The exact projected Einstein equation

The exact projected Einstein equation

Israel junction condition

Let the brane surface stress tensor be $$S_{\mu\nu}=-\lambda q_{\mu\nu}+T_{\mu\nu}, \qquad S\equiv q^{\mu\nu}S_{\mu\nu}=-4\lambda+T .\tag{22}$$ The Israel condition [3] for a thin shell in five dimensions is $$[K_{\mu\nu}]\equiv K^+_{\mu\nu}-K^-_{\mu\nu} =-\kappa_5^2\Bigl(S_{\mu\nu}-\tfrac13q_{\mu\nu}S\Bigr),\tag{23}$$ the factor $\tfrac13$ being $1/(D-2)$ with $D=5$. Imposing the RS2 $Z_2$ reflection symmetry gives $K^+_{\mu\nu}=-K^-_{\mu\nu}$, so $[K_{\mu\nu}]=2K_{\mu\nu}$ evaluated with the outward normal on one side, and hence $$\boxed{\; K_{\mu\nu} =-\frac{\kappa_5^2}{2} \Bigl[\,T_{\mu\nu}+\tfrac13(\lambda-T)q_{\mu\nu}\,\Bigr]. \;}\tag{24}$$ Indeed, $S_{\mu\nu}-\tfrac13q_{\mu\nu}S =T_{\mu\nu}-\lambda q_{\mu\nu} -\tfrac13q_{\mu\nu}(T-4\lambda) =T_{\mu\nu}+\tfrac13q_{\mu\nu}(\lambda-T)$, which gives (24) on halving. Traces: $$K=q^{\mu\nu}K_{\mu\nu} =-\frac{\kappa_5^2}{2}\Bigl[T+\tfrac43(\lambda-T)\Bigr] =-\frac{\kappa_5^2}{2}\Bigl[\tfrac43\lambda-\tfrac13T\Bigr] =-\frac{\kappa_5^2}{6}\bigl(4\lambda-T\bigr).\tag{25}$$

Thus, under $Z_2$ symmetry, the extrinsic curvature is fixed algebraically by the brane stress tensor. This is precisely why the quadratic correction below descends to brane data while the projected Weyl tensor does not.

Gauss–Codazzi projection

The Gauss equation for the embedding is $${}^{(4)}R_{\mu\nu\rho\sigma} ={}^{(5)}R_{ABCD}\, e^A{}_\mu e^B{}_\nu e^C{}_\rho e^D{}_\sigma +K_{\mu\rho}K_{\nu\sigma}-K_{\mu\sigma}K_{\nu\rho}.\tag{26}$$ Decompose the five-dimensional Riemann tensor into Weyl and Ricci parts, $${}^{(5)}R_{ABCD} ={}^{(5)}C_{ABCD} +\tfrac23\bigl(g_{A[C}{}^{(5)}R_{D]B} -g_{B[C}{}^{(5)}R_{D]A}\bigr) -\tfrac16\,{}^{(5)}R\,g_{A[C}g_{D]B},\tag{27}$$ and use (5), which gives ${}^{(5)}R_{AB}=\tfrac23\Lambda_5g_{AB}$ and ${}^{(5)}R=\tfrac{10}{3}\Lambda_5$ away from the brane. Define the electric projection of the bulk Weyl tensor, $$E_{\mu\nu}:={}^{(5)}C_{ABCD}\, n^Ae^B{}_\mu\,n^Ce^D{}_\nu ,\tag{28}$$ which is symmetric and traceless, $q^{\mu\nu}E_{\mu\nu}=0$, by the trace identities of the Weyl tensor and $n\cdot n=1$.

Contracting (26) twice and substituting (27) yields the well-known intermediate identity $$G_{\mu\nu} =-\tfrac12\Lambda_5q_{\mu\nu} +Q_{\mu\nu}(K)-E_{\mu\nu},\tag{29}$$ where the complete extrinsic-curvature contribution is $$Q_{\mu\nu}(K) =KK_{\mu\nu}-K_{\mu\beta}K_\nu{}^\beta -\tfrac12q_{\mu\nu}\bigl(K^2-K_{\alpha\beta}K^{\alpha\beta}\bigr).\tag{30}$$

