Exact tree-level bulk elimination in the linear weak-field sector

Exact tree-level bulk elimination in the linear weak-field sector

The homogeneous sector evaluates the nonlinear local response but hides the spectral structure of the eliminated bulk gravitons. That structure is explicit at linear order [1, 6, 7].

Bulk perturbation equation, brane bending, and the boundary condition

In Randall–Sundrum gauge, a TT perturbation of the background (11) obeys $$\Bigl[ e^{2|y|/\ell}\,\Box_4 +\partial_y^2-\frac{4}{\ell^2} +\frac{4}{\ell}\,\delta(y) \Bigr]h^{\mathrm{TT}}_{\mu\nu} =-2\kappa_5^2\,\Sigma^{\mathrm{TT}}_{\mu\nu}\,\delta(y).\tag{51}$$ The source $\Sigma_{\mu\nu}$ contains the brane stress tensor and the brane-bending contribution required by the junction condition. In the Garriga–Tanaka convention [6], $$\boxed{\; \Sigma_{\mu\nu} =T_{\mu\nu}-\tfrac13\eta_{\mu\nu}T +\frac{2}{\kappa_5^2}\,\partial_\mu\partial_\nu\hat\xi^5, \qquad \Box_4\hat\xi^5=\frac{\kappa_5^2}{6}\,T . \;}\tag{52}$$ The bending equation makes $\Sigma^\mu{}_\mu=0$: indeed $\eta^{\mu\nu}\Sigma_{\mu\nu} =T-\tfrac43T+\tfrac{2}{\kappa_5^2}\Box_4\hat\xi^5 =-\tfrac13T+\tfrac13T=0$. Together with $\partial^\mu T_{\mu\nu}=0$ it also gives $\partial^\mu\Sigma_{\mu\nu}=0$, as RS gauge requires.

Brane bending is not an optional refinement. It is the statement that the brane, being a dynamical surface, is displaced from $y=0$ in RS gauge by the presence of matter; restoring the Gaussian-normal frame adapted to the physical brane produces, up to a tangential gauge transformation, $$\bar h_{\mu\nu} =h_{\mu\nu} +\ell\,\partial_\mu\partial_\nu\hat\xi^5 +\frac{2}{\ell}\,\eta_{\mu\nu}\hat\xi^5 -\partial_{(\mu}\hat\xi_{\nu)} .\tag{53}$$ Decomposing $h_{\mu\nu}=h^{(m)}_{\mu\nu}+h^{(\xi)}_{\mu\nu}$ into the part sourced by $T_{\mu\nu}-\tfrac13\eta_{\mu\nu}T$ and the part sourced by the bending term, and choosing the tangential gauge to cancel the pure gradient pieces, the brane value reduces to $$\boxed{\; \bar h_{\mu\nu} =h^{(m)}_{\mu\nu} +\frac{2}{\ell}\,\eta_{\mu\nu}\,\hat\xi^5 . \;}\tag{54}$$ This single term converts the apparent $\tfrac13$ trace coefficient of the RS-gauge source into the four-dimensional massless $\tfrac12$ coefficient in the zero-mode response , as Section 6.3 makes explicit. Omitting brane bending, or using the wrong trace coefficient, produces a graviton that is not the four-dimensional graviton and gives the wrong light-bending prediction.

The retarded Green function imposes causal response; regularity, i.e. absence of incoming radiation from the AdS horizon, selects the forced solution. If arbitrary incoming bulk radiation is admitted, a homogeneous solution must be added; that homogeneous term is exactly the initial-state contribution associated with the same-readout fiber of Section 3. The two constructions are the same physics seen twice.

