Use the arc length $s$ and the unit tangent $\mathbf{t}={\partial }_s\mathbf{r}$ of an inextensible filament. In the high-force state choose its positive longitudinal branch. The constraint givesThe extension is $z={\int }_0^L{t}_{z}ds=L-\frac{1}{2}{\int }_0^L|{\mathbf{t}}_{\perp }{|}^{2}ds+\cdots$. The curvature of a space curve obeys ${\kappa }^2=|{\partial }_s\mathbf{t}{|}^2$. Since ${\partial }_s{t}_z=-{\mathbf{t}}_{\perp }\cdot {\partial }_s{\mathbf{t}}_{\perp }+\cdots$ is second order in small fluctuations, its squared contribution is fourth order. Thus the worm-like chain energy through quadratic order isPositive force suppresses transverse orientation fluctuations, while the filament bending modulus penalizes their spatial variation.

To apply the equipartition theorem without missing either component or a mode-counting factor, expand each component in a real orthonormal basis of eigenfunctions of $-{\partial }_s^2$, $t_{\alpha }(s)={\sum }_q{a}_{\alpha q}{w}_q(s)$, with $\int w_q{w}_{q^{\prime }}ds={\delta }_{q{q}^{\prime }}$ and $\alpha =x,y$. The quadratic energy is $\frac{1}{2}{\sum }_{\alpha ,q}(A{q}^2+f)a_{\alpha q}^2$. Each real quadratic degree of freedom contributes $k_{B}T/2$, soFor a chain long compared with the force-dependent correlation length, the interior is translation invariant and the mode sum per length becomes $\int dq/(2\pi )$. Equivalently, for the Fourier transform ${\tilde{t}}_{\alpha }(q)=\int e^{-iqs}{t}_{\alpha }(s)ds$,The negative and positive wavenumbers here represent the real-field mode count consistently, rather than two independent complex modes. Summing both transverse components gives the high-force worm-like chain elasticitywhere the integral follows by setting $q=\sqrt{f/A}u$ and using ${\int }_{-\infty }^{\infty }(1+u^2{)}^{-1}du=\pi$. ThereforeHere $L_p$ is the three-dimensional zero-force persistence length. The result requires $f{L}_p\gg k_{B}T$, negligible backbone stretching, and a long-chain bulk approximation. It does not describe the eventual bond-stretching regime. For a continuum polymer description, the force-dependent length should also exceed the microscopic cutoff.

For comparison, a three-dimensional freely jointed chain has contour length $L=Nb$ and independent link orientations. At force $f$, a link of polar angle $\vartheta$ has Boltzmann factor $e^{x\cos \vartheta }$ with $x=fb/(k_{B}T)$. Its angular canonical partition function isThe total canonical partition function is $Z_1^N$. Differentiating with respect to force gives the force-extension of a three-dimensional freely jointed chain:ConsequentlyThe extension deficits scale as $f^{-1}$ and $f^{-1/2}$ respectively. Inverting them gives force divergences proportional to the first and second inverse powers of the deficit. Freely jointed link rotations cost no bending energy; the worm-like chain instead has a continuum of coupled bending modes whose contributing wavelengths change with force. Setting a long-wavelength coarse-grained link length $b=2L_p$ can match weak-force statistics, but does not make these high-force laws identical.

For the transverse tangent correlation of a stretched worm-like chain, use a separation $r$ consistently: the source writes $C(y)$ but integrates a separation denoted $r$. In the translation-invariant bulk the spatial average is redundant andTo evaluate the integral without an assumed decay law, let $G(r)$ be the inverse Fourier transform of $(A{q}^2+f{)}^{-1}$. Then $(-A{\partial }_r^2+f)G=\delta (r)$. Away from zero, decay at both infinities requires $G=B{e}^{-|r|/\xi }$ with $\xi =\sqrt{A/f}$. Integrating across zero yields $-A[G^{\prime }(0^{+})-G^{\prime }(0^{-})]=1$, so $B=1/(2\sqrt{Af})$. ThusAt $r=0$ this agrees with the two-component equipartition theorem calculation. The correlation length decreases with force and is distinct from the zero-force persistence length. The full tangent correlation has a nonzero aligned background; the exponentially decaying quantity here is specifically the transverse correlation.

