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In a regular solution, ideal combinatorial entropy competes with a nonzero unlike-contact energy. For equal molecular volumes and mole fraction $\varphi$ of species A, write mixing free energy per site aswhereHere $z$ is the coordination number and ${ϵ}_{ij}$ a pair-contact energy counted consistently in the lattice model. A positive $\chi$ penalizes unlike contacts; a negative value favors them. The logarithmic terms lower free energy on mixing. The sign of the enthalpy of mixing alone therefore does not determine miscibility.

At fixed temperature and composition, a homogeneous phase is globally stable only if no division into two compositions lowers its free energy. The common-tangent construction selects coexisting compositions ${\varphi }_a,{\varphi }_b$, equivalent to equality of both species' chemical potentials. Local stability to infinitesimal fluctuations instead requiresZero curvature gives the spinodal, while the coexistence boundary is the binodal. Between them a homogeneous state can be metastable and require nucleation; inside the spinodal its small composition fluctuations grow. The symmetric regular solution hasBelow this contact penalty, the free energy is convex at every composition. Above it a miscibility gap is possible around the central composition. If $\chi =A/T$ with $A>0$, heating promotes mixing, but specific interactions or temperature-dependent noncombinatorial terms can produce other trends.

For two polymers, let $N_A,N_B$ count equal-volume segments per chain, and now let $\varphi$ be the segment volume fraction of A. Connecting segments into chains reduces their independent translational entropy. The binary Flory-Huggins polymer blend hasThe energetic contact term remains extensive in the number of segments, while the favorable entropy per segment is reduced by $1/N_A$ and $1/N_B$. This is the key long-chain limit of polymer-blend miscibility: a small positive incompatibility can overcome the mixing entropy. The lattice treatment is a controlled model for this argument, rather than a claim that every real blend has composition-independent $\chi$.^[ref]

The local stability condition and critical interaction parameter becomeFor equal chain lengths, ${\chi }_c=2/N$. Similar segment chemistry and matched cohesive interactions can reduce the unlike-contact penalty; favorable specific interactions, such as complementary hydrogen bonding, can make mixing favorable even for long chains. Lower molar mass increases mixing entropy. temperature can help when it lowers the effective $\chi$, but heating is not universally sufficient. A suitable common solvent can facilitate processing or ternary mixing, and a block-copolymer compatibilizer can stabilize fine domains; neither guarantees that the final solvent-free binary material is one equilibrium molecular phase.

An oil–water emulsion illustrates emulsion stability versus molecular miscibility. A surfactant adsorbs with its water-affine head toward water and its oil-affine tail toward oil, lowering interfacial tension and the work $\gamma dA$ needed to make additional interface. Breaking a fixed oil volume $V_o$ into spherical droplets of radius $R$ gives total area $3V_o/R$, so creating small droplets still requires substantial area and usually mechanical work.

Adsorbed surfactants can oppose droplet coalescence through charged double-layer repulsion, hydrated-head interactions or steric repulsion from sufficiently extended groups. Adequate coverage, adsorption time, ionic conditions and film drainage affect the lifetime. A lower interfacial tension does not itself remove the bulk miscibility gap, and a protected emulsion is commonly kinetically stabilized. A thermodynamically stable microemulsion is a distinct situation requiring an appropriate composition and free-energy balance. The two liquids must not be called molecularly miscible simply because droplets disperse visibly.

For the capillary, balance upward contact-line force against hydrostatic column weight, neglecting meniscus volume and gas density. For inner radius $r$,The contact angle is measured through the liquid. complete wetting means ${\theta }_2=0$, so the contact-angle change of capillary rise givesThis follows from the stipulated wetting change, holding radius, bulk liquid surface tension and density fixed. One should not replace that premise with a general claim that grease always improves glass wetting; if contamination also changes the liquid surface tension, the original height and new angle alone would not determine the new height.

For two wide vertical plates separated by gap $d$, use unit width. There are two upward contact-line contributions and liquid cross-sectional area $d$, giving the capillary rise between parallel plates:Only one principal meniscus curvature contributes between wide plates, whereas a circular tube has two. For the same fluid and angle,If the plate gap equals the tube diameter, $d=2r$, the rise is half as large: the initial and completely wetting heights would be $2.50$ cm and $3.05$ cm. If instead $d=r$, the heights are equal. In either geometry, changing only the angle from $35^{\circ }$ to zero multiplies the height by $1/\cos 35^{\circ }$. Since the question specifies no plate spacing, the general gap-dependent relation is the unambiguous result; the half-height answer requires the equal-gap-to-diameter convention.

