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Assume both densities are nonnegative and ${\Omega }_{\mathrm{DE},0}<1$, so a nonzero pressureless matter component exists. Their ratio obeysThus matter domination givesAt fixed $H_0$, the early expansion rate does depend on the present dark energy fraction: increasing it reduces ${\rho }_{M,0}=(1-{\Omega }_{\mathrm{DE},0})3H_0^2/(8\pi G)$. There is no contradiction with negligible early dark energy, since the leading expression is simply $H^2\sim 8\pi G{\rho }_M/3$. At fixed physical matter density, that leading rate is independent of the negligible component.

The endpoint ${\Omega }_{\mathrm{DE},0}=1$ has no matter, so the ratio argument does not apply. Its Hubble parameter is $H=H_0{a}^{-3(1+w)/2}$: it diverges for $-1<w<0$, is constant for $w=-1$, and tends to zero for $w<-1$.

Normalize the scale factor by $x=a/a_0$. Constant-equation-of-state density scaling for pressureless matter gives $\rho ={\rho }_0{x}^{-3}$. Multiplying the Friedmann equation by $x^2$ and substituting the expressions from part (i),Take the expanding branch $\dot{x}>0$ and set the origin of cosmic time at the Big Bang, $x=0$. The age of a dust universe without a cosmological constant is consequently

For pressureless matter and zero cosmological constant, the Friedmann acceleration equation gives $\ddot{a}/a=-4\pi G\rho /3$. The deceleration parameter therefore satisfies the dust relation between deceleration and density, $q=4\pi G\rho /(3H^2)$. Evaluating today,Substitute this into the present Friedmann equation to obtain $H_0^2+k{c}^2/a_0^2=2q_0{H}_0^2$, orIn particular, $q_0>1/2$, $q_0=1/2$, and $q_0<1/2$ correspond to positive, zero, and negative spatial curvature.

Use conformal time, $dt=ad\tau$, and let primes mean $d/d\tau$. Then $a^{\prime }=a\dot{a}$ and $a^{\prime \prime }=a{\dot{a}}^2+a^2\ddot{a}$. For $k=1$, $\Lambda =0$, the Friedmann equation and acceleration equation giveSubstitution yieldsUse the supplied solution without deriving it. Its first maximum has sine equal to one, so its amplitude must equal the physical turnaround scale in part (a):Choose the expanding branch with $B=0$ and the first zero at $\tau =0$.

For pressureless matter, $w=0$, the closed matter-dominated Friedmann solution isThe Big Crunch is at ${\tau }_c=2\pi$, or proper time $t_c=\pi A=\pi {\rho }_0/3$, measured from the Big Bang.

For radiation, $w=1/3$, the closed radiation-dominated Friedmann solution isThe Big Crunch is at ${\tau }_c=\pi$, or $t_c=2A$.

A radial null geodesic in FLRW spacetime obeys $d\chi /d\tau =\pm 1$. The spatial slice is a unit three-sphere times $a$, so a great circle has comoving circumference $2\pi$. The dust lifetime supplies conformal distance $2\pi$: one circumference, with the return occurring only in the crunch limit. Strictly before the singular endpoint no full return is completed. The radiation lifetime supplies conformal distance $\pi$: the photon reaches the antipode, half a great circle, in the crunch limit. These are limiting null rays from near the initial singularity, not photons at a regular event on the singular surface.

Normalize the present scale factor to one. For $0<\Omega <1$, the parametric solution givesThus the age of a flat matter-coasting-fluid universe isA direct age integral proves the bound more transparently than manipulating inverse hyperbolic functions. Since $\dot{a}=H_0\sqrt{\Omega /a+b}$,For $0\le a\le 1$ and $0\le \Omega \le 1$, one has $a\le \Omega +ba\le 1$, henceIntegrating givesEquality on the left is the pure pressureless matter universe, $\Omega =1$; equality on the right is the pure coasting fluid, $\Omega =0$. These are all physically allowed density fractions for the nonnegative two-fluid model. The integrand decreases with $\Omega$, so the coasting-fluid age bound is approached continuously between those endpoints. In the closed expression the apparent $b=0$ singularities cancel using $\mathrm{arsinh}x=x-x^3/6+\cdots$; as $\Omega \to 0$, the term $\Omega \log (1/\Omega )$ vanishes. The limits are respectively $H_0{t}_0=2/3$ and $1$.

