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When the self-gravity source in the pressureless linear cosmological density perturbation equation is negligible, $\ddot{\delta }+2(\dot{a}/a)\dot{\delta }=0$. Multiplication by $a^2$ makes it $d(a^2\dot{\delta })/dt=0$, so $\delta =C_1+C_2\int dt/a^2(t)$. This explains how Hubble friction changes growth for an arbitrary positive scale factor. For $a\propto t^p$, the nonconstant mode is $t^{1-2p}$ when $p\ne 1/2$ and $\log t$ when $p=1/2$.

Gravitational instability is the growth of a perturbation because its increased self-gravity overcomes the opposing support or restoring response. For a uniform nonexpanding gas, linearization gives the Jeans instability dispersion relation ${\omega }^2=c_s^2{k}^2-4\pi G\rho$: sufficiently long wavelengths grow rather than oscillate. In an expanding universe the background and Hubble friction modify the growth rate; a pressureless growing density mode can eventually become nonlinear. Rotation, magnetic stress, pressure and geometry can all alter the stability threshold in other systems.

For a homogeneous canonical inflaton in an expanding FLRW metric,The expansion contributes the damping term called Hubble friction.

Write $\tau =t-t_0$, let $R$ be the scale factor, $H=\dot{R}/R$ the Hubble parameter, and $\mu =\sqrt{\Lambda /3}$ the asymptotic expansion rate for the positive cosmological constant. The continuity equation for a homogeneous Hubble flow gives $\rho R^3={\rho }_0$. Isotropy and the Poisson equation for Newtonian gravity give a relative acceleration proportional to displacement from any chosen origin:This uses the local relative gravitational acceleration, rather than trying to sum an absolutely convergent potential over an infinite universe. A fluid element with constant comoving coordinate $\mathbf{x}$ has $\mathbf{r}=R\mathbf{x}$. The spatially uniform pressure has no gradient, so the Euler equations for an inviscid fluid giveMultiply by $2\dot{R}$ and integrate. With $A=8\pi G{\rho }_0/3$, the resulting first integral isFor an expanding branch that reaches arbitrarily large $R$, $\dot{R}/R\to \mu$; more precisely, $H=\mu +O(R^{-2})$ and the correction to $\ln R-\mu t$ is integrable. Therefore $R\sim C{e}^{\mu t}$. Positive $\Lambda$ alone does not make every initially expanding branch escape: sufficiently large positive $\kappa$ can give a turning point or a limiting static radius. Under the stated vacuum-dominance inequalities at $R=1$, however, $A/R$ and $|\kappa |$ remain small compared with ${\mu }^2{R}^2$ for the future expanding branch, and $R\simeq e^{\mu \tau }$ is the required approximation.

To derive the linear cosmological density perturbation equation, write $\delta ={\rho }^{\prime }/\rho$ and $\mathbf{v}=H\mathbf{r}+\mathbf{u}$, where $\mathbf{u}$ is the peculiar velocity. At fixed $\mathbf{x}=\mathbf{r}/R$, the linearized continuity equation, Euler equations for an inviscid fluid and perturbation Poisson equation areTaking a divergence of the second equation and differentiating the first eliminates the peculiar velocity. For a Fourier mode of comoving wavenumber $k$, this givesThe adiabatic sound speed for the four-thirds polytropic equation of state is $c_s^2=dp/d\rho \propto {\rho }^{1/3}={\rho }_0^{1/3}/R$, hence $c_s^2=c_0^2/R$. Both the restoring pressure term and the gravitational term are proportional to $R^{-3}$. Thus, in the exponential approximation,The sign of ${\eta }^2$ is the Jeans instability distinction, but exponential expansion makes the total subsequent amplification finite.

Remove Hubble friction by setting $\epsilon =e^{-\mu \tau }y$. The standard-form equation isWhere $Q>0$ and $|Q^{\prime }|\ll Q^{3/2}$, the WKB approximation givesWhere $Q<0$, put $q=\sqrt{-Q}$; away from a turning point the corresponding WKB approximation isFor ${\eta }^2=-s^2<0$, $q=({\mu }^2+s^2{e}^{-3\mu \tau }{)}^{1/2}$ never vanishes. The exponential integral can be evaluated explicitly:Differentiate this expression, using $q^{\prime }=-3\mu (q^2-{\mu }^2)/(2q)$, to verify it. Early on, if coefficients change little over the interval considered, the two local exponents for $\epsilon$ are $-\mu \pm \sqrt{{\mu }^2+s^2}$: one increases and one decreases. At late times $q\to \mu$ and the displayed integral is $\mu \tau +O(1)$. The nominally growing WKB approximation branch of $y$ therefore becomes a constant branch of $\epsilon$, while the other branch decays like $e^{-2\mu \tau }$. The WKB approximation is not automatically accurate at all intermediate times; its stated derivative criterion must be checked.