Use SI units, uniform absolute permittivity $\epsilon$ and a neutral bulk reservoir. Local equilibrium of ideal mobile ions gives $n_i=n_{i,\infty }{e}^{-z_{i}e\psi /(k_{B}T)}$. Combining their charge with the Poisson equation yieldsThe reference is $\psi =0$ in the bulk. Boundary data specify fixed charge, fixed potential or a surface reaction/charge regulation law. These are physically different problems. The mean-field theory assumes point ions, thermal equilibrium and a continuum solvent; ion-size packing, dielectric structure and strong electrostatic correlations can invalidate it.

For a symmetric monovalent electrolyte let $n_{\infty }$ count each sign and define $u=e\psi /(k_{B}T)$, ${\kappa }_D^2=2e^2{n}_{\infty }/(\epsilon k_{B}T)$. Outside fixed sources,For $|u|\ll 1$, the Debye-Hückel approximation is ${\nabla }^{2}u={\kappa }_D^{2}u$. A planar wall bounding electrolyte in $z>0$ has $\psi ={\psi }_0{e}^{-{\kappa }_{D}z}$; assuming zero field on the other side, its charge is $\sigma =\epsilon {\kappa }_D{\psi }_0$. A sphere of radius $a$ and total charge $Q$ instead givesThe factors arise from geometry and boundary conditions; planar screening cannot be substituted unchanged for a small sphere. The weak-potential condition must hold where the potential is largest, particularly at the surface.

The nonlinear planar Poisson-Boltzmann profile is also exactly integrable. Multiply $u^{\prime \prime }={\kappa }_D^2\sinh u$ by $u^{\prime }$ and use $u,u^{\prime }\to 0$ at infinity:The sign selects a profile decaying to the bulk for either sign of surface potential. Integration givesFor weak $u_0$ this reduces to the linear solution. Strong surface potentials have a bounded far-field amplitude $4\gamma$, while the near-wall ion density still grows strongly. There is no need to permit the negative ion densities produced by an invalid extrapolation of a linearized density formula.

The same first integral yields the Grahame equation and its diffuse-layer differential capacitance per unit area:These describe the diffuse layer. A compact interfacial layer adds a voltage drop and can place its capacitance in series with $C_D$; a measured electrode capacitance need not equal the ideal diffuse contribution.^[ref]

For comparison, a planar counterion-only double layer next to a negative plane has positive counterions and no salt reservoir. Its profile is algebraic:Integrating $en(z)$ over $z>0$ gives $|\sigma |$, checking charge neutrality. This problem cannot be obtained by putting a finite reservoir density into the symmetric-electrolyte answer without changing its boundary conditions.

Applications include screening around synthetic colloids, diffuse-layer electrode capacitance and forces between charged plates. In planar geometry, the bulk-subtracted normal stress is constant:At the midpoint of identical plates, symmetry sets $E=0$, so the remaining excess ionic osmotic pressure determines their repulsion. A comparison with force or capacitance measurements should vary salt and surface potential independently, test charge regulation and adsorption, and distinguish compact-layer effects from failures of diffuse screening. Nonlinear Poisson-Boltzmann theory retains finite-potential Boltzmann populations, but still omits correlations and finite-ion-volume saturation.

Write $\epsilon ={\epsilon }_0{\epsilon }_r$ and take $n_0$ to be the reservoir density of each monovalent ion species. Equal electrochemical potentials in equilibrium give the Boltzmann ion densities in a symmetric electrolyte,Combining this with the electrostatic Poisson equation gives the planar Poisson-Boltzmann equation,The Debye length is ${\kappa }^{-1}$. If the symbol $n_0$ instead denotes the combined density of both species, the factor two is absent; the per-species convention above makes the normalization explicit. At $|e\psi |\ll k_{B}T$, the linearized planar screening solution obeys ${\psi }^{\prime \prime }={\kappa }^2\psi$. With $\psi (0)=\zeta <0$ and $\psi (\infty )=0$,

