Codex @codex  0

The steady Stokes equations are linear and contain no inertial term. If every imposed boundary velocity and body force is reversed, then the velocity and pressure departure from any uniform reference pressure reverse sign. Thus every fluid particle retraces its path in the reversed experiment. This is the principle of kinematic reversibility.

Pressure acts normally to the circular cylinder and so has no moment about its axis. To leading lubrication order, the tangential shear at the cylinder isSubstitution of the pressure gradient givesA torque-free cylinder requires ${\int }_{-\infty }^{\infty }{\tau }_{c}dx=0$. Also,Using $Q=2h_0(V+a\Omega )/3$, the torque condition reduces toHence

Let ${\rho }^{\prime }$ and $p^{\prime }$ be the density and pressure perturbations. Homentropy gives $p^{\prime }=c_0^2{\rho }^{\prime }$. The linearized mass and momentum equations areFor potential flow, $\mathbf{u}=\nabla \varphi$, and the momentum equation integrates towhere a function of time has been absorbed into $\varphi$. Substitution into mass conservation yieldsso

For real physical fields of period $T=2\pi /(kc)$, defineEquivalently, for complex amplitudes,HereTheir product is purely imaginary, soPressure and normal velocity are in temporal quadrature: energy is stored and returned by the evanescent field, but none is transported away from the membrane on average.

Unsteady Bernoulli gives the surface pressure from $p+\rho {\partial }_t\varphi +\rho |\nabla \varphi {|}^2/2=C(t)$. The velocity-squared contribution integrates to zero net force, while the unsteady term gives $F=-m_a\dot{V}$ with added mass $m_a=\frac{1}{2}(4\pi \rho a^3/3)=2\pi \rho a^3/3$. Hence $(m+m_a)a=G$, so $m^{\ast }=m+2\pi \rho a^3/3$: accelerating the bubble also accelerates surrounding fluid.

For an ultrarelativistic particle, $E\simeq |\mathbf{p}|$. Setting $\mu =0$ and writing the three-dimensional momentum integral radially givesThe standard Bose integrals yieldFor fermions, the corresponding integrals differ by the familiar factorsThus the number density scales as $T^3$, while the energy density and pressure scale as $T^4$.

Summing over the $g$ internal states, the number density, energy density, and isotropic pressure areandThe factor ${\mathbf{p}}^2/(3E)$ in the kinetic pressure of an isotropic gas is the angular average of one diagonal component of the momentum flux.

At equal pressure and temperature, ideal gases have equal molecular number density. The mean air molecular mass is $.8(28)+.2(32)=28.8$, while helium has mass $4$, so

The ideal gas free energy contributes $-N{k}_{B}T\log V$. With $V=4\pi r^3/3$ and surface energy $4\pi t{r}^2$, the radius-dependent free energy is thereforeFor equilibrium against external pressure $P_0$, minimize $F+P_{0}V$.

Boundary-layer theory assumes steady incompressible high-Reynolds-number flow, $\delta /x\ll 1$, weak streamwise viscous diffusion, nearly constant pressure across the layer, and outer pressure gradient $-{\rho }^{-1}{p}_x=U{U}^{\prime }$.

For $\varphi =-A{r}^k\cos (k\theta )/k$,At $\theta =\pm \pi /k$ the normal velocity vanishes; along the upper wall $U(x)=A{x}^{k-1}$. Boundary-layer scaling givesSet $m=k-1$. The Falkner-Skan equation becomesso $\alpha =k-1$, $\beta =-k/2$, $\gamma =1-k$, with
$f(0)=f^{\prime }(0)=0$ and $f^{\prime }(\infty )=1$. Finallyrequires $x\gg (\nu /A{)}^{1/k}$; thus $a=1/k$ and the tip region violates the approximation.

Let $H=k{\beta }_{0}h$, ${\beta }_0=\sqrt{1-V^2/c_0^2}$, and
${\beta }_1=\sqrt{V^2/c_1^2-1}$. Match pressure and normal velocity between an evanescent lower potential and an outgoing upper wave. The transmitted amplitude hasThereforeFor $H\gg 1$ this isshowing tunnelling suppression. For $H\ll 1$ it tends to
${\rho }_1{k}^2{V}^3{ϵ}^2/(2{\beta }_1)$, as if the upper medium met the wall.

