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A membrane-mediated interaction potential is the separation-dependent energy produced when deformation fields around two membrane-bound objects overlap. Its sign and range depend on membrane tension, bending stiffness, inclusion geometry, and which side of the membrane each object occupies.

The tension term $T{w}_{xx}$ is the transverse restoring force from a nearly straight, tensioned rod. The bending term $-EI{w}_{xxxx}$ is the force from Euler-Bernoulli flexure, where $E$ is Young's modulus and $I$ the second moment of area. The right side is transverse inertia. In the displayed one-dimensional equation $\rho$ denotes linear mass density; if it denotes volumetric density instead, the inertial coefficient must include cross-sectional area.

For a Fourier mode, $w_{xx}=-k^{2}w$, $w_{xxxx}=k^{4}w$ and $w_{tt}=-{\omega }^{2}w$, givingFor $k>0$ the phase speed is $c_p=\omega /k=\sqrt{T/\rho +EI{k}^2/\rho }$, strictly increasing with $k$ when $EI>0$ and $T\ge 0$. Thus shorter wavelengths propagate faster in phase. The group velocity $c_g=(T+2EI{k}^2)/\sqrt{\rho (T+EI{k}^2)}$ is also increasing, so the statement is consistent with wave-packet transport.

At a simple support the displacement and bending moment vanish: $w=0$ and $EI{w}_{xx}=0$ at $x=0,L$. The eigenmodes are $\sin (n\pi x/L)$ for $n=1,2,\dots$. Setting $T=0$ in the dispersion relation therefore yieldsThe integer is positive for a nonzero mode; $n=0$ gives the identically zero sine function, and negative integers repeat the same modes. Bending stiffness makes the frequencies quadratic in mode number, unlike a purely tensioned string.

Let $s$ be arc length, so inextensibility makes $r^{\prime }=d_3$ a unit tangent vector. The curvature is $\kappa ={\theta }^{\prime }$, hence $m^{\prime }=EI{\theta }^{\prime \prime }{e}_z$ for uniform bending stiffness. Balance of linear momentum in static equilibrium gives a constant resultant $n=F{e}_x+G{e}_y$.

Taking the cross product directly,The scalar component of conservation of angular momentum therefore becomesTogether with $x^{\prime }=\cos \theta$ and $y^{\prime }=\sin \theta$, this Euler elastica equation determines the centerline once the appropriate boundary conditions are supplied.

Solutions to all three questions of the University of Oxford Part A fluids and waves paper, sat on 16 June 2023. Source: original released PDF, bundle pages 242–245; paper pages 1–4. Question statements are omitted. Throughout, the liquid has constant positive mass density $\rho$; a nonzero doublet and positive bending stiffness are understood in the corresponding questions.

Linearize about the resting state $\eta =0$, $\varphi =0$. Introduce a dimensionless small parameter $ϵ$ and writeholding the depth, bending stiffness and variation scales fixed. Thus both boundary displacement and induced velocity are small. For any first-order derivative $D\varphi$, a Taylor expansion at the moving surface givesThe displacement correction to the leading field is ${ϵ}^2{\eta }_1{\partial }_z(D{\varphi }_1)(x,0,t)$, so evaluating it on $z=0$ changes it only at second order. Also ${\varphi }_x{\eta }_x=O({ϵ}^2)$ and $|\nabla \varphi {|}^2=O({ϵ}^2)$. The first-order kinematic boundary condition and dynamic boundary condition are thereforeDropping the subscript, the linearized boundary condition pair isThe fixed-bed condition remains ${\varphi }_z(x,-h,t)=0$. The source's shorthand $|\eta |\ll 1$ requires a length scale: for the subsequent plane wave, small surface slope $k|\alpha |\ll 1$ and displacement small compared with depth are sufficient conditions. Small height alone, with arbitrarily rapid spatial variation or a finite background disturbance, does not justify discarding the products.

The irrotational flow has a velocity potential $\varphi$ with $(u,w)=({\varphi }_x,{\varphi }_z)$. The incompressible flow satisfies $u_x+w_z=0$, hence the interior Laplace equation isThe ice supplies a boundary load through its bending stiffness; it does not alter this interior consequence of mass conservation.

Define the dimensionless parameterThe stipulated hierarchy $kh\ll \delta \ll 1$ implies both $kh\ll 1$ andThus the finite-depth correction uses $\tanh (kh)\sim kh$, while the restoring effect of bending stiffness is small compared with gravity. The dispersion relation becomesHence, to leading order, these surface gravity waves are nondispersive waves withFor clarity, the first corrections to the phase speed areThey exhibit the weak residual wave dispersion at finite wavenumber. The claimed nondispersive behavior is the asymptotic leading-order result, not exact equality for the complete relation.

