The Randall-Sundrum Model, a Brane-Singularity Problem and AdS-Compactification

By George Shiber Tuesday, March 15, 2016Saturday, August 17, 2019 In Cosmology, Grand Unification Physics, M Theory, Mathematical Physics, Mathematics, Philosophy Of Physics, Philosophy Of Science, Physics, Super String Theory, Supersymmetry AdS/CFT, Branes

Let no man who is not a mathematician read the elements of my work ~ Leonardo da Vinci

In my last post, I analyzed a relationship between the AdS/CFT correspondence and the Randall-Sundrum brane-world ‘model‘, where the brane-world five-dimensional action is

$\begin{array}{c}S = \int {{d^5}} x\sqrt { - {g_{\left( 5 \right)}}} \left[ {{M^3}{R_{\left( 5 \right)}} - \Lambda } \right]\\ + \int {{d^4}x} \sqrt { - {g_{\left( 4 \right)}}} \widetilde {\not L}_{{\rm{BRANE}}}^{{\rm{Lag}}}\end{array}$

with $M$ being the five-dimensional Planck mass, ${M^3} = 1/\left( {16\pi {G_5}} \right)$ , and $\Lambda$ is the cosmological constant in the bulk, yielded us, via s mall fluctuations of the metric on the brane

$\begin{array}{c}d{s^2} = {e^{ - 2\left| y \right|/L}}\left[ {{\eta _{\mu \nu }} + {h_{\mu \nu }}\left( {x,y} \right)} \right] \cdot \\d{x^\mu }d{x^\nu } + d{y^2}\end{array}$

with $L$ being the radius of AdS, defined by

$R_{MNPQ}^{\left( 5 \right)} = - \frac{1}{{{L^2}}}\left( {g_{MP}^{\left( 5 \right)}g_{NQ}^{\left( 5 \right)} - g_{MQ}^{\left( 5 \right)}g_{NP}^{\left( 5 \right)}} \right)$

and then deduced the crucial relation $\Lambda = - 12{M^3}/{L^2}$ , which yields the CFT-brane relation

$\begin{array}{l}{h_{\mu \nu }}\left( p \right) = - \frac{2}{{L{M^3}}}{\Delta _3}\left[ {{T_{\mu \nu }}\left( p \right) - \frac{1}{2}{\eta _{\mu \nu }}{T^\alpha }_\alpha \left( p \right)} \right]\\ - \frac{1}{{{M^3}}}{\Delta _{KK}}\left( p \right)\left[ {{T_{\mu \nu }}\left( p \right) - \frac{1}{3}{\eta _{\mu \nu }}{T^\alpha }_\alpha \left( p \right)} \right]\end{array}$

possessing real physical meaning, as can be seen by

the fact that the four-dimensional massless propagator ${\Delta _4}\left( p \right)$ corresponds exactly to the zero-mode graviton, localized on the brane, while

${\Delta _{KK}}\left( p \right) = - \frac{1}{p}\frac{{{K_0}\left( {pL} \right)}}{{{K_1}\left( {pL} \right)}}$

being the propagator for the continuum Kaluza–Klein graviton modes, gives us an analytic relation between four-and-five-dimensional Newton’s constants. So, at large distances, corresponding to $pL \ll 1$ , a small-argument expansion for Bessel functions gives us

$\begin{array}{c}{\Delta _{KK}}\left( p \right) = \frac{L}{2}\left( {{\rm{In}}\frac{{{p^2}{L^2}}}{4} + 2y} \right) + \\\vartheta {\left( p \right)^2}\end{array}$

Thus, by evaluating the Fourier transform for $r \gg L$ we get the linearized metric

$\left\{ {\begin{array}{*{20}{c}}{{h_{00}} = \frac{{2Gm}}{r}\left( {1 + \frac{{2{L^2}}}{{3{r^2}}} + ...} \right)}\\{{h_{ij}} = \frac{{2Gm}}{r}\left( {1 + \frac{{{L^2}}}{{3{r^2}}} + ...} \right){\delta _{ij}}}\end{array}} \right.$

giving ‘real’ physical meaning to the Newtonian potential: hence the $1/{r^3}$ –corrections to Newton’s law are identical between the Maldacena ‘model’ and Randall–Sundrum ‘model’ which entails a deep classical/quantum duality expressed by