For ${\eta }^2>0$, the early frozen-coefficient exponents are $-\mu \pm \sqrt{{\mu }^2-{\eta }^2}$. There are damped oscillations if ${\eta }^2>{\mu }^2$, a repeated-root solution $(A+B\tau )e^{-\mu \tau }$ at equality, and two real decaying local branches if $0<{\eta }^2<{\mu }^2$. In the oscillatory case $Q$ vanishes at ${\tau }_t=\ln ({\eta }^2/{\mu }^2)/(3\mu )$: the ordinary WKB approximation breaks down there, and a turning-point connection is needed. After that transition the solutions again approach a constant or decay. A positive pressure-minus-gravity restoring term does not imply indefinitely continuing oscillation; its physical frequency is redshifted away.

The limiting conclusion does not depend on the accuracy of WKB approximation. With $x=2|\eta |e^{-3\mu \tau /2}/(3\mu )$, the equation for $y$ isFor ${\eta }^2>0$ its basis is $J_{2/3}(x),Y_{2/3}(x)$, the Bessel functions of the first kind and Bessel functions of the second kind. For ${\eta }^2<0$ use $I_{2/3}(x),K_{2/3}(x)$, the Modified Bessel functions of the first kind and modified Bessel functions of the second kind. Their small-$x$ branches are proportional to $x^{2/3}$ and $x^{-2/3}$. Multiplication by $e^{-\mu \tau }$ gives the decaying and constant branches respectively. At ${\eta }^2=0$ the exact solution is $A+B{e}^{-2\mu \tau }$. Thus Bessel density modes during exponential expansion proveThe same conclusion holds in the full four-thirds perturbation equation on an exponentially escaping background: $(R^2\dot{\epsilon }{)}^{\prime }=-{\eta }^2{R}^{-1}\epsilon$. Its integrated equation has an absolutely integrable Volterra kernel, bounded by a constant times $e^{-3\mu \tau }$. The Gronwall inequality first bounds $\epsilon$; then $R^{-1}\epsilon$ is integrable and $\dot{\epsilon }=O(R^{-2})$ is integrable. A finite limiting amplitude follows without setting $H$ exactly constant. This argument requires positive $\Lambda$ and the escaping branch; it is not a claim about $\Lambda =0$ or recollapse.

The requested strict increase of that limit needs an initial-velocity qualification. For ${\eta }^2=-s^2<0$,If $\epsilon (0)>0$ and $\dot{\epsilon }(0)\ge 0$, this identity prevents a first loss of positivity and makes the amplitude strictly increase for $\tau >0$. Integrating it twice givesThis is monotone Jeans growth with positive cosmological constant. Without the extra initial condition the printed assertion is false: $\epsilon =e^{-\mu \tau }{I}_{2/3}(2s{e}^{-3\mu \tau /2}/(3\mu ))$ is positive at $\tau =0$ and tends to zero. An arbitrarily small multiple remains an infinitesimal linear cosmological density perturbation, so linearity does not remove the counterexample.

For constant $a$ and constant ${\rho }_m>0$, Hubble friction vanishes. Put ${\omega }_J=\sqrt{4\pi G{\rho }_m}$. The constant-coefficient ordinary differential equation is $\ddot{\delta }-{\omega }_J^2\delta =0$, so its characteristic polynomial has the real roots of a polynomial $\pm {\omega }_J$:These are the pressureless special case of the Static-universe Jeans modes. A positive growing-mode coefficient gives exponential gravitational instability, with time scale $(4\pi G{\rho }_m{)}^{-1/2}$. For initial data $\delta (t_{\ast })={\delta }_{\ast }$ and $\dot{\delta }(t_{\ast })=v_{\ast }$, the equivalent answer is ${\delta }_{\ast }\cosh [{\omega }_J(t-t_{\ast })]+(v_{\ast }/{\omega }_J)\sinh [{\omega }_J(t-t_{\ast })]$. This solves the stated perturbation model; a globally static dust-only FRW metric would require additional background support.