For a finite open chain, the source's integral samples $s+r$ beyond $L$ and supplies no endpoint tangent conditions. An exact finite-length correlation is therefore not uniquely specified. The result above is the intended bulk form when $L\gg \xi$. As one explicit finite convention, periodically extending the tangent fluctuations gives the periodic finite-length tangent correlation of a stretched worm-like chainwhere $d_L(r)$ is the distance from zero around the periodic interval. Then $C_L(0)=(k_{B}T/\sqrt{Af})\coth [L/(2\xi )]$, and the extension formula acquires this same finite-length factor. The bulk law is recovered as $L/\xi \to \infty$. This makes the endpoint assumption explicit rather than claiming an unspecified finite-chain correlation is exactly translationally invariant.

Set $\zeta =\gamma /m>0$ and $\mathbf{A}={\mathbf{A}}^{\prime }/m$. Multiplying the Langevin equation by $e^{\zeta t}$ and integrating givesThe expected value of $\mathbf{U}$ is zero. For the printed dot-product force covariance, write $C_A(s-s^{\prime })=\varphi (|s-s^{\prime }|)/m^2$, so $\int C_A(y)dy=\tau$. Without any white-noise approximation the trace variance isThe intended short-correlation approximation replaces $C_A(y)$ by $\tau \delta (y)$, where $\delta$ is the Dirac delta. It requires the force correlation time to be much shorter than ${\zeta }^{-1}$ and the observation time. ThenAt thermal equilibrium, the equipartition theorem gives $m⟨|\mathbf{u}{|}^2⟩/2=3k_{B}T/2$. Taking the long-time limit determines the vector noise normalization in inertial Langevin dynamics:This factor six corresponds to the dot-product covariance across three components. If $\varphi$ were defined for one component instead, its integrated strength would be $2\gamma k_{B}T$ before dividing by $m^2$, and the acceleration strength would be $2\zeta k_{B}T/m$.

Finite rapid correlation is not literally a Dirac delta. For a stationary force and local drag, its exact limiting velocity trace isThis follows by splitting the double convolution into $s\ge s^{\prime }$ and $s^{\prime }\ge s$. Thus equipartition fixes the weighted integral, not just $\tau$, unless the short-correlation approximation is made. For example, $C_A(y)=\tau e^{-|y|/t_c}/(2t_c)$ gives $\tau /[2\zeta (1+\zeta t_c)]$. The remaining explicit formulae use the intended continuous-time Gaussian white noise limit.

Assume isotropic jointly Gaussian noise, as appropriate to an isotropic bath. The component covariance isThe scalar moment identities in the question must now be applied to a fixed Cartesian component $X=U_i$, not to the magnitude of the vector. Its variance isIn the $2n$-fold integral for $⟨X^{2n}⟩$, the Isserlis theorem replaces the force product by a sum of complete pairings. Each pairing yields $n$ identical two-time integrals, hence $(s_t^2{)}^n$. Pair the first labelled factor in $2n-1$ ways, then the first unpaired factor in $2n-3$ ways, and continue. The Gaussian even-moment pairing count isOdd force products vanish, soFor completeness, summing the scalar moment-generating function givesThe same Gaussian process argument for every linear combination of components gives the vector characteristic function $⟨e^{i\mathbf{k}\cdot \mathbf{U}}⟩=e^{-s_t^2|\mathbf{k}{|}^2/2}$. Inverting this Fourier transform yields the normalized multivariate Gaussian distributionThe initial condition is a velocity Dirac delta as $t\downarrow 0$, and the long-time limit is the thermal velocity density. The converted TeX's exponent is damaged; the PDF has the squared norm displayed here. The scalar pairing formula cannot be interpreted as a vector-norm formula: the radial moments of an isotropic Gaussian vector instead give $⟨|\mathbf{U}{|}^{2n}⟩=(2n+1)!!(s_t^2{)}^n$ in three dimensions.