$\prod {_2} \left( p \right) + \vartheta \left( {{G^2}} \right) = \frac{L}{4}{\Delta _{KK}}\left( p \right)$

However, let me describe a serious problem related to a type of singularity and solve it by RR-compactification on an AdS minus a CFT-brane, with no loss of metaplectic content. As was shown, the AdS/CFT correspondence implies that the RS model is isomorphic to a four dimensional gravity coupled to a strongly interacting CFT. First note that on the RS side, the singular brane in the bulk is very crucial for the two massless modes and two towers of continuum-massive-modes on AdS, and so the field equations pick up a delta function source and the RS background constitutes a solution: the problem now is that we lose the delta function because the brane lives on the boundary, and it would follow that the brane-action cannot modify the field equations which are deduced by fixing the metric on the boundary thus one cannot include the degrees of freedom corresponding to the path integral. I will try a solution based on replacing the brane-action in the bulk with a term proportional to the area of the boundary of the compactified AdS space.

Before proceeding, keep this following relation in mind, it is as deep as it gets

$\begin{array}{c}Z\left[ \gamma \right]{e^{i{S_1}\left[ \gamma \right]}}\exp \left[ { - \frac{i}{{8\pi {G_{d + 1}}}}} \right.\int_{\partial M} {\sqrt { - \gamma } } \\\left( {\frac{{d - 1}}{l}} \right. + \frac{l}{{2d - 4}}{R_{icci}} + \left. {\left. {...} \right)} \right] \cdot \\{\left\langle {{e^{i{\gamma _{\mu \nu {T^{\mu \nu }}}}}}} \right\rangle _{CFT}}\end{array}$

In d + 1-dimensions, the field equations for gravity with a negative cosmological constant Λ are

${R_{AB}} - \frac{1}{2}{G_{AB}}R = \Lambda {G_{AB}}$

that admit the AdS space

$d{s^2} = \frac{{{l^2}}}{{{z^2}}}\left( {d{z^2} + {\eta _{\mu \nu }}d{x^\mu }d{x^\nu }} \right)$

as a solution with ${l^2} = - d\left( {d - 1} \right)/\left( {2\Lambda } \right)$ and taking $z$ as the radial coordinate, AdS space can be interpreted as the metric of a domain wall that models our observable universe located at an arbitrary radial position, and spanned by the coordinates ${x^\mu }$ . It is clear that without matter, the metric on the domain wall is flat. With matter on the wall, one would replace the flat metric ${\eta _{\mu \nu }}$ with a curved metric ${g_{\mu \nu }}(x)$ , hence recovering the two basic properties of gravity without a cosmological constant. Now, from an effective action point of view, in d + 1- dimensions the action is

$\begin{array}{c}S = \frac{1}{{16\pi {G_{d + 1}}}}\int_M {\sqrt { - G} } \left( {R - 2\Lambda } \right)\\ + \frac{1}{{8\pi {G_{d + 1}}}}\int_{\partial M} {\sqrt { - \gamma } } K\end{array}$

with ${G_{d + 1}}$ the Newton’s constant and ${\gamma _{\mu \nu }}$ the boundary-metric, yielding a Chern-Simons constant $a$ , giving us the modified action

${S_1} = \frac{a}{{16\pi {G_{d + 1}}}}\int_{\partial M} {\sqrt { - \gamma } }$

which is proportional to the area of the boundary. The condition for the metric is

$d{s^2} = \frac{{{l^2}}}{{{z^2}}}\left( {d{z^2} + {g_{\mu \nu }}(x)d{x^\mu }d{x^\nu }} \right)$

and even though the boundary is located at $z = 0$ , a solution is gotten by cutting the asymptotic region at the surface $z = \varepsilon$ and let the space be denoted by $Ad{S_\varepsilon }$ . By inserting

