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For a single barotropic perfect fluid in general relativity, the perturbations obey $\delta P=w\delta \rho =w\bar{\rho }\delta$. This adiabatic closure is needed: constant background $P/\rho$ alone would not eliminate an independent entropy perturbation. The absence of scalar anisotropic stress allows the common potential used in Newtonian gauge in cosmology.

Substituting the density constraint into the pressure equation gives the gravitational potential evolution of a barotropic fluidSince $a\propto {\tau }^{2/(1+3w)}$, the conformal Hubble parameter is $\mathcal{H}=2/[(1+3w)\tau ]$. The bracket cancels identically, leavingFor a Fourier transform mode, ${\nabla }^2$ becomes $-k^2$.

During radiation domination, $w=1/3$ and ${\Phi }_k^{\prime \prime }+4{\Phi }_k^{\prime }/\tau +(k^2/3){\Phi }_k=0$. Set $x=k\tau /\sqrt{3}$ and write $\Phi =u/x$. The resulting equation is $u_{xx}+2u_x/x+(1-2/x^2)u=0$, so the two Spherical Bessel functions in the hint giveAt $x\ll 1$, the two solutions approach a constant and a mode proportional to ${\tau }^{-3}$. The regular adiabatic mode, normalized to its primordial potential, isAfter entry into the sound horizon, $x\gg 1$, the potential oscillates at cosmological sound speed $1/\sqrt{3}$ with envelope $x^{-2}\propto a^{-2}$. The Hubble radius and sound horizon differ by the sound-speed factor; outside the Hubble radius the regular potential is constant, while well inside it radiation supports acoustic oscillations.

During matter domination, $w=0$ gives ${\Phi }_k^{\prime \prime }+6{\Phi }_k^{\prime }/\tau =0$ at every wavenumber. HenceThe growing density mode has a constant potential both outside and inside the Hubble radius; the other potential mode decays. Pressureless matter has zero cosmological sound speed, so horizon entry does not produce the radiation acoustic decay. These formulas cover both independent solutions, while the subsequent sketches select the regular adiabatic growing mode.

Put $\Omega ={\Omega }_{m,0}$ and $b=1-\Omega$. Integrating the cosmological perfect-fluid continuity equation for the separately conserved cosmological fluids givesIn particular, ${\rho }_X{a}^2$ is constant. At the present epoch the flat Friedmann equation implies ${\Omega }_{X,0}=b$. Nonnegative fluid densities therefore require $0\le \Omega \le 1$.

The conformal time relation $dt=ad\tau$ gives the conformal Hubble parameter $\mathcal{H}=aH$. ConsequentlyOn the expanding branch, divide by $\mathcal{H}$ and use $a^{\prime }=a\mathcal{H}$:Eliminating the matter term yields the conformal Riccati equation for matter and a coasting fluid,The nonnegative square root fixes the convenient parameter convention; only ${\alpha }^2$ enters the differential equation.

The same invariant interval can be written as $g_{\mu \nu }d{x}^{\mu }d{x}^{\nu }={\tilde{g}}_{\alpha \beta }d{\tilde{x}}^{\alpha }d{\tilde{x}}^{\beta }$. Applying the chain rule to the differentials proves the metric transformation lawFor first-order scalar cosmological perturbations and ${\xi }^{\mu }=(T,{\partial }^{i}L)$, compare perturbations at the same background coordinate label. Expanding the Jacobians and the shifted background gives the linear metric gauge-transformation lawFor the background ${\bar{g}}_{00}=-a^2$, ${\bar{g}}_{ij}=a^2{\delta }_{ij}$, the Lie derivative components areHere primes denote conformal time and $\mathcal{H}$ is the conformal Hubble parameter. These expressions establish all four alternatives, including any chosen pair.

The $00$ component is $\delta g_{00}=-2a^{2}A$, immediately giving the lapse function perturbation transformation. The $0i$ component is $a^2{\partial }_{i}B$, giving the shift vector scalar-potential transformation. The trace of the spatial perturbation is $6a^{2}C$, so tracing the spatial Lie derivative gives the transformation of $C$. Its trace-free scalar part is $2a^2({\partial }_i{\partial }_j-{\delta }_{ij}{\nabla }^2/3)E$, giving the transformation of $E$. ThusAs usual, identifying scalar potentials from their derivatives uses the standard boundary conditions, or nonzero Fourier modes, to remove homogeneous ambiguities.