Integrate the velocity once more. Interchanging the time integrals gives the integrated Ornstein-Uhlenbeck displacementFor the stated fixed initial position and velocity,Let $\mathbf{R}=\mathbf{r}-⟨\mathbf{r}⟩$ and $D=k_{B}T/\gamma$. Its component covariance comes from integrating the squared noise kernel:Therefore the total centered variance isThe mean-square displacement from ${\mathbf{r}}_0$ additionally contains $|{\mathbf{u}}_0{|}^2(1-e^{-\zeta t}{)}^2/{\zeta }^2$. Confusing this deterministic drift contribution with centered variance changes the short-time conclusion.

For times longer than the force correlation time but $\zeta t\ll 1$, the mean motion is ${\mathbf{r}}_0+{\mathbf{u}}_{0}t+O(t^2)$ and the centered trace is $2\zeta (k_{B}T/m)t^3+O(t^4)$. Thus a nonzero fixed initial velocity produces a leading ballistic mean-square displacement $|{\mathbf{u}}_0{|}^2{t}^2$; conditioned on ${\mathbf{u}}_0=0$, the white-noise displacement variance instead starts as $t^3$. If one averages over an independent equilibrium initial velocity, its covariance adds $3(k_{B}T/m)(1-e^{-\zeta t}{)}^2/{\zeta }^2$, givingThis is the usual thermal ballistic regime, before velocity memory has relaxed. At $\zeta t\gg 1$, the conditional mean approaches ${\mathbf{r}}_0+{\mathbf{u}}_0/\zeta$, while the centered trace is $6Dt-9D/\zeta +O(e^{-\zeta t})$. Its leading behavior is diffusion with $⟨|\mathbf{R}{|}^2⟩\sim 6Dt$ and $D=k_{B}T/\gamma$, as successive displacement increments lose inertial memory. At times below a finite force correlation time the white-noise short-time expansion is not applicable; for a regular force covariance and fixed initial velocity the leading noise-induced displacement variance is of order $t^4$.

The three contributions to the Frank elastic energy penalize distinct slow distortions of the nematic director. Nematic splay measures local spreading or convergence and is governed by $K_1(\nabla \cdot \mathbf{n}{)}^2/2$. Nematic twist measures rotation around an axis perpendicular to the director, with cost $K_2(\mathbf{n}\cdot \nabla \times \mathbf{n}{)}^2/2$. Nematic bend measures curvature of director integral lines; for a unit vector, $(\mathbf{n}\cdot \nabla )\mathbf{n}=-\mathbf{n}\times (\nabla \times \mathbf{n})$, giving cost $K_3|\mathbf{n}\times \nabla \times \mathbf{n}{|}^2/2$. These terms respect the head-tail symmetry of a nematic liquid crystal. Positive elastic constants oppose these distortions.

For the one-dimensional reduction assume lateral uniformity, so $\theta$ and $\varphi$ depend only on $z$. A prime below means $d/dz$. Directly differentiating the nematic director givesThe squared curl is ${\sin }^2\theta {\theta }^{\prime 2}+{\cos }^2\theta {\varphi }^{\prime 2}$. Subtracting the squared twist invariant givesConsequently the one-dimensional twisted nematic energy has elastic coefficientsA normal electric field has $\mathbf{E}\cdot \mathbf{n}=E_0\sin \theta$. There is a genuine factor-of-two inconsistency in the source: its initial functional places the electric term inside the overall factor $1/2$, whereas its subsequent density and threshold use the full electric term. For the requested density and threshold, use the usual orientational coupling $-ϵE_0^2{\sin }^2\theta /(8\pi )$. Here $ϵ$ is the dielectric anisotropy of a nematic, in the printed Gaussian-unit convention. If the first functional is instead read literally, this coupling is halved and the resulting threshold voltage is multiplied by $\sqrt{2}$.

