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A Klein-Gordon solution is second order in time: both $\Psi$ and ${\partial }_t\Psi$ are needed as initial data. Its free modes have both signs of frequency,This differs from the first-order Schrödinger evolution of a fixed one-particle wavefunction.

Moreover, $|\Psi {|}^2$ does not generally obey the required probability continuity equation. The conserved Klein-Gordon density is proportional toup to a positive conventional constant. A pure positive-frequency mode gives $+2\omega |\Psi {|}^2$ and a negative-frequency mode gives $-2\omega |\Psi {|}^2$. This density is not positive definite and cannot be a universal position probability density.

For a free field, one can choose a positive-frequency subspace with a positive integrated inner product. Time-dependent backgrounds generally mix it with the negative-frequency subspace, so simply deleting the latter is not dynamically consistent. In quantum field theory the complex Klein-Gordon field instead has particle and antiparticle sectors; its conserved current is interpreted as charge current, and creation or annihilation of pairs need not fit a fixed-particle-number wavefunction. These are limitations of the naive single-particle interpretation, not a failure of the relativistic field equation.

Farther than a few edge lengths from the root, replace the radial coordinate $r$ by a physical coordinate $x=ar$ and smooth the master equation. It has the same constant-coefficient Fokker-Planck equation as in (b). Under the conventional first-time-derivative truncation,The radial drift $v=-C=a(z-2)/(z\tau )$ is outward. For $z=2$ the tree is the infinite line, its radial distance is an unbiased reflected walk, and $C=0$.

The root cannot absorb probability or allow negative radial distance. Write the equation as a continuity equation, using the Fokker-Planck probability currentThe appropriate continuum boundary condition is reflecting, or zero flux:It is not generally ${\partial }_{x}P(0,t)=0$: that simpler condition is correct only when $C=0$. With the finite-bias fixed-step Gaussian convention from (b), use $D=D_{\mathrm{fixed}}=2a^2(z-1)/(z^2\tau )$ instead; the drift sign and zero-current boundary condition are unchanged. A continuum boundary condition represents the microscopic root region after coarse-graining, rather than reproducing each discrete root probability at individual step times.

Choose $x$ vertically upward. Let $G(x,t)$ be the gas-occupied part of a pipe cross-section, of area $\alpha A$. Define the phase mean mass density, mass-weighted velocity, second moment of velocity, and phase mean pressure byThe void fraction $\alpha$ is a volume fraction, not a mass fraction. The gas momentum in a slice of length $dx$ is $A\alpha {\bar{\rho }}_g\bar{v}dx$; its advected momentum flux is $A\alpha {\bar{\rho }}_g\overline{v^2}$. The two end pressures give the net force $-A{\partial }_x(\alpha {\bar{p}}_g)dx$. Include every other vertical force on the gas, including interfacial forces, in $M_{g}dx$. Newton's second law for this fixed slice, followed by division by $Adx$, therefore givesThis is a cross-section averaged phase momentum balance. In particular, mass weighting makes the momentum and its flux exact averages; one must not replace $\overline{v^2}$ by ${\bar{v}}^2$ without an additional closure. With liquid mean velocity $\bar{u}$, constant liquid mass density ${\rho }_{\ell }$, and liquid second moment $\overline{u^2}$, the same calculation givesWith the common pressure $p={\bar{p}}_g={\bar{p}}_{\ell }$, the specified pressure contributions to $M_g,M_{\ell }$ cancel the phase-fraction derivatives from the product rule. The remaining pressure forces per unit mixture volume are $-\alpha p_x$ and $-(1-\alpha )p_x$.