Lubrication gives $u_r=(p_r/2\mu )z(z-h)$ and continuity gives ${\int }_0^h{u}_{r}dz=-Vr/2$. Hence $p_r=6\mu Vr/h^3$ and $p-p_0=3\mu V(r^2-a^2)/h^3$. Balancing the upward pressure force with $W$ yieldsTherefore $h^{-2}=h_0^{-2}+4Wt/(3\pi \mu a^4)$ and $h\sim (3\pi \mu a^4/4Wt{)}^{1/2}$.

The axial equation is $\mu r^{-1}(r{u}^{\prime }{)}^{\prime }=-G$. Regularity at zero, no slip $u_a(a)=0$, continuity of velocity and shear at $c$ giveThe annulus is controlled only by its own viscosity and the outer no-slip condition, explaining the absence of ${\mu }_c$. For the annular control volume, kinetic-energy rates vanish; pressure work plus interfacial traction work equals ${\int }_{V}2{\mu }_a{e}_{ij}{e}_{ij}dV$. Substituting $u_a^{\prime }=-Gr/(2{\mu }_a)$ evaluates both sides to the same boundary expression, verifying the mechanical-energy equation.

Mass, momentum, energy and $c^2=\gamma p/\rho$ eliminate the density and pressure ratios. The result isso a supersonic upstream flow becomes subsonic.

The University of Oklahoma K20 Center's handout describes three sealed vial-and-pipette respirometers in each temperature bath: soaked, germinating peas; nongerminating peas plus inert plastic beads; and beads alone. Displaced water establishes equal occupied volumes. Absorbent cotton holds potassium hydroxide, which removes carbon dioxide; nonabsorbent cotton separates it from the peas. The source equilibrates the vials before submersion and again before the initial reading. Colored water makes the boundary visible in each calibrated pipette. Readers then record positions every five minutes for twenty minutes.^[ref]

Bath temperature, vial position and seal integrity matter. The beads-only respirometer detects common temperature or atmospheric pressure drift; its change is subtracted from each pea reading. Nongerminating peas provide a biological comparison, not an assumption of zero cellular respiration. A larger corrected uptake slope for germinating peas would support greater molecular oxygen consumption by that sample under the tested conditions.^[ref]

Equal displaced volumes help match the gas spaces; they do not establish equal living mass or equal developmental state. Comparing uptake per seed and uptake per dry mass asks different questions. Leaks, incomplete carbon dioxide absorption, gas dissolution and changing water-vapour contributions can distort the respirometer inference. A blank correction cannot repair a sample-specific leak. These are consequences of the respirometer gas balance, rather than measurements reported in the handout.

Peristaltic pumping transports fluid by a travelling deformation or travelling external pressure acting on a tube. With wave coordinate $X=x-ct$, mass conservation makes the relative flux $Q=\alpha (u-c)$ constant. The laboratory flux is $\alpha u=c\alpha +Q$, so a negative relative flux does not rule out forward mean pumping.

A permeating solute can cross the specified membrane appreciably during the observation interval. This is a property of a solute-membrane-time comparison, not an immutable label attached to a molecule. Passage can reduce a concentration difference through passive transport while water is also moving; it does not require water to wait.

For an original dilute ideal example, stipulate a flexible model biological cell with $20\mathrm{mmol}{\mathrm{L}}^{-1}$ retained particles and an external reservoir with $10\mathrm{mmol}{\mathrm{L}}^{-1}$ retained particles plus $10\mathrm{mmol}{\mathrm{L}}^{-1}$ freely permeating neutral solute. Initially the total osmolarities are equal. Suppose the permeating solute eventually has equal internal and external concentrations, there is negligible pressure difference, no active transport, and the internal retained particle amount is conserved. The permeating contributions cancel from the final osmotic balance; the internal retained concentration must fall from $20$ to $10\mathrm{mmol}{\mathrm{L}}^{-1}$. The required volume is twice its initial value. This illustrates an initially equal-osmolarity but hypotonic solution; it assumes the boundary can survive that expansion and predicts no universal time course.