Put $c=ϵE_0^2/(8\pi )$ and $\mathcal{L}=f{\theta }^{\prime 2}/2+g{\varphi }^{\prime 2}/2-c{\sin }^2\theta$. Apply the Euler-Lagrange equations for two fields. Variation in $\varphi$, which does not appear explicitly, givesVariation in $\theta$ gives $(f{\theta }^{\prime }{)}^{\prime }-f_{\theta }{\theta }^{\prime 2}/2-g_{\theta }{\varphi }^{\prime 2}/2+2c\sin \theta \cos \theta =0$, orAt $\theta =0$, $f=K_1$, $g=K_2$, and $f_{\theta }=g_{\theta }=0$. With zero field, ${\theta }^{\prime }=0$ and ${\varphi }^{\prime }=\pi /(2L)$ satisfy both equations and the imposed azimuthal endpoints. Thus the quarter-turn planar texture is a stationary solution. Stationarity alone is not a claim of stability for arbitrary elastic constants.

To find the Fréedericksz transition, impose the additional strong planar anchoring $\theta (0)=\theta (L)=0$. Write $\alpha =\pi /(2L)$. The small-angle expansions areThe conserved $g{\varphi }^{\prime }$ and fixed total twist imply ${\varphi }^{\prime }=\alpha +O({\theta }^2)$ for a pure tilt perturbation. Linearizing the tilt Euler-Lagrange equation therefore givesSubstitute $\theta ={\theta }_0\sin (\pi z/L)$. A nonzero neutral amplitude requiresSince $V_c=E_{c}L$, the quarter-turn nematic instability threshold isThis is the first zero-endpoint mode: higher sine modes replace $K_1$ by $n^2{K}_1$ and cost more elastic energy. The quadratic energy change is $\frac{1}{2}\int [K_1{\theta }^{\prime 2}+\{(K_3-2K_2){\alpha }^2-ϵE_0^2/(4\pi )\}{\theta }^2]dz$, so the first mode has negative stiffness above this voltage. The result assumes $ϵ>0$ and $K_1+(K_3-2K_2)/4>0$. If the latter is negative the planar state is already unstable at zero field; for negative dielectric anisotropy of a nematic, a normal field opposes this tilt. The literal initial electric prefactor would give $V_{c,\mathrm{literal}}=\sqrt{2}{V}_c$.

For reflective operation of the twisted nematic field effect, orient each polarizer's transmission axis along the adjacent planar nematic director. Light entering through the top polarizer initially has $y$ linear polarization. At low voltage, adiabatic optical following in a twisted nematic rotates it through a quarter turn to $x$ as it travels downward, so it passes the bottom polarizer and reaches the mirror. On returning, it follows the reverse twist back to $y$ and exits the top polarizer: the pixel appears bright. At sufficiently high voltage, positive dielectric anisotropy of a nematic aligns most of the director with the normal field. The bulk no longer produces the quarter-turn optical rotation, so the crossed bottom polarizer blocks the incident light and the pixel appears dark. Elastic relaxation restores the bright twisted state after voltage is removed. The adiabatic requirement is more precisely $2\pi |\Delta n|/\lambda \gg |{\varphi }^{\prime }|$, involving birefringence, not thickness alone. Voltage switches director orientation and hence the reflected brightness; it does not require the liquid crystal to emit light.

Retaining the nonlinear factor $f$ determines the order-$ϵ$ forcing and the matching constants. Write $\delta =\sqrt{ϵ}$. The inner leading equation and its boundary condition at $r=1$ give $f_0=C(1-r^{-1/2})$. Matching to the far-field value fixes $C=1$.

The tail $-r^{-1/2}$ becomes $-\delta x^{-1/2}$ under $x=ϵr$, so a regular expansion in integer powers of $ϵ$ cannot suffice. This is a matched expansion with square-root and logarithmic corrections. DefineThe exponential integral here is normalized by the displayed definite integral. Integration by parts also gives $E(x)=2e^{-x}/\sqrt{x}-2\sqrt{\pi }\mathrm{erfc}\sqrt{x}$, using the complementary error function.

For the outer asymptotic expansion, put $F(x)=f(x/ϵ)$ andThe transformed equation gives $L{F}_1=0$ and $L{F}_2=-F_1{F}_1^{\prime }$, with both corrections vanishing at infinity. Matching the leading inner tail determines $F_1=-E/2$. Multiplying the equation for $F_2$ by its integrating factor yieldsFix the particular solution $G$ unambiguously byThen $(x^{3/2}{e}^x{G}^{\prime }{)}^{\prime }=E$, $G(\infty )=0$, and $G$ has no $x^{-1/2}$ term at zero. The general outer correction is $F_2=\beta E+G/4$, whereThe logarithm in this asymptotic expansion explains why the inner expansion needs an $ϵ\log ϵ$ term.