$d{s^2} = \frac{{{l^2}}}{{{z^2}}}\left( {d{z^2} + {g_{\mu \nu }}(x)d{x^\mu }d{x^\nu }} \right)$

into the total action $S + {S_1}$ and picking $al = - 2\left( {d - 1} \right)$ , one gets

$S + {S_1} = \frac{1}{{16\pi {G_d}}}\int {{d^d}} \_\sqrt { - \gamma } {R_{icci}}$

with the d-dimensional Newton’s constant is given by

${G_d} = \left( {s - 2} \right)\frac{{{\varepsilon ^{d - 1}}}}{{{l^{d - 1}}}}{G_{d + 1}}$

To get an effective action that describes pure gravity without a cosmological constant, we start with the total action $S + {S_1}$ where the free parameter $a$ in ${S_1}$ is fixed by $al = - 2\left( {d - 1} \right)$ .

The central point now is that in the RS model, ${S_1}$ has been replaced with a singular brane action

$S = \tau \int_\Sigma {\sqrt { - \gamma } }$

coupled to the gravity, where $\Sigma$ is the world-volume embedded in the bulk, and the key point is that the tension $\tau$ of the brane would be fine-tuned, pushing the cosmological constant problem into the brane! A solution suggests itself by noting that ${S_1}$ with coefficient in $al = - 2\left( {d - 1} \right)$

come from an AdS-compactification of string/M theory, thus getting a conformally invariant graviton 2-point function using AdS/CFT duality, in turn solving the cosmological constant problem not by fine tuning but by the presence of a critical boundary action induced by such compactification.

Hence, from

$d{s^2} = \frac{{{l^2}}}{{{z^2}}}\left( {d{z^2} + {g_{\mu \nu }}(x)d{x^\mu }d{x^\nu }} \right)$

we get

$\begin{array}{c}{R_{ABCD}}{R^{ABCD}} = \frac{{2d\left( {d + 1} \right)}}{{{l^4}}} + \\\frac{{{z^4}}}{{{l^4}}}{R_{\mu \nu \sigma \rho }}{R^{\mu \nu \sigma \rho }}\end{array}$

with ${R_{\mu \nu \sigma \rho }}$ the curvature tensor of ${g_{\mu \nu }}$ ,

which indicates a generic curvature singularity on the ‘horizon’

at $z = \infty$ . A solution can be achieved by insersion of a wall located at a finite $z$ value, giving rise to a radial coordinate compactified on a line, which is quasi-morphic to a Horava-Witten construction in 11-dimensions, where one coordinate of the flat Minkowski space has been compactified by two 10-branes. So, let me determine the spectrum of modes on $Ad{S_\varepsilon }$ , implying $\delta {n_A} = 0$ and $\delta {n^A} = 0$ on $\partial M$ , where ${n_A}$ is the unit normal vector to the boundary. We now vary the action $S + {S_1}$ by treating the boundary terms coming from integration-by-parts as Picard-Lefschetz terms, to obtain

$\begin{array}{c}\delta \left( {S + {S_1}} \right) = \frac{1}{{16\pi {G_{d + 1}}}}\int_M {\sqrt { - G} } \cdot \\\left( {{R_{AB}}} \right. - \frac{1}{2}{G_{AB}}R - \Lambda \left. {{G_{AB}}} \right)\delta {G^{AB}}\\ + \frac{1}{{16\pi {G_{d + 1}}}}\int_{\partial M} {\sqrt { - \gamma } } \left( {{K_{AB}}} \right. - \\{\gamma _{AB}}K - \frac{a}{2}\left. {{\gamma _{AB}}} \right)\delta {\gamma ^{AB}} = 0\end{array}$

with ${\gamma _{AB}}$ the induced metric on $\partial M$ , ${K_{AB}}$ the extrinsic curvature of the boundary defined by

