Past exam of the general relativity and cosmology course of the University of Oxford / 2017 / 1

The covariant derivative differentiates a vector field while accounting for how its coordinate basis changes between neighboring points. For the Levi-Civita connection,The Christoffel symbol term makes this a tensor with one contravariant and one covariant index; ordinary partial derivatives of vector components do not generally have that transformation law. Equivalently, one compares the neighboring vector with the original after parallel transport.

For a curve $x^{\gamma }(\lambda )$ with tangent $U^{\gamma }=d{x}^{\gamma }/d\lambda$, contraction gives the covariant derivative along a curveIt is the rate of change measured relative to a parallel transport frame along the curve. It vanishes precisely when $v$ undergoes parallel transport along that curve. If $v=U$, it is the covariant acceleration, and its vanishing is the geodesic equation with an affine parameter.

Use metric signature $(-,+,+,+)$ and a bulk four-velocity normalized by $u^{\alpha }{u}_{\alpha }=-c^2$. A perfect fluid in general relativity is isotropic in its local rest frame, with no rest-frame energy flux or anisotropic stress. In an orthonormal frame in spacetime comoving with that fluid,Here $\rho$ is rest-frame energy density divided by $c^2$, including random kinetic energy, and $p$ is isotropic pressure. For a particle gas, $p=\frac{1}{3}\int \mathbf{p}\cdot \mathbf{v}dn$ and $\rho c^2=\int Edn$, with $dn$ the rest-frame particle distribution. Isotropy makes the off-diagonal spatial stresses and net energy flux vanish.

Introduce the spatial projector of a timelike congruenceThe only isotropic decomposition with the preceding components isSince both sides are tensors, equality in the comoving orthonormal frame in spacetime establishesCoordinate-independent definitions of the two scalar fields areThus $\rho$ is not, in general, only the sum of the particle rest masses per volume.

The microscopic averaging notation needs a normalization qualification. The normalization of an averaged four-velocity matters because a literal average of normalized microscopic four-velocities is not normally a normalized bulk four-velocity. For example, six equally weighted streams with the same ordinary speed $v$, directed along the positive and negative coordinate axes, have an isotropic velocity distribution andThe physical rest-frame bulk four-velocity is $(c,0,0,0)$, not this unnormalized average. Even taking the source's schematic kinetic expression literally with prefactor ${\rho }_s$ givesSubstituting $u=⟨v⟩$ and the same ${\rho }_s$ into the claimed fluid expression would instead increase $T^{00}$ by $p({\gamma }^2-1)$. The standard result therefore requires a normalized bulk velocity and a carefully defined rest-frame total energy density; it does not follow with both microscopic identifications interpreted literally. More generally the bulk velocity can be defined as the future-directed timelike energy eigenvector of the stress-energy tensor.

The equation ${\nabla }_{\alpha }{T}^{\alpha \beta }=0$ is local stress-energy conservation. In a weak gravitational field and the nonrelativistic limit its leading energy component gives mass conservation,while its spatial components give momentum conservation, or the inviscid Euler equationsThe next energy order gives conservation of internal plus kinetic energy, including pressure work and gravitational work. These are local conservation laws, not an assertion that arbitrary curved spacetimes possess a global conserved total energy.

For relativistic dust, $T^{\alpha \beta }=\rho u^{\alpha }{u}^{\beta }$, with $u^{\alpha }{u}_{\alpha }=-c^2$. Expanding stress-energy conservation givesContract with $u_{\beta }$. Metric compatibility and the fixed normalization implyConsequentlyThis is the covariant mass conservation equation. Substituting it back, wherever $\rho >0$, givesThus the dust's fluid elements follow timelike geodesics. With $u^{\alpha }=d{x}^{\alpha }/d\tau$, the proper time $\tau$ is an affine parameter, and explicitlyThe absence of pressure removes the force produced by pressure gradients; gravity is encoded by the Levi-Civita connection. At a point with $\rho =0$ this calculation makes no claim about a fluid velocity.

For nonrelativistic particles the energy density is dominated by rest energy, while their isotropic kinetic pressure is approximately $p_m=nm⟨v^2⟩/3$. ThereforeThe matter pressure is negligible in this approximation, including in the gravitational combination ${\rho }_m+3p_m/c^2$. For isotropic radiation in cosmology, $E=pc$ implies one third of the energy density in each diagonal spatial stress:

Because the two fluids do not exchange energy or momentum, they are separately conserved cosmological fluids: ${\nabla }_{\alpha }{T}_i^{\alpha \beta }=0$ for each, rather than merely conservation of their sum. In a comoving volume proportional to $a^3$, the first law of thermodynamics givesEquivalently this follows by projecting the individual stress-energy conservation equation along the common comoving four-velocity. Differentiating and dividing by ${\rho }_i{c}^2{a}^3$ yieldsFor constant equation-of-state parameter $w_i$,The extra power for radiation in cosmology represents the cosmological redshift of photon energies in addition to number dilution. Differentiating the total Friedmann equation gives only the total continuity equation; the separate equations also use the stated absence of interactions between components.

Both positive-density components have nonnegative pressure, so the Friedmann acceleration equation gives $\ddot{a}<0$ whenever $a>0$. Suppose $a(t_1)=a_1>0$ and $\dot{a}(t_1)=v_1>0$. Going backward while the solution is regular, $\dot{a}(t)\ge v_1$. Integrating,and thereforeA positive scale factor cannot persist indefinitely into the past: the right side vanishes at the finite time $t_1-a_1/v_1$.

For completeness, in this model the past endpoint must actually have $a\to 0$, rather than another singularity at positive $a$. From part (d) write ${\rho }_m=C_m/a^3$ and ${\rho }_r=C_r/a^4$ with $C_m,C_r>0$. ThenThe right side strictly decreases with $a$ and is positive at $a_1$, so it remains positive throughout $0<a<a_1$. There is no earlier bounce. It is smooth at every positive $a$, permitting continuation there. The lookback integral isNear $a=0$ its integrand is $a/\sqrt{K_r}$, which is integrable. ThusThe early asymptotic behavior is $a\simeq [2\sqrt{K_r}(t-t_0){]}^{1/2}$. Both densities diverge at this Big Bang boundary. Writing $a(t_0)=0$ denotes the limiting singular endpoint of the classical solution, not an ordinary regular spacetime slice.