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First supply the introductory scaling and geometry. The leading tangential Stokes equation in a thin gap balances $P/L$ against $\mu U/H^2$. The shear stress scale is $\tau \sim \mu U/H$, so lubrication pressure dominates shear stress:The outer circle, viewed along a ray $r(\cos \theta ,\sin \theta )$ from the inner centre, satisfies $r^2-2\alpha \Delta r\sin \theta +{\alpha }^2{\Delta }^2=(a+\Delta {)}^2$. Expanding its positive root to first order in $\Delta /a$ gives $r=a+\Delta +\alpha \Delta \sin \theta +O({\Delta }^2/a)$, hence $h=\Delta (1+\alpha \sin \theta )$.

Take positive $\Omega$ to mean increasing $\theta$, and put $s=a\theta$, $y=r-a$, $U=\Omega a$. With no-slip boundary conditions at $y=0,h$, lubrication theory givesSteady mass conservation makes $q$ independent of $\theta$. Single-valued pressure requires ${\int }_0^{2\pi }{p}_{\theta }d\theta =0$. Write $I_n={\int }_0^{2\pi }(1+\alpha \sin \theta {)}^{-n}d\theta$ and $D_{\alpha }=1+{\alpha }^2/2$. It follows thatThe tangential fluid tractions on the inner and outer solids, measured in the same increasing-$\theta$ direction, are respectivelyThis sign convention distinguishes the traction on a wall from the corresponding stress component in the fluid.

To leading order the holding force balances the pressure traction on the inner circle. Integration by parts givesUse $p_{\theta }=a{p}_s$ and $\int \sin \theta /(1+\alpha \sin \theta {)}^{n}d\theta =(I_{n-1}-I_n)/\alpha$, with the continuous limit at $\alpha =0$. Substituting the boxed flux yieldsThe zero vertical component also follows from reflection in mathematics in the vertical line: this leaves the geometry and vertical force unchanged, reverses $\Omega$, and Linearity of Stokes flow then requires $F_z=0$.

Pressure gives no moment about either circular wall's own centre. Integrating the viscous shear torque with lever arm $a$ to leading order gives the fluid couplesIndeed $C_i=a^2\int {\tau }_{i}d\theta$ and $C_o=a^2\int {\tau }_{o}d\theta$ at this order. Applied holding or driving couples are the negatives of these fluid couples. They need not cancel because they are moments about different axes. The fluid force on the outer solid equals the holding force $F_x$ on the inner one. Transferring its moment to the inner axis adds $-\alpha \Delta F_x$, so angular momentum balance about a common axis isFor $\alpha =0$ the usual equal and opposite concentric couples are recovered. All these formulas are leading thin-gap expressions at fixed $|\alpha |<1$; taking the contact limit requires maintaining the local slender-gap assumptions.

This is a squeezing motion, not the rotational motion of the previous part. The gap changes at rate $h_t=V\sin \theta$. The Reynolds lubrication equation therefore givesThe outer wall also has tangential speed $V\cos \theta$, but its local Couette flow flux is $O(V\Delta )$ compared with the squeezing flux $O(Va)$ and does not affect the leading force. Thus $p_{\theta }=-12\mu aq/h^3$ at leading order. Periodic pressure gives $C=0$, because the integral of $\cos \theta /h^3$ vanishes over a full circle. Balancing pressure traction on the fixed inner cylinder and using integration by parts,The supplied integral gives the squeeze resistance of an eccentric journal bearing:The fluid tends to pull the inner cylinder upwards with the moving outer wall, so a downward holding force is required when $V>0$. At concentricity the translation resistance is $R_0=12\pi \mu a^3/{\Delta }^3$. The omitted wall-tangential flux and shear traction give smaller thin-gap contributions.

Put the fixed outer centre at the origin, and let the moving inner centre be $\mathbf{d}=(x,z)$. This reverses the sign of the vertical offset used in the previous parts: now $h=\Delta -\mathbf{d}\cdot \mathbf{n}$. At concentricity rotation produces no force and translation has isotropic resistance $R_0=12\pi \mu a^3/{\Delta }^3$. Balancing the external load givesThere is no inertial acceleration in this Stokes flow model; velocity is determined instantaneously by force balance.

