# Derivation of Gauss-Codazzi type equation (Ricci relation)

I am following Padmanabhan's book Gravitation for the particular derivation. The derivation goes as follows, \begin{align} R_{abst}n^t&=\nabla_a\nabla_b n_s-\nabla_b\nabla_a n_s=\nabla_a(-K_{bs}-n_b a_s)-\nabla_b(-K_{as}-n_a a_s)\nonumber\\ &=-\nabla_a K_{bs}-n_b\nabla_a a_s-a_s\nabla_a n_b+\nabla_b K_{as}+n_a \nabla_b a_s+a_s\nabla_b n_a \end{align} Contracting again with a normal vector, \begin{align} R_{abst}n^b n^t &=-n^b \nabla_a K_{bs}+\nabla_a a_s+n^b \nabla_b K_{as}+n_a n^b \nabla_b a_s+a_s n^b\nabla_b n_a\nonumber\\ &=-n^b \nabla_a K_{bs}+(\delta^r_a+n_an^r)\nabla_r a_s+n^b \nabla_b K_{as}+a_s n^b\nabla_b n_a\nonumber\\ &=-n^b \nabla_a K_{bs}+h^r_a \nabla_r a_s+n^b \nabla_b K_{as}+a_s a_a \end{align} Now we project the free indices onto $$\Sigma(t)$$, \begin{align} R_{abst}h^a_m n^b h^s_n n^t &=-n^b h^a_m h^s_n \nabla_a K_{bs}+D_m a_n+n^b h^a_m h^s_n \nabla_b K_{as}+a_m a_n \end{align} Here we have used the fact that $$n_a a^a=0\implies h^a_b a_a=a_b$$. Moreover, from $$K_{ab}n^b=0$$ we have, $$n^b\nabla_m K_{ab}=-K_{ab}\nabla_m n^b$$, thus we have, \begin{align} R_{abst}h^a_m n^b h^s_n n^t &=K_{bn} h^a_m \nabla_a n^b+D_m a_n+n^b h^a_m h^s_n \nabla_b K_{as}+a_m a_n\nonumber\\ &=-K_{bn} K^b_m+D_m a_n+n^b h^a_m h^s_n \nabla_b K_{as}+a_m a_n\nonumber \end{align} Now, \begin{align} \mathcal{L}_{\mathbf{n}}K_{as}=n^b \nabla_b K_{as}+K_{al}\nabla_s n^l+K_{ls}\nabla_a n^l. \end{align} \begin{align}\label{eq:curavaturetwonormal} R_{abst}h^a_m n^b h^s_n n^t &=K_{bn} h^a_m \nabla_a n^b+D_m a_n+n^b h^a_m h^s_n \nabla_b K_{as}+a_m a_n\nonumber\\ &=-K_{bn} K^b_m+D_m a_n+ h^a_m h^s_n (\mathcal{L}_{\mathbf{n}}K_{as}-K_{al}\nabla_s n^l-K_{ls}\nabla_a n^l)+a_m a_n\nonumber\\ &=-K_{bn} K^b_m+D_m a_n+ h^a_m h^s_n (\mathcal{L}_{\mathbf{n}}K_{as})-K_{ml} h^s_n\nabla_s n^l-K_{ln}h^a_m \nabla_a n^l+a_m a_n\nonumber\\ &=-K_{bn} K^b_m+D_m a_n+ h^a_m h^s_n (\mathcal{L}_{\mathbf{n}}K_{as})+K_{ml} K^l_n+K_{nl}K^l_m+a_m a_n\nonumber\\ &=-K_{bn} K^b_m+D_m a_n+ h^a_m h^s_n (\mathcal{L}_{\mathbf{n}}K_{as})+2K_{bn}K^b_m+a_m a_n\nonumber\\ &=K_{bn} K^b_m+D_m a_n+ h^a_m h^s_n (\mathcal{L}_{\mathbf{n}}K_{as})+a_m a_n \end{align} The extrinsic curvature is defined as $$-K_{mn}=h^a_m\nabla_an_n$$. \begin{align} \mathcal{L}_{\mathbf{n}}K_{mn}&=\mathcal{L}_{\mathbf{n}}(h^a_m h^a_n K_{as})\nonumber\\ &=h^a_m h^s_n (\mathcal{L}_{\mathbf{n}} K_{as})+h^a_m K_{as} (\mathcal{L}_{\mathbf{n}} h^s_n)+h^s_n K_{as} (\mathcal{L}_{\mathbf{n}}h^a_m)\nonumber\\ &=h^a_m h^s_n (\mathcal{L}_{\mathbf{n}} K_{as})-2h^a_m K_{as} K^s_n-2h^s_n K_{as} K^a_m\nonumber\\ &=h^a_m h^s_n (\mathcal{L}_{\mathbf{n}} K_{as})-4 K_{nb} K^b_m \end{align} Thus, we have, \begin{align} h^a_m h^s_n (\mathcal{L}_{\mathbf{n}} K_{as})=\mathcal{L}_{\mathbf{n}}K_{mn}+4 K_{nb} K^b_m \end{align} Putting this into the original equation, we obtain, \begin{align} R_{abst}h^a_m n^b h^s_n n^t &=5K_{bn} K^b_m+D_m a_n+\mathcal{L}_{\mathbf{n}}K_{mn}+a_m a_n \end{align} Which doesn't seem right.

