Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 64 / 2

This question uses proper cosmological time $t$. The future-directed unit normal isThe non-null hypersurface projection removes all normal components. The only relevant Christoffel symbols are ${\Gamma }_{ij}^0=a\dot{a}{\delta }_{ij}$. Since the components of $n_{\mu }$ are constant,ConsequentlyWith the printed positive extrinsic curvature convention, this is the expansion of the normal comoving congruence. A fixed coordinate cell has proper volume proportional to $a^3$, so $K=d\log (a^3)/dt={\nabla }_{\mu }{n}^{\mu }$: the fractional proper-volume expansion rate.

Write $q_{\alpha \beta }=g_{\alpha \beta }+n_{\alpha }{n}_{\beta }$ for the induced metric. By definition of the spatial covariant derivative,The spacetime Levi-Civita connection is metric compatible, so $\nabla g=0$. The remaining terms contain either $P_{\alpha }{}^{\sigma }{n}_{\sigma }$ or $P_{\beta }{}^{\lambda }{n}_{\lambda }$, each zero because $n^{\mu }{n}_{\mu }=-1$. ThereforeThis verifies metric compatibility of the spatial covariant derivative directly, without assuming that the unprojected derivative of $q$ vanishes.

For tangent vectors $X,Y$ on a slice, the extrinsic curvature is $K(X,Y)=g({\nabla }_{X}n,Y)$. Differentiating $g(n,Y)=0$ and using metric compatibility gives $K(X,Y)=-g(n,{\nabla }_{X}Y)$. Since the connection is torsion-free,The final equality holds because tangent fields on a hypersurface have a tangent Lie bracket. All normal components are projected out, so $K_{\mu \nu }$ is symmetric. This is precisely hypersurface orthogonality implies symmetric extrinsic curvature; without an integrable tangent distribution the corresponding antisymmetric vorticity need not vanish.

Double contraction of the projected Gauss equation, with the timelike-normal sign and curvature convention in the question, givesHere $q^{\mu \nu }=g^{\mu \nu }+n^{\mu }{n}^{\nu }$. Expanding the two projectors givesThe four-normal contraction vanishes by Riemann antisymmetry. Because $g_{\mu \nu }{n}^{\mu }{n}^{\nu }=-1$, the right side is $2G_{\mu \nu }^{(4)}{n}^{\mu }{n}^{\nu }$. Rearranging proves the Hamiltonian constraint identity:For flat FRW, $R^{(3)}=0$, $K^2=9H^2$ and $K_{\mu \nu }{K}^{\mu \nu }=3H^2$, so $G_{nn}=3H^2$. Einstein's equation is therefore the first Friedmann equation, $3H^2=8\pi G\rho$, or $3H^2=8\pi G\rho +\Lambda$ if a cosmological constant is placed separately in Einstein's equation.

The scalar $R^{(3)}$ measures intrinsic spatial curvature, $K^2$ measures squared volume expansion, and the norm of $K_{ij}$ measures directional expansion including shear. More generally $K_{ij}=K{q}_{ij}/3+s_{ij}$ with trace-free $s_{ij}$ gives $K_{ij}{K}^{ij}=K^2/3+s_{ij}{s}^{ij}$. The shear vanishes in FRW, leaving the positive $3H^2$ contribution. This is an instantaneous constraint on initial data, not the separate acceleration equation.