At first order, tensor cosmological perturbations are spatial transverse-traceless tensors. The coordinate-generated spatial perturbation consists of a trace term and symmetrized derivatives of the displacement. In Fourier space, the latter terms carry a factor $k_i$ or $k_j$. The transverse-traceless projector removes these longitudinal terms and the trace, including the derivative of a transverse vector displacement. Therefore the tensor perturbation is gauge invariant at linear order around the homogeneous background. This is a first-order statement, not a claim of automatic invariance at arbitrary perturbative order.

The tight-coupling approximation requires the positive scattering rate $\kappa =-\dot{\tau }$ to exceed both $k$ and the conformal Hubble parameter. The photon-baryon velocity slip and photon quadrupole are then small, and the leading fluid motion has $v_b=v_{\gamma }=v$. One must combine the equations before setting the slip to zero: a small slip times a large collision rate can exert a finite force.

Add the photon Euler equation to $R$ times the baryon Euler equation with Thomson drag. The collisions cancel and, neglecting ${\Theta }_2$ at this order, the result isSince $\dot{R}=\mathcal{H}R$, the common velocity obeysThe inertia of the baryons reduces the photon-baryon sound speed to $c_s^2=1/[3(1+R)]$.

Differentiate the photon continuity equation and substitute this velocity equation:Thus the photon-baryon acoustic oscillator isThe terms on the right drive the oscillation gravitationally; the first-derivative term comes from changing baryon inertia. Diffusion damping enters only beyond this leading tight-coupling approximation.

Throughout this question dots denote conformal time derivatives, and $\mathcal{H}=\dot{a}/a$ is the conformal Hubble parameter. The monopole of the photon Boltzmann hierarchy gives ${\dot{\Theta }}_0+k{\Theta }_1/3=\dot{\varphi }$. With the specified velocity convention, the photon continuity equation is thereforeThe dipole equation is ${\dot{\Theta }}_1+k(2{\Theta }_2/5-{\Theta }_0)=\dot{\tau }({\Theta }_1+v_b)+k\psi$. Thus the photon Euler equation isIn particular, $\dot{\tau }<0$ makes the scattering term damp the photon-baryon velocity slip; it does not amplify it.

The linear momentum density of the photon-baryon fluid is weighted by enthalpy, not merely energy density:Elastic Thomson scattering exchanges momentum internally. Its photon force density is $(4{\bar{\rho }}_{\gamma }/3)\dot{\tau }(v_{\gamma }-v_b)$, so the baryon force density is its negative. Dividing by ${\bar{\rho }}_b$ gives the baryon Euler equation with Thomson dragThis $R$ is the baryon loading parameter, the baryon-to-photon inertia ratio. The two collision terms cancel exactly when the equations are weighted by their enthalpies. Background conservation gives ${\bar{\rho }}_b\propto a^{-3}$ and ${\bar{\rho }}_{\gamma }\propto a^{-4}$, so $\dot{R}=\mathcal{H}R$.

The printed $\mathcal{H}$ is the conformal Hubble parameter, $\mathcal{H}=aH=\dot{a}$; ${\mathcal{H}}^{-1}$ is the comoving Hubble radius, rather than the physical radius $H^{-1}$. Since the derivative requested is with respect to cosmic time,Thus the cosmic-time derivative of the comoving Hubble radius hasFor an expanding universe, the same factor $1+3\omega$ determines whether a small departure from flatness grows and whether the comoving Hubble radius grows. Ordinary matter with $\omega >-1/3$ therefore has both signs associated with the conventional Flatness problem and Horizon problem. Scales whose physical wavelengths now exceed the Hubble radius were even farther outside it, in relative terms, earlier in such an era, rather than being brought inside for causal equilibration. Accelerated expansion with $\omega <-1/3$ reverses both signs.

There is a qualification to the word “always”: an actual Horizon problem depends on the complete past history, not just this local sign. In a flat, constant-$\omega$ hot Big Bang model with $\omega >-1/3$, $a\propto (t-t_B{)}^p$ with $p=2/[3(1+\omega )]<1$, so the comoving particle horizon ${\int }_{t_B}^{t}d{t}^{\prime }/a(t^{\prime })$ is finite. The conventional causal problem then accompanies the flatness instability. An earlier accelerated era or a different past boundary can change the causal conclusion even when the present-era comoving Hubble radius is growing. Thus the requested correspondence is valid in that usual expanding hot Big Bang setting, not a universal logical equivalence between global horizons and local stability. At $\omega =-1/3$ both local effects are marginal.