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Write $A=8\pi G/(3c^2)$ and $H=\dot{a}/a$. Differentiating the Friedmann equation givesThe cosmological perfect-fluid continuity equation says $\dot{\rho }=-3H(\rho +P)$, so, first where $H\ne 0$ and then everywhere by continuity,Using $K{c}^2/a^2=A\rho -H^2$ from the original equation givesSince $\dot{H}+H^2=\ddot{a}/a$ for the Hubble parameter, this is the Friedmann acceleration equation.

Put $K=8\pi G/(3c^2)$. Differentiating $Q=K\rho a^2-{\dot{a}}^2$ givesThe continuity equation and the Friedmann acceleration equation implyandThese terms cancel, soThis is the First integral of the Friedmann acceleration equation.

The Flatness problem is that the observed cosmological density parameter is close to $\Omega =1$, although in an ordinary decelerating universe any departure from one grows with time. Extrapolation backward therefore requires implausibly precise early cancellation of the curvature term.

The cosmological perfect-fluid continuity equation isConsequentlyFor an expanding universe $H>0$. Ifthen $\rho a^2$ increases. The Friedmann equation impliesIt therefore decreases during this period, driving $\Omega$ toward one. By the Friedmann acceleration equation, the same condition gives $\ddot{a}>0$ when the cosmological constant is included in the effective fluid. This is the inflationary solution of the flatness problem.

For the scalar field, the slow-roll approximation meansThe equations reduce toTaking the positive-field branch gives $H\simeq m\varphi /\sqrt{3}$, and thereforeWith $\varphi (0)={\varphi }_i$,Now integrate $H=\dot{a}/a\simeq m\varphi /\sqrt{3}$:Thus the quadratic-potential slow-roll solution isFinally,so equivalently

Differentiate the Friedmann equation and useAfter cancelling the curvature term with the original Friedmann equation, one obtains the Friedmann acceleration equation

The Newtonian derivation does not explain why pressure gravitates through the $3P$ term; that input enters only through a relativistic-looking continuity law. It also treats space and time as Euclidean and absolute, so it cannot supply the spacetime meaning of curvature $k$, horizons, or the global geometry of the cosmological model.

For radiation in cosmology with zero cosmological constant, $p=\rho /3$. The Friedmann acceleration equation gives $\ddot{a}/a=-8\pi G\rho /3$, so the deceleration parameter equals the cosmological density parameter, regardless of spatial curvature. This is a relation for a pure radiation fluid, not a general identity for a mixture.

A closed universe containing only a cosmic string network can remain static. Its Friedmann acceleration equation vanishes identically because $P_s=-{\rho }_s/3$, and its Friedmann equation requires ${\rho }_s{a}^2=3K/(8\pi G)$. Once this conserved normalization is tuned, every positive radius gives a static solution. The usual Hubble-normalized cosmological density parameter is undefined at a static epoch.