Past exam of the cosmology course of the University of Oxford / 2025 / 2

The Helmholtz decomposition separates the peculiar velocity into a gradient field and a transverse vector field:With decaying or periodic spatial boundary conditions this is equivalently the Fourier transform projection, for $k\ne 0$,The transverse projection operator annihilates a gradient, because $({\delta }_{ij}-k_i{k}_j/k^2)k_j=0$. Both forces in the linear cosmological Euler equation are gradients, with spatially homogeneous coefficients. Applying the transverse projection operator therefore givesSince $H=\dot{a}/a$, this is $d(a{\mathbf{v}}_{\perp })/dt=0$, and henceAt this order the scalar pressure and gravitational forces cannot generate vorticity; an initially vanishing transverse vector field remains zero. Its nonzero initial value is damped by cosmic expansion.

Use the source's $+---$ metric signature and $x^0=\eta$, with conformal time derivatives denoted by primes. For this vector cosmological perturbation, the metric tensor and inverse metric through first order areSpatial indices on perturbations are raised with the background Kronecker delta. The Christoffel symbol formula isBecause $a=a(\eta )$ only, it givesFor the other requested Christoffel symbols, only the time derivative of the spatial metric tensor contributes:ThusFinally contract the first upper and first lower spatial indices, summing over $i$:Since $J_{ii}=2{\partial }_i{F}_i=0$ by the transverse vector field condition on $\mathbf{F}$,The useful Christoffel symbol traces are consequently ${\Gamma }_{\mu 0}^{\mu }=4\mathcal{H}$ and ${\Gamma }_{\mu j}^{\mu }=0$ to first order: the first uses $J_{ii}^{\prime }=0$ and the second uses the contraction just proved.

Define the momentum density $q_i=(\bar{\rho }+\bar{P})v_{\perp i}$, lowering its spatial label with ${\delta }_{ij}$, so the mixed stress-energy tensor obeys $T^i{}_0=q_i$ and $T^0{}_i=-q_i$. All products of $J_{ij}$ or its derivatives with $q_i$ are second order and must be dropped in linear cosmological perturbation theory.

For $\nu =0$, the three terms in stress-energy conservation areHere the trace $J_{ii}^{\prime }$ vanishes, ${\Gamma }_{00}^i={\Gamma }_{i0}^0=0$, and ${\partial }_i{q}_i=(\bar{\rho }+\bar{P}){\partial }_i{v}_{\perp }^i=0$. ThusFor $\nu =i$, spatial pressure gradients vanish because this is a purely vector cosmological perturbation. The first two terms areIn the remaining contraction, the two background-connection times momentum density terms cancel:The other terms are ${\Gamma }_{0i}^0\bar{\rho }=0$ and $-\bar{P}{\Gamma }_{ji}^j=0$. ConsequentlyExpanding this conservation law and using the background cosmological continuity equation gives the equivalent peculiar velocity equation, when $\bar{\rho }+\bar{P}\ne 0$,

The $\nu =0$ result is the background cosmological perfect-fluid continuity equation. For a comoving region of physical volume proportional to $a^3$, it is preciselyexpressing local conservation of energy including the pressure-volume work done during cosmic expansion. A transverse vector field has no velocity divergence, and the vector spatial metric tensor has no trace perturbation, so neither produces a first-order compression or density change.

The spatial result is vector-fluid momentum dilution and integrates immediately:Thus, for any specified background perfect fluid,A statement about $\mathbf{q}$ alone does not fix the velocity scaling independently of the equation of state. For constant equation-of-state parameter $w\ne -1$, the cosmological continuity equation gives $\bar{\rho }\propto a^{-3(1+w)}$ and henceFor pressureless matter, $w=0$ and ${\mathbf{v}}_{\perp }\propto a^{-1}$, exactly as in the Newtonian calculation in part (a). For radiation in cosmology, $w=1/3$ and the vector peculiar velocity remains constant, although its momentum density still falls as $a^{-4}$. More generally, for a barotropic perfect fluid with cosmological adiabatic sound speed $c_a^2={\bar{P}}^{\prime }/{\bar{\rho }}^{\prime }$, the equivalent equation is ${\mathbf{v}}_{\perp }^{\prime }+\mathcal{H}(1-3c_a^2){\mathbf{v}}_{\perp }=0$. The Newtonian damping law describes nonrelativistic inertia; it need not apply to a relativistic perfect fluid. A pure cosmological constant has $\bar{\rho }+\bar{P}=0$ and supplies no such fluid velocity degree of freedom.