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For the open matter-dominated Friedmann solution, write $\Omega ={\Omega }_{m,0}$, so $0<\Omega <1$. The present Friedmann equation gives $-K=H_0^2(1-\Omega )$. With $A=H_0^2\Omega$ and $B=H_0^2(1-\Omega )>0$, the expanding equation becomesChoose the hyperbolic substitutionThen $da=(A/(2B))\sinh \eta d\eta$ and $\sqrt{a/(a+A/B)}=\tanh (\eta /2)$. Their product givesTake the Big Bang to be $t=0$, $\eta =0$, and integrate. The parametric solution isThe parameter is determined explicitly byThus $\eta$ is conformal time multiplied by $\sqrt{-K}$, with its origin at the Big Bang; it is not equal to unscaled cosmic time. In particular today ${\eta }_0=\mathrm{arcosh}(2/\Omega -1)$. Small $\eta$ gives $a\propto {\eta }^2$ and $t\propto {\eta }^3$, hence $a\propto t^{2/3}$. Large $\eta$ gives $a/t\to H_0\sqrt{1-\Omega }=\sqrt{-K}$, in agreement with the curvature-dominated universe limit.

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.

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 $\dot{R}=\mathcal{H}R$.

Let $q^i=e^i-\frac{1}{2}{h}^i{}_j{e}^j$, so $p^0=ϵ/a^2$ and $p^i=(ϵ/a^2)q^i$. The orthonormal tetrad gives ${\delta }_{ij}{e}^i{e}^j=1$. To first order,Hence the time-index Christoffel symbols contract toThe $\mathcal{H}{h}_{ij}$ term cancels the perturbation in the coordinate direction norm. This cancellation is why the local tetrad direction must not be mistaken for the coordinate direction.

Along the photon null geodesic, $d/d\lambda =p^{0}d/d\eta$, andThe time component of the geodesic equation now gives the photon energy redshift from a tensor metric perturbation:The derivative on the left follows the ray; the dot on $h$ is a partial conformal time derivative. The background redshift has already been removed by using comoving energy. The calculation only needs ${\Gamma }^0{}_{00}$ and ${\Gamma }^0{}_{ij}$; the unrelated printed ${\Gamma }^i{}_{j0}$ should contain $\frac{1}{2}{\dot{h}}^i{}_j$, rather than the full derivative.

Distinguish the slow-roll approximation parameter $ϵ$ from the infinitesimal positive contour regulator $\epsilon$. In the interaction picture, the ordered unequal-time Wick contraction isA differentiated vertex field instead supplies $u_k^{\ast \prime }(\tau )$. These conjugates are fixed by the annihilation operators and creation operators and cannot be dropped while retaining the same contour.

Convert the interaction time integral to conformal time. Since $dt=ad\tau$ and each cosmic-time derivative is $a^{-1}{\partial }_{\tau }$, the integrated cubic vertex isThe in-in formalism supplies $-2i$ multiplying the expectation of the Hamiltonian, so its negative sign gives $+2i$ multiplying this vertex. At nonzero external momenta, the connected contractions join each of the three external fields to one vertex field. There are six bijections, not three. Choosing which external field meets the undifferentiated vertex gives three cyclic choices; swapping the two differentiated fields gives a further factor of two. This is the Wick-pairing multiplicity for a zeta zeta-prime-squared vertex.

For example, before the internal momenta are integrated, one cyclic contribution includesThe factors of $2\pi$ cancel to leave one overall $(2\pi {)}^3$ and the external momentum delta. Thus the properly normalized reduced expression isThe vacuum prescription for inflationary in-in integrals damps the early-time oscillations and selects the Bunch-Davies vacuum. Internal self-contractions correspond to disconnected zero-momentum tadpole terms and are excluded from this connected three-point function.

The printed intermediate expression has unconjugated modes and only three literal cyclic terms with coefficient $-2i$. The conjugate form of the result above would use $-4i$, together with the conjugated contour. A literal $-2i$ with only three terms gives half the final printed answer. The six-contraction expression above is the consistent reduction of the supplied Hamiltonian and leads to that final answer.

