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For separately conserved cosmological fluids, the prerequisite is absence of energy or momentum exchange between components. Each matter action then gives its own local conservation equation ${\nabla }_{\mu }{T}_i^{\mu \nu }=0$. In the homogeneous perfect fluid in general relativity, its time component isThis is the cosmological perfect-fluid continuity equation. Equivalently, a physical volume $V\propto a^3$ following the fluid satisfies $d({\rho }_{i}V)=-P_{i}dV$.

The Einstein field equations instead have the single shared metric on its left and ${\sum }_i{T}_i^{\mu \nu }$ on its right. The Friedmann equation and Friedmann acceleration equation therefore involve total density and total pressure; the same geometry cannot be sourced separately by each component's density alone.

The PDF's separate-conservation statement needs this noninteraction qualification. In an interacting system,where $Q_i$ is the energy-transfer rate into component $i$. Total conservation remains true, but the source-free equation does not hold individually during processes such as annihilation or particle decay.

For a single positive-density component, staticity requires $\rho +3P=0$ from the Friedmann acceleration equation. Among the listed components, only the cosmic string network has $w=-1/3$. Put $a(t)=a_E>0$ and use the Friedmann equation to obtainAll other components are absent. This is the static closed cosmic-string universe.

There is a normalization condition hidden in this expression. String conservation fixes $C_s={\rho }_s{a}^2$, and a static solution exists only whenOnce this condition is imposed, every positive radius $a_E$ is a static solution, with its density determined as above. One cannot independently prescribe an arbitrary string normalization and then choose a radius to compensate: both the string density term and curvature term scale as $a^{-2}$.

Choose a reference epoch with $H_0=\dot{a}(t_0)/a_0\ne 0$ and let $x=a/a_0$. Define the critical density and each cosmological density parameter byFor a closed universe ${\Omega }_{K,0}<0$. Take the dark-energy component to be a cosmological constant, as indicated by its $\Lambda$ label. The constant-equation-of-state density scaling givesSubstitution into the Friedmann equation givesAt the reference epoch, ${\Omega }_{r,0}+{\Omega }_{m,0}+{\Omega }_{s,0}+{\Omega }_{w,0}+{\Omega }_{\Lambda ,0}+{\Omega }_{K,0}=1$. The Friedmann acceleration equation isStrings have ${\rho }_s+3P_s=0$ and make no direct acceleration contribution, although they affect the Friedmann constraint. The normalized density parameters are undefined at a static epoch, where $H_0=0$; the static solution must therefore be described directly with dimensional densities.

Increase the density at the initial radius by a small fraction $ϵ>0$, maintaining the string equation of state. Then $C_s=C_E(1+ϵ)$ and the Friedmann equation fixesBecause $\ddot{a}=0$, the exact expanding and contracting branches areThe expanding branch grows indefinitely; the contracting branch reaches zero scale factor after $a_E/\sqrt{Kϵ}$. Thus a positive density-normalization perturbation destroys staticity. The departure speed is proportional to $\sqrt{ϵ}$, reflecting the fact that the Friedmann constraint is quadratic in the velocity at the static point. This is another aspect of homogeneous stability of a static cosmic-string universe.

The distinction between the evolution equation and the constraint is essential. For strings alone, the exact equations reduce toLinearizing the acceleration equation gives $\delta a=A+Bt$. At the exactly tuned static $C_s$, however, the full Friedmann equation requires ${\dot{a}}^2=0$, so $B=0$ for an admissible exact perturbation.

A small nonzero velocity $v$ can be admissible if its initial density also changes:Arbitrarily small $v$ and the corresponding second-order density change therefore produce a secular departure. There is no Lyapunov stability against all nearby constrained initial data for the static closed cosmic-string universe. Neutral radius shifts at fixed normalization and drift when the normalization changes are different perturbation classes.

First perturb the radius consistently with the conserved string normalization. If $a_E\mapsto a_E(1+ϵ)$ while $C_s={\rho }_s{a}^2$ stays at its static value, the density becomesThe Friedmann equation still has an exact cancellation between the string and curvature terms, and the Friedmann acceleration equation still vanishes. ThusThe homogeneous stability of a static cosmic-string universe is neutral in this restricted sense: there is no restoring force and no exponentially growing displacement. Holding the density fixed while changing the radius would instead change $C_s$ and would not be this pure displacement perturbation.