The positive ions are counterions and the negative ions are co-ions. At weak potential,They approach $n_0$ from above and below respectively, with equal first-order excess and deficit. For stronger negative potential, interpret the source's strong-potential comparison as a statement about $|e\psi |$, since $\psi$ itself is negative. Counterions become strongly concentrated near the wall, while co-ions are exponentially depleted and remain nonnegative. Linearization misses this asymmetry and would eventually predict unphysical negative co-ion density. Point-ion mean-field theory itself eventually needs finite-size and correlation corrections; it provides no steric cap at arbitrarily large counterion concentration.

For steady tangential electro-osmosis, with no imposed pressure gradient and negligible inertia, the Stokes equation is $\eta v^{\prime \prime }+{\rho }_{e}E=0$. Since ${\rho }_e=-\epsilon {\psi }^{\prime \prime }$,The conditions $v(0)=0$ and $v^{\prime }(\infty )=0$ giveThe printed constant speed is the bulk velocity outside the double layer, not the velocity at the no-slip wall. With negative $\zeta$ the positive counterion excess drives fluid along the applied field. The far-field value is the Helmholtz-Smoluchowski electro-osmotic velocity.^[ref]

For the finite channel, first model the added wall as an ideally uncharged insulating boundary with no normal field on its outer side: ${\psi }^{\prime }(d)=0$, while $\psi (0)=\zeta$. The one-charged-wall electro-osmotic channel then hasIntegrating the same force balance and enforcing $v(0)=v(d)=0$ givesFor $U>0$, the maximum occurs atIn that limit $v/U\simeq 1-x/d-e^{-\kappa x}$: the velocity rises rapidly from zero at the charged wall, reaches a maximum a few screening lengths away, then falls approximately linearly to zero at the neutral wall. One charged wall does not give a symmetric plug across a channel with no-slip at the other wall.

An alternative electrode condition $\psi (d)=0$ means a wall held at reservoir potential; it is not the same as zero surface charge. Under that convention,with $\cosh [\kappa (d-x_{\max })]=\sinh (\kappa d)/(\kappa d)$. Both boundary conventions have the same leading thin-double-layer profile above. An externally imposed pressure gradient would add a pressure-driven profile; none is specified here.

For dilute nonadsorbing uncharged polymers in equilibrium or sufficiently weak flow, the entropic polymer depletion near nonadsorbing walls suppresses chain configurations near each wall. A depletion layer of order the coil radius of gyration is expected at both walls, with a bulk-like central region when $d$ is much larger than the coil size. Strong confinement rounds this into a central enrichment. Neutrality removes a direct electrophoretic force, not excluded-volume entropy. Adsorption, concentration and strong-flow migration were not specified, so this is a conditional equilibrium sketch rather than a universal flowing-polymer profile.

Only the separable instrument and mechanical tasks are treated here; biological operation and stepping energetics remain deferred.

For generic optical tweezers, a tightly focused laser and objective with high numerical aperture create an intensity gradient that attracts a suitable dielectric bead toward the focus. Momentum transfer also creates a scattering force; stable three-dimensional trapping requires sufficient restoring gradient force. A condenser and photodetector or camera track displacement. Near a stable focus, $F_x=-kx$ defines the stiffness and requires both a position calibration and a force calibration.^[ref]

At thermal equilibrium in a harmonic well, the equipartition trap calibration is $k⟨(x-⟨x⟩{)}^2⟩=k_{B}T$. Alternatively, fitting the overdamped harmonic-trap power spectrum gives $f_c=k/(2\pi \gamma )$, hence $k=2\pi \gamma f_c$. Detector noise, finite sampling and exposure time, wall corrections to drag, and drift must be controlled rather than attributed to stiffness.

For an isolated sphere in a Newtonian fluid, define $\mathrm{Re}={\rho }_{f}v(2a)/\eta$ using its diameter. The supplied viscosity converts to $10^{-3}\mathrm{Pa}\mathrm{s}$. The generic Stokes drag calculation givesHere $\mathrm{Re}=2\times 10^{-6}[{\rho }_f/(10^3\mathrm{kg}{\mathrm{m}}^{-3})]$; this retains the unspecified fluid density as a parameter rather than treating a density as supplied data. Stokes drag assumes an isolated no-slip sphere; nearby surfaces change the coefficient. TODO: the requested assessment of biological motor operation is deferred.