Here primes mean conformal time derivatives, $d\tau =dt/a$, and spatial derivatives use comoving coordinates in Newtonian gauge in cosmology. Set ${\theta }_m=\nabla \cdot {\mathbf{v}}_m$. The linear cosmological Euler equation gives ${\theta }_m^{\prime }=-\mathcal{H}{\theta }_m-{\nabla }^2\Phi$, while the continuity equation gives ${\theta }_m=-{\delta }_m^{\prime }+3{\Phi }^{\prime }$.

Differentiate the continuity equation and substitute:Thus the linear matter density equation in Newtonian gauge isThe common potential in these supplied equations assumes that the relevant scalar anisotropic stress is neglected. No Fourier-sign convention is needed for the derivation in position space.

Use the comoving curvature perturbation mode normalization. Take $H,ϵ,c_s$ to be effectively constant at leading slow-roll approximation order, with $H,ϵ,c_s>0$, and use the Bunch-Davies vacuum selected by the early-time $i\epsilon$ contour. The regulator $\epsilon$ is distinct from the Hubble slow-roll parameter $ϵ$.

Let $C=ϵ+1-c_s^2$ and use the Fourier transform convention $\zeta (\mathbf{x})=\int d^{3}k\zeta (\mathbf{k})e^{i\mathbf{k}\cdot \mathbf{x}}/(2\pi {)}^3$. Conversion to conformal time givesThe last sign follows from substituting $a=-1/(H\tau )$, which is positive for $\tau <0$; dropping the explicit minus sign in that substitution would reverse the final primordial bispectrum.

At the order needed by the in-in formalism, the unequal-time Wick contraction isIt uses the Gaussian quantum vacuum two-point function with the displayed operator ordering; an arbitrary classical unequal-time covariance cannot replace it. For the connected three-point function, each external field contracts with one of the three fields at the vertex. There are $3!=6$ such Wick contractions, all equal. The spatial integral conserves total momentum and leaves one factor $(2\pi {)}^3{\delta }^{(3)}({\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3)$. The other nine pairings are disconnected tadpole terms supported on a zero external momentum; they are excluded from the connected primordial bispectrum, equivalently by subtracting the one-point function or normal ordering the vertex.

The free Gaussian odd correlator is zero. Applying the stated first-order in-in formalism therefore givesBy definition, the full connected correlator is $(2\pi {)}^3{\delta }^{(3)}({\sum }_r{\mathbf{k}}_r)B$. The assumptions are a Gaussian Bunch-Davies vacuum, the given leading cubic vertex, tree order, the late-time limit $\tau \to 0^{-}$, effectively constant background parameters, and connected nonzero external modes. No symmetry factor $1/3!$ is inserted into the given Hamiltonian.

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.

Put $\Gamma (\eta )=a{\bar{n}}_e{\sigma }_T\ge 0$, the Thomson scattering rate per unit conformal time. The cosmological optical depth is the integrated scattering rate from emission time to observation, so a Poisson scattering process has survival probabilitySince $\dot{\tau }=-\Gamma$, the cosmological visibility function isA photon scatters in an interval $d\eta$ with rate factor $\Gamma d\eta$ and then reaches us without another scattering with probability $E$, so $gd\eta$ is its probability of last scattering in that interval. For an optically thick initial epoch,If the initial optical depth is finite, the missing weight represents photons that have already stopped scattering before ${\eta }_i$.

Before cosmological recombination, the optical depth to the present is enormous: $E\simeq 0$, and $g$ is also small because almost every scattering is followed by another. Through photon decoupling, $E$ rises rapidly and $g$ has its dominant positive peak. After decoupling, without reionization, $E$ is close to one and reaches exactly one at ${\eta }_0$, while $g$ is small because the remaining electron density and scattering rate are small. Residual ionization can give a weak tail; there is no second reionization peak. The peak marks the Cosmic microwave background last-scattering surface.

Let $\mathcal{H}=a^{\prime }/a=\dot{a}$, where a prime denotes differentiation with respect to conformal time $d\tau =dt/a$. Multiplying the Friedmann equation by $a^2$ givesThe critical density is the value obtained when $k=0$, so

Since ${\mathcal{H}}^{\prime }=a\ddot{a}$, the pressure-free Raychaudhuri equation givesorDifferentiating $\Omega =1+k{c}^2/{\mathcal{H}}^2$ and substituting this equation yields

For an expanding universe $\mathcal{H}>0$. The Einstein-de Sitter universe $\Omega =1$ is an unstable constant solution: curves starting with $0<\Omega <1$ decrease towards $0$, while curves starting with $\Omega >1$ rise away from $1$ and diverge as the closed model reaches maximum expansion. This is the Flatness problem.