Use units $c=\hslash =k_B=1$, and write $H=\dot{a}/a$ for the Hubble parameter, with dots denoting cosmic time. Differentiating the Friedmann equation givesOn an interval where $H\ne 0$, use the Hubble parameter identity $\dot{H}=\ddot{a}/a-H^2$ and the Friedmann acceleration equation to findSubstitution cancels the curvature term and gives the continuity equationThe cosmological perfect-fluid continuity equation extends by continuity through a regular isolated turning point. Physically it states that the change of energy in a comoving volume is the negative of the pressure work: $d(\rho a^3)=-Pd(a^3)$.

For separately conserved cosmological fluids, each component obeys this equation individually. A constant equation-of-state parameter $w_i=P_i/{\rho }_i$ therefore giveswhere the present scale factor is normalized to $a_0=1$. Thus the component density laws areThe extra factor for radiation in cosmology is the cosmological redshift of each photon's energy; pressureless matter has only number dilution, while the specified dark energy is a cosmological constant.

The critical density at a given expansion rate is the total density that makes the spatial curvature vanish:These cosmological density parameters use the critical density at that same time. Put $E(a)=H(a)/H_0$ and ${\Omega }_{K,0}=-K/H_0^2$. The Friedmann equation yieldsConsequently the fractional-density evolution, rather than just the component-density evolution, isIn particular ${\sum }_i{\Omega }_i=1+K/(a^2{H}^2)$. The simple powers of $a$ alone apply to ${\rho }_i$, or to ${\rho }_i/{\rho }_{\mathrm{crit},0}$, not to the instantaneous cosmological density parameters. At a recollapse turning point $H=0$, those instantaneous ratios are undefined even though the component densities remain finite.

Insert the component-density laws into the Friedmann equation, using ${\rho }_{i,0}=3H_0^2{\Omega }_{i,0}/(8\pi G)$ and $a_0=1$. Multiplying by $a^2$ givesThe radiation in cosmology and pressureless matter contributions add; the printed first formula has lost the plus sign between them. In the Friedmann acceleration equation, $\rho +3P=2{\rho }_r+{\rho }_m-2{\rho }_{\Lambda }$, soEvaluating the first equation today gives $K=H_0^2({\Omega }_{r,0}+{\Omega }_{m,0}+{\Omega }_{\Lambda ,0}-1)$.

Now remove dark energy and assume a positive pressureless matter or radiation in cosmology density, starting on an expanding branch. Let $f(a)=H_0^2({\Omega }_{r,0}/a^2+{\Omega }_{m,0}/a)$. It decreases from infinity to zero, and ${\dot{a}}^2=f(a)-K$, while $\ddot{a}<0$. This immediately distinguishes all three cases of spatial curvature of an FLRW universe.

The early small-$a$ solution has $a\propto t^{1/2}$ if radiation in cosmology is present, and $a\propto t^{2/3}$ for pressureless matter alone. The figure isolates the effect of curvature in three matter-only examples, rather than comparing models with the same present density parameters.

Put $\Omega =1+\delta$ with $|\delta |\ll 1$. Linearizing the cosmological density parameter flow gives $d\delta /d\log a=(1+3\omega )\delta +O({\delta }^2)$. For constant equation-of-state parameter,For variable $\omega$, the correct replacement is $\delta (a)=\delta (a_i)\exp [{\int }_{a_i}^a(1+3\omega (\tilde{a}))d\log \tilde{a}]$. The printed power assumes constant $\omega$.

For pressureless matter, the departure from flatness grows as $a$; for radiation in cosmology, it grows as $a^2$. Hence in a conventional decelerating expanding history, maintaining $|\Omega -1|\ll 1$ at a late epoch requires extremely small departures at early times. This is the flatness problem, not an assertion that exact flatness is inconsistent: $\Omega =1$ is an exact solution, but it is unstable to small departures in spatial curvature of an FLRW universe during these eras. With $\omega <-1/3$ the perturbation instead decays, which supplies the inflationary solution of the flatness problem.

Normalize the scale factor so that $a_0=1$. With separately conserved pressureless matter and radiation in cosmology, their energy densities scale as $a^{-3}$ and $a^{-4}$. Hence the flat Friedmann equation isEquality of the two densities defines matter-radiation equality, with $a_{\mathrm{eq}}={\Omega }_{r,0}/{\Omega }_{m,0}$. Define the dimensionless variableSubstituting $a=a_{\mathrm{eq}}y$ givesBoth reference density fractions are positive. In the two-component flat model their sum at the reference epoch is one; no dark-energy component is being included in this background calculation.