For the subsequent motion, the full-film journal bearing assumption matters: pressure is single-valued around the whole annulus, with no specified cavitation boundary. Let $r=|\mathbf{d}|$, $e=r/\Delta$, $\varphi =\arg (x+iz)$, and ${\mathbf{e}}_r,{\mathbf{e}}_{\varphi }$ be its radial and tangential unit vectors. The squeeze resistance of an eccentric journal bearing has radial and tangential coefficientsFor completeness, rotate coordinates until $\mathbf{d}$ points along $x$, so $h=\Delta (1-e\cos \theta )$. Radial translation gives squeezing flux $q=a{v}_r\sin \theta$ and resistance proportional to $\int {\sin }^2\theta /(1-e\cos \theta {)}^3=\pi /(1-e^2{)}^{3/2}$. Tangential translation gives $q=a{v}_t[-\cos \theta +3e/(2+e^2)]$; the constant enforces periodic pressure. Its resistance usesThese expressions supply both translation directions; retaining only the vertical squeezing formula would miss the moving axle's horizontal dynamics.

The rotational fluid force from the earlier calculation points along ${\mathbf{e}}_{\varphi }$, with magnitude $R_t\Omega r/2$. Superposing rotation and translation by Linearity of Stokes flow, and resolving the load, givesFor positive $\Omega$ the axle initially moves down, then to the right because the rotation-generated force is perpendicular to the downward eccentricity. The equilibrium is displaced horizontally to the right, at the same height as the outer axis. Its eccentricity $0<e_{\ast }<1$ is the unique solutionThe rotational force then points upwards and balances the load. For negative $\Omega$ the displacement is to the left, using $|\Omega |$ in the magnitude equation. Uniqueness follows because the logarithmic derivative of $e/[\sqrt{1-e^2}(1+e^2/2)]$ has positive numerator $1-e^2/2+e^4$ after a common positive denominator. If $\Omega =0$ and $F>0$, no separated static equilibrium exists in this ideal model. If $F=0$, concentricity remains a solution.

One should not assert that this full-film model necessarily settles to the equilibrium. It predicts full-film journal-bearing whirl. For $F,\Omega >0$, eliminate time from the two boxed evolution equations and put $K=6\pi \mu \Omega a^3/{\Delta }^2$. The orbit starting at concentricity obeysso, with its zero initial horizontal displacement,The orbit begins on the lower branch, turns at $z=0$ before contact, and returns along the upper branch, forming a closed orbit through concentricity. The ratio of the right-hand side to $e$ increases from zero to infinity, giving a unique turning eccentricity below one. Linearizing at equilibrium likewise gives $\delta \dot{r}=-(F/R_{r,\ast })\delta \varphi$ and $\delta \dot{\varphi }=c_{\ast }\delta r$ with $c_{\ast }>0$: the eigenvalues are imaginary, not damped. Thus the equilibrium position exists, but the initially concentric axle circulates around it in the stated ideal full-film approximation. Cavitation or additional mechanical constraints would change the model; none were specified.

The normalized centre trajectory below illustrates the full-film prediction for $K/F=1$. The dashed circle is the contact locus of the axle centre, not either cylinder wall.

Take $h$ positive downwards and write $B$ for the filament bending modulus. Under the material-modulus interpretation of the printed $A$ as Young's modulus, the circular cross-section gives the second moment of area and bending modulusIndeed a small curvature $\kappa$ gives axial strain $-y\kappa$, so integrating $A{y}^2{\kappa }^2/2$ over the section gives bending energy $B{\kappa }^2/2$ per unit length. Keeping this distinction is essential to the powers of radius in the answer.

After allowing for buoyancy, the effective downward force per unit length isThe filament loses potential energy as it moves down; the displaced fluid's hydrostatic contribution is included through the density difference. For an inextensible filament with arc-length coordinate $s$, the geometric bending and gravitational energy is ${\int }_0^L[B\kappa (s{)}^2/2-wh(s)]ds$, up to a constant. For the small-slope Monge representation this reduces toThe upward-positive convention reverses the sign of $h$ and the gravitational term together. There is no stretching contribution at this order when bending a slender filament without imposed axial tension. The wording does not explicitly distinguish a material elastic modulus from a flexural modulus: if $A$ instead denotes the filament bending modulus in the intended convention, set $B=A$ throughout; then the $\pi a^4/4$ factor is already included in $A$.

