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For a homogeneous canonical scalar field, the cosmological perfect-fluid continuity equation reduces to $\dot{\varphi }(\ddot{\varphi }+3H\dot{\varphi }+V^{\prime })=0$. It implies the inflaton equation of motion wherever $\dot{\varphi }\ne 0$, and by continuity at isolated turning points. It cannot by itself exclude a frozen field on a time interval. Indeed, choose $\varphi ={\varphi }_0$ with $V({\varphi }_0)>0$ but $V^{\prime }({\varphi }_0)\ne 0$; a de Sitter scale factor in flat slicing with $H^2=V({\varphi }_0)/(3M_{\mathrm{Pl}}^2)$ satisfies the homogeneous Friedmann equation and Friedmann acceleration equation, but not the scalar equation. Varying the inflaton action in an expanding universe supplies the missing equation also for a frozen field, requiring $V^{\prime }({\varphi }_0)=0$.

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.

Write the action density asUnder a compactly supported variation $\varphi \mapsto \varphi +ϵ\eta$,Integrating the first term by parts in time and the second in space, and discarding the boundary terms, givesThe principle of stationary action requires this to vanish for every $\eta$. Sincedivision by $-a^3$ yields the field Euler-Lagrange equationThis is the equation generated by the inflaton action in an expanding universe.