Brane-to-brane propagator: zero mode and KK continuum

The brane-to-brane retarded Green function of (51) has the spectral resolution $$\boxed{\; \begin{aligned} G_R(p;0,0) &=G_R^{(0)}(p)+G_R^{(\mathrm{KK})}(p),\\ G_R^{(0)}(p) &=-\frac{1}{\ell}\, \frac{1}{-(\omega+i0^+)^2+\mathbf k^2},\\ G_R^{(\mathrm{KK})}(p) &=-\int_0^\infty dm\, \frac{u_m(0)^2}{m^2-(\omega+i0^+)^2+\mathbf k^2} . \end{aligned} \;}\tag{55}$$ The normalized continuum modes of the transverse operator are Bessel combinations, and their brane value gives the spectral weight $$\boxed{\; u_m(0)^2 =\frac{2}{\pi^2 m\ell\bigl[J_1^2(m\ell)+Y_1^2(m\ell)\bigr]} . \;}\tag{56}$$ Using the small-argument behavior $J_1(x)\simeq x/2$ and $Y_1(x)\simeq-2/(\pi x)$, one has $J_1^2+Y_1^2\simeq 4/(\pi^2x^2)$, hence $$u_m(0)^2 =\frac{m\ell}{2} +O\bigl((m\ell)^3\log(m\ell)\bigr), \qquad m\ell\ll1 .\tag{57}$$

Two structural facts follow. The first term in (55) is the normalizable four-dimensional graviton zero mode: an isolated massless pole. The second is a continuum of physical KK modes with nonnegative spectral density $u_m(0)^2\ge0$, producing a branch cut beginning at threshold $m=0$.

Tensor projectors and the response operator

At linear order the bulk gravitational action is quadratic. Integration over bulk modes is therefore Gaussian, and the resulting Schur complement is exact at tree level. We first retain the tensor structure that distinguishes the massless graviton from the massive continuum.

For four-momentum $p$, define the transverse tensor $$\theta_{\mu\nu} =\eta_{\mu\nu}-\frac{p_\mu p_\nu}{p^2+i0^+},\tag{58}$$ and the Barnes–Rivers spin projectors on symmetric tensors [20], $$\begin{aligned} P^{(2)}_{\mu\nu\rho\sigma} &=\tfrac12\bigl(\theta_{\mu\rho}\theta_{\nu\sigma} +\theta_{\mu\sigma}\theta_{\nu\rho}\bigr) -\tfrac13\theta_{\mu\nu}\theta_{\rho\sigma},\\ P^{(0)}_{\mu\nu\rho\sigma} &=\tfrac13\theta_{\mu\nu}\theta_{\rho\sigma}. \end{aligned}\tag{59}$$ (The $\tfrac13$ is $1/(D-1)$ with $D=4$, appropriate to the three-dimensional transverse subspace.)

The brane-bending field of Section 6.1 is precisely what converts the scalar coefficient of the massless part from $\tfrac13$ to $\tfrac12$: the term $\tfrac{2}{\ell}\eta_{\mu\nu}\hat\xi^5$ in (54) supplies a trace piece proportional to $\Box_4^{-1}T$, i.e. $-\tfrac16\theta_{\mu\nu}T$ times the zero-mode propagator. Consequently the complete gauge-invariant response to a conserved brane source, modulo longitudinal pure-gauge terms, is $$\boxed{\; \bar h_{\mu\nu}(p) =-\mathbb R_{R\,\mu\nu}{}^{\rho\sigma}(p)\,T_{\rho\sigma}(p), \;}\tag{62}$$ with $$\boxed{\; \mathbb R_R(p) =2\kappa_5^2\Bigl[ G_R^{(0)}(p)\Bigl(P^{(2)}-\tfrac12P^{(0)}\Bigr) +G_R^{(\mathrm{KK})}(p)\,P^{(2)} \Bigr]. \;}\tag{63}$$ Equation (63) fixes both the normalization and the relative tensor coefficients. It is equivalent to the RS-gauge source equation (51) together with the bending equation (52) and the transformation (54). It displays directly why the zero mode carries the four-dimensional massless trace coefficient while the KK continuum carries the massive spin-two coefficient , the physical content of the fact that RS2 has no vDVZ discontinuity in the observable weak-field limit.

