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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

During matter domination, $a\propto t^{2/3}$, $H=2/(3t)$, and $4\pi G{\bar{\rho }}_m=2/(3t^2)$. Substituting ${\Delta }_m\propto t^s$ into the linear growth equation giveswhose roots are $s=2/3$ and $s=-1$. ThereforeThe growing one of the matter-dominated density-perturbation modes is

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.

On long wavelengths the gradient term is negligible. The Friedmann equation impliesDuring matter domination, $a\propto t^{2/3}$ and $H=2/(3t)$, so the linear growth equation becomesThe power-law ansatz $\delta \propto t^s$ gives $s=2/3$ or $s=-1$. Since $a\propto t^{2/3}$, the growing and decaying modes are

Write physical position as $\mathbf{r}=a(t)\mathbf{x}$ and physical velocity asSubtract the homogeneous expanding background from the pressureless Euler equation. To linear order in peculiar velocity and density contrast, the convective term is negligible andorHere $\nabla$ differentiates with respect to comoving position and $\Phi$ is the peculiar gravitational potential.

The linear continuity and Poisson equations areFor the growing mode $\delta (\mathbf{x},t)=D(a){\delta }_i(\mathbf{x})$, matter conservation gives ${\bar{\rho }}_m\propto a^{-3}$, so the potential scales asafter absorbing the fiducial normalization into ${\Phi }_i$. Integrating the linear Euler equation with a negligible decaying mode gives

Taking the divergence of Euler and using continuity and Poisson yields the linear growth equationIn the fiducial normalization used in the question this can be rearranged asThe cosmological Lagrangian displacement obeys $\dot{\mathbf{\psi }}=\mathbf{v}/a$. Combining the last two relations and choosing the growing displacement gives the Zeldovich approximation

DefineThen $\mathbf{x}={\mathbf{x}}_i-b\nabla {\Phi }_i$ and $\mathbf{v}=-a\dot{b}\nabla {\Phi }_i$. Keeping the lowest nonvanishing order in displacement in the halo angular momentum givesThe angular momentum vanishes for a spherical Lagrangian patch, and also whenever the patch's inertia principal axes align with the local tidal-field principal axes. It likewise vanishes in a locally isotropic tidal field.

Taylor-expandThe constant-gradient term vanishes by the barycentre definition. Withone obtains the tidal torque theory result$I$ is the protohalo inertia tensor and $T$ is the local tidal, or gravitational Hessian, tensor. Their eigenframe misalignment produces the torque.

In an Einstein-de Sitter universe, $D\propto a\propto t^{2/3}$. Hence $\dot{b}\propto \dot{a}\propto t^{-1/3}$ andThis linear estimate is normally stopped near turnaround. Simulated haloes subsequently gain angular momentum through nonlinear torques, anisotropic accretion, and mergers, including the orbital angular momentum of infalling subhaloes, so the early linear estimate underpredicts the final value.