# Understanding boundary conditions on slices of AdS5

This is a thing Iïve seen on many papers dealing with Warped Extra Dimensions, specifically on slices of AdS5. But the one where it appears more clearly is a lecture by Tony Gherghetta:

http://arxiv.org/abs/hep-ph/0601213

Essentially what is done is that one builds a 5-dimensional theory in a slice of a Anti-De-Sitter space, the slice meaning that the fifth dimension has a small size and ends with a 4D brane at each extreme.

The principle of least action applied to the 5D action leads us to two terms (eq. 7 in the paper above), one on the bulk (the vanishing of this one leads us to the bulk equation of motion) and one on the branes. The vanishing of this second term can be accomplished in two ways:

1) the variation of the field on the branes is zero

2) the term multiplying the variation is zero on the branes (let?s call it B)

Now, all authors affirm that, in absence of extra terms on the brane, this two conditions lead to Neumann or Dirichlet conditions for the field on the branes (in the paper above this is said just after equation 10). My question is: why is that?

For scalar fields it can be shown that B equals the derivative of the field, so I can see the Neumann condition there (am I wrong?). But the 1st condition says the variation of the field is zero not the field itself. Clearly I am misunderstanding something here...

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## 3 Answers

The answer is well-known from the analysis of open strings in perturbative string theory: at the end points, the coordinates $X(\sigma)$ must also satisfy either Neumann or Dirichlet boundary conditions for the very same reason that you mentioned.

In fact, I think that you have almost answered your own question.

1. The variation of the field is zero. But the very point of the variational calculus - or the principle of the least action - is that the variations may change an allowed configuration to any other nearby allowed configuration. So if you want a constraint of the kind $\delta X(0)=0$ - or, in your AdS case, $\delta \phi|_{y*} = 0$, the only reason to guarantee it is to impose the condition $X(0)=const$ or $\phi|_{y*}=const$ for the allowed configurations. You may imagine that the constant is zero - it's just a convention. But because it's a constant, the variation - the difference of the value of $X(0)$ between two nearby allowed configuration - vanishes. Setting a scalar field at the boundary equal to a constant is called the Dirichlet boundary condition.

2. Alternatively, the boundary term in the variation - the second term in equation 7 - may vanish because the arbitrary nonzero $\delta \phi$ is multiplied by a vanishing coefficient. The coefficient is called $B_{curly}\phi$ in the equation but $B_{curly}$ is an operator proportional to $\partial_5$. So this method to make the boundary term vanish says that $\partial_5 \phi=0$ at the boundary, and this condition is called the Neumann boundary condition. Just to be sure, the $D_{curly}$ in the first term of equation 7 is also an operator - and it contains second derivatives. Note that the first term has second derivatives integrated over the 5D space, while the second term has first derivatives integrated over the 4D space, so the dimensions agree.

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In the case of the string, saying $X(0) = 0$ is indeed a convention, because there's a translational symmetry. But for more general fields, there's physics in which constant you choose. – Matt Reece Jan 21 '11 at 21:59
I kind of agree even though moduli (scalar fields with no potential) could have any value of the constant, too. And on the contrary, the translational symmetry in string theory may also be broken, and even branes could perhaps be stabilized at some points. So to summarize, I kind of disagree that there is any difference. $X(\sigma,\tau)$ is just a field in a quantum field theory - a rather general two-dimensional CFT - so you can't possibly find any universal "qualitative difference" from other QFTs. – Luboš Motl Jan 21 '11 at 22:16

"Dirichlet" in general means you specify the field value on the boundary (and don't allow it to fluctuate, setting the variation to zero), not necessarily that that boundary value is zero. But if it's nonzero, it corresponds in AdS/CFT to turning on a source for some operator, i.e. deforming the Lagrangian by a term like $J(x){\cal O}(x)$. So the minimal thing to do, without changing the theory, is to set this source to zero.

(Fair warning: this is off-the-cuff, without thinking carefully about the $\epsilon \to 0$ limit that replaces a UV brane with conditions on the field behavior at the AdS conformal boundary. Saying that this is setting a source to zero might be a slightly wrong interpretation at finite but small $\epsilon$.)

The interpretation of the case with an IR brane is more subtle, but you might imagine justifying any choice $\phi = \phi_0$ by adding a boundary-localized potential $V(\phi)$ with minimum at $\phi_0$ and taking the limit as it becomes steep.

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The commonly stated claim that we can write down any old effective brane action on the IR brane in Randall-Sundrum models compatible with the orbifold projections because, well, anything can come out from a strongly coupled CFT in the dual picture has always struck me as dubious. There ought to be a swampland criteria against that. For instance, in the Hořava-Witten model, consistency conditions force the boundaries of space to admit an $E_8$ gauge group. – QGR Jan 22 '11 at 10:00
@QGR: I absolutely agree. Generally the IR brane isn't really meant to be a "brane" at all, but is serving as a proxy for something like the dynamics that caps off the throat in a solution like Klebanov-Strassler. From the AdS/CFT perspective you would definitely expect that the dynamics you specify in the UV determine what happens there. I was just trying to give an answer within the usual 5d EFT rules that people follow when building these models.... – Matt Reece Jan 22 '11 at 15:19

The point is that you can satisfy the boundary condition $\delta\phi = 0$ by taking $\phi = const$. This is all that is required to satisfy the variational principle. In other words $\phi = const \implies \delta \phi = 0$ but not vice versa.

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