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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.

Take downward velocity as positive. Lubrication theory reduces the axial momentum equation to a constant pressure gradient, and the no-slip conditions are $u(0)=0$, $u(h)=U$. Twice integrating givesThe flux per unit circumference and the total gap flux areThe shear at the cylinder iswith the fluid traction on the falling cylinder opposing its motion.

The descending solid displaces volume at rate $\pi a^{2}U$, so closed-container mass conservation givesSince $h\ll a$, the Couette-flux term is smaller, andThe pressure or form drag isUsing the pressure-dominated part of ${\tau }_h$, the side shear force is of orderso $F_p/F_{\mu }\simeq a/h\gg 1$. Balancing form drag against the excess weight $\Delta \rho g\pi a^{2}L$ givesThis is annular lubrication drag on a settling cylinder.

For the outgoing wave, the pressure and velocity amplitudes are $-i\omega {\rho }_{+}T$ and $-i{k}_{+}T$. Its time-averaged acoustic energy flux is thereforeIf $\lambda =n\pi$, then $|T|=ϵc_{+}$ andIf $\lambda =(n+\frac{1}{2})\pi$, thenandWhen ${\rho }_{+}\ll {\rho }_{-}$ and $c_{+}\simeq c_{-}$, the half-integer family is strongly resonantly enhanced relative to the integer family by approximately $({\rho }_{-}/{\rho }_{+}{)}^2$ at equal $\omega$.

Use complex fields with time factor $e^{i\omega t}$ and put $k_{\pm }=\omega /c_{\pm }$. The outgoing potential in $x>0$ isContinuity of pressure and velocity at $x=0$ givesHenceThe piston condition ${\Phi }_{-}^{\prime }(-L)=i\omega ϵ$, with $\lambda =k_{-}L$, givesThis is the transmitted velocity potential from a piston through an acoustic interface.

An intensive quantity is unchanged when the size of a homogeneous system is scaled; temperature and pressure are examples. An extensive quantity scales in proportion to system size; volume and particle number are examples. Energy and entropy are also extensive, while chemical potential is intensive. These are intensive and extensive thermodynamic quantities.

Since $P=P(\rho )$ and $\rho =\bar{\rho }(1+\delta )$,To first order, continuity and the divergence of Euler giveEliminating $\theta$ and using $\nabla \mapsto i\mathbf{k}/a$ yieldsThe second term is Hubble damping, gravity drives growth, and pressure produces oscillatory restoration. Their balance definesModes longer than the Jeans length are gravitationally unstable; shorter modes are pressure supported.

The tensor ${\sigma }_{ij}$ is the flux of the $j$th component of electromagnetic momentum across a surface normal to the $i$ direction, with the sign convention used in the conservation law. Its diagonal components are normal momentum fluxes, interpreted as electromagnetic pressures or tensions. Its off-diagonal components are tangential momentum fluxes, interpreted as shear stresses.

With streamwise and normal velocities $u,v$, zero exterior pressure gradient gives the steady boundary-layer equationsAt the organism,and matching to the quiescent exterior requires $u(x,y)\to 0$ as $y\to \infty$.

If $k>N/U$, then $m=i\alpha$ with $\alpha >0$ chosen so that the disturbance decays as $z\to \infty$. The pressure amplitude is then ninety degrees out of phase with $w$, since $\hat{p}/\hat{w}$ is purely imaginary. Hence

In the Dieterici equation, $b$ is the excluded volume per particle caused by short-range repulsion. The positive constant $a$ measures cohesive attraction, which lowers the pressure through the Boltzmann factor.

The interface is the material level set $F=r-R(\theta ,t)=0$. The condition $DF/Dt=0$ giveswhich is the stated equation.

Let the perturbation potentials inside and outside beLinearizing the kinematic condition at $r=a$ givesThe linearized interior Euler equations givewhereas the stationary exterior gives $p_o^{\prime }=-\rho \sigma {\varphi }_o$. Since the basic interior pressure has $d{p}_i/dr=\rho {\omega }^{2}r$, pressure continuity on the displaced boundary isEliminating $A$ and $B$ gives the circular vortex-sheet mode relationsoThe $k=1$ mode is a neutral displacement. Every $k>1$ mode has one exponentially growing branch, so the circular interface rolls up through a Kelvin--Helmholtz instability. Its pattern angular velocity is $\omega (k-1)/(2k)$, in the direction of the gyre but slower than the solid-body motion; the disturbances therefore propagate upstream relative to the rotating water.

Take the downstream tissue pressure as gauge zero. The two transverse momentum fluxes cancel. The force of the fluid on the junction is axial and equalswhere $U=q/A$, $v=q/(2a)$ and the pressure difference is that in part (i).

Take $x$ vertically downward and $y$ outward from the wall, with the film occupying $0<y<h$. The ambient pressure isAt $y=h$, normal stress continuity gives $p=p_a$ and zero ambient shear gives $\mu U^{\prime }(h)=0$; at the wall, no slip gives $U(0)=0$.

For steady parallel flow $u=(U(y),0)$, incompressibility is automatic and Navier–Stokes reduces toSince $p_x={\rho }_{a}g$, integration yields

In the fluid region $y>\eta (x)$, incompressibility and potential flow giveNo penetration through the exact surface gives

The condition $hk\ll 1$ says the hill slope is small; $|\nabla \varphi |\ll U$ says the disturbance velocity is small compared with the background wind. Dropping the product ${\varphi }_x{\eta }_x$ and Taylor-shifting the boundary from $y=\eta$ to $y=0$ therefore givesThe decaying solution is

Bernoulli's equation isTo first order on the surface,ThusEquivalently, crest minus trough is the negative of this. For $k{U}^2/g\ll 1$, hydrostatic elevation dominates; for $k{U}^2/g\gg 1$, the Bernoulli pressure drop caused by faster crest flow dominates.

The two binodal compositions ${\varphi }_1<0<{\varphi }_2$ are the contact points of the common-tangent construction. Equivalently, they solveThe first equality is equality of chemical potential; the second is equality of pressure. Heretogether with the intercept equation determines the two densities. As $C\downarrow a$, both contact points approach zero continuously with scale $|{\varphi }_{1,2}|\propto \sqrt{C-a}$, so this mean-field onset of phase separation is continuous.

The normalized one-particle height density is exponential:ConsequentlyThe number density and local ideal-gas pressure areThis also obeys hydrostatic balance $p^{\prime }(z)=-mgn(z)$.