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Write $A_k=H/\sqrt{4ϵM_{\mathrm{Pl}}^2{k}^3}$, so $u_k(0)=A_k$. For the cyclic term with the first field undifferentiated, insertion into the in-in formalism giveswhere $K=k_1+k_2+k_3>0$. The two differentiated modes contribute ${\tau }^2$, cancelling the ${\tau }^{-2}$ in $a^2$.

Along the vacuum-selected contour, the lower boundary vanishes. The elementary integrals areand thereforeMultiplying by $4i{M}_{\mathrm{Pl}}^2{ϵ}^2$, taking the real part and summing the three cyclic choices gives the curvature bispectrum from a zeta zeta-prime-squared interaction:This matches the final normalization. The coefficient requires all six connected Wick contractions.

The calculation assumes the Gaussian Bunch-Davies vacuum, tree order in the specified cubic interaction, effectively constant $H$ and $ϵ$ during the integral, the approximate de Sitter scale factor, and observation after the modes freeze as $\tau \to 0^{-}$. The contour tilt is kept until the early boundary has been eliminated; an undamped real-axis integral cannot simply discard that boundary. The reduced Planck mass convention is the one in the supplied De Sitter curvature mode function. Other cubic interactions and nonlinear field redefinitions are outside the specified model, so this is not by itself the complete bispectrum of a general single-field action.

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 $\dot{\varphi }$.

Substitution of these expressions into the cosmological perfect-fluid continuity equation givesContinuity through a turning point therefore yields the homogeneous Klein-Gordon equationDuring slow-roll inflation, the acceleration and kinetic-energy terms are negligible. In terms of the reduced Planck mass, the approximate evolution is

For an unwarped product metric with fixed compact volume $V_n$, integrating the higher-dimensional Einstein-Hilbert action gives $M_{\mathrm{Pl}}^2=M_{\ast }^{n+2}{V}_n$ when the action coefficient is $M_D^{D-2}/2$. The reduced Planck mass convention must be distinguished from the unreduced one, and stabilizing the internal metric is a separate dynamical issue.

Planck units use fundamental constants to set natural units of mass, length and time. The displayed convention sets the ordinary Planck mass to one. The related reduced Planck units convention instead normalizes the reduced Planck mass.

Near a flat quartic maximum, $V\simeq \alpha {\sigma }^4-3\alpha {\varphi }^4$. With reduced Planck mass $M$, the slow-roll approximation gives $N\simeq {\sigma }^4/(24M^2{\varphi }_N^2)$, ${\eta }_V\simeq -3/(2N)$ and ${ϵ}_V\simeq {\sigma }^4/(192M^4{N}^3)$. Hence $n_s\simeq 1-3/N$, $n_T\simeq -2{ϵ}_V$, and ${\mathcal{P}}_{\mathcal{R}}\simeq 8\alpha N^3/{\pi }^2$. The result is dominated by the near-maximum part of the trajectory; extending slow roll to a distant minimum is unnecessary.

// Target: cosmic-inflation.bigb

Here the reduced Planck mass $M_{\mathrm{Pl}}=(8\pi G{)}^{-1/2}$ is one, so the Friedmann equation for an energy density $\rho$ is $3H^2=\rho$. This differs by a factor of $\sqrt{8\pi }$ from setting the ordinary Planck mass to one; the convention must be stated in inflationary amplitudes.