Use the downward-positive displacement and the filament bending modulus $B$ and load density $w$ defined above. Two applications of integration by parts give the first variationThus the elastic-plus-gravitational force density is $-\delta E/\delta h=-B{h}_{xxxx}+w$. In resistive-force theory, transverse motion has drag per unit length $-{\zeta }_{\perp }{h}_t$. Neglecting filament and fluid inertia, instantaneous force balance gives the forced small-slope elastohydrodynamic filament equationClamping fixes position and tangent, while the free end has zero bending moment and zero transverse shear force. ThereforeThe last two conditions also follow as natural boundary conditions from the variation, since the free-end values of $\delta h$ and $\delta h_x$ are arbitrary. Distributed weight does not add a concentrated force or moment at the tip.

At steady state, $B{h}_s^{\prime \prime \prime \prime }=w$. Integrating first from the free end gives $h_s^{\prime \prime \prime }=(w/B)(x-L)$ and $h_s^{\prime \prime }=(w/2B)(x-L{)}^2$. Integrating twice more with the clamp conditions gives the uniform-load bending of a clamped filament:In particular the free-end displacement is $\delta =h_s(L)=w{L}^4/(8B)$ and the total buoyancy-corrected gravitational load is $F=wL$. Thus the tip stiffness of a uniformly loaded cantilever isThe last expression uses $A$ as Young's modulus; if $A$ itself is flexural rigidity, the same result reads $k_{\mathrm{eff}}=8A/L^3$. This is Hooke's law for the specified loading pattern. It is not the stiffness for a point force at the tip, which would be $3B/L^3$; replacing a distributed force by its resultant does not preserve its bending moment distribution.

For dimensional analysis, $[A]=$ force/length squared, $[I]=$ length to the fourth power, hence $[B]=$ force times length squared. Consequently $[B/L^3]=$ force/length, as required for a spring constant. Also $[{\zeta }_{\perp }]=$ force times time/length squared makes ${\zeta }_{\perp }{h}_t$, $B{h}_{xxxx}$ and $w$ all forces per unit length. Neither viscosity nor density enters the stiffness, although viscous drag controls the rate of approach to the steady shape. The small-slope approximation requires $w{L}^3/B\ll 1$, since $h_s^{\prime }(L)=w{L}^3/(6B)$.

Take the current to be dilute, vertically well mixed and Boussinesq approximation, with constant channel width $w$, a horizontal absorbing bed, negligible ambient fluid entrainment, negligible bed friction, and no resuspension. The shallow-water approximation neglects vertical inertia and gives hydrostatic pressure. Particle loss alters reduced gravity but does not alter the leading-order carrier-fluid volume or inertia; otherwise additional mass and momentum terms would be necessary. The current occupies depth $h(x,t)$, has depth-uniform speed $u(x,t)$, and has particle volume fraction $c(x,t)$, with $g^{\prime }=g_{o}c$ and constant $g_o>0$.

Volume conservation of a short channel segment gives $h_t+(uh{)}_x=0$. Its suspended particle amount per unit bed area is $hc$, and the downward particle deposition flux is $v_{s}c$. Thus $(hc{)}_t+(uhc{)}_x=-v_{s}c$, which becomes $(h{g}^{\prime }{)}_t+(uh{g}^{\prime }{)}_x=-v_s{g}^{\prime }$ on multiplying by $g_o$.

At vertical coordinate $y$ above the bed, excess hydrostatic pressure divided by reference mass density is $g^{\prime }(h-y)$. Integrating across the depth gives the depth-integrated force flux $g^{\prime }{h}^2/2$. The horizontal momentum per unit bed area is $hu$ and the advected kinematic momentum flux is $h{u}^2$. The resulting variable-buoyancy shallow water equations areThe derivative of the pressure term acts on both $h$ and $g^{\prime }$; treating reduced gravity as horizontally constant while particles settle would lose part of the force.