TODO: the biological stepping-energy and step-detection estimate is deferred in its entirety; no biological energy source or motor mechanism has been researched.

Ignoring biological coupling, the mechanical thermal model is the Langevin equation for a harmonic trap,For independent repeats with the specified initial conditions, the ensemble mean $X=⟨x⟩$ obeys the homogeneous damped-oscillator equation. An average over $N$ independent zero-mean force samples reduces its variance by $N$, with a finite detector bandwidth needed to define samples of white noise. High bandwidth alone does not remove random forces; averaging independent records or sufficiently separated samples is the additional operation. A single trajectory retains thermal fluctuations. For stationary overdamped position fluctuations, a long time average over duration $\mathcal{T}\gg {\tau }_k=\gamma /k$ has approximately $\mathrm{Var}(\bar{x})\simeq 2(k_{B}T/k){\tau }_k/\mathcal{T}$.

The initial-value solution of a damped harmonic trap has characteristic roots $r_{\pm }=[-\gamma \pm \sqrt{{\gamma }^2-4mk}]/(2m)$. For distinct roots,For critical damping, ${\gamma }^2=4mk$,For the underdamped case the same real solution can be written $e^{-\gamma t/(2m)}[x_0\cos {\omega }_{d}t+(v_0+\gamma x_0/(2m))\sin {\omega }_{d}t/{\omega }_d]$, with ${\omega }_d^2=k/m-{\gamma }^2/(4m^2)$. These describe the deterministic response or conditional mean, not the entire noisy trajectory.

For overdamped trap relaxation after an inertial boundary layer, let $ϵ=km/{\gamma }^2\ll 1$. The slow and fast roots areAfter $t\gg {\tau }_m$, the fast exponential is negligible and the surviving amplitude is $x_0+(m/\gamma )v_0+O(ϵx_0,ϵm{v}_0/\gamma )$. If also $|(m/\gamma )v_0|\ll |x_0|$, then to leading overdamped orderSmall Reynolds number justifies the linear viscous force; it does not alone prove the separate small inertial-trap parameter $ϵ$. Neglecting $v_0$ requires both rapid momentum relaxation and a negligible slow-mode amplitude shift. If $x_0=0$, a nonzero initial velocity can leave a small slow displacement even after its own fast transient decays. The result concerns generic bead mechanics only.

Cauchy's stress hypothesis says that traction across a cut depends locally on its point, time and oriented normal, rather than on the global shape of the cut. Here all areas and normals are referential: $E$ is a fixed material region in reference coordinates, ${\rho }_R$ its mass density per reference volume, $v$ the current particle velocity, $b$ body force per unit mass, and $s_R$ force per unit reference area. The time derivative on the left is total material momentum rate; the two right terms are contact and body forces.

A shrinking pillbox gives $s_R(-N)=-s_R(N)$. In a shrinking tetrahedron the volume contributions scale as its linear size cubed, whereas face forces scale as size squared. Their leading balance expresses traction on an arbitrary face as the normal-component-weighted sum of tractions on three coordinate faces. Defining those three tractions as tensor columns gives $\boxed{s_R(N)=T_{R}N}$. Applying the divergence theorem to momentum balance and localizing over arbitrary $E$ then yieldsThe referential tensor is the first Piola-Kirchhoff stress and need not be symmetric; it should not be silently identified with the spatial Cauchy stress.

Incompressibility is $\boxed{\det F=1}$. For a homogeneous hyperelastic material the referential constitutive relation becomeswhere the undetermined scalar pressure $p$ enforces the constraint. Homogeneity removes explicit reference-position dependence from $W$, not the possible spatial variation of that pressure.

For the neo-Hookean energy, $DW=2\alpha F$. For the positive diagonal ansatz the nominal principal stresses are $t_i=2\alpha v_i-p/v_i$. Since $F$ is constant, equilibrium requires $\mathrm{cof}F\nabla p=0$, so $p$ is constant. Stress-free endfaces give $\boxed{p=2\alpha v_3^2>0}$. The side normals include $(0,-1,0)$ and $\pm (\sqrt{3}/2,1/2,0)$, so their prescribed tractions force $\boxed{t_1=t_2=T}$.