Causal influence functional

The same result follows from a causal influence functional, which is the formulation that makes the initial-state dependence explicit [12]. Causality requires the closed-time-path fields to remain doubled. Set $$h_c=\tfrac12(h_++h_-), \qquad h_\Delta=h_+-h_- .\tag{64}$$ On the physical conserved-source subspace let $\mathbb K_R=\mathbb R_R^{-1}$. For equal physical sources on the two contour branches, the quadratic closed-time-path functional is $$\boxed{\; \Gamma^{(2)}_{\mathrm{CTP}}[h_c,h_\Delta;T] =\tfrac12\bigl\langle h_\Delta,\mathbb K_R h_c\bigr\rangle +\tfrac12\bigl\langle h_\Delta,T\bigr\rangle +\tfrac{i}{2}\bigl\langle h_\Delta,\mathbb N h_\Delta\bigr\rangle , \;}\tag{65}$$ where the pairing includes momentum integration and contraction on the conserved tensor subspace. The factor $\tfrac12$ in the source term is the standard linear metric coupling $\delta S_m=\tfrac12\int T^{\mu\nu}\delta h_{\mu\nu}$. The symmetric kernel $\mathbb N$ is fixed by the chosen bulk density matrix; it controls fluctuations and does not enter the mean classical field. Varying (65) with respect to $h_\Delta$ and then taking the physical limit $h_\Delta=0$ gives $$\tfrac12\,\mathbb K_R h_c+\tfrac12\,T=0 \qquad\Longrightarrow\qquad h_c=-\mathbb R_R\,T ,\tag{66}$$ reproducing (62). Restricted to the TT source in RS gauge this reduces to $$h_c^{\mathrm{TT}}=-2\kappa_5^2\,\mathcal G_R\,\Sigma^{\mathrm{TT}},\tag{67}$$ where $\mathcal G_R$ denotes the retarded convolution operator whose momentum-space kernel is the brane-to-brane Green function $G_R(p;0,0)=G_R^{(0)}(p)+G_R^{(\mathrm{KK})}(p)$ of (55), acting as $$\bigl(\mathcal G_R\,\Sigma^{\mathrm{TT}}\bigr)(x) =\int d^4x'\;\mathcal G_R(x-x')\,\Sigma^{\mathrm{TT}}(x'), \qquad \mathcal G_R(x-x') =\int\frac{d^4p}{(2\pi)^4}\;e^{ip\cdot(x-x')}\,G_R(p;0,0).\tag{68}$$ This recovers the normalization used in (51). Thus the standard $\tfrac12 h_{\mu\nu}T^{\mu\nu}$ matter coupling reproduces the response normalization rather than fixing it by convention.

The retarded prescription fixes causality. In the positive-tension model the KK spectral density (56) is positive, so the linear continuum introduces no negative-norm graviton. The negative-tension branch instead gives the wrong sign of the four-dimensional Newton constant and is not used.

Evaluated long-distance response

For a static source and $r\gg\ell$, the three-dimensional massive Green function is $-e^{-mr}/(4\pi r)$. The small-$m$ spectral weight (57) therefore evaluates the KK contribution rather than leaving it as a formal kernel: $$G^{(\mathrm{KK})}(r) =-\frac{1}{4\pi r}\int_0^\infty dm\,u_m(0)^2\,e^{-mr} =-\frac{1}{4\pi r}\cdot\frac{\ell}{2}\int_0^\infty dm\,m\,e^{-mr} +\cdots =-\frac{\ell}{8\pi r^3} +O\!\Bigl(\frac{\ell^3}{r^5}\log\frac r\ell\Bigr),\tag{69}$$ using $\int_0^\infty m e^{-mr}dm=r^{-2}$. Adding the zero mode $G^{(0)}(r)=-1/(4\pi\ell r)$ gives the brane scalar Green function $$G(\mathbf x,0;\mathbf x',0) =-\frac{1}{4\pi\ell r} \Bigl[1+\frac{\ell^2}{2r^2} +O\!\Bigl(\frac{\ell^4}{r^4}\log\frac r\ell\Bigr)\Bigr], \qquad r=|\mathbf x-\mathbf x'| .\tag{70}$$