The term $2H\dot{\delta }$ is Hubble friction in the equation for the density contrast. During expansion, $H>0$, it damps the peculiar motion producing density growth. It therefore slows gravitational instability relative to the otherwise identical static model; it does not necessarily stop growth.

Multiplying by $a^2$ shows the mechanism without confusing it with microscopic dissipative friction:When the self-gravity term is negligible, the expansion-weighted derivative of a passive density perturbation is conserved, so $\dot{\delta }\propto a^{-2}$. Expansion also reduces ${\rho }_m\propto a^{-3}$ and hence weakens the source on the right. A contracting background has $H<0$ and reverses the sign of this effect.

Continue to choose the CDM-comoving synchronous frame. For nonrelativistic baryons, neglect $O(c_s^2)$ corrections to gravitational density weights and Hubble friction, but retain $c_s^2{k}^2$, since gradients may compensate its small coefficient. ThenTogether with the metric tensor equation these giveThe friction terms contain first derivatives, as the original PDF confirms; they are missing in the TeX's transcription of these equations.

When ${\Omega }_B\ll {\Omega }_C$, Cold dark matter sets the gravitational forcing. Its matter-era growing mode is ${\delta }_C\propto a\propto {\tau }^2$. Compare pressure with the gravitational timescale:Converting the comoving wavenumber to a physical wavelength givesFor $k\ll k_J$, pressure is negligible and baryons can follow the growing Cold dark matter mode. In that limit subtracting the two equations gives $D^{\prime \prime }+2D^{\prime }/\tau =0$ for $D={\delta }_B-{\delta }_C$, so $D=D_0+D_1/\tau$. Relative to ${\delta }_C\propto {\tau }^2$, this offset vanishes: baryons catch up.

The phrase “only grow” needs this gravitational-growth interpretation. For a small baryon fraction the baryon equation is forced, not an autonomous Jeans-instability equation. At $k\gg k_J$, its slowly varying response isplus acoustic transients. It is suppressed rather than identically absent. For example, with constant sound speed, $\mathcal{H}=2/\tau$ and ${\delta }_C=A{\tau }^2$, the equation is ${\delta }_B^{\prime \prime }+2{\delta }_B^{\prime }/\tau +c_s^2{k}^2{\delta }_B=6A$. Its particular solution $6A/(c_s^2{k}^2)$ is constant, and its oscillatory homogeneous part decays as $1/\tau$. This demonstrates Jeans support versus forced baryon response; the scale above is not a theorem forbidding every forced or transient sub-Jeans change.

Before decoupling, tight photon coupling makes the effective sound speed large, about $1/\sqrt{3}$. The small-$c_s$ approximation just derived is not quantitative in that regime; the coupled photon-baryon fluid must be used. Its large pressure supports acoustic oscillations, and baryon clustering below the corresponding large Jeans/sound scale is inhibited while Cold dark matter wells grow. After decoupling the supplied sound speed is tiny and decreases further as the gas cools. The Jeans length drops sharply, so many previously supported modes fall into Cold dark matter wells and approach its growing density contrast. Modes still below the instantaneous pressure-support scale remain suppressed. The subsequent detailed thermal scaling depends on the gas's temperature evolution; the given sound-speed history suffices for this qualitative conclusion.

Use natural units and the unreduced Planck mass $M_P=G^{-1/2}$; it is $\sqrt{8\pi }$ times the reduced Planck mass. A homogeneous canonical inflaton has ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. Vacuum domination by a scalar field requires ${\dot{\varphi }}_i^2/2\ll V({\varphi }_i)$, with other matter and the curvature term negligible in the Friedmann equation. The equation-of-state parameter is then near $-1$, and the Friedmann acceleration equation gives accelerating expansion.

For sustained slow-roll inflation, Hubble friction must dominate the inflaton acceleration: $|\ddot{\varphi }|\ll 3H|\dot{\varphi }|$, with the scalar potential changing only slightly in one Hubble time. In terms of the potential slow-roll parameter and second potential slow-roll parameter, this requiresFor the quartic scalar potential,Thus $|{\varphi }_i|\gg M_P$ supplies a broad slow-roll range when $\lambda \ll 1$. At an initial density of order the Planck density, potential domination givesso both potential slow-roll parameters are small for small $\lambda$. Small initial kinetic energy is a separate condition; the height of the scalar potential alone does not enforce it. These classical equations describe the post-Planck evolution under the model assumptions, rather than resolving quantum gravity at the initial endpoint.