${K_{AB}} = {\gamma _A}^C{\gamma _B}^D{\nabla _C}{n_D}$

and $K = {\gamma ^{AB}}{K_{AB}}$

Hence, the metric near $\partial M$ is

$d{s^2} = \frac{{{l^2}}}{{{z^2}}}d{z^2} + {\gamma _{\mu \nu }}\left( {x,z} \right)d{x^\mu }d{x^\nu }$

where the boundary is located at $z = \varepsilon$ and $n = - \left( {z/l} \right){\not \partial _z}$ is the unit normal vector. So the following holds

$\left\{ {\begin{array}{*{20}{c}}{\delta {G_{AB}} \to \delta {\gamma _{\mu \nu }}}\\{\delta {G^{AB}}{n_B} = 0}\end{array}} \right.\quad \quad {\rm{on }}\partial M$

and the extrinsic curvature is given by

${K_{\mu \nu }} = - \varepsilon /\left( {2l} \right){\not \partial _z}\gamma \left| {_{z = \varepsilon }} \right.$

so the field equations following from $\delta \left( {S + {S_1}} \right)$ are

$\left\{ {\begin{array}{*{20}{c}}{{R_{AB}} - \frac{1}{2}{G_{AB}}R - \Lambda {G_{AB}} = 0}\\{\left( {{K_{AB}} - {\gamma _{AB}}K - \frac{a}{2}{\gamma _{AB}}} \right)\left| {_{\partial M}} \right. = 0}\end{array}} \right.$

One can thus see that $Ad{S_\varepsilon }$ satisfies the immediate-above condition two only if the free parameter $a$ is fixed by $al = - 2\left( {d - 1} \right)$ .

The ‘magic’ now is, the same fine tuning imposed to cancel the induced cosmological constant on the boundary is now required to have $Ad{S_\varepsilon }$ as a solution of the theory. The metric of the wall in

$d{s^2} = \frac{{{l^2}}}{{{z^2}}}\left( {d{z^2} + {\eta _{\mu \nu }}d{x^\mu }d{x^\nu }} \right)$

is now

${g_{\mu \nu }} = {\eta _{\mu \nu }} + {h_{\mu \nu }}\left( {x,z} \right)$

and with the following gauges constraints

$\left\{ {\begin{array}{*{20}{c}}{{{\not \partial }_\mu }{h^\mu }_\nu = 0}\\{{h^\mu }_\mu = 0}\end{array}} \right.$

the linearized equations become

$\begin{array}{c}{z^2}\not \partial _z^2{\widehat h_{\mu \nu }} - \left( {d - 1} \right)z{{\not \partial }_z}{\widehat h_{\mu \nu }} + \\{m^2}{z^2}{\widehat h_{\mu \nu }} = 0\end{array}$

and

${\not \partial _z}{\widehat h_{\mu \nu }}\left| {_{z = \varepsilon }} \right. = 0$

with ${h_{\mu \nu }} = {e^{ipx}}{\widehat h_{\mu \nu }}\left( z \right)$ and $m$ being the d-dimensional mass given by ${m^2} = - {p^2}$ , hence, the solutions are

$\widehat h = \left\{ {\begin{array}{*{20}{c}}{{\rm{constant , }}m = 0}\\{{N_m}{{\left( {z/l} \right)}^{d/2}}\left( {{A_m}{J_{d/2}}\left( {mz} \right) + {B_m}{Y_{d/2}}\left( {mz} \right)} \right)}\end{array}} \right.{\rm{ , }}m \ne 0$

with ${N_m}$ the normalization constant and ${A_m} = {Y_{d/2 - 1}}\left( {m\varepsilon } \right)$ , ${B_m} = - {J_{d/2}}\left( {m\varepsilon } \right)$ . Let us check the regularity and the normalizability conditions, which are given by