For a nonrelativistic point source, $T_{00}=M\delta^3(\mathbf x)$ and $T=-M\delta^3(\mathbf x)$, so the bending equation (52) gives $$\nabla^2\hat\xi^5=-\frac{\kappa_5^2M}{6}\,\delta^3(\mathbf x), \qquad \hat\xi^5(r)=\frac{\kappa_5^2M}{24\pi r} .\tag{71}$$ Substituting (70) into $h^{(m)}=-2\kappa_5^2\,G*\bigl(T_{\mu\nu}-\tfrac13\eta_{\mu\nu}T\bigr)$, adding $2\eta_{\mu\nu}\hat\xi^5/\ell$ as in (54), and using $G_4=\kappa_5^2/(8\pi\ell)$ (Section 6.6), one obtains, outside a nonrelativistic spherical source of mass $M$, $$\boxed{\; \bar h_{00} =\frac{2G_4M}{r} \Bigl[1+\frac{2\ell^2}{3r^2} +O\!\Bigl(\frac{\ell^4}{r^4}\log\frac r\ell\Bigr)\Bigr], \;}\tag{72}$$ $$\boxed{\; \bar h_{ij} =\frac{2G_4M}{r} \Bigl[1+\frac{\ell^2}{3r^2} +O\!\Bigl(\frac{\ell^4}{r^4}\log\frac r\ell\Bigr)\Bigr]\delta_{ij}. \;}\tag{73}$$ Equivalently, in asymptotically Minkowski Newtonian gauge, $$ds^2=-(1+2\Phi)\,dt^2+(1-2\Psi)\,d\mathbf x^2,\tag{74}$$ the gauge-invariant static potentials are $$\boxed{\; \Phi(r)=-\frac{G_4M}{r} \Bigl(1+\frac{2\ell^2}{3r^2}+\cdots\Bigr), \qquad \Psi(r)=-\frac{G_4M}{r} \Bigl(1+\frac{\ell^2}{3r^2}+\cdots\Bigr). \;}\tag{75}$$ This reproduces the Garriga–Tanaka result [6].

Three observations. First, the correction is not a mass renormalization: it has a different radial scaling and cannot be absorbed into $G_4M$. Second, the coefficients in $\Phi$ and $\Psi$ are unequal, which correlates nonrelativistic acceleration with light deflection: the former probes $\Phi$, the latter $\Phi+\Psi$. Third, in momentum space the same physics is the KK branch cut of (55), not an extra isolated pole.

The result is controlled by $$\frac{G_4M}{r}\ll1, \qquad \frac{\ell}{r}\ll1 ,\tag{76}$$ and assumes the minimal two-derivative bulk action and the regular retarded bulk state. It is not an expansion valid at $r\lesssim\ell$, where the full spectral integral (56) must be used.

One-sided versus two-sided normalization

Two conventions occur in the literature and must not be mixed. In the two-sided covering-space convention used throughout this paper, $$y\in(-\infty,\infty), \qquad a(y)=e^{-|y|/\ell},\tag{77}$$ and the zero-mode reduction gives $$\frac{1}{\kappa_4^2} =\frac{1}{\kappa_5^2}\int_{-\infty}^{\infty}dy\;a(y)^2 =\frac{1}{\kappa_5^2}\int_{-\infty}^{\infty}dy\;e^{-2|y|/\ell} =\frac{\ell}{\kappa_5^2},\tag{78}$$ so that $$\boxed{\; \kappa_4^2=\frac{\kappa_5^2}{\ell}, \qquad G_4=\frac{G_5}{\ell}. \;}\tag{79}$$ If instead one works on a single AdS patch $y\ge0$ with the same undoubled bulk normalization, the integral is $\ell/2$ and $$\kappa_4^2=\frac{2\kappa_{5,\mathrm{one}}^2}{\ell}, \qquad G_4=\frac{2G_{5,\mathrm{one}}}{\ell}.\tag{80}$$ Equivalently, the $Z_2$ orbifold fundamental domain may be assigned a doubled bulk action, reproducing the two-sided convention. The weak-field coefficients $2\ell^2/3$ and $\ell^2/3$, expressed in terms of the measured $G_4$, are unchanged provided the gravitational coupling and the number of AdS patches are converted consistently.

The same distinction appears in the four-dimensional dual description (Section 10). One AdS patch corresponds to one cutoff CFT contribution; a two-sided geometry has two identical bulk regions and hence a doubled CFT polarization contribution. Duff and Liu [9] present the one-patch normalization explicitly; doubling both the bulk contribution and its CFT counterpart gives the two-sided convention used here. Their equality of the nonanalytic $1/r^3$ residue therefore remains valid, but only after this conversion. No equation in this paper combines a one-sided Newton-constant relation with a two-sided junction condition.