For the inner asymptotic expansion, all corrections larger than order $ϵ$ solve the homogeneous leading equation and vanish at $r=1$. At order $ϵ$, the forcing isIntegration gives the particular solution $1-\sqrt{r}+\log r$, which vanishes at $r=1$. Thus writeExpress this inner expansion in $x=ϵr$ and compare the terms that survive in the overlap $ϵ\ll x\ll 1$. The order-$\delta$ constant gives $A=\sqrt{\pi }$. At order $ϵ$, the term $C\log ϵ+\log r$ becomes $\log x+(C-1)\log ϵ$, so $C=1$. The singular term $-ϵA/\sqrt{x}$ matches $2ϵ\beta /\sqrt{x}$, giving $\beta =-\sqrt{\pi }/2$. Finally, the regular constants give $B+1=\pi +\mu /4$.

The requested expansions, including every correction through order $ϵ$, are thereforeThe inner formula satisfies the boundary condition at $r=1$ term by term; the outer formula satisfies the condition at infinity. Each formula has its own stated region of validity, and the matched asymptotic expansion relates them in their overlap.

The constant can also be made explicit. Integration by parts in the definition of $G$ gives $G=-EI-{\int }_x^{\infty }E(t{)}^{2}dt$, and $E(x)I(x)\to 8$. Squaring the complementary error function expression for $E$ separates the integral into three elementary pieces:The first piece is $-4\log x-4\gamma -4\log 2+o(1)$, where $\gamma$ is the Euler--Mascheroni constant. Substitution $u=\sqrt{t}$ makes the second integral tend to $\sqrt{\pi }/2$, because $(\mathrm{erfc}u{)}^{\prime }=-2e^{-u^2}/\sqrt{\pi }$. Integration by parts in the third integral gives $1/2-1/\pi$. ConsequentlyEquivalently, defining $G$ by its convergent integral already determines the same matching constant without needing this closed form.

The geostrophic material derivative isIt advects horizontally with the geostrophic flow. The exact material derivative additionally includes horizontal ageostrophic flow and vertical advection $w{\partial }_z$. The leading balances are geostrophic balance and the hydrostatic approximation:The conventional quasi-geostrophic streamfunction is $p=f_0\psi$, after fixing the horizontally uniform reference pressure consistently with the basic buoyancy profile. ThereforeHere and below pressure is divided by reference density, and $\sigma$ is the anomaly about the prescribed background buoyancy.

For the zonal average assume periodicity or vanishing boundary terms, so that averages of $x$ derivatives vanish and averaging commutes with $y,z$ derivatives. Put $S(z)=f_0^2/N^2(z)$. Since $\overline{{\psi }_x^{\prime }}=0$, the disturbance quasi-geostrophic potential vorticity is $Q^{\prime }={\psi }_{xx}^{\prime }+{\psi }_{yy}^{\prime }+{\partial }_z(S{\psi }_z^{\prime })$. The zonal part of $\overline{{\psi }_x^{\prime }{Q}^{\prime }}$ vanishes as $\frac{1}{2}\overline{{\partial }_x({\psi }_x^{\prime }{)}^2}$. For the other parts, integration by parts givesAs $u^{\prime }=-{\psi }_y^{\prime }$, $v^{\prime }={\psi }_x^{\prime }$, and ${\sigma }^{\prime }=f_0{\psi }_z^{\prime }$, the Eliassen–Palm flux isWe have therefore proved the exact Taylor identity for quasi-geostrophic flux,without imposing small amplitude. This identity is algebraic within the quasi-geostrophic approximation; it does not assert validity of that approximation for arbitrary physical flows.