We now insert (24) into (30). Write $K_{\mu\nu}=-\tfrac{\kappa_5^2}{2}\bigl(T_{\mu\nu}+\alpha q_{\mu\nu}\bigr)$ with $\alpha\equiv\tfrac13(\lambda-T)$, so that $K=-\tfrac{\kappa_5^2}{2}(T+4\alpha)$. Then $$\begin{aligned} KK_{\mu\nu} &=\frac{\kappa_5^4}{4}\bigl(T+4\alpha\bigr) \bigl(T_{\mu\nu}+\alpha q_{\mu\nu}\bigr), \qquad\text{(31)}\\ K_{\mu\beta}K_\nu{}^\beta &=\frac{\kappa_5^4}{4} \bigl(T_{\mu\beta}T_\nu{}^\beta +2\alpha T_{\mu\nu}+\alpha^2q_{\mu\nu}\bigr), \qquad\text{(32)}\\ K^2 &=\frac{\kappa_5^4}{4}\bigl(T+4\alpha\bigr)^2, \qquad K_{\alpha\beta}K^{\alpha\beta} =\frac{\kappa_5^4}{4} \bigl(T_{\alpha\beta}T^{\alpha\beta} +2\alpha T+4\alpha^2\bigr). \qquad\text{(33)} \end{aligned}$$ Substituting (31)–(33) directly into (30), without yet using the value of $\alpha$, gives the unsimplified expression $$\begin{aligned} Q_{\mu\nu} =\frac{\kappa_5^4}{4}\Bigl[ &\bigl(T+4\alpha\bigr)\bigl(T_{\mu\nu}+\alpha q_{\mu\nu}\bigr) -\bigl(T_{\mu\beta}T_\nu{}^\beta +2\alpha T_{\mu\nu}+\alpha^2q_{\mu\nu}\bigr)\\ &-\tfrac12 q_{\mu\nu} \bigl(\bigl(T+4\alpha\bigr)^2 -T_{\alpha\beta}T^{\alpha\beta} -2\alpha T-4\alpha^2\bigr) \Bigr]. \end{aligned}\tag{34}$$ Expanding and collecting on the independent structures $T_{\mu\nu}$, $T_{\mu\beta}T_\nu{}^\beta$, and $q_{\mu\nu}$: $$Q_{\mu\nu} =\frac{\kappa_5^4}{4}\Bigl[ \bigl(T+2\alpha\bigr)T_{\mu\nu} -T_{\mu\beta}T_\nu{}^\beta +q_{\mu\nu}\Bigl( \tfrac12 T_{\alpha\beta}T^{\alpha\beta} -\tfrac12 T^2 -2\alpha T -3\alpha^2 \Bigr)\Bigr].\tag{35}$$ Indeed, the $q_{\mu\nu}$ coefficient collects $\alpha(T+4\alpha)-\alpha^2 -\tfrac12\bigl(T^2+8\alpha T+16\alpha^2 -T_{\alpha\beta}T^{\alpha\beta}-2\alpha T-4\alpha^2\bigr) =\tfrac12 T_{\alpha\beta}T^{\alpha\beta}-\tfrac12T^2-2\alpha T-3\alpha^2$, while the $T_{\mu\nu}$ coefficient collects $(T+4\alpha)-2\alpha=T+2\alpha$.