$\left\{ {\begin{array}{*{20}{c}}{\delta {G_{zz}} = 0}\\{\delta {G_{z\mu }} = 0}\\{\delta {G_{\mu \nu }} = \frac{{{l^2}}}{{{z^2}}}{h_{\mu \nu }}}\end{array}} \right.$

viewed in d + 1-dimensional space, and note the above regularity and normalizability conditions indicate that, contrary to the gravitational waves propagating along the boundary, the ones moving through the horizon are irregular. A similar singularity arises in the Randall–Sundrum model. Let me concentrate on the problem in Euclidean signature. For normalizability, I must introduce an inner-product in the space of linearized perturbations. Noting that

$\begin{array}{c}{z^2}\not \partial _z^2{\widehat h_{\mu \nu }} - \left( {d - 1} \right)z{{\not \partial }_z}{\widehat h_{\mu \nu }} + \\{m^2}{z^2}{\widehat h_{\mu \nu }} = 0\end{array}$

is the Laplacian acting on the scalars, let me define

$\begin{array}{c}\left\langle {h,h'} \right\rangle = i\int_\Sigma {\left( {h{{\not \partial }_\mu }{{h'}^ * } - {{h'}^ * }} \right.} {{\not \partial }_\mu }\left. h \right) \cdot \\{n^\mu }\not \partial \Sigma \end{array}$

with where $\omega \equiv {p_0}$ , and

$I\left( {m,m'} \right) = \int_\varepsilon ^\infty {dz} \frac{{{l^{d - 1}}}}{{{z^{d - 1}}}}\widehat h{\widehat h^\prime }$

and from it, the massless mode is normalizable if and only if $\varepsilon \ne 0$ and for massive modes, via the Bessel’s differential equation, we get

$\begin{array}{c}I\left( {m,m'} \right) = \frac{{{N_m}{N_{m'}}}}{{{m^2} - {{m'}^2}}}\left[ {\frac{z}{l}} \right.\left( {{f_m}} \right.\frac{d}{{dz}}{f_{m'}}\\ - {f_{m'}}\frac{d}{{dz}}\left. {\left. {{f_m}} \right)} \right]_\varepsilon ^\infty \end{array}$

Analyzing the $z = 0$ limit using the asymptotic forms of the Bessel functions, one finds

$\begin{array}{c}I\left( {m,m'} \right) = \frac{{\sqrt {2\pi } }}{{ml}}\left( {A_m^2 + B_m^2} \right) \cdot \\\delta \left( {m - m'} \right)\end{array}$

now picking

${N_m} = {\left( {ml/\left[ {2\omega \left( {A_m^2 + B_m^2} \right)} \right]} \right)^{1/2}}$

the massive modes are normalized as

$\left\langle {h,h'} \right\rangle = {\left( {2\pi } \right)^{d - 1}}\sqrt {2\pi } \delta \left( {\vec p - \vec p'} \right)\delta \left( {m - m'} \right)$

and since the product of the massless mode with a massive KK mode turns out to be zero since

$I\left( {m,m'} \right) = \int_\varepsilon ^\infty {dz} \frac{{{l^{d - 1}}}}{{{z^{d - 1}}}}\widehat h{\widehat h^\prime }$

reduces to

$\begin{array}{c}I\left( {m,0} \right) \sim \int_\varepsilon ^\infty {zdz} {f_m}{z^{ - d/2}} \sim \\\frac{1}{{{m^2}}}\left[ {z\left( {{f_m}} \right.} \right.\frac{d}{{dz}} - {z^{ - d/2}}\frac{d}{{dz}}\left. {\left. {{f_m}} \right)} \right]_\varepsilon ^\infty = 0\end{array}$

so now we see, from

$\left\langle {h,h'} \right\rangle = {\left( {2\pi } \right)^{d - 1}}\sqrt {2\pi } \delta \left( {\vec p - \vec p'} \right)\delta \left( {m - m'} \right)$

that the measure on the set of continuum eigenvalues is simply $dm$ . Since the dimensionless coordinate on $Ad{S_\varepsilon }$ is $z/l$ , the dimensionless eigenvalues and the measure are $lm$ and $ldm$ , respectively. Therefore, the graviton propagator can be gotten by superposing the modes

given their completeness. Hence, must solve

${\overline {\not \Theta } _L}\Delta \left( {x,z;x',z'} \right) = \frac{{{\delta ^d}\left( {x - x'} \right)\delta \left( {z - z'} \right)}}{{\sqrt { - G} }}$

where ${\overline {\not \Theta } _L}$ being the brane-bulk Laplacian and the boundary condition