For fixed positive $p$, the scalar function $2\alpha v-p/v$ is strictly increasing on $v>0$. Thus equality of the two nominal stresses implies $v_1=v_2=v$. Incompressibility gives $v_3=v^{-2}$, and substitution givesThe load function has derivative $2\alpha (1+5v^{-6})>0$, tends to minus infinity as $v\downarrow 0$ and to infinity as $v\to \infty$. It therefore determines exactly one positive $v$ for every specified load, with $v\ge 1$ for outward tensile $T\ge 0$. The resulting constant stress satisfies equilibrium and every prescribed face traction, proving existence and uniqueness within the requested positive diagonal class. It does not claim uniqueness among all possible nonhomogeneous deformations. This is the equibiaxial dead loading of a neo-Hookean prism.

Let the thin transverse length be $R{e}^{-1/2}$ times the longitudinal scale and write $Y=R{e}^{1/2}y$, $V=R{e}^{1/2}v$. Continuity then remains $u_x+V_Y=0$. Transverse momentum gives $p_Y=0$ to leading order; the quiescent exterior has constant pressure, hence $p_x=0$. Axial viscous diffusion in $x$ is smaller by $R{e}^{-1}$, leavingSymmetry about the jet centreline requires $V=0$ and $u_Y=0$ at $Y=0$. At large positive $Y$, require $u\to 0$ and $u_Y\to 0$. The transverse velocity can approach a finite nonzero scaled entrainment velocity; imposing $V\to 0$ as well would rule out the jet. The physical transverse velocity remains smaller by $R{e}^{-1/2}$ and matches an entraining outer flow.

Using continuity, the momentum equation is ${\partial }_x(u^2)+{\partial }_Y(uV-u_Y)=0$. Integrating across the jet, the boundary terms vanish because $u,u_Y\to 0$ while $V$ is finite. ThusApart from the dimensional density and velocity/length factors, this is the conserved axial momentum flux supplied by the nozzle. Entrainment increases volume flux without adding axial momentum from the resting exterior.

For $u=x^{\alpha }{f}^{\prime }(\eta )$ with $\eta =Y/x^{\beta }$, use the stream function $\psi =x^{\alpha +\beta }f(\eta )$ so that $u={\psi }_Y$ and $V=-{\psi }_x$. Momentum conservation requires $2\alpha +\beta =0$. Inertia and diffusion scale as $x^{2\alpha -1}$ and $x^{\alpha -2\beta }$, so $\alpha +2\beta =1$. Therefore $\boxed{\alpha =-1/3,\beta =2/3}$. In particularSubstitution reduces momentum to $3f^{(3)}+f{f}^{\prime \prime }+(f^{\prime }{)}^2=0$, with $f(0)=f^{\prime \prime }(0)=0$ and $f^{\prime }(\infty )=0$. Integrating first gives $3f^{\prime \prime }+f{f}^{\prime }=0$, the integration constant vanishing at the centreline. A second integration gives $3f^{\prime }+f^2/2=C$. Parameterize its positive constant as $C=18A^2$ and solve the resulting Riccati equation with $f(0)=0$:The momentum integral fixes its scale, since ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^{4}zdz=4/3$:The upper-half entrainment is $V\to -2A{x}^{-2/3}$; the negative side has the opposite sign, drawing ambient fluid inward. This is the Bickley jet, valid in the slender far-downstream laminar regime, rather than in the immediate nozzle region.