Assume that the bubbles at a given height have the stated radius $a$. Their number per unit mixture volume is $N=3\alpha /(4\pi a^3)$. The drag force on a rising bubble opposes its relative velocity, so multiplication of the single-bubble law by $N$ gives the signed magnitudeGravity acts downward; the drag force on the liquid is the opposite of that on the gas. With no wall force,A bubble retains its mass in the absence of transfer between phases. Consequently $a^3{\bar{\rho }}_g=a_0^3{\rho }_0$ andFor clarity suppress bars on $p,{\rho }_g,v$ in the steady calculation. On a macroscopic variation length $\ell$, gas inertia per unit volume scales like $\alpha {\rho }_g{v}^2/\ell$, whereas the quadratic drag force scales like $\alpha {\rho }_{\ell }{v}^2/a$. Their ratio is $O[({\rho }_g/{\rho }_{\ell })(a/\ell )]$. Thus ${\rho }_g=O({\rho }_0)\ll {\rho }_{\ell }$, together with the scale separation $a\ll \ell$ inherent in this averaging model, justifies the leading-order neglect of gas inertia. Small reference density alone does not justify it in an arbitrarily short or strongly compressed region. Neglect gas weight at the same order, with the liquid fraction not vanishing. For a stationary liquid, summing the two momentum equations givesThe leading gas balance is $0=-\alpha p_x-D_v$. Substitution of the bubble-radius law yieldsThe gas continuity equation, with constant pipe area and no mass exchange, is ${\partial }_t(\alpha {\rho }_g)+{\partial }_x(\alpha {\rho }_{g}v)=0$. Hence the steady gas mass flux $J=\alpha {\rho }_{g}v$ is constant. For $0<\alpha <1$ the pressure equation selects $v>0$, andEvaluate the latter constant at a reference point ${\rho }_g={\rho }_0$, $\alpha ={\alpha }_0$ and divide. The terminal-drift density law for compressible bubbles isThe exponent $3/5$ comes from combining mass conservation with the density dependence of bubble radius; no gas equation of state was needed.

Take $z$ positive away from the wall into the liquid, and write $p_a$ for atmospheric pressure. For a constant-mass density Newtonian fluid, the two-dimensional Navier-Stokes equation and continuity equation areThe stationary wall imposes the no-slip boundary condition and no-penetration boundary condition, $u=w=0$ at $z=0$. For a flat free surface with negligible air shear stress, the stress boundary condition is $u_z=0$, $p=p_a$ at $z=h_0$; the kinematic boundary condition is $w=0$ there.

For steady flow independent of $x$, these equations give $w=0$, $\mu u_{zz}=-\rho g\sin \alpha$ and $p_z=-\rho g\cos \alpha$. Two integrations and the boundary conditions give the uniform gravity-driven Newtonian film:Its surface velocity is $U_0=\rho g{h}_0^2\sin \alpha /(2\mu )$, while its mean velocity is $2U_0/3$.

Use an asymptotic expansion at fixed $z$, evaluating the boundary conditions at the instantaneous $z=h(x,t)$. At leading order, the downslope equation gives ${\hat{u}}_{0,zz}=0$. The wall condition and the surface condition ${\bar{u}}_z(h)+{\hat{u}}_{0,z}(h)=0$ implyHere $u^{(0)}$ denotes the total leading velocity, to distinguish it from the dimensional scale $U_0$. The continuity equation and wall impermeability then giveThe transverse momentum equation gives ${\hat{p}}_{0,z}=0$. The normal stress condition therefore yieldsThe leading volume flux per unit width is $q^{(0)}={\int }_0^h{u}^{(0)}dz=2h^3/3$. Consequently mass conservation gives $h_t=-2h^2{h}_x+O(\epsilon )$. This relation eliminates $h_t$ inside the first inertial correction:Set $G=2h_x\cot \alpha -3S{h}_{xxx}$. The $O(\epsilon )$ downslope equation and boundary conditions areTwo integrations giveThe pressure correction ${\hat{p}}_1$ is not needed for this calculation because its downslope pressure gradient enters only at $O({\epsilon }^2)$.