Take ${\varphi }_i>0$; the negative branch follows by the symmetry $\varphi \mapsto -\varphi$. Dropping kinetic energy from the Friedmann equation and acceleration from the inflaton equation of motion givesWriting $b=M_P\sqrt{\lambda /(6\pi )}$, the quartic-potential slow-roll solution isIndeed, $\dot{\varphi }\simeq -b\varphi$, and integrating $H=\dot{a}/a$ gives $\log (a/a_i)=\pi ({\varphi }_i^2-{\varphi }^2)/M_P^2$. These expressions apply during slow-roll inflation; their formal late-time saturation must not be used beyond the end of that approximation.

The exact end of accelerating expansion is ${ϵ}_H=-\dot{H}/H^2=1$, equivalently ${\dot{\varphi }}^2=V$ for the scalar-dominated flat model. The quartic-inflation endpoint estimate uses the leading potential condition ${ϵ}_V\simeq 1$, givingThe numerical coefficient is an endpoint estimate: the slow-roll approximation breaks down there. Imposing $|{\eta }_V|=1$ estimates when slow roll loses accuracy and gives a different order-one coefficient, not a different exact acceleration condition.

The slow-roll e-fold count isThe quartic-inflation expansion from Planck density combines this initial estimate with the conventional endpoint to giveFor small $\lambda$ the dominant term is enormous; the order-one endpoint uncertainty does not affect that conclusion.

To estimate the reheating temperature, assume prompt, efficient reheating converts the end-of-inflation energy into a thermal radiation bath. With the same endpoint convention, $V_e\simeq \lambda M_P^4/(4{\pi }^2)$, and equating this to the radiation energy density givesThe quartic-inflation prompt-reheating estimate fixes the scaling; the endpoint kinetic energy changes the coefficient by order one. Without an inflaton coupling and its thermalization history, $V$ alone cannot specify the actual reheating temperature.

Finally, during cosmic inflation the comoving Hubble radius decreases:An initially causally connected patch is enlarged enough that the present observable region can lie within it. Its portions may subsequently be far outside one another's Hubble radii while retaining their common earlier thermal history. This solves the horizon problem if the number of e-folds is sufficiently large, conventionally of order sixty for the relevant thermal history; $N\sim 2\pi /\sqrt{\lambda }$ can readily exceed that. The necessary count depends on reheating and later expansion. The causal mechanism and the conversion back to a hot universe are discussed in David Tong's account of inflation.

Vary the scalar field and integrate by parts, taking the variation to vanish at the boundary:Thus the curved-spacetime Klein-Gordon equation is $\square \varphi -V_{,\varphi }=0$. In the flat FRW metric, $\sqrt{-g}=a^3$ and $g^{00}=-1$. A spatially homogeneous field therefore hasConsequentlyDots now denote cosmic time derivatives. The $3H\dot{\varphi }$ term is Hubble friction caused by expansion of the comoving volume.

Assume $V>0$ and that its derivatives do not change substantially over the field interval traversed in the estimate. With initially negligible kinetic energy, $H^2\simeq V/(3M_{\mathrm{Pl}}^2)$. The force $-V_{,\varphi }$ changes the field velocity on the Hubble friction time. Keeping $H$ and $V_{,\varphi }$ approximately constant for that local estimate, and starting at rest, solves the field equation explicitly:After one Hubble time,Writing $A=M_{\mathrm{Pl}}^2{V}_{,\varphi }^2/V^2$, the condition $A<1$ keeps this kinetic-to-potential ratio below about $0.31$, comfortably inside the inflationary bound. Integrating once more gives the fractional potential change estimateover the same interval. Thus the characteristic time for an order-one potential change is $H^{-1}/A$, up to constants of order one. The requested local Hubble-timescale criterion isUsing a strict inequality with one as the order-one threshold is conventional; its coefficient is not an exact universal end-of-inflation bound. Much smaller $A$ gives the controlled slow-roll regime and many Hubble times of cosmic inflation if the bounds persist.