${\not \partial _z}\Delta \left| {_{z = \varepsilon }} \right. = 0$

is entailed by

${\not \partial _z}{\widehat h_{\mu \nu }}\left| {_{z = \varepsilon }} \right. = 0$

This Green function is localized near the boundary, and ${\overline {\not \Theta } _L}$ can be separated into the standard d-dimensional propagator plus a contribution coming from the exchange of KK modes, which in momentum space reads

$\Delta \left( p \right) = \frac{{d - 2}}{\varepsilon }\frac{1}{{{p^2}}} + {\Delta _{KK}}\left( p \right)$

where

${\Delta _{KK}} = - \frac{1}{p}\frac{{H_{d/2 - 2}^{\left( 1 \right)}\left( {p\varepsilon } \right)}}{{H_{d/2 - 1}^{\left( 1 \right)}\left( {p\varepsilon } \right)}}$

and ${H^{\left( 1 \right)}}$ is the first Hankel function given by ${H^{\left( 1 \right)}} = J + iY$ . To proceed, let me start from the following path integral

$Z = \int {\left[ {dG} \right]} \,{e^{iS + {S_1}}}$

summing over all metrics on M. It can be evaluated using saddle point approximation when there is an extremum of the functional $S + {S_1}$ under these conditions

The path integral above allows us to connect the existence of a complementary picture implied by the AdS/CFT correspondence

yielding

$Z = \int {\left[ {d\gamma } \right]} \,Z\left[ \gamma \right]$

with

$Z\left[ \gamma \right] = {e^{i{S_1}\left[ \gamma \right]}}\int {\left[ {dG} \right]} G\left| {_{\partial M = \gamma }} \right.{e^{iS}}$

holding. And by the AdS/CFT duality, we get

For a minimally coupled massless scalar field on $Ad{S_\varepsilon }$ , one starts from the action

${S_S} = \int_M {\sqrt { - G} } {\nabla _A}\phi {\nabla ^A}\phi$

with the scalar allowed to vary on the boundary, getting

$\left\{ {\begin{array}{*{20}{c}}{{{\overline {\not \Theta } }_L}\phi = )}\\{{n^A}{\nabla _A}\phi \left| {_{\partial M} = 0} \right.}\end{array}} \right.$

and in the AdS/CFT picture, the partition function

$Z = \int {\left[ {d\phi } \right]} \,{e^{i{S_S}}}$

which can be calculated as thus

$Z = \int {\left[ {d{\phi _0}} \right]} {\left[ {d\phi } \right]_{\phi \left| {_{\partial M = {\phi _0}}} \right.}}{e^{i{S_S}}}$

summing is over all bulk fields. By the AdS/CFT correspondence, one derives

$\begin{array}{c}Z = \int {\left[ {d{\phi _0}} \right]} \exp \left( {i\int {{d^d}} } \right.x\frac{1}{{2\left( {d - 2} \right)}}\\\left( {{\phi _0}\not \partial _x^2{\phi _0} + \left. {...} \right)} \right.{\left\langle {{e^{i{\phi _{0\not {\rm O}}}}}} \right\rangle _{CFT}}\end{array}$

$\not {\rm O}$ being the dual CFT operator. And we have a solution to the RS-singularity problem via AdS-compactification

${\Delta _{KK}} = - \frac{1}{p}\frac{{H_{d/2 - 2}^{\left( 1 \right)}\left( {p\varepsilon } \right)}}{{H_{d/2 - 1}^{\left( 1 \right)}\left( {p\varepsilon } \right)}}$