If the initial density is decreased by $ϵ<0$ at the same radius, the string-only Friedmann equation would requirewhich has no real solution. It is not an allowed initial perturbation of a closed string-only universe at fixed curvature normalization. In particular, it should not be interpreted as an oscillating or automatically recollapsing solution.

Likewise, increasing the density while keeping both $a$ and $\dot{a}=0$ fixed violates the constraint; the admissible velocity is the one obtained above. The homogeneous stability of a static cosmic-string universe must always be assessed using both the acceleration equation and the Friedmann constraint. Allowing extra components or changing the spatial curvature would define a different perturbation problem.

For one Fourier mode, the trace-free differential operator in the Einstein field equations is $-k^2{Q}_{ij}$, with $Q_{ij}={\hat{k}}_i{\hat{k}}_j-{\delta }_{ij}/3$. Part (ii) gives ${\Pi }_{ij}=(4/5){\bar{\rho }}_{\nu }{\Theta }_2{Q}_{ij}$. The two minus signs in the field equation therefore implyInsert ${\Theta }_2={\psi }_0{k}^2{\eta }^2/6$. During radiation domination, $a\propto \eta$, $\mathcal{H}=1/\eta$, and the Friedmann equation gives $8\pi G{a}^2\bar{\rho }=3{\mathcal{H}}^2$. ThusThese are the radiation-era adiabatic potentials with free-streaming neutrinos; photons are taken to have negligible scalar anisotropic stress.

To normalize the potentials, use $\bar{P}=\bar{\rho }/3$. The denominator in the comoving curvature perturbation becomesWith $\dot{\varphi }=0$, the conserved curvature therefore obeys $\mathcal{R}=-{\varphi }_0-{\psi }_0/2$. Combining it with the potential ratio givesAs a sign and normalization check, $f_{\nu }=0$ gives ${\varphi }_0={\psi }_0=-2\mathcal{R}/3$. Increasing the free-streaming radiation fraction increases ${\varphi }_0/{\psi }_0$ while preserving the supplied curvature convention. The TeX's malformed definition of $\mathcal{H}$ is restored by the PDF: $\mathcal{H}=\dot{a}/a$.

Differentiate the Friedmann equation:The scalar-field equation gives $\ddot{\varphi }+c^2{V}^{\prime }=-3H\dot{\varphi }$, henceBy continuity this also holds through isolated zeros of $H$, so the Raychaudhuri equation is

Put $u=\tanh (\lambda t)=e^{\varphi /{\varphi }_0}$. The hyperbolic cotangent double-angle identity givessoAlsoSolving the Friedmann equation for the scalar potential,therefore gives

Let $\mathcal{H}=a^{\prime }/a=\dot{a}$, where a prime denotes differentiation with respect to conformal time $d\tau =dt/a$. Multiplying the Friedmann equation by $a^2$ givesThe critical density is the value obtained when $k=0$, so

Since ${\mathcal{H}}^{\prime }=a\ddot{a}$, the pressure-free Raychaudhuri equation givesorDifferentiating $\Omega =1+k{c}^2/{\mathcal{H}}^2$ and substituting this equation yields

For an expanding universe $\mathcal{H}>0$. The Einstein-de Sitter universe $\Omega =1$ is an unstable constant solution: curves starting with $0<\Omega <1$ decrease towards $0$, while curves starting with $\Omega >1$ rise away from $1$ and diverge as the closed model reaches maximum expansion. This is the Flatness problem.

For pressureless matter, $w=0$ and part (a) gives $\rho =3\mu /(8\pi a^3)$. With $\Lambda =0$, the Friedmann equation becomesorThis is the $\Lambda =0$ case of the Friedmann effective potential for pressureless matter. The potential rises monotonically from $-\infty$ at $a=0$ to zero from below as $a\to \infty$.

For $k=+1$, $C=-1$ intersects the potential at $a_{\max }=\mu$. An initially expanding closed matter-dominated Friedmann solution reaches this turning point and recollapses to a Big Crunch. For $k=0$, $C=0$ gives expansion for all time with $\dot{a}\to 0$ and $a\propto t^{2/3}$. For $k=-1$, $C=1$ gives eternal expansion with $\dot{a}\to 1$ in these units. Thus positive curvature recollapses, whereas zero and negative curvature expand forever.