Take a fixed comoving region, whose physical volume is $V=V_0{a}^3$. Its energy is $E=\rho V$, so the first-law relation $dE=-PdV$ givesDifferentiating with respect to cosmic time and using the Hubble parameter $H=\dot{a}/a$ yields the cosmological perfect-fluid continuity equation

For the barotropic equation of state $P=w\rho$, separation gives the constant-equation-of-state density scalingSubstituting this into the flat Friedmann equation and integrating the expanding branch gives, for $w\ne -1$,For $w=-1$, the density and Hubble parameter are constant and instead $a(t)\propto e^{Ht}$.

Conformal time is defined byAt radiation–string equality, let each component have density ${\rho }_{\mathrm{eq}}$ at $a=a_{\mathrm{eq}}$. Since radiation in cosmology has $w=1/3$ and a cosmic string network has $w=-1/3$,ConsequentlyThusOn the expanding branch, integration givesChoosing the Big Bang to occur at $\tau =0$ gives the radiation--cosmic-string Friedmann solution

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.

Conformal time is defined byFor $a=a_0{e}^{H_{\inf }t}$, choose $\tau \to 0$ as $t\to +\infty$. Integration givesas recorded by Conformal time during de Sitter expansion.

Writing a prime for $d/d\tau$, the relationsturn the Fourier-mode equation intoSetA direct substitution givesHere $a^{\prime \prime }/a=2/{\tau }^2$, so the Canonically rescaled de Sitter scalar mode obeys

For successive wavecrests from the same comoving source, $1+z=a(t_0)/a(t_s)$. ThereforeThe same radial null path implies equality of the elapsed conformal time,Expanding both scale factors to first order gives the redshift drift

The Euler-Lagrange equation isAfter a spatial Fourier transform and the change to conformal time, ${\partial }_t=a^{-1}{\partial }_{\tau }$, this becomesFor the exact de Sitter spacetime scale factor $a=-1/(H\tau )$, one has $a^{\prime \prime }/a=2a^2{H}^2$. Setting $u_{\mathbf{k}}=a{\phi }_{\mathbf{k}}$ therefore cancels both the friction term and the effective mass term, leavingThis is the special cancellation for a conformally coupled scalar field.

Conformal time obeys $dt=ad\eta$, so $\dot{\varphi }=a^{-1}{\varphi }^{\prime }$. ThereforeFor expanding de Sitter spacetime, $a=-1/(H\eta )$ with $-\infty <\eta <0$.

Substitution of $\delta N=C\phi$ and conversion to conformal time turn the interaction intoTreat $C$ and $H$ as constant at leading slow-roll order and use the massless de Sitter modeFor $K=k_1+k_2+k_3$, the in-in formalism time integral at a vertex with leg $i$ undifferentiated is proportional toSumming the three choices of undifferentiated leg gives the primordial bispectrumThe sign follows from the interaction sign displayed in the question and $H_I=-L_{\mathrm{int}}$ at this order.

Define conformal time by $d\tau =dt/a(t)$. The radial coordinate and the angular-radius function areSubstitution yields the stated conformal metric. A radial null ray satisfiesso $d\chi /d\tau =\pm c$ and $\chi ={\chi }_0\pm c(\tau -{\tau }_0)$.

Consider a spatially flat universe containing radiation and a cosmic string network. If their densities are both ${\rho }_{\mathrm{eq}}$ at scale factor $a_{\mathrm{eq}}$, thenIn conformal time $d\tau =dt/a$, the flat Friedmann equation becomesChoosing the Big Bang at $\tau =0$, its expanding solution is

In a flat universe containing separately conserved radiation in cosmology and pressureless matter, the Friedmann equation becomes $(a^{\prime }{)}^2=H_0^2({\Omega }_{r,0}+{\Omega }_{m,0}a)$ in conformal time. Setting $a=0$ at $\tau =0$ gives the stated quadratic-plus-linear scale factor. The linear term controls radiation domination and the quadratic term controls matter domination. The formula presumes $a_0=1$, ${\Omega }_{r,0}+{\Omega }_{m,0}=1$ and no cosmological constant.