Integrating the total velocity across the film givesIndeed the three inertial polynomial terms integrate to $\mathrm{Re}{h}^6{h}_x(1/30-1/6+2/3)=8\mathrm{Re}{h}^6{h}_x/15$. Integrating the continuity equation and using the kinematic boundary condition gives $h_t+q_x=0$. Hence the Benney equation isNormal gravity smooths height variations on the upper side of the slope, while the inertial correction can amplify them. Surface tension suppresses sufficiently short wavelengths.

Write $u_i=U_i+u_i^{\prime }$ and $p=P+p^{\prime }$, with $\overline{u_i^{\prime }}=\overline{p^{\prime }}=0$. Assume a statistically stationary mean and sufficient regularity for Reynolds averaging to commute with spatial derivatives. Thenprovided the endpoint difference grows sublinearly. Averaging the continuity equation givesIn Einstein notation, the nonlinear term expands asThus the Reynolds-averaged Navier-Stokes equation isThe extra term survives because the average of a product of zero-mean fluctuations need not vanish; it represents turbulent transport of mean linear momentum.

Let $\delta \ll L$ be a representative mixing-layer thickness, ${ϵ}_m=\delta /L$, and let $U_s$ be a representative streamwise velocity. Use the scalesThe transverse velocity scale follows from the mean continuity equation. For turbulent transverse transport to balance streamwise inertia,With this ordering, the streamwise eddy-stress divergence isThus transverse diffusion of streamwise linear momentum is retained, while streamwise diffusion and the $V_x$ contribution are smaller by ${ϵ}_m^2$. Molecular kinematic viscosity is neglected where ${\nu }_T\gg \nu$.

In the transverse momentum equation, inertia and the retained eddy stresses have scale ${ϵ}_m{U}_s^2/L$, whereas an order-one transverse pressure variation would give $P_y/\rho \sim U_s^2/({ϵ}_{m}L)$. Hence the leading dimensionless transverse balance is $P_y=0$. The resulting boundary layer equations for the planar turbulent mixing layer areHere ${\nu }_T={\mu }_T/\rho$. The effective pressure convention explained in part (a)(iii) is understood. Since the outer streams have uniform horizontal velocities, their inviscid momentum equations give $P_x=0$; choosing the far-field gauge pressure as zero gives $P=0$ throughout this approximation.

For an arbitrary material volume $V(t)$, Newton's second law equates the rate of change of linear momentum to the resultant contact force, since there is no body force:Use the Reynolds transport theorem on the left. Expanding the product and using the continuity equation leavesOn the right, the divergence theorem applied to each row of the Cauchy stress tensor givesConsequently the volume integral of $\rho D{u}_i/Dt-{\partial }_j{\sigma }_{ij}$ vanishes on every material volume. Localizing as in part (i) gives the Cauchy momentum equation:The left side is mass density times material acceleration; the right side is the divergence of the Cauchy stress tensor.

Let $V(t)$ have a boundary moving with velocity ${\mathbf{v}}_b$, and let $\mathbf{n}$ be its outward unit normal vector. For a smooth scalar field $q$, the Reynolds transport theorem isIn the last expression, extend ${\mathbf{v}}_b$ smoothly into $V(t)$ and use the divergence theorem. For a material volume, its boundary moves with the fluid velocity $\mathbf{u}$, so ${\mathbf{v}}_b=\mathbf{u}$. Equivalently, the Reynolds transport theorem for a material volume readswhere $D/Dt$ is the material derivative.

For an arbitrary moving region, mass conservation includes the relative mass flow rate across its boundary:Substitution of the Reynolds transport theorem and the divergence theorem givesSince the region is arbitrary, a continuous integrand must vanish: if it had a positive or negative value at a point, a sufficiently small surrounding region would have an integral of the same sign. This gives the continuity equation. With constant positive mass density $\rho$, it reduces toThus the flow is incompressible. For a material volume, the same argument starts directly from conservation of its total mass, with zero relative boundary flux.