For a steady interior, let $q=Q/w=uh>0$ denote volume flux per unit width. Then the particle balance is a first-order linear differential equation,Here $g_0^{\prime }=g^{\prime }(0)>0$, $v_s\ge 0$ is constant, and deposited particles are not returned to the current. In particular the decay length is $q/v_s=Q/(w{v}_s)$ for $v_s>0$, and the cancellation of depth follows from $uh=q$: dilution and transit time compensate in the deposition balance.

The steady momentum equation also givesThe inlet depth selects $K=q^2/h_0+g_0^{\prime }{h}_0^2/2$. Depth and speed at every downstream position are determined by the continuous positive root of a polynomial inThis implicit expression also makes clear why the inlet condition is needed: there are generally two possible positive depths.

The depth branches of a steady depositing gravity current are distinguished by the Froude number ${\mathrm{Fr}}^2=q^2/(g^{\prime }{h}^3)$. Differentiating the momentum invariant, and using the exponential decay of reduced gravity, givesFor $v_s>0$, supercritical flow has $\mathrm{Fr}>1$ and decreasing depth, with $h\to q^2/K$ and $u\to K/q$ as $x\to \infty$. Subcritical flow has $\mathrm{Fr}<1$ and increasing depth, with $h\sim \sqrt{2K/g^{\prime }(x)}\propto e^{v_{s}x/(2q)}$. This indefinitely deepening branch eventually violates the assumed shallow or confined-channel geometry. Hence the depth does not universally decrease: that conclusion needs a supercritical inlet.

For fixed $g^{\prime }>0$, $q^2/h+g^{\prime }{h}^2/2$ has its unique minimum at $h_c=(q^2/g^{\prime }{)}^{1/3}$, with $K_c=3q^{4/3}(g^{\prime }{)}^{1/3}/2$. The two positive roots when $K>K_c$ lie below and above $h_c$. Since $K_c$ decreases downstream while $K$ is constant, an initially strict branch remains separate from the other branch and cannot cross a critical point. A critical inlet with $v_s>0$ has no finite depth derivative: at $\mathrm{Fr}=1$ the first differentiated equation would equate zero to a positive value. It requires a singular entrance adjustment or physics beyond this steady model. If $v_s=0$, both reduced gravity and the continuous depth branch are constant. These steady solutions describe the supplied current interior, not a moving finite-release front.

For two particle populations, write $g^{\prime }=g_1^{\prime }+g_2^{\prime }$ and $g_i^{\prime }=g_{0i}{c}_i$, allowing different constant particle-density conversion factors. The bidisperse particle-laden gravity current equations areFor the steady case,Thus the depth is obtained from the same cubic polynomial invariant with the sum of the two exponentially decreasing reduced gravity contributions.

The local deposit consists of the two particle deposition fluxes $v_1{c}_1$ and $v_2{c}_2$. For a positive total deposition rate, the bidisperse gravity-current deposition fraction by particle volume isIf the particles have a common material density, this is also their local mass fraction. If they have common $g_o$, it can be written in the source's buoyancy notation asFor unequal $g_{0i}$ replace each $g_{i0}^{\prime }$ by $g_{i0}^{\prime }/g_{0i}$. The exponent sign matters: with both populations present and $v_1>v_2>0$, the faster-settling population has decreasing local deposit fraction downstream, since $f_1^{\prime }=(v_2-v_1)f_1(1-f_1)/q<0$. Equal positive settling speeds give a constant fraction. A nonsettling population contributes no deposit; if the entire local deposition rate is zero the deposit fraction is undefined.

If “fraction of particles” means number fraction rather than particle-volume or mass fraction, let $a_i$ be a single particle's volume in population $i$. The correct weights are then $v_i{c}_i/a_i$, soSettling speeds alone do not determine $a_i$ without particle properties and a drag law. Finally, if the intended deposit is cumulative over the bed from $0$ to $x$, rather than local at $x$, integrating particle deposition flux gives the species volumes per unit channel width $D_i(x)=q{c}_{i0}(1-e^{-v_{i}x/q})$. The cumulative particle-volume fraction isFor both $v_i>0$, its limit at infinity is the incoming particle-volume fraction, whereas its limit as $x\downarrow 0$ is the local deposition fraction at the inlet. Local sorting and the cumulative inventory are different observables.