Differentiate the Friedmann equation and use energy conservation to obtain $\dot{H}=-4\pi G(\rho +P/c^2)$ on the expanding branch, where $H\ne 0$. Since $\ddot{a}/a=\dot{H}+H^2$,For pressureless matter, conservation gives ${\rho }_M={\rho }_{M0}{a}^{-3}$. In terms of the specified cosmological density parameters, flatness implies ${\Omega }_M+{\Omega }_{\Lambda }=1$, andSet $b=a^{3/2}$ and take ${\Omega }_M>0$. Then $\dot{b}=\frac{3}{2}{H}_0\sqrt{{\Omega }_M+{\Omega }_{\Lambda }{b}^2}$. Integrating with the Big Bang at $t=0$ givesFor another time origin replace $t$ by $t-t_B$. The cosmological constant has effective density ${\rho }_{\Lambda }=3H_0^2{\Omega }_{\Lambda }/(8\pi G)$. Equality with the matter density means $a^3={\Omega }_M/{\Omega }_{\Lambda }$, soIf ${\Omega }_M=0$, the expanding solution is pure exponential de Sitter scale factor in flat slicing and there is no finite matter-density equality time or dust Big Bang; the stated hyperbolic-sine formula presumes a nonzero dust component.

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.

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

Assume nonrelativistic pressureless matter, an initial irrotational vector field of growing-mode linear cosmological density perturbations, a homogeneous expanding background, and no shell crossing. All gradients below refer to initial comoving coordinates. Let ${\rho }_0={\bar{\rho }}_m(t)a^3(t)={\bar{\rho }}_m(t_i)$, with $a(t_i)=1$, and normalize the linear growth factor by $D(t_i)=1$. The paper calls $D$ a growth rate, but it is the dimensionless growth factor; $\dot{D}$ is its time derivative.

Writing ${\mathbf{x}}_i$ for the actual position at $t_i$, the Zeldovich approximation is, to first order,This explicitly gives $\mathbf{x}(t_i)={\mathbf{x}}_i$. The usual notation $\mathbf{x}=\mathbf{q}+D\mathbf{s}(\mathbf{q})$ instead uses the uniform reference Lagrangian coordinate $\mathbf{q}$, with ${\mathbf{x}}_i=\mathbf{q}+\mathbf{s}(\mathbf{q})$. The two labels can be interchanged inside a first-order displacement, but not in the unperturbed position without the initial displacement correction.

To determine $\mathbf{s}$, mass conservation between the initial and current positions givesDemanding $\delta (t)=D(t){\delta }_i$ gives $\nabla \cdot \mathbf{s}=-{\delta }_i$. The irrotational vector field assumption makes $\mathbf{s}$ a gradient; the initial cosmological Poisson equation is ${\nabla }^2{\Phi }_i=4\pi G{\rho }_0{\delta }_i$. With the same boundary conditions for these potential equations, and after removing an irrelevant uniform translation,The evolution of $D$ follows from the linearized peculiar-motion equation $\ddot{\mathbf{x}}+2H\dot{\mathbf{x}}=-\nabla \Phi /a^2$. Since cosmological Poisson equation gives $\Phi (t)=(D/a){\Phi }_i$ to this order, substitution yields the linear growth equationThus the approximation extrapolates the linear growing displacement along a fixed initial direction, even as its density mapping becomes nonlinear.

The early spin of a dark matter halo comes from an external gravitational torque on its nonspherical initial mass region. A uniform external acceleration moves its centre of mass without spinning it; the spatially varying tidal tensor exerts different forces on different parts. The tidal torque theory requires misalignment of the region's shape and the surrounding tidal field. A local irrotational vector field of velocity does not imply zero integrated angular momentum for a nonspherical region.

Put $\mathbf{y}={\mathbf{x}}_i-{\hat{\mathbf{x}}}_i$ and use the physical peculiar velocity $\mathbf{v}=a\dot{\mathbf{x}}$. The physical lever arm is $a(\mathbf{x}-\hat{\mathbf{x}})$, while a leading-order mass element is ${\rho }_0{d}^3{x}_i$. Substituting the velocity formula into the supplied definition of angular momentum about the centre of mass givesTerms from the perturbed mass measure are higher order. Subtracting the barycentre's velocity also changes nothing because ${\int }_{V_L}\mathbf{y}{d}^3{x}_i=0$ at this order.