The flat plate moves tangentially and the curved plate is stationary. The no-slip boundary condition and no-penetration boundary condition giveAt the curved plate, impermeability can also be written $w=u{h}_x$; together with zero tangential velocity, this gives both components zero. The native PDF writes $v^{\prime }=0$ in the scaled wall conditions although the normal component is defined as $w^{\prime }$. We use $w^{\prime }$ consistently.

The parabolic lubrication gap changes by order $h_0$ when $x^2/R$ is order $h_0$. Its natural horizontal scale is thereforeThe proposed normal velocity scale $h_{0}U/L$ then makes the two terms of the continuity equation comparable. Balancing the horizontal pressure gradient against the transverse viscous term in the Navier-Stokes equation requires $P/L=\mu U/h_0^2$, soHere pressure is measured from a common reference value, which is chosen at the outer fluid in part (b).

For clarity, define the dimensionless material derivativeSubstitution of all the scalings gives the exact dimensionless equationsFor example, the factor in the last left side is obtained by dividing the vertical acceleration scale $\rho h_0{U}^2/L^2$ by the vertical pressure gradient scale $P/h_0$.

Taking $\delta \ll 1$ and ${\delta }^2\mathrm{Re}\ll 1$ leaves the lubrication approximationThe scaled gap and boundary conditions areThe Reynolds number itself need not be small: it is ${\delta }^2\mathrm{Re}$ that measures inertia relative to transverse viscous resistance in this thin geometry.

In Newtonian gauge in cosmology, the Free-streaming photon Boltzmann equation is $\dot{\Theta }+\mathbf{e}\cdot \nabla \Theta =\dot{\varphi }-\mathbf{e}\cdot \nabla \psi$. Angular averaging and the photon angular temperature moments give the displayed continuity equation. Its dipole yields the photon Euler equation, with the sign of the anisotropic stress term determined by the stress convention.

In a nonzero steady one-dimensional flow with an isothermal equation of state with $p=c_s^2\rho$ and constant cross section, the continuity equation gives $\rho u=\text{constant}$. Eliminating density from momentum gives $(u-c_s^2/u)u^{\prime }=-{\Phi }^{\prime }$. With Mach number $\mathcal{M}=|u|/c_s$, integration yields ${\mathcal{M}}^2/2-\ln \mathcal{M}+\Phi /c_s^2=C$. The function $f(\mathcal{M})={\mathcal{M}}^2/2-\ln \mathcal{M}$ has minimum $1/2$ at $\mathcal{M}=1$, so a smooth transonic crossing at $x_{\ast }$ requires a local maximum of $\Phi$ and $C=1/2+\Phi (x_{\ast })/c_s^2$. At a nondegenerate maximum, $f=1/2+(\mathcal{M}-1{)}^2+\cdots$ gives ${\mathcal{M}}^{\prime }(x_{\ast })=\pm \sqrt{-{\Phi }^{\prime \prime }(x_{\ast })/(2c_s^2)}$.

With the normal fluid at rest, linearize about the unit-amplitude plane wave $e^{i(\mathbf{k}\cdot \mathbf{x}-k^{2}t/2)}$. For amplitude and phase perturbations proportional to $e^{i\mathbf{l}\cdot \mathbf{x}+\sigma t}$, the continuity equation gives $(\sigma +i\mathbf{k}\cdot \mathbf{l})R=l^2\Theta /2$, and the phase equation gives $(\sigma +i\mathbf{k}\cdot \mathbf{l})\Theta =-(2+l^2/2+2\zeta \sigma )R$. Their determinant is the displayed dispersion relation. The damping contains the laboratory exponent $\sigma$, not the Doppler-shifted exponent, because the reference flow is the normal component.