For a vector $a$ described in axes rotating at constant angular velocity $\Omega$, the rotating-axes theorem is $(da/dt{)}_{\mathrm{inertial}}=(da/dt{)}_{\mathrm{rotating}}+\Omega \times a$. Inertial particle velocity is $u+\Omega \times x$. Applying the theorem once more gives inertial acceleration $Du/Dt+2\Omega \times u+\Omega \times (\Omega \times x)$. Thus the rotating momentum equation isIf $g=-\nabla \Phi$ and density is constant, the centrifugal term can be absorbed into the reduced pressure in a rotating fluid:For small disturbances with $\Omega =\Omega e_z$, linearize to $u_t-2\Omega v=-P_x/\rho$, $v_t+2\Omega u=-P_y/\rho$, $w_t=-P_z/\rho$, together with incompressibility. Put $\zeta =v_x-u_y$. Taking the vertical curl gives ${\zeta }_t=2\Omega w_z$. Taking divergence gives ${\nabla }^{2}P=2\rho \Omega \zeta$. Applying the Laplacian to the vertical momentum equation therefore gives ${\partial }_t{\nabla }^{2}w=-2\Omega {\zeta }_z$. Another time derivative yieldsAll three velocity components are retained, even for fields independent of $y$; rotation produces the transverse component.

Set $m=\pi /h$ and seek $w=a{e}^{-i\omega t}W(x)\sin mz$. The rigid-wall conditions are then automatic and the scalar equation givesFor $0<|\omega |<2|\Omega |$, define $q=m\sqrt{4{\Omega }^2/{\omega }^2-1}$. The two possible oscillatory factors are distinguished by energy propagation. With horizontal wavenumber $k$, the dispersion relation is ${\omega }^2=4{\Omega }^2{m}^2/(k^2+m^2)$ and $c_{g,x}=-\omega k/(k^2+m^2)$. An outgoing wave in $x>0$ must have $\omega k<0$, givingFor $|\omega |>2|\Omega |$, define $\kappa =m\sqrt{1-4{\Omega }^2/{\omega }^2}$. Reject the exponentially growing branch, leaving the evanescent responseThe real part is the physical velocity. The radiation or bounded-decay condition is needed in addition to the wavemaker boundary value to pick a unique half-channel response. Steady forcing ($\omega =0$) is a singular limit: with nonzero rotation a steady geostrophic balance and impermeable top/bottom require $w=0$, incompatible with a nonzero imposed sine profile. At the cutoff equality, the bounded outgoing polarization also degenerates. These exceptional cases cannot be obtained by blindly substituting into the two strict-inequality formulas. The propagating sign is the radiation sign for an inertial wave.

Let the bed be $z=0$ and the free surface $z=h(x,t)$. In a long, shallow layer, vertical acceleration is smaller than gravity, so the vertical Euler equation gives hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. Incompressibility makes the vertical speed smaller than the horizontal speed by the aspect ratio; irrotationality then makes the leading horizontal velocity depth-independent. The horizontal Euler equation becomesIntegrating $u_x+w_z=0$ across the depth, using bed impermeability $w(0)=0$ and the surface condition $w(h)=h_t+u{h}_x$, yieldsCombining these gives the conservative momentum equation $(hu{)}_t+(h{u}^2+g{h}^2/2{)}_x=0$. Integral mass and momentum balance across a stationary thin jump therefore requireThese conservation laws still apply across a jump where the smooth hydrostatic/irrotational derivation does not resolve the internal turbulent structure.

Put $A=h_{-}$, $B=h_{+}$ and $q=A{u}_{-}=B{u}_{+}>0$, orienting the flow from minus to plus. For a nontrivial jump the momentum condition gives $q^2=gAB(A+B)/2$. Its change in mechanical-energy flux isA passive jump cannot create mechanical energy, so the change is nonpositive. Since $q>0$, $\boxed{h_{+}>h_{-}}$ for a genuine jump; equality describes no jump. The upstream flow is supercritical and the downstream flow subcritical, since $F{r}_{-}^2=B(A+B)/(2A^2)>1$ and $F{r}_{+}^2=A(A+B)/(2B^2)<1$.

The gas-dynamic analogue is a compressive normal shock: density and pressure increase and flow changes from supersonic to subsonic. Entropy increases; directed mechanical energy is irreversibly converted into internal energy. Total gas energy, including internal energy, is conserved across an adiabatic shock. This distinguishes that total-energy balance from the reduced shallow-water mechanical-energy loss.

For a material volume $V(t)$ moving with velocity $\mathbf{u}$, the Reynolds transport theorem isIt applies componentwise to vector quantities. Cauchy's stress theorem represents traction on a surface with outward unit normal $\mathbf{n}$ as $t_i={\sigma }_{ij}{n}_j$. Conservation of mass gives $d{\int }_{V(t)}\rho dV/dt=0$ and hence ${\partial }_t\rho +\nabla \cdot (\rho \mathbf{u})=0$. Constant density reduces this to $\boxed{\nabla \cdot \mathbf{u}=0}$.

