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Use the Einstein-de Sitter universe relations $H=2/(3t)$ and $\bar{\rho }\propto t^{-2}$. Write $H_{\mathrm{ta}}=H(t_{\mathrm{ta}})$ and $R_{\mathrm{ta}}=R(t_{\mathrm{ta}})$. Since $M=(4\pi /3){\Delta }_{\mathrm{ta}}{\bar{\rho }}_{\mathrm{ta}}{R}_{\mathrm{ta}}^3$,The dimensionless first integral with zero velocity at $y=1$ is $(dy/d\tau {)}^2={\Delta }_{\mathrm{ta}}(1/y-1)$. On the expanding branch this givesEvaluating the integral gives the inverse-trigonometric expression in the source. Alternatively, differentiate it: $d\tau /dy={\Delta }_{\mathrm{ta}}^{-1/2}\sqrt{y/(1-y)}$, so differentiation of the first integral gives exactly the required equation of motion. At $y=1$, the integral equals $\pi /2$. Thus ${\tau }_{\mathrm{ta}}=2/3$ fixes $\sqrt{{\Delta }_{\mathrm{ta}}}=3\pi /4$. The collapsing branch is $\tau =4/3-{\tau }_{\mathrm{expand}}(y)$, not the same increasing inverse function.

For the early-time density contrast, expand the integral directly:The leading coefficient in the printed small-radius approximation needs correction to $8/(9\pi )$. The printed coefficient is inconsistent with both the exact integral and approach to the homogeneous background. Indeed, the nonlinear density contrast satisfiesHence ${\delta }_{\mathrm{L}}\sim 3y/5$. In the Einstein-de Sitter universe the growing linear cosmological density perturbation is proportional to $t^{2/3}$, so its extrapolation isThe formal collapse occurs at ${\tau }_{\mathrm{coll}}=4/3$, giving the linear spherical-collapse thresholdThe actual nonlinear density contrast diverges at collapse; the finite number is the extrapolated linear amplitude. Equivalently, the full spherical-collapse model has $y=(1-\cos \theta )/2$ and $\tau =2(\theta -\sin \theta )/(3\pi )$, with turnaround at $\theta =\pi$ and formal collapse at $2\pi$. Combining the half-radius estimate with the background-density decrease up to $2t_{\mathrm{ta}}$ gives ${\rho }_{\mathrm{vir}}/\bar{\rho }(t_{\mathrm{coll}})=32{\Delta }_{\mathrm{ta}}=18{\pi }^2$.

The flat matter-dominated Friedmann equation gives $\rho =1/(6\pi G{t}^2)$ when $a=t^{2/3}$. With $P=C{\rho }^{4/3}$,For linear density modes with polytropic pressure, substitution gives the Euler-Cauchy equationA power law has exponent satisfying $n(n-1)+4n/3+{\bar{c}}_s^2{k}^2-2/3=0$. Thus, for distinct roots,For $k\ll k_J$, the two modes approach $t^{2/3}$ and $t^{-1}$: gravity produces a growing linear cosmological density perturbation and a decaying mode. For $k\gg k_J$, put $\nu =\sqrt{{\bar{c}}_s^2{k}^2-25/36}$; the real solution isThese are pressure-supported oscillations with a decaying envelope. For the repeated-root cosmological density mode $k=k_J$, the complete solution is $t^{-1/6}(A_k+B_k\log t)$. The specified $k_J=5/(6{\bar{c}}_s)$ separates real exponents from logarithmic oscillations; the growing exponent actually crosses zero at $k=\sqrt{2/3}/{\bar{c}}_s$, a slightly smaller value.

Relativistic pressure gives the radiation fluid a sound speed $c_s^2=1/3$. Well inside the sound horizon, $k\gg \mathcal{H}$, radiation linear cosmological density perturbations undergo rapid pressure-supported oscillations instead of sustained gravitational growth. Their potential decays. For an ideal radiation fluid with negligible scalar anisotropic stress, the regular Newtonian gauge in cosmology solution during radiation domination illustrates this explicitly:For $u\gg 1$ its amplitude is $O(u^{-2})$. Thus radiation in cosmology remains essential to the background Friedmann equation, while its rapidly oscillating gravitational forcing can be smoothed out when following the slowly varying cold dark matter density mode. This is the approximation behind the Mészáros equation.

The important limitation is that “subhorizon” alone is not a pointwise theorem that $|{\Phi }_r|\ll |{\Phi }_m|$. Radiation can dominate the instantaneous potential during early radiation domination, and the forcing near horizon entry seeds the later logarithmic matter mode. We neglect that transient and rapidly oscillating forcing in the subsequent coarse-grained growth equation, not radiation's contribution to the expansion rate. Close to horizon entry, or for accurate oscillatory evolution, the coupled radiation and matter perturbation equations must be retained. For the slow matter potential, the metric time derivatives in part (a) are also suppressed relative to its spatial Laplacian by $O((\mathcal{H}/k{)}^2)$. The approximation smooths radiation perturbations while retaining radiation in cosmology in the background cosmic expansion.

For small $\theta$,Put $u=(6t/B{)}^{1/3}$. Inverting the second series givesSubstitution into the first series yieldsUsing the definitions of $A$ and $B$ simplifies its leading factor toTherefore

The homogeneous sphere containing the same mass has radius $\bar{R}=R_0{\Omega }_{m,0}^{1/3}a(t)$. By mass conservation, its nonlinear density contrast isThe leading, linear cosmological density perturbation is consequentlyIn the parametric solution, complete spherical collapse occurs at $\theta =2\pi$, when $R=0$ and $t=B(2\pi -\sin 2\pi )=2\pi B$. Extrapolating the linear growing mode to that time gives the standard linear spherical-collapse threshold

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 linearized cosmological continuity equation, the divergence of the cosmological Euler equation, and the cosmological Poisson equation giveEliminating $\theta =-{\delta }^{\prime }$ yields the equation for a linear cosmological density perturbation:In an Einstein-de Sitter universe, ${\Omega }_m=1$ and $\mathcal{H}=2/\tau$, soSubstitution of the power-law ansatz $\delta \propto {\tau }^p$ gives $(p-2)(p+3)=0$. HenceThe leading one of the matter-era growing and decaying density modes is $\delta \propto a\propto {\tau }^2$, as direct substitution confirms, and its velocity divergence is

During matter domination, pressureless matter has ${\bar{P}}_m=\delta P_m=0$. On subhorizon scales the time derivative of the Newtonian gauge in cosmology potential is negligible, so the continuity and Euler equations reduce toTaking the divergence of the second equation, differentiating the first, and using the Poisson equation givesSince $d/d\tau =a,d/dt$, one hasDivision by $a^2$ yields the standard linear cosmological density perturbation equation

During matter domination, $a\propto t^{2/3}$, $H=2/(3t)$ and $4\pi G{\bar{\rho }}_m=2/(3t^2)$. Trying ${\delta }_m=t^p$ giveswith roots $p=2/3,-1$. The growing linear cosmological density perturbation is therefore