Now restore $\alpha=\tfrac13(\lambda-T)$ in (35) and collect by powers of $\lambda$. The $T_{\mu\nu}$ coefficient becomes $T+\tfrac23(\lambda-T)=\tfrac13T+\tfrac23\lambda$, contributing $\tfrac{\kappa_5^4}{4}\bigl(\tfrac13 T T_{\mu\nu} +\tfrac23\lambda T_{\mu\nu}\bigr) =\tfrac{\kappa_5^4}{12}TT_{\mu\nu}+\tfrac{\kappa_5^4\lambda}{6}T_{\mu\nu}$. The $q_{\mu\nu}$ coefficient becomes $\tfrac12 T_{\alpha\beta}T^{\alpha\beta}-\tfrac12T^2 -\tfrac23(\lambda-T)T-\tfrac13(\lambda-T)^2 =\tfrac12T_{\alpha\beta}T^{\alpha\beta}-\tfrac16T^2-\tfrac13\lambda^2$, contributing $\tfrac{\kappa_5^4}{8}q_{\mu\nu}T_{\alpha\beta}T^{\alpha\beta} -\tfrac{\kappa_5^4}{24}q_{\mu\nu}T^2 -\tfrac{\kappa_5^4\lambda^2}{12}q_{\mu\nu}$. Together with the $-\tfrac{\kappa_5^4}{4}T_{\mu\beta}T_\nu{}^\beta$ term this is exactly $$\boxed{\; Q_{\mu\nu} =-\frac{\kappa_5^4\lambda^2}{12}\,q_{\mu\nu} +\frac{\kappa_5^4\lambda}{6}\,T_{\mu\nu} +\kappa_5^4\,\Pi_{\mu\nu}, \;}\tag{36}$$ with $$\boxed{\; \Pi_{\mu\nu} =-\tfrac14 T_{\mu\beta}T_\nu{}^\beta +\tfrac1{12}\,T\,T_{\mu\nu} +\tfrac18\,q_{\mu\nu}T_{\alpha\beta}T^{\alpha\beta} -\tfrac1{24}\,q_{\mu\nu}T^2 . \;}\tag{37}$$ No cross terms are dropped and no $\lambda$-dependence is hidden in the purely quadratic part: all $\lambda^2$ dependence enters through $-\tfrac13\lambda^2$ in the $q_{\mu\nu}$ coefficient, and all $\lambda T$ dependence through $\tfrac23\lambda$ in the $T_{\mu\nu}$ coefficient.

Substituting (36) into (29) gives the Shiromizu–Maeda–Sasaki equation [2]: $$\boxed{\; G_{\mu\nu}[q] =-\Lambda_4\,q_{\mu\nu} +\kappa_4^2\,T_{\mu\nu} +\kappa_5^4\,\Pi_{\mu\nu} -E_{\mu\nu}, \;}\tag{38}$$ with $$\boxed{\; \Lambda_4=\tfrac12\Bigl(\Lambda_5+\tfrac16\kappa_5^4\lambda^2\Bigr), \qquad \kappa_4^2=8\pi G_4=\tfrac16\kappa_5^4\lambda . \;}\tag{39}$$ Equivalently, writing the equation in the form $G_{\mu\nu}+\Lambda_4q_{\mu\nu}=\kappa_4^2T_{\mu\nu} +\mathcal C^{\mathrm{RS}}_{\mu\nu}$, the RS2 realization of the source-induced metric response is $$\boxed{\; \mathcal C^{\mathrm{RS}}_{\mu\nu} =\kappa_5^4\,\Pi_{\mu\nu}-E_{\mu\nu}. \;}\tag{40}$$

Every coefficient in (38)–(40) follows from the action (3), the junction condition (23), and the Gauss identity (26). Dimensions are consistent: $[\kappa_5^4]=L^6$, $[T_{\mu\nu}T^{\mu\nu}]=L^{-8}$, so $[\kappa_5^4\Pi]=L^{-2}=[E_{\mu\nu}]=[G_{\mu\nu}]$.

Conservation, Bianchi, and nonclosure

With no exchange of matter energy into the bulk, the Codazzi equation $D^\mu\bigl(K_{\mu\nu}-Kq_{\mu\nu}\bigr) ={}^{(5)}R_{AB}n^Ae^B{}_\nu=\tfrac23\Lambda_5\,n_Ae^A{}_\nu=0$ combined with (24) gives $$\nabla^\mu T_{\mu\nu}=0 .\tag{41}$$ Applying the four-dimensional contracted Bianchi identity $\nabla^\mu G_{\mu\nu}=0$ to (38) then yields $$\boxed{\; \nabla^\mu E_{\mu\nu} =\kappa_5^4\,\nabla^\mu\Pi_{\mu\nu}. \;}\tag{42}$$

This is the exact nonclosure statement in local form. Equation (42) fixes the divergence of $E_{\mu\nu}$, which is four of its components’ worth of information, but leaves its transverse-traceless content , precisely the free bulk gravitational response , undetermined by brane data. The projected system (38)–(42) is covariantly consistent but is not in general a closed four-dimensional initial-value system. This is the tensorial counterpart of Proposition 3.1: the wave packet of Section 3 supplies two solutions with identical instantaneous brane data and different later $E_{\mu\nu}$.