Now linearize about a steady zonal basic flow $\bar{u}=U(y,z)$, with ${\bar{Q}}_y\ne 0$ and no mean meridional velocity. The linear quasi-geostrophic potential-vorticity equation is $Q_t^{\prime }+U{Q}_x^{\prime }+v^{\prime }{\bar{Q}}_y=0$. Multiply by $Q^{\prime }/{\bar{Q}}_y$ and average. The advection term is a zonal derivative and vanishes, givingThis is quasi-geostrophic wave-activity conservation law. The basic-state denominator is time-independent at the retained order; its change caused by the waves is higher order. The wave activity is signed if ${\bar{Q}}_y<0$, and the formula cannot be used where that gradient vanishes. The original PDF includes the zonal average over the squared disturbance in this definition.

For the resonant solution it is helpful to avoid squaring an imaginary coefficient as though it were real. Writewhere $B_0$ is real. Assume $f_0\ne 0$, $\beta \ne 0$, and $k\ne 0$. Define $c(z)=\cos (mz)$ and $r(z)=(z-H)\sin (mz)$. The complex coefficient of the physical streamfunction disturbance isThe original PDF contains $Bt\cos (mz)$: the converted TeX's missing $t$ would incorrectly remove the resonance and fails the differential equations.

For the vertical potential-vorticity inversion operator, $c^{\prime \prime }=-m^{2}c$ and $r^{\prime \prime }=-m^{2}r+2mc$. ConsequentlySubstitution in the linear potential-vorticity equation givessince ${\kappa }^2{B}_0=2k{f}_{0}U$. Thus this is a genuine resonant topographic quasi-geostrophic wave, with linear amplitude growth rather than an assumed stationary wave.

At the flat upper boundary $h_z(H,t)=0$, so $w^{\prime }(H)=0$. At the lower boundary, linearization of the impermeability condition gives $w^{\prime }(0)=U{\partial }_x{h}_b$, where $h_b=\mathrm{Re}\{\epsilon H{e}^{ikx}\sin (\ell y)\}$ is the bottom corrugation. Here $h_z(0,t)=-CHm$ is time-independent, so the specified vertical velocity givesusing $f_{0}Cm/N^2=1$. Both boundary conditions are verified. The linear solution is valid while its growing displacement and velocity remain small; it is not a large-time nonlinear solution.

To evaluate all fluxes, use $\overline{\mathrm{Re}(a{e}^{ikx})\mathrm{Re}(b{e}^{ikx})}=\frac{1}{2}\mathrm{Re}(a{b}^{\ast })$. The coefficients of $u^{\prime },v^{\prime },{\sigma }^{\prime }$ are respectively $-\epsilon \ell \cos (\ell y)h$, $i\epsilon k\sin (\ell y)h$, and $\epsilon f_0\sin (\ell y)h_z$. HenceThe products of the $B_{0}t$ terms alone and the $C$ terms alone are imaginary in the vertical flux and contribute nothing. Only the cross terms remain. Sincewe obtainDifferentiating $R$ gives $R^{\prime }=m(1+\cos (2mz))=2m{\cos }^2(mz)$. As $SmC=f_0$,For an independent evaluation of the right side of the Taylor identity for quasi-geostrophic flux, write $q=(2f_{0}c-{\kappa }^{2}Cr)-i{\kappa }^2{B}_{0}tc$. Multiplication of $ikh$ by $q^{\ast }$ gives real part $-2k{f}_0{B}_{0}t{c}^2$, because the remaining real cross terms cancel. Thereforewhich verifies the flux identity pointwise in $y,z$.

The basic potential-vorticity gradient is ${\bar{Q}}_y=\beta$, so the wave activity and its derivative areHere ${\kappa }^4{B}_0^2/(2\beta )=k{f}_0{B}_0$ follows from $U=\beta /{\kappa }^2$. Thus ${\mathcal{A}}_t+F_y+G_z=0$. The growth term involves $B_0^2=|B{|}^2=-B^2$; the distinction matters because the printed complex coefficient $B$ is imaginary. These formulas evaluate both flux components, their divergence, the buoyancy and PV covariances, and wave activity. In particular $G(H)=0$ and $G(0)={\epsilon }^{2}k{f}_0{B}_{0}Ht{\sin }^2(\ell y)/2$, so for $\beta >0$ the growing activity is supplied by the positive flux from the lower boundary.