Eliminating momentum between the continuity equation and conservation of momentum gives ${\rho }_{tt}={\partial }_i{\partial }_j(\rho u_i{u}_j-{\sigma }_{ij})+{\nabla }^2(p+\chi )$ when the force density is $-\nabla \chi$. For $q=p-(P-\chi )$ and time-independent reference fields,No energy equation is needed for this identity. The arbitrary reference fields should match the stationary medium where the outgoing sound is evaluated.

A uniform spherical cloud with negligible pressure and velocity $\mathbf{u}=(\dot{R}/R)\mathbf{r}$ remains uniform: the continuity equation gives $\rho R^3={\rho }_{00}{R}_0^3$. The shell theorem gives interior gravitational acceleration $-(4\pi G\rho /3)\mathbf{r}$, so its outer radius satisfies the displayed equation, with fixed mass $M=4\pi {\rho }_{00}{R}_0^3/3$. Collapse from rest has ${\dot{R}}^2=2GM(1/R-1/R_0)$. Writing $C^2=2GM/R_0$, its cycloid parametrization is $R=R_0(1+\cos \varphi )/2$ and $Ct=R_0(\varphi +\sin \varphi )/2$, for $0\le \varphi \le \pi$. Complete collapse occurs after $\sqrt{3\pi /(32G{\rho }_{00})}$. This Newtonian dust solution ceases to be a regular fluid configuration at zero radius.

For a wavefunction satisfying the Schrödinger equation with a real potential,This continuity equation expresses local conservation of probability.

The ensemble density of actual configurations equals the squared norm of the guiding wave function. For spinless particles with $v_k=j_k/|\Psi {|}^2$, the Schrödinger continuity equation is ${\partial }_t|\Psi {|}^2+{\sum }_k{\nabla }_k\cdot (|\Psi {|}^2{v}_k)=0$. A density transported by the same guiding field satisfies the same equation, so initial equality is preserved. This equivariance is a consistency property, not a proof that every possible initial ensemble is in equilibrium. The physical account of the equilibrium assumption and subsystem statistics requires an additional argument.

A steady two-dimensional purely radial incompressible Newtonian fluid flow has $u_r=g(\theta )/r$ by the continuity equation. The polar Navier-Stokes equation givesEquality of the mixed pressure derivatives yields the displayed third-order equation. On fixed walls $\theta =\pm \alpha$, no-slip boundary conditions require $g(\pm \alpha )=0$. A source with volume flux $Q$ per unit out-of-plane length requires ${\int }_{-\alpha }^{\alpha }gd\theta =Q$. Consequently $g=QG$ leaves the single inertial parameter $ϵ=Q/\nu$ in $G^{(3)}+4G^{\prime }+2ϵG{G}^{\prime }=0$. The origin is a source singularity; the radial ansatz describes the wedge away from that point, not the detailed source apparatus.

For tightly wound perturbations in a locally uniform, razor-thin gas disk, continuity equation and rotating-fluid momentum equations couple the surface-density perturbation to radial and azimuthal velocities. The sheet potential is $\delta \Phi =-2\pi G\delta \Sigma /|k|$. Eliminating these perturbations gives the displayed dispersion relation. radial epicyclic frequency supplies the rotation term, the sound-speed term resists short waves, and self-gravity supplies the negative term. For an axisymmetric mode $n=0$, use ${\omega }^2$ directly; the pattern speed $\omega /n$ is undefined.

A thin dense gravity current on a rigid no-slip substrate can be carried downstream by imposed upper shear stress $\tau$. For density contrast $\Delta \rho$, lubrication theory gives $\mathbf{q}=A{h}^2{\mathbf{e}}_x-D{h}^3\nabla h$, where $A=\tau /(2\mu )$ and $D=\Delta \rho g/(3\mu )$. The continuity equation gives the displayed equation. Besides small slopes and reduced inertia, the hydrostatic reduction requires $\tau /(\Delta \rho gL)\ll 1$ to keep shear-related normal stress small. Finite zero-thickness fronts are formal outer profiles and need a separate local description when their slopes are large.