1
$\begingroup$

I am trying to reproduce the results in the paper Holographic Dual of BCFT (cf. https://arxiv.org/abs/1105.5165) in which, performing a variation of the metric of the Hilbert action with the Gibbons-Hawking boundary term (eq. 1) $$ I = \frac{1}{16\pi G_N} \int_N \sqrt{-g} (R-2\Lambda) + \frac{1}{8\pi G_N}\int_Q \sqrt{-h} K $$ where $h$ is the metric of the manifold Q, $K = h^{ab} K_{ab}$ is the trace of the extrinsic curvature $K_{ab}$, defined as $K_{ab} = \nabla_a n_b$, and $n$ is the unit vector normal to Q with a projection of indices onto Q from N.

They then arrive "after a partial integration" to eq. 2 $$ \delta I = \frac{1}{16\pi G_N} \int_Q \sqrt{-h} (K_{ab}\delta h^{ab} - K h_{ab}\delta h^{ab}) $$ From that they derive the boundary condition (eq. 3) $$ K_{ab} - h_{ab} K = 8\pi G_N T^Q_{ab} $$ where $T^Q_{ab}:= \frac{2}{\sqrt{-h}}\frac{\delta I_Q}{\delta h_{ab}}$ and $I_Q$ is the action of some matter fields localized on Q.

  • Is the variation carried out with respect of the metric $g$ or the metric $h$? I'm confused since the result is expressed in terms of $\delta h^{ab}$ and I don't understand how both variations would be related.
  • Could anyone provide the explicit calculation that takes us from eq. 1 to eq. 2? How does that "partial integration" actually take place?
  • How do we arrive to the final boundary conditions? (eq. 3)

I have read different sources and treatments of the GH term, but I am getting extremely confused at this point, even though I think it must be a quite simple calculation. This is in part due to the fact that I am not familiar with the concept of an induced metric, so any information about this would be extremely appreciated, specially in order to arrive to eq. 2.

$\endgroup$
3
  • 1
    $\begingroup$ Wikipedia $\endgroup$ Commented Oct 30, 2023 at 10:28
  • $\begingroup$ You should explain what $K_{\alpha\beta}$ and its possible relation with $K$ (trace?) means. Additionally you should explain what $T^Q_{ab}$ is in particular what the index $Q$ means. And all the other symbols which might be more evident what they mean though. $\endgroup$ Commented Oct 30, 2023 at 14:34
  • $\begingroup$ Sorry, I just edited the original post to add the additional information. $\endgroup$ Commented Oct 30, 2023 at 15:10

1 Answer 1

0
$\begingroup$

This derivation is not particularly simple. In this answer, let greek indices $\mu,\nu,\dots$ take the values $0,1,\dots,n$ and latin indices $a,b,\dots$ take the values $1,2,\dots,n$. Spacetime $M$ is assumed to be $n+1$ dimensional with its boundary $\Sigma:=\partial M$ $n$ dimensional. We will also suppose that $\Sigma$ consists of a single timelike or spacelike piece with $\varepsilon=\pm 1$ with $\varepsilon=+1$ for a timelike boundary and $\varepsilon = -1$ for a spacelike boundary.


Let $\mathfrak g:=|\det(g_{\mu\nu})|$, then $$ \delta\left(R\sqrt{\mathfrak g}\right)=G_{\mu\nu}\delta g^{\mu\nu}\sqrt{\mathfrak g}+\nabla_\rho Q^\rho\sqrt{\mathfrak g}, $$ where $$ Q^\rho=g^{\mu\nu}\delta\Gamma^\rho_{\ \mu\nu}-g^{\mu\rho}\delta\Gamma^\nu_{\ \nu\mu}. $$

Then Stokes' theorem gives $$ \int_M\nabla_\rho Q^\rho\sqrt{\mathfrak g}\,\mathrm d^{n+1}x=\int_\Sigma\varepsilon n_\rho Q^\rho\sqrt{\mathfrak h}\,\mathrm d^ny. $$

We have $$ Q(n):=n_\rho Q^\rho=g^{\mu\nu}n_\rho\delta\Gamma^\rho_{\ \mu\nu}-n^\mu\delta\Gamma^\nu_{\ \mu\nu}. $$

We want to find the variation of the normal vector, for which we get $$ \delta n_\mu = -\frac{\varepsilon}{2}n_\alpha n_\beta\delta g^{\alpha\beta} n_\mu. $$ Then $$ \delta(\nabla_\mu n_\nu)=\nabla_\mu\delta n_\nu-\delta\Gamma_{\ \mu\nu}^\rho n_\rho $$and $$ \delta(\nabla_\mu n^\mu)=\delta\Gamma^\nu_{\ \nu\mu}n^\mu+\nabla_\mu\delta n^\mu $$(note that $\delta n^\mu\neq g^{\mu\nu}\delta n_\nu$), which nets us $$ Q(n)=g^{\mu\nu}\nabla_\mu\delta n_\nu-g^{\mu\nu}\delta(\nabla_\mu n_\nu)+\nabla_\mu\delta n^\mu-\delta(\nabla_\mu n^\mu). $$ Let $\delta n^\mu_{\parallel}:=\delta n^\mu+g^{\mu\nu}\delta n_\nu$, then $$ Q(n)=\nabla_\mu\delta n^\mu_\parallel +\nabla_\mu n_\nu \delta g^{\mu\nu}-\delta(2\nabla_\mu n^\mu). $$

