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Expression for extrinsic curvature

Physics Asked on April 2, 2021

In Padmanabhan’s book Gravitation Foundations and Frontiers, the following equation can regarding the extrinsic curvature of a hypersurface can be found in section 12.2 (see just above equation 12.19 in that book),

begin{align}
K_{alphabeta}=-nabla_alpha n_beta=-NGamma^0_{alphabeta}.
end{align}

According to the book’s convention greek indices run for spatial coordinates ($alpha=1,2,3$) and latin indices run for space-time coordinates ($a=0,1,2,3$). Thus the above equation gives an expression for the spatial components of the extrinsic curvature, $K_{alphabeta}$. Here, $n^a$ is the vector field normal to the hypersurface and $N$ is the lapse function. Now the book claims if we expand the Christoffel symbol, we shall get the following expression (see equation 12.19 in the book),

$$K_{alphabeta}=frac{1}{2N}left(D_alpha N_{beta}+D_beta N_{alpha}-partial_0 h_{alphabeta}right)$$

Here, $N^alpha$ is the shift vector, $h_{alphabeta}$ is the induced spatial metric on the hypersurface, and $D_m$ is the intrinsic covariant derivative on the hypersurface with its action on the purely spatial vectors $X_s$, which satisfies a constraint like $X_sn^s=0$, defined as

$$D_mX_s=h^a_mh^b_snabla_aX_b,$$

where, $h^a_b=delta^a_b+n^an_b$ are the projection tensor on the hypersurface, and $nabla_a$ is the usual covariant derivative for the spacetime.

I have failed to derive the equation 12.19 giving the expression for $K_{alphabeta}$. Below I show, how I tried to do it. The Christoffel symbol can be expanded as,
begin{align}
Gamma^0_{alphabeta}&=frac{1}{2}g^{0a}left(partial_alpha g_{beta a}+partial_beta g_{alpha a}-partial_a g_{alphabeta}right)nonumber
&=frac{1}{2}g^{00}left(partial_alpha g_{beta 0}+partial_beta g_{alpha 0}-partial_0 g_{alphabeta}right)+frac{1}{2}g^{0gamma}left(partial_alpha g_{beta gamma}+partial_beta g_{alpha gamma}-partial_gamma g_{alphabeta}right)nonumber
&=frac{-1}{2}N^{-2}left(partial_alpha N_{beta}+partial_beta N_{alpha}-partial_0 h_{alphabeta}right)+frac{1}{2}N^{-2}N^{gamma}left(partial_alpha h_{beta gamma}+partial_beta h_{alpha gamma}-partial_gamma h_{alphabeta}right)nonumber
&=frac{-1}{2}N^{-2}left(D_alpha N_{beta}+D_beta N_{alpha}+2Gamma^gamma_{alphabeta}N_{gamma}-partial_0 h_{alphabeta}right)+frac{1}{2}N^{-2}N^{gamma}left(partial_alpha h_{beta gamma}+partial_beta h_{alpha gamma}-partial_gamma h_{alphabeta}right)nonumber
&=frac{-1}{2}N^{-2}left(D_alpha N_{beta}+D_beta N_{alpha}-partial_0 h_{alphabeta}right)-N^{-2}Gamma^gamma_{alphabeta}N_{gamma}+frac{1}{2}N^{-2}N^{gamma}left(partial_alpha h_{beta gamma}+partial_beta h_{alpha gamma}-partial_gamma h_{alphabeta}right)
end{align}

In the above, I have used the facts that, $$n_0=-N,quad n_alpha=0,$$
$$D_alpha N_beta=h^a_alpha h^b_betanabla_a N_b=partial_alpha N_beta-Gamma^gamma_{alphabeta}N_gamma,$$
$$h_{00}=N^gamma N_gamma,quad h_{0alpha}=N_alpha,quad h_{alphabeta}=g_{alphabeta}$$