In cosmic time, all three Friedmann-Lemaitre-Robertson-Walker metrics can be writtenwhereThe closed spatial slice is a three-sphere of radius $a(t)$, soThe complete flat and open spatial slices are noncompact and have

For a homogeneous fluid in the Spatially flat FLRW metric, the equations in part (a) reduce to the cosmological perfect-fluid continuity equationwhile homogeneity makes the spatial relativistic Euler equation automatic. The Einstein field equations give the flat Friedmann equation and Friedmann acceleration equation,equivalently $\ddot{a}/a=-4\pi G(\rho +3P)/3$.

Normalize the present scale factor to $a(t_0)=1$. In a flat universe containing radiation in cosmology, pressureless matter, and a cosmological constant, the Friedmann equation iswhere ${\Omega }_{r,0}+{\Omega }_{m,0}+{\Omega }_{\Lambda ,0}=1$. Since $dt=da/(aH)$, the age of an FLRW universe isFor a matter-only flat Einstein-de Sitter universe,Taking the present value $H_0\simeq 70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$, for which the Hubble time is $H_0^{-1}\simeq 14.0\mathrm{Gyr}$, gives

For each Fourier mode, the cosmological Poisson equation givesBecause nonrelativistic matter has ${\bar{\rho }}_m\propto a^{-3}$, this implies $\Phi \propto {\Delta }_m/a$. Write $\Phi =C{\Delta }_m/a$ and use $\mathcal{H}=a^{\prime }/a$. Direct differentiation in the stated potential equation givesThe supplied Friedmann equation identity therefore yieldsSince $d/d\tau =ad/dt$, this becomes the linear growth equation

A larger cosmological constant changes both the expansion history and the growth history: it alters cosmological distances, the baryon acoustic oscillation scale as observed in angle and redshift, and the time at which growth begins to freeze. The new interaction in part (d) leaves the background Friedmann equation unchanged but changes clustering abruptly from $t_f$ onward.

Redshift-resolved measurements of the linear growth factor, weak gravitational lensing, galaxy clustering, and cosmological distance measures can therefore compare geometry with growth. Equal present amplitudes do not imply equal $D(z)$, and only the higher-dark-energy model changes the distance-redshift relation. This breaks the degeneracy.

Insert the result for the neutrino tensor anisotropic stress into the tensor wave equation and divide out $m_{ij}{e}^{i\mathbf{k}\cdot \mathbf{x}}/\sqrt{2}$. Since the Fourier transform replaces $-{\nabla }^2$ by $k^2$, the source coefficient is $-64\pi G{a}^2{\bar{\rho }}_{\nu }$. The flat Friedmann equation givesso $64\pi G{a}^2{\bar{\rho }}_{\nu }=24{\mathcal{H}}^2{\bar{\rho }}_{\nu }/{\bar{\rho }}_{\mathrm{tot}}$. HenceThis memory term transfers coherent tensor motion into directional neutrino perturbations. Free-streaming neutrinos damp the tensor oscillation amplitude after horizon entry, suppressing the gravitational-wave contribution to cosmic microwave background temperature and polarization spectra. The kernel acts on $\dot{h}$, so a constant superhorizon tensor mode is unaffected at leading order.

Since $H=\dot{a}/a$, differentiation gives the Hubble parameter identitySubtracting the given Friedmann equation from the Raychaudhuri equation therefore yieldsOn the other hand, differentiating the Friedmann equation givesSubstitute $\dot{H}=-4\pi G{\dot{\varphi }}^2$ and divide by $(8\pi G/3)\dot{\varphi }$, which is permitted by the assumption $\dot{\varphi }\ne 0$. The result is the homogeneous inflaton equation of motionThe middle term is Hubble friction.

For an infinitesimal test mass $\delta m$ on the surface of a homogeneous sphere, its Newtonian mechanical energy isThe spatially flat case corresponds to zero specific energy. Usingand dividing by $\delta m{R}^2/2$ givesSince $R$ is proportional to the scale factor $a$, this is the Newtonian derivation of the flat Friedmann equation

Now take $a(t_0)=1$ and $\rho ={\rho }_0{a}^{-4}$. PutThe contracting branch of the Friedmann equation isIntegrating $ada=-Cdt$ from $t_0$ givesThereforeon the contracting interval, reaching $a=0$ at $t=t_0+1/(2C)$.

Differentiate the Friedmann equation $H^2=8\pi G\rho /3$ and use the continuity equation $\dot{\rho }=-3H(\rho +P)$:so $\dot{H}=-4\pi G(\rho +P)$. Since $H=\dot{a}/a$,Hence the Raychaudhuri equation is