Apply the divergence theorem componentwise:The antisymmetry of the Levi-Civita symbol kills the last term. Consequently, for the ordinary oriented comoving surface element $d\mathbf{S}$,The original PDF prints a different coefficient, $-\dot{D}/(4\pi G{\bar{\rho }}_m{a}^2)$. That coefficient does not follow from its own mass measure and velocity equation and has the wrong dimensions for total angular momentum. The corrected coefficient above also gives the time dependence consistent with the paper's later tensor expression; this is a source typo, not a TeX transcription issue.

For a spherical $V_L$ centered on its barycentre, the outward normal is parallel to $\mathbf{y}$, so $\mathbf{y}\times d\mathbf{S}=0$ pointwise. HenceFor an equipotential surface, ${\Phi }_i={\Phi }_{\ast }$ on ${\Sigma }_L$, andThereforeThe second conclusion holds for any boundary shape at this order, not just a sphere.

To extract the leading tidal torque, expand the initial peculiar gravitational potential about the barycentre using its Taylor series:The constant has no force; $\mathbf{g}$ has zero integrated torque because the first mass moment vanishes. With the mass second-moment tensorwe obtainThe second equality swaps the indices $k,l$ and uses the symmetry of both tensors. Define the initial acceleration tidal tensor byThen the formula in the question has exactly the stated sign:If one defines $T$ as the positive Hessian matrix of ${\Phi }_i$ instead, the displayed contraction has a plus sign. Defining the convention is essential. Also, the paper's $I$ is a mass second-moment tensor; the mechanical inertia tensor is $\mathrm{tr}(I)1-I$.

Only the anisotropic parts contribute: a multiple of the identity produces zero contraction with the Levi-Civita symbol. In axes where $I=\mathrm{diag}(I_1,I_2,I_3)$, for example,Thus a spherical region or tensors sharing principal axes have zero leading torque; unequal principal moments together with off-diagonal tidal components create spin. The Taylor series truncation assumes that higher spatial derivatives of the tidal field are sufficiently small across the region.

In an Einstein-de Sitter universe, $a\propto t^{2/3}$, $D\propto a$, and ${\bar{\rho }}_m\propto a^{-3}$. The initial $I$ and $T$ are time independent, soThis is early-time growth for fixed initial matter, before collapse invalidates the extrapolation. It does not predict indefinite linear spin growth for a virialized halo.

Two reasons for only approximate agreement with simulations are that nonlinear collapse and shell crossing change the trajectories and the tidal field, invalidating the first-order displacement and fixed leading tidal tensor; and that real halos undergo dark-matter halo mergers, anisotropic accretion, and exchange of material and angular momentum, so the halo identified at a later time need not be the same isolated collection of initial particles. These processes can change both the magnitude and the direction of the spin.

The Friedmann acceleration equation and $H=\dot{a}/a$ give the deceleration parameterAn Einstein-de Sitter universe contains only pressureless matter, so $q_0=1/2$ and cannot explain the observed negative value. For the stated Lambda-CDM model,so its cosmological constant produces the required accelerating universe.

Near the minimum, the Taylor expansion of the potential isNeglecting Hubble friction during one short oscillation reduces the Klein-Gordon equation to the harmonic oscillator equationThe virial theorem gives $⟨{\dot{\varphi }}^2/2⟩=⟨V⟩$, so $⟨P_{\varphi }⟩=0$. Restoring the slow cosmological damping in the averaged cosmological perfect-fluid continuity equation givesThe coherently oscillating inflaton therefore behaves as pressureless matter.

For example, choose $w=-5/3$. A schematic solution can join the pressureless matter law $a\propto t^{2/3}$ to the phantom-dominated law $a\propto (t^{\ast }-t{)}^{-1}$. It begins at the Big Bang, bends down during matter domination, changes to accelerated expansion when phantom energy takes over, and has a vertical asymptote at the finite Big Rip time $t^{\ast }$.

Separate stress-energy conservation and the constant equation-of-state parameter $w$ give the constant-equation-of-state density scalingBecause $w<-1$, the phantom energy density grows with the scale factor, whereas the pressureless matter density decays as $a^{-3}$. Phantom energy therefore dominates at late times. With $\Lambda =0$, the Friedmann equation becomesIntegrating and absorbing the integration constant into the finite time $t^{\ast }$ givesThe exponent is negative, so $a(t)\to \infty$ as $t\to t^{\ast -}$. This is a Big Rip.