mathcal{k}\right] $$

where $g_1$,$g_2$ are the metrics of two regions and $R_1,\Lambda_1$,$R_2,\Lambda_2$ are the ricci curvature and cosmological constants of the two regions. Then $h$ is the metric on the interface and $K_1,K_2$ are the extrinsic curvatures of the two regions on the interface while $\mathcal{k}$ is the tension parameter of the interface.

When we vary the action we get the einstein field equations from the first two terms(each one giving the equations of motion for the respective region). When we vary the last two terms we should get the relation $$ K_{1ab}-K_{2ab}=\mathcal{k}h_{ab} $$

When I try to do this I am stuck when I vary $\delta K_1$ for example where I would set $$ \delta K_1 = \delta(K_{1\mu\nu}h^{\mu\nu})= \delta K_{1\mu\nu}h^{\mu\nu}+K_{1\mu\nu}\delta h^{\mu\nu} $$ from here I don't understand how to proceeed with the $\delta K_{1\mu\nu}$ term.

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In Gaussian normal coordinates (with the surface at constant $\lambda$ $$ ds^2 = \sigma d\lambda^2 + h_{ij}(\lambda,y) dy^i dy^j$$ $K_{\mu\nu}$ may be written as $$ K_{ij} = \frac12 \partial_\lambda h_{ij} \,.$$

From this you should be able to go on (if you still intend to do so, this is a late answer).

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