2 Answers

The OP's calculation seems alright. If we proceed along that line the required expression can be achieved rather easily. First, I note that, $$D_alpha N_beta=partial_alpha N_beta-{}^{(3)}Gamma^gamma_{alphabeta}N_gammaneq partial_alpha N_beta-{}^{(4)}Gamma^gamma_{alphabeta}N_gamma=partial_alpha N_beta-Gamma^gamma_{alphabeta}N_gamma,$$ Perhaps this substitution is what was confusing in OP's calculation. If we correct that then it follows, begin{align} &frac{1}{2}g^{0a}left(partial_alpha g_{beta a}+partial_beta g_{alpha a}-partial_a g_{alphabeta}right)nonumber &=-frac{1}{2}N^{-2}left(D_alpha N_{beta}+D_beta N_{alpha}+2{}^{(3)}Gamma^gamma_{alphabeta}N_{gamma}-partial_0 h_{alphabeta}right)+frac{1}{2}N^{-2}N^{gamma}left(partial_alpha h_{beta gamma}+partial_beta h_{alpha gamma}-partial_gamma h_{alphabeta}right)nonumber &=-frac{1}{2}N^{-2}left(D_alpha N_{beta}+D_beta N_{alpha}-partial_0 h_{alphabeta}right)-N^{-2}{}^{(3)}Gamma^gamma_{alphabeta}N_{gamma}+frac{1}{2}N^{-2}N^{gamma}left(partial_alpha h_{beta gamma}+partial_beta h_{alpha gamma}-partial_gamma h_{alphabeta}right)nonumber &=-frac{1}{2}N^{-2}left(D_alpha N_{beta}+D_beta N_{alpha}-partial_0 h_{alphabeta}right)-N^{-2}{}^{(3)}Gamma^gamma_{alphabeta}N_{gamma}nonumber &qquad+frac{1}{2}N^{-2}N_{sigma}h^{gammasigma}left(partial_alpha h_{beta gamma}+partial_beta h_{alpha gamma}-partial_gamma h_{alphabeta}right)nonumber &=-frac{1}{2}N^{-2}left(D_alpha N_{beta}+D_beta N_{alpha}-partial_0 h_{alphabeta}right)-N^{-2}{}^{(3)}Gamma^gamma_{alphabeta}N_{gamma}+N^{-2}N_{sigma}{}^{(3)}Gamma^{sigma}_{alphabeta}nonumber &=-frac{1}{2}N^{-2}left(D_alpha N_{beta}+D_beta N_{alpha}-partial_0 h_{alphabeta}right) end{align} Therefore, $$K_{alphabeta}=-NGamma^0_{alphabeta}=frac{1}{2N}left[D_alpha N_{beta}+D_beta N_{alpha}-partial_0 h_{alphabeta}right].$$

Correct answer by VacuuM on April 2, 2021

  1. The extrinsic curvature is defined in the ambient spacetime (rather than on the hypersurface) as $n_a$: $$K_{ab} = -P_perp{}^c{}_{a}P_perp{}^d{}_b nabla_c n_d,$$ with $P_perp$ the projection tensor on the hypersurface. Notice that by construction the extrinsic curvature is spatial and symmetric in its two indices.
  2. Use the symmetry to write $K_{ab}$ as a Lie derivative:$$K_{ab} ={-scriptsizefrac{1}{2}} P_perp{}^c{}_{a}P_perp{}^d{}_b mathcal{L}_n ,g_{cd}.$$
  3. Use the orthogonal decomposition of the metric and the adapted coordinate system $t^a = Nn^a + N^a$ for the lapse function and shift vector to arrive at $$K_{ab} = {scriptsizefrac{1}{2}}N^{-1}mathcal{L}_{(N-t)}h_{ab}.$$

References:

  • T. Thiemann, Introduction to Modern Canonical Quantum General Relativity, subsection I.1.1

Answered by haelewiin on April 2, 2021

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