After the transient in relaxation to the slow-roll attractor, $3H\dot{\varphi }\simeq -V_{,\varphi }$. Differentiating this approximate relation givesTherefore the standard sufficient conditions for negligible acceleration throughout slow roll are ${ϵ}_V\ll 1$ andThe absolute value matters: large negative curvature also drives rapid evolution. The more precise local acceleration criterion is $|{\eta }_V-{ϵ}_V|\ll 1$, with the usual separate smallness conditions avoiding a tuned cancellation. If a field starts exactly at rest on a nonzero slope, its initial acceleration is not negligible; the displayed exponential transient describes its approach to the attractor.

These are local timescale and slow-roll conditions, rather than a global duration theorem from derivatives at a single point. A smooth potential can have a narrow flat shoulder followed by a steep drop, ending cosmic inflation soon despite small starting derivatives. The conditions must hold over the traversed interval. This is the additional smooth-evolution assumption implicit in the estimate.

Vary the scalar field with compactly supported variation, leaving the metric tensor fixed. The action variation isIntegration by parts moves the derivative from $\delta \varphi$ and gives the Euler-Lagrange equationFor the FRW metric, $\sqrt{-g}=a^3$, so the covariant wave operator becomesRearranging gives the requested scalar field evolution:The damping term is Hubble friction; it comes from the expanding volume factor $a^3$, not from a microscopic dissipative force.

Let $M_{\mathrm{Pl}}$ be the reduced Planck mass, $M_{\mathrm{Pl}}^{-2}=8\pi G$ in units $\hslash =c=1$. Combining that equation with the Friedmann acceleration equation givesDifferentiate $3M_{\mathrm{Pl}}^2{H}^2=\rho$ to obtain $\dot{\rho }=-3H(\rho +P)=-3H{\dot{\varphi }}^2$. For the homogeneous canonical scalar field, $\dot{\rho }=\dot{\varphi }\ddot{\varphi }+V_{,\varphi }\dot{\varphi }$, soWhere $\dot{\varphi }\ne 0$, division gives the inflaton equation of motion; continuity extends it to isolated turning points:The Hubble friction term describes loss of kinetic energy into expansion.

There is a small logical caveat if the field is frozen on an entire interval: the homogeneous Einstein constraints alone only give the product equation and do not force $V_{,\varphi }=0$. The frozen scalar fields and the Friedmann constraint counterexample makes this explicit. For a canonical scalar, the field equation also follows directly from its inflaton action in an expanding universe, whose homogeneous Lagrangian is $a^3[{\dot{\varphi }}^2/2-V(\varphi )]$. The Euler-Lagrange equation gives $d(a^3\dot{\varphi })/dt+a^3{V}_{,\varphi }=0$, including the frozen case, and yields the same boxed equation without division by $\dot{\varphi }$.

In conformal time, $\sqrt{-g}=a^4$ and $g^{\mu \nu }=a^{-2}\mathrm{diag}(-1,1,1,1)$. The inflaton action in an expanding universe becomesVary $\varphi$ with compact support, integrate the derivative terms by parts, and discard the boundary term. The Euler-Lagrange field equation isDividing by $a^2$ provesThe factor two is the conformal-time form of Hubble friction. Using $d\tau =dt/a$ converts it to the usual inflaton equation of motion $\ddot{\varphi }+3H\dot{\varphi }-a^{-2}{\nabla }^2\varphi +V_{,\varphi }=0$, providing a sign and normalization check.

Since $H=\dot{a}/a$, differentiation gives the Hubble parameter identitySubtracting the given Friedmann equation from the Raychaudhuri equation therefore yieldsOn the other hand, differentiating the Friedmann equation givesSubstitute $\dot{H}=-4\pi G{\dot{\varphi }}^2$ and divide by $(8\pi G/3)\dot{\varphi }$, which is permitted by the assumption $\dot{\varphi }\ne 0$. The result is the homogeneous inflaton equation of motionThe middle term is Hubble friction.

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 a barotropic equation of state, $\delta P=c_s^2\delta \rho =c_s^2\bar{\rho }\delta$. A spatial Fourier mode of comoving wavenumber $k$ therefore satisfiesThe Jeans wavenumber is defined by equality of the pressure and self-gravity terms:For $k\gg k_J$, pressure dominates and produces acoustic oscillations whose amplitude is affected by Hubble friction. For $k\ll k_J$, self-gravity dominates and a growing mode develops: this is Jeans instability.