In the book, however, we have something like this

Could someone please explain what I am doing wrong and how to obtain the correct results?

• Up until the sentence The extrinsic curvature is defined as..." everything is certainly good, compare e.g with eq. (3.43). in 3+1 formalism and Bases of Numerical Relativity by ́Eric Gourgoulhon. There the terms with the 4-acceleration are expressed slightly different, using the relation $a_{\mu}=D_{\mu}\ln N$.
– K.T.
Sep 15, 2021 at 19:18
• @K.T. Thanks for checking. Then probably the relation $h^a_m h^s_n (\mathcal{L}_{\mathbf{n}} K_{as})=\mathcal{L}_{\mathbf{n}}K_{mn}+4 K_{nb} K^b_m$ is wrong! But I don't understand why. Sep 15, 2021 at 19:25
• Also, as for the differently expressed term with the Lie derivative, it can be similarly proven that $\mathcal{L}_{\mathbf{m}}K_{\mu\nu}=\frac{1}{N}\mathcal{L}_{\mathbf{n}}K_{\mu\nu}$, where $\mathbf{m}=N\mathbf{n}$ in a derivation very similar to (3.25) and using the fact that the extrinsic curvature (projected on the hypersurface) is a spatial tensor. Do you use different indices for making a distinction between the tensor field as living on the entire manifold versus its spatial projection?
– K.T.
Sep 15, 2021 at 19:31
• Professor T. Padmanabhan author of the book Gravitation: Foundations and Frontiers has unfortunately passed away on 17th September 2021. Sep 18, 2021 at 14:56