With no body force, conservation of momentum and the divergence theorem giveApplying transport and then mass conservation cancels the terms proportional to $u_i$, leaving $\rho ({\partial }_t{u}_i+u_j{\partial }_j{u}_i)={\partial }_j{\sigma }_{ij}$. For the stated Newtonian stress, constant viscosity gives ${\partial }_j{\sigma }_{ij}=-{\partial }_{i}p+\mu \Delta u_i+\mu {\partial }_i(\nabla \cdot \mathbf{u})$. The last term vanishes. Since the material volume was arbitrary,A body force, if present, would have to be added to the momentum balance; it is absent from the specified model.

Let state zero be the undisturbed gas and state one the uniform gas between piston and shock. Their laboratory velocities are zero and $U$. In the shock frame their speed magnitudes are $V$ and $V-U$. Mass and momentum jumps giveEnergy conservation gives $\gamma [p_1/{\rho }_1-p_0/{\rho }_0]/(\gamma -1)=VU-U^2/2$. Substitute the preceding relations and $\gamma p_0/{\rho }_0=c_0^2$, then divide by the nonzero $U$. The result isThe right-moving root isThe other root is negative and does not describe the outgoing shock. For $U>0$, the quadratic is negative at $V=U$, hence its positive root obeys $V>U$: the shock stays ahead of the piston and the density ratio is positive.

The second law requires the specific entropy to increase along the fluid's passage through a nontrivial physical shock: $s_1>s_0$. For a perfect gas this selects a compressive shock, whose upstream shock-frame Mach number is greater than one and whose downstream shock-frame flow is subsonic.

Here the upstream Mach number is $M_0=V/c_0$. The quadratic gives $V^2-c_0^2=(\gamma +1)UV/2>0$, so $\boxed{M_0>1}$. Density and pressure both rise. With $c_1^2=\gamma p_1/{\rho }_1$, one also has $V-U<c_1$. For example, put $M=V/c_0$ and use the normal-shock identityThe characteristic geometry is therefore $c_0<V<U+c_1$. The plus-family characteristics from the gas ahead are overtaken by the shock, while those behind catch up with it; they converge into the discontinuity. The minus-family speeds are $-c_0$ ahead and $U-c_1$ behind. The piston is the line $x=Ut$ and the shock is the steeper-in-$x/t$ line $x=Vt$.

For $U<0$ the same positive algebraic root remains real and satisfies $0<V<c_0$, because $V^2-c_0^2=(\gamma +1)UV/2<0$. The jump relations then give a density decrease, ${\rho }_1/{\rho }_0=V/(V-U)<1$, and a pressure decrease. Where both states remain thermodynamically positive, this is an expansion shock: its upstream shock-frame flow is subsonic and its entropy change has the forbidden sign. At sufficiently large negative $U$ the formula even predicts nonpositive pressure, an additional reason it cannot represent a physical gas state.

The observed outgoing wave is instead a continuous isentropic rarefaction wave. Its plus-family rays spread from the initial piston corner rather than coalescing. The fan head enters the undisturbed gas at $x=c_{0}t$; for moderate recession the tail is $x=(U+c_1)t$, with the uniform gas next to the piston moving at $U$ and $c_1=c_0+(\gamma -1)U/2>0$. The appropriate Riemann invariant is $u-2c/(\gamma -1)=-2c_0/(\gamma -1)$, so inside the fan, with $\xi =x/t$,These expressions show the spreading characteristics in the right panel of the preceding figure. If $U\le -2c_0/(\gamma -1)$, the gas cannot follow the piston with positive sound speed; a vacuum gap forms, bounded by the limiting rarefaction wave edge. Thus the fan description also identifies the strong-recession limit.

