Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 57 / 1

Use $\mathcal{H}=a^{\prime }/a$ for the conformal expansion rate and $H=a^{\prime }/a^2$ for the physical Hubble parameter. A future unit normal to a constant-$\tau$ slice isIt has norm minus one, and the spatial projection tensor is $P^{\mu }{}_{\nu }=\mathrm{diag}(0,1,1,1)$. The nonzero connections needed here are ${\Gamma }^0{}_{00}=\mathcal{H}$, ${\Gamma }^0{}_{ij}=\mathcal{H}{\delta }_{ij}$ and ${\Gamma }^i{}_{0j}=\mathcal{H}{\delta }^i{}_j$. Thus ${\nabla }_i{n}_j=-{\Gamma }^0{}_{ij}{n}_0=a^{\prime }{\delta }_{ij}$. Projection removes every normal component, yieldingThis uses the positive-expansion convention for extrinsic curvature in the question. With the alternative convention that inserts an overall minus sign, all these extrinsic-curvature quantities change sign.

For a comoving cell, the physical volume is proportional to $a^3$. Its logarithmic derivative in proper time is $3\dot{a}/a=3H=K$. Thus the trace is the expansion rate of the normal congruence: positive for expansion and negative for contraction. The isotropic tensor $K^i{}_j=H{\delta }^i{}_j$ has vanishing trace-free shear. It measures homogeneous expansion, not the intrinsic spatial curvature, which is zero in this flat model.

Write $q_{\alpha \beta }=g_{\alpha \beta }+n_{\alpha }{n}_{\beta }$ for the induced metric. The spatial covariant derivative projects its derivative index and both tensor indices:Four-dimensional metric compatibility removes the derivative of $g$. Each remaining term contains an undifferentiated normal on a projected index and hence vanishes. ThereforeThis establishes metric compatibility of the spatial covariant derivative directly from its projected definition.

Differentiating the normal's normalization givesso $n^{\mu }{\nabla }_{\nu }{n}_{\mu }=0$. Consequently the second projection in the extrinsic curvature definition is redundant:Finally differentiate $P^{\lambda }{}_{\nu }={\delta }^{\lambda }{}_{\nu }+n^{\lambda }{n}_{\nu }$ and project:The first term vanishes by $P^{\nu }{}_{\beta }{n}_{\nu }=0$, and the second is precisely the defining curvature. The positive sign follows from the specified positive-expansion convention, rather than from an independent projector convention.

Let $a_{\mu }=n^{\nu }{\nabla }_{\nu }{n}_{\mu }$ be the normal acceleration. It is tangential by normalization. Resolving the derivative's first index into normal and tangential parts givesThe extrinsic curvature is symmetric because the normal is hypersurface orthogonal: locally $n_{\mu }=-N{\partial }_{\mu }t$, and the antisymmetric derivative of $n$ vanishes after projecting both indices tangentially.

Apply the supplied Ricci commutator to $n^{\nu }$ and project its free index. With the positive-expansion convention it givesTo identify the first two terms, expand the fully projected divergence:Tangency gives $n_{\lambda }{K}^{\lambda }{}_{\mu }=0$. Differentiating it shows that the last term is $-P^{\mu }{}_{\alpha }{K}^{\lambda }{}_{\mu }{n}^{\rho }{\nabla }_{\rho }{n}_{\lambda }=-K_{\alpha \lambda }{a}^{\lambda }$. Substituting into the commutator proves the contracted Codazzi equation with positive expansion:No assumption of geodesic normals was needed; the acceleration terms cancel rather than being discarded.

The metric-trace term in the Einstein field equations drops out of a tangential-normal projection, since $P^{\mu }{}_{\alpha }{g}_{\mu \nu }{n}^{\nu }=0$. Define the normal-frame momentum density $j_{\alpha }=-P^{\mu }{}_{\alpha }{T}_{\mu \nu }{n}^{\nu }$. The equation becomesThus Codazzi supplies the momentum constraint on each spatial slice. In flat homogeneous FRW, $K^i{}_j=H{\delta }^i{}_j$ and $H$ has no spatial gradient, so its left side vanishes. A comoving homogeneous perfect fluid likewise has $j_i=0$. The background satisfies the constraint automatically; the Friedmann expansion law is instead obtained from the normal-normal constraint.

Suitable scalar variables are lapse and shift potentials $A,B$, spatial curvature and scalar shear potentials $\psi ,E$, and a matter momentum potential $J$ with $j_i={\partial }_{i}J$. For example, chooseA density contrast and pressure perturbation can be added for the remaining scalar matter equations. With this positive-shift convention, the linear extrinsic-curvature perturbation isIts scalar shear cancels out of the momentum-constraint combination, leavingThese are the scalar momentum constraint with positive extrinsic curvature variables; a specified gauge can then remove two of the four scalar metric potentials.