\begin{align} R_{abst}n^t&=\nabla_a\nabla_b n_s-\nabla_b\nabla_a n_s=\nabla_a(-K_{bs}-n_b a_s)-\nabla_b(-K_{as}-n_a a_s)\nonumber\\ &=-\nabla_a K_{bs}-n_b\nabla_a a_s-a_s\nabla_a n_b+\nabla_b K_{as}+n_a \nabla_b a_s+a_s\nabla_b n_a \end{align} Contracting with a normal vector, \begin{align} R_{abst}n^b n^t &=-n^b \nabla_a K_{bs}+\nabla_a a_s+n^b \nabla_b K_{as}+n_a n^b \nabla_b a_s+a_s n^b\nabla_b n_a\nonumber\\ &=-n^b \nabla_a K_{bs}+(\delta^r_a+n_an^r)\nabla_r a_s+n^b \nabla_b K_{as}+a_s n^b\nabla_b n_a\nonumber\\ &=-n^b \nabla_a K_{bs}+h^r_a \nabla_r a_s+n^b \nabla_b K_{as}+a_s a_a \end{align} Projecting onto the hypersurface,
\begin{align} R_{abst}h^a_m n^b h^s_n n^t &=-n^b h^a_m h^s_n \nabla_a K_{bs}+D_m a_n+n^b h^a_m h^s_n \nabla_b K_{as}+a_m a_n \end{align} Using the fact, $$n_a a^a=0\implies h^a_b a_a=a_b$$. Moreover, $$K_{ab}n^b=0\implies n^b\nabla_m K_{ab}=-K_{ab}\nabla_m n^b$$, gives, \begin{align} R_{abst}h^a_m n^b h^s_n n^t &=K_{bn} h^a_m \nabla_a n^b+D_m a_n+n^b h^a_m h^s_n \nabla_b K_{as}+a_m a_n\nonumber\\ &=-K_{bn} K^b_m+D_m a_n+ h^a_m h^s_n (n^b\nabla_b K_{as})+a_m a_n \end{align} Now, \begin{align} h^a_m h^s_n\mathcal{L}_{\mathbf{n}}K_{as}&=h^a_m h^s_n\left(n^b\nabla_b K_{as}+K_{bs}\nabla_a n^b+K_{ab}\nabla_s n^b\right)\nonumber\\ &=h^a_m h^s_n\left(n^b\nabla_b K_{as}\right)-K_{bn}K_m^b-K_{mb}K^b_n \end{align} Also, \begin{align} \mathcal{L}_{\mathbf{n}}h^a_b&=\mathcal{L}_{\mathbf{n}}\left(\delta^a_b+n^a n_b\right)\nonumber\\ &=\mathcal{L}_{\mathbf{n}}\delta^a_b+\mathcal{L}_{\mathbf{n}}(n^a n_b)\nonumber\\ &=n^m\nabla_m \delta^a_b-\delta^m_b\nabla_m n^a+\delta^a_m\nabla_b n^m+n^m\nabla_m(n^a n_b)-n^m n_b\nabla_m n^a+n^a n_m\nabla_b n^m\nonumber\\ &=n^m n^a\nabla_m n_b\nonumber\\ &=n^a a_b \end{align} Then, \begin{align} \mathcal{L}_{\mathbf{n}}K_{mn}&=\mathcal{L}_{\mathbf{n}}(h^a_m h^s_n K_{as})\nonumber\\ &=h^a_m h^s_n (\mathcal{L}_{\mathbf{n}} K_{as})+h^a_m K_{as} (\mathcal{L}_{\mathbf{n}} h^s_n)+h^s_n K_{as} (\mathcal{L}_{\mathbf{n}}h^a_m)\nonumber\\ &=h^a_m h^s_n\left(n^b\nabla_b K_{as}\right)-K_{bn}K_m^b-K_{mb}K^b_n-h^a_m K_{as} n^s a_n-h^s_n K_{as} n^a a_m\nonumber\\ &=h^a_m h^s_n\left(n^b\nabla_b K_{as}\right)-K_{bn}K_m^b-K_{mb}K^b_n\nonumber \end{align} Putting all these together, we have, \begin{align} R_{abst}h^a_m n^b h^s_n n^t &=-K_{bn} K^b_m+D_m a_n+ h^a_m h^s_n (n^b\nabla_b K_{as})+a_m a_n\nonumber\\ &=-K_{bn} K^b_m+D_m a_n+ \mathcal{L}_{\mathbf{n}}K_{mn}+2K_{bn}K^b_m+a_m a_n\nonumber\\ &=K_{bn} K^b_m+D_m a_n+ \mathcal{L}_{\mathbf{n}}K_{mn}+a_m a_n\nonumber \end{align} This is the desired expression.