The Maxwell equations with vacuum permittivity and permeability give $\rho ={ϵ}_0\nabla \cdot E$ and $J={\mu }_0^{-1}\nabla \times B-{ϵ}_0{\partial }_{t}E$. The current identity is the Ampère-Maxwell equation, and Faraday's law gives ${\partial }_{t}B=-\nabla \times E$. With $g={ϵ}_{0}E\times B$,The identity $(\nabla \times B)\times B=(B\cdot \nabla )B-\frac{1}{2}\nabla B^2$, and its electric analogue, turn this into the divergence of the Maxwell stress tensor $T$. Thus the tensor in the question is its negative:The vector $g$ is electromagnetic momentum density, and ${\sigma }_{ij}$ is the outward flux of its $i$th component across a surface normal in the $j$ direction. In static conditions the total force on matter is $F_i=-\oint {\sigma }_{ij}{n}_{j}dS$; generally subtract the time derivative of the enclosed electromagnetic field momentum as well.

For the two infinite sheets, superposition and the electromagnetic field jumps give $E=(\sigma /{ϵ}_0)e_x$, $B={\mu }_{0}K{e}_y$ between them and zero outside. Substitution yields exactlyA thin pillbox enclosing the $x=0$ sheet has only its $+x$ face in the interior electromagnetic field. ThereforeDirectly, omit the sheet's self-electromagnetic field: the other sheet supplies $E_{\mathrm{other}}=\sigma e_x/(2{ϵ}_0)$ and $B_{\mathrm{other}}={\mu }_{0}K{e}_y/2$. The Lorentz force per area is $\sigma E_{\mathrm{other}}+K{e}_z\times B_{\mathrm{other}}$, giving the same result. The electric contribution is attractive and the magnetic contribution repulsive.

If $K=\sigma v$ for a single moving charge population, use ${\mu }_0{ϵ}_0=c^{-2}$ to obtain $F/A={\sigma }^2(1-v^2/c^2)e_x/(2{ϵ}_0)$. For physical $|v|<c$ this is attractive, so motion of these charges cannot make the force repulsive. Independently imposed currents need not satisfy that restriction on $K/\sigma$.

At the change in buoyancy frequency, continuity of vertical velocity and fluid pressure givesThe background mass density is continuous, so there is no interfacial density jump or separate surface restoring force. Since $P=i\omega W^{\prime }/k^2$, these internal-wave transmission across a stratification step conditions are continuity of $W$ and $W^{\prime }$. The buoyancy perturbation need not be continuous, because the background density gradient jumps.

Let $W_0=-i\omega {\eta }_0$ and $\ell =k\sqrt{N_2^2/{\omega }^2-1}$. The lower unstratified layer obeys $W^{\prime \prime }-k^{2}W=0$ and the upper radiation condition givesPropagating these matching data downward givesThe kinematic boundary condition at the moving lower boundary therefore givesThe upper vertical-displacement complex amplitude at $H$ isIn particular, $|{\eta }_H/{\eta }_0|=[{\cosh }^2(kH)+(\ell /k{)}^2{\sinh }^2(kH){]}^{-1/2}$. The unstratified layer attenuates and phase-shifts the transmitted internal gravity wave. The matching height is $H$ in the original PDF; the TeX transcription's $H^{\prime }$ is a typo.

For a neutrally buoyant horizontal turbulent round jet, a segment's horizontal momentum flux changes only if an external horizontal force or ambient momentum input exists. Neither is present, soPositive buoyancy adds a vertical force, not a horizontal one. Therefore $d{M}_x/ds=0$ remains true for an inclined forced plume even when $F_0\ne 0$; in general $M_x=M_0\cos {\theta }_0$.

The net upward force on a segment is $g({\rho }_0-\rho )\pi b^{2}ds$. Since $F=g({\rho }_0-\rho )\pi b^{2}V$ and $V=|\mathbf{M}|/Q$, vertical momentum balance givesIn kinematic plume fluxes, the equations are $m_x^{\prime }=0$, $m_z^{\prime }=fq/m$, where $m=(m_x^2+m_z^2{)}^{1/2}$. The centreline geometry satisfiesFor $F_0>0$, $Q>0$ beyond the source, and an initially nonvertical turbulent round jet pointing right, $M_z$ increases and the centreline turns toward the vertical. The magnitude of momentum flux is not conserved once the tangent has a vertical component.