We recall some facts whose proof is left to OP as an exercise:

  • We have $K=\nabla_\mu n^\mu$, assuming that the unit normal $n^\mu$ has been extended off $\Sigma$ the following way. We have a (local) foliation by surfaces such that $\Sigma$ is a particular leaf in the foliation, and $n^\mu$ is a unit normal to all of the leaves in the foliation. This will be assumed throughout this calculation.
  • Then also $K_{\mu\nu}=h^\rho_{\ \mu}\nabla_\rho n_\nu=\nabla_\mu n_\nu - \varepsilon n_\mu a_\nu$, where $a_\nu =n^\mu\nabla_\mu n_\nu$ is the "acceleration" of the unit normal.
  • If $S^\mu$ is a vector field tangent to $\Sigma$ (extended off $\Sigma$ in a way that it remains tangent to all leaves), then $\nabla_\mu S^\mu=D_a S^a-\varepsilon a_\mu S^\mu$.
  • The vector field $\delta n^\mu_\parallel$ is tangent to $\Sigma$ and satisfies $\delta n^\mu_\parallel a_\mu=n_\mu a_\nu\delta g^{\mu\nu}$.

Putting these together, we get $$ Q(n)=K_{\mu\nu}\delta g^{\mu\nu}-\delta(2K)+D_a\delta n^a_\parallel. $$ Multiplying this by $\sqrt{\mathfrak h}$, we get $$ Q(n)\sqrt{\mathfrak h}=K_{\mu\nu}\delta g^{\mu\nu}\sqrt{\mathfrak h}- Kh_{ab}\delta h^{ab}\sqrt{\mathfrak h}-\delta(2K\sqrt{\mathfrak h})+ D_a\delta n^a_\parallel\sqrt{\mathfrak h}. $$ For the first term, we have $$ K_{\mu\nu}\delta g^{\mu\nu}=-K^{\mu\nu}\delta g_{\mu\nu}=-K^{ab}e^\mu_{\ a}e^\nu_{\ b}\delta g_{\mu\nu}=-K^{ab}\delta h_{ab}=K_{ab}\delta h^{ab}. $$ Thus $$ Q(n)\sqrt{\mathfrak h}=\pi_{ab}\delta h^{ab}\sqrt{\mathfrak h}-\delta(2K\sqrt{\mathfrak h})+D_a\delta n^a_\parallel\sqrt{\mathfrak h}, $$ where $$ \pi_{ab}=K_{ab}-Kh_{ab}. $$


Let $$ S[g]=\frac{1}{2\kappa}\int_M R\sqrt{\mathfrak g}\,\mathrm d^{n+1}x+\frac{\varepsilon}{\kappa}\int_\Sigma K\sqrt{\mathfrak h}\,\mathrm d^ny. $$ Then $$ 2\kappa\delta S[g]=\int_M G_{\mu\nu}\delta g^{\mu\nu}\sqrt{\mathfrak g}\,\mathrm d^{n+1}x+\int_\Sigma\varepsilon\left(Q(n)\sqrt{\mathfrak h}+\delta(2K\sqrt{\mathfrak h})\right)\mathrm d^ny \\ = \int_M G_{\mu\nu}\delta g^{\mu\nu}\sqrt{\mathfrak g}\,\mathrm d^{n+1}x+\int_\Sigma\varepsilon\pi_{ab}\delta h^{ab}\sqrt{\mathfrak h}\,\mathrm d^ny. $$


The above derivation essentially answers the first and second question of OP. For the third question, suppose also that there is an action $S_\Sigma$ concentrated on the boundary that depends on the metric (it can depend on either the full metric or only the induced metric, it does not actually matter; for simplicity, let us suppose it depends on the induced metric). Then the combined action $S+S_\Sigma$ has variation $$ \delta\left(S+S_\Sigma\right)=\int_M\frac{1}{2\kappa}G_{\mu\nu}\delta g^{\mu\nu}\sqrt{\mathfrak g}\,\mathrm d^{n+1}x+\int_\Sigma\left(\frac{\varepsilon}{2\kappa}\pi_{ab}\sqrt{\mathfrak h}+\frac{\delta S_\Sigma}{\delta h^{ab}}\right)\delta h^{ab}\,\mathrm d^ny. $$

The bulk and boundary terms must vanish separately for $\delta(S+S_\Sigma)=0$, so we obtain the equation $$ 0=\frac{\varepsilon}{2\kappa}\pi_{ab}\sqrt{\mathfrak h}+\frac{\delta S_\Sigma}{\delta h^{ab}}. $$ If we set $$ T_{ab}^\Sigma:=-\frac{2}{\sqrt{\mathfrak h}}\frac{\delta S_\Sigma}{\delta h^{ab}}, $$ then this results in the equation $$ \pi_{ab}=\varepsilon\kappa T^\Sigma_{ab}. $$

Remark: It appears that OP's source either assumes that the boundary is timelike or it uses a different convention for the unit normal (i.e. inward pointing when the boundary is spacelike), hence why in the results contained in this answer, there are factors of $\varepsilon$ that do not appear in the OP.

$\endgroup$

Your Answer

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