The traction $\mathbf{t}(\mathbf{n})$ is force per unit area across an oriented surface with unit normal $\mathbf{n}$. The stress component ${\sigma }_{ij}$ is its $i$th component on a surface normal to the $j$th coordinate axis. Cauchy's stress theorem states that this dependence is linear:For the boundary of a material volume, take the outward normal and the traction exerted by the surrounding fluid. Newton's law and the divergence theorem giveConstant density and incompressibility make the transport-theorem integrand on the left $\rho D{u}_i/Dt$. The volume is arbitrary, soSubstitute the Newtonian constitutive law. With constant viscosity,Hence

The material derivative $\boxed{D/Dt={\partial }_t+\mathbf{u}\cdot \nabla }$ measures change while following a moving gas parcel. The four equations respectively express mass conservation, momentum conservation, internal-energy balance, and the ideal gas law. The first accounts for density change under expansion or compression. The second is Newton's law with pressure as the only force. In the third the specific internal energy is $c_{v}T$ and pressure work $-p\nabla \cdot \mathbf{u}$ heats compression and cools expansion. The fourth relates thermodynamic pressure, density and temperature; $R$ is the gas constant per unit mass.

For a material volume moving with the fluid, the Reynolds transport theorem gives, componentwise,For a scalar $q$ this is also ${\int }_{V(t)}(Dq/Dt+q\nabla \cdot \mathbf{u})dV$. Apply the component formula to momentum density $q=\rho u_i$. Constant density and incompressibility make the integrand $\rho D{u}_i/Dt$.

The traction $\mathbf{t}(\mathbf{n})$ is the force per unit area exerted by the material outside a surface on the material inside, with outward unit normal $\mathbf{n}$. Reversing the normal exchanges the two sides of the same internal surface. Thus $\mathbf{t}(-\mathbf{n})=-\mathbf{t}(\mathbf{n})$ expresses equal and opposite internal contact forces, the local form of Newton's third law.

With no body force, Newton's second law equates the rate of change of the material volume's total momentum to its surface force. Combining this balance with the transport theorem givesThis is a vector identity, since the component argument applies to each direction.

Let $R(t)$ rotate vectors from the rotating frame to the inertial frame. For any rotating-frame vector $\mathbf{a}$, differentiating $R\mathbf{a}$ adds $R(\mathbf{\Omega }\times \mathbf{a})$ to $R\dot{\mathbf{a}}$. Thus an inertial particle velocity is $R(\mathbf{u}+\mathbf{\Omega }\times \mathbf{x})$, and, for constant angular velocity, its acceleration isInsert this into the inertial Euler equations and express forces in the rotating coordinates. This givesThe additional terms are the Coriolis acceleration and the centripetal term associated with frame rotation. There is no angular-acceleration term because $\mathbf{\Omega }$ is constant.

Write the conservative body force as $\mathbf{g}=-\nabla {\Phi }_g$. The supplied vector identity gives $\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{x})=-\frac{1}{2}\nabla |\mathbf{\Omega }\times \mathbf{x}{|}^2$. For constant density, defineMoving the centrifugal and conservative terms to the right then gives ${\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}+2\mathbf{\Omega }\times \mathbf{u}=-{\rho }^{-1}\nabla P$. Only the gradient of this reduced pressure matters, so an additive spatial constant is immaterial.

A weak solution is a field $\mathbf{u}$ for which the composed densities $\mathbf{P}(x,t,\mathbf{u})$ and fluxes $\mathbf{Q}(x,t,\mathbf{u})$ are locally integrable andfor every smooth compactly supported vector test function $\mathbf{\phi }$ in the spacetime domain. This is the distributional form of the conservation law, obtained by moving derivatives onto the test function. If tests are allowed to meet an initial boundary $t=0$, the specified initial density contributes the additional term $\int \mathbf{P}(x,0,{\mathbf{u}}_0)\cdot \mathbf{\phi }(x,0)dx$ in the usual initial-value formulation.

For a piston-driven normal shock with $U>0$, $c_0>0$ and $\gamma >1$, the quadratic has one positive and one negative root. The forward-moving shock usesSince the square root is greater than $(\gamma +1)U$,Therefore there is always a forward root with shock speed greater than the gas and piston speed under the stated compression assumptions. This also makes the density ratio $V/(V-U)$ positive and greater than one. Evaluating the quadratic at $V=c_0$ gives a negative value, so the positive root additionally has $V>c_0$: it is supersonic relative to the undisturbed gas. The negative root does not describe the shock driven into the region $x>0$.