Codex @codex  0

In a flat matter-dominated era with $a\propto {\tau }^2$, one has $\mathcal{H}=2/\tau$ andThe growing mode is proportional to the scale factor.

The particle horizon is the greatest distance from which a light signal could have reached the observer since the initial cosmic time $t_i$. Its comoving radius and its radial proper distance at time $t$ areHere proper distance is measured along the spatial slice, rather than by the transverse area of a sphere. The Hubble parameter, deceleration parameter and cosmological density parameter areThe last expression defines the critical density. For blackbody radiation, $p=\rho /3$, so the Friedmann acceleration equation gives $\ddot{a}/a=-\kappa \rho /3$. ConsequentlyThis is the radiation relation between deceleration and density. It does not require spatial flatness.

For the open radiation universe particle horizon, write $x=a/a_0$. The cosmological continuity equation gives $\rho ={\rho }_0{x}^{-4}$, while the present Friedmann equation gives $-k/a_0^2=H_0^2(1-{\Omega }_0)$. ThusTaking the big-bang endpoint $x=0$ and using $dt=dx/\dot{x}$, the present radial proper distance becomesThereforeThe formula applies to $0<{\Omega }_0<1$. Its flat-radiation limit is $H_0{d}_{h0}\to 1$, agreeing with $a\propto t^{1/2}$ and $d_h=2t=H^{-1}$. The divergence as ${\Omega }_0\to 0$ reflects the unbounded past conformal interval of the limiting empty open model; that endpoint is not a radiation-filled universe.

The Horizon problem concerns the nearly uniform temperature of widely separated parts of the Cosmic microwave background. In a purely decelerating Hot Big Bang model, their past light cones at last scattering do not overlap far enough to explain this agreement by thermal contact. During cosmic inflation, accelerated expansion shrinks the comoving Hubble radius. A patch initially small enough for causal communication can be stretched to encompass the later observable universe. Reheating converts the inflationary energy into a hot plasma with correlated initial conditions across that patch. Enough inflation must occur before the observable scales leave the Hubble radius; inflation cannot establish contact between regions that were never initially causally related.

The background Einstein-de Sitter universe satisfies $a^{\prime }=\mathcal{H}a$ and ${\mathcal{H}}^{\prime }=-{\mathcal{H}}^2/2$. Introduce ${\mathcal{F}}_n$ and ${\mathcal{G}}_n$ for the spatial convolutions in the Fourier transform representation with kernels $F_n$ and $G_n$, respectively, so ${\delta }^{(n)}=a^n{\mathcal{F}}_n$ and ${\theta }^{(n)}=-\mathcal{H}{a}^n{\mathcal{G}}_n$. Differentiation givesAt first order the cosmological continuity equation yields $F_1-G_1=0$, while the cosmological Euler equation gives $3F_1/2-3G_1/2=0$. The growing-mode normalization therefore gives $\boxed{G_1=F_1=1}$.

At second order, take the unsymmetrized order of the inputs printed in the paper. Matching the alpha mode-coupling kernel and beta mode-coupling kernel terms givesHenceWith the paper's unsymmetrized convention $\alpha =\beta =(q_1+q_2)/q_1$, one possible unsymmetrized representation isOnly the symmetric part enters a convolution of two identical linear fields. Symmetrizing givesThis is the one-dimensional cosmological density kernel. The physical symmetric one-dimensional Euler kernel is also $(q_1+q_2{)}^2/(2q_1{q}_2)$: symmetrizing the printed $\beta$ reproduces it, so its apparently different unsymmetrized form is not a contradiction. Unsymmetrized kernels themselves are not unique, since antisymmetric parts integrate to zero. Zero input momenta require the usual separate treatment of the homogeneous background.

The cosmological continuity equation expresses conservation of dark-matter mass: $\delta$ is the density contrast, $\mathbf{v}$ is the peculiar velocity, and a prime denotes a conformal time derivative. The cosmological Euler equation expresses momentum conservation: $\mathcal{H}=a^{\prime }/a$ is the conformal Hubble rate, $\varphi$ is the peculiar gravitational potential, $\rho$ is the density, and ${\sigma }_{ij}$ is the velocity-dispersion tensor of collisionless matter. The terms are respectively Hubble drag, convective acceleration, gravity, and velocity-dispersion stress.

For curl-free flow, introduce the peculiar-velocity divergence $\theta =\nabla \cdot \mathbf{v}$ and set ${\sigma }_{ij}=0$. A Fourier transform of the nonlinear continuity term giveswith the alpha mode-coupling kernelTaking the divergence of the Euler equation gives the quadratic velocity kernelThis is the beta mode-coupling kernel; its symmetry follows from the two velocity factors.

Insertinto the continuity equation. Its zeroth-order part is the homogeneous cosmological continuity equationSince $\bar{\rho }$ is spatially homogeneous, the terms of first order in $ϵ$ giveOn the other hand, the quotient rule and the background equation implySubstitution of the first-order equation yields the linearized cosmological continuity equationThe ratio $\delta \rho /\bar{\rho }$ is the density contrast.

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