The second moment of area about an axis weights each element of cross-sectional area by its squared perpendicular distance from that axis. The ellipse extends farther along ${\mathbf{d}}_2$ than along ${\mathbf{d}}_1$, so its material lies, on balance, farther from the ${\mathbf{d}}_1$ axis. ThereforeWith zero intrinsic curvature of an elastic filament, the respective bending moments required by the constitutive equation are $E{I}_1\kappa {\mathbf{d}}_1$ and $E{I}_2\kappa {\mathbf{d}}_2$. Thus their magnitudes satisfyfor $\kappa \ne 0$: bending about the minor axis is harder. Here $E$ is Young's modulus, and $E{I}_i$ is the corresponding bending stiffness.

Write $w(x,t)$ for the vertical displacement of the sea ice, and take the undisturbed water surface as $z=0$, with water occupying $z<0$. The elastic plate has areal mass $m={\rho }_{i}h$ and bending stiffnessHere $E$ is Young's modulus and $\eta$ is Poisson's ratio. We neglect in-plane prestress, viscosity and plate shear deformation, and linearize about hydrostatic equilibrium. These are important assumptions: perfect elasticity alone does not specify every term in a floating-plate model.

For a plane wave $w=\hat{w}{e}^{i(kx-\omega t)}$, $k>0$, potential flow in deep water has velocity potential $\varphi =\hat{\varphi }{e}^{kz}{e}^{i(kx-\omega t)}$. This solves Laplace's equation and decays downwards. The kinematic boundary condition gives $k\hat{\varphi }=-i\omega \hat{w}$. Linearizing the water pressure at the displaced interface gives an upward excess loadThe elastic plate equation is $m{w}_{tt}+D{w}_{xxxx}=p$. Substitution and multiplication by $k$ therefore give the flexural-gravity wave dispersion relationEquivalently, with $\alpha ={\rho }_{i}h/{\rho }_w$ and $\beta =D/{\rho }_w$,This determines the positive wavenumber implicitly for prescribed positive angular frequency. It is unambiguous: the derivative of ${\omega }^2$ is $(g+5\beta k^4+4\alpha \beta k^5)/(1+\alpha k{)}^2>0$, while ${\omega }^2$ runs from zero to infinity.

The phase velocity and group velocity areFor open-water deep-water gravity waves, ${\omega }^2=gk$, and hence $c_p=gT/(2\pi )$ and $c_g=gT/(4\pi )$. At large period the ice-covered curves approach these straight lines. At shorter period, plate bending raises the speeds, so both ice-covered curves turn upward as period decreases. In the bending regime with negligible plate inertia, $\omega \propto k^{5/2}$ and $c_g\simeq (5/2)c_p$; in the formal plate-inertia-dominated limit, $\omega \propto k^2$ and $c_g\simeq 2c_p$. The latter extrapolation eventually leaves thin-plate validity and should not be read as a prediction at arbitrarily small wavelength.

The plotted parameters are illustrative rather than measured at the observation site. They give a group-velocity minimum of a flexural-gravity wave near $15.1\mathrm{m}{\mathrm{s}}^{-1}$ at period $17.6\mathrm{s}$. There is no arbitrarily slow wave-energy branch under a continuous elastic sheet. Energy put into a localized disturbance travels away at at least this minimum group velocity; a slowly moving wind system cannot retain a wave packet indefinitely beneath itself. This reduces the opportunity for sustained local growth compared with slow, short open-water waves, and the continuous cover also prevents direct wind forcing of an exposed water surface. Incoming long swell can still propagate.

A group-velocity minimum of a flexural-gravity wave is not by itself a universal minimum wind speed for wave generation. A steadily translating forcing pattern requires a phase velocity matching its translation speed, so the minimum of $c_p$, not $c_g$, supplies the corresponding resonance threshold. Random wind forcing, dissipation and aerodynamic coupling must be specified before making an absolute generation claim.

It is important to distinguish loss of forward-going surface-gravity-wave energy from conversion of mechanical energy into heat. Scattering attenuation by ice floes redirects energy; it can attenuate a coherent transmitted wave without dissipating the total energy.

For fixed floe geometry, increasing frequency usually increases attenuation over the relevant swell range: shorter wavelengths respond more strongly to the contrast between water and the elastic plate, and to repeated floe edges. Long surface gravity waves have weak curvature and often penetrate much farther. This is a trend over a specified frequency range, not a theorem excluding resonances.

The diameter dependence is governed by $d/\lambda$. An ice floe much smaller than the wavelength moves nearly with the water and scatters weakly. Scattering becomes appreciable when floe size is comparable with the wavelength, and interference between its two edges can give maxima and minima. At fixed ice concentration, larger ice floes also mean fewer edges per unit propagation distance, roughly proportional to $p/d$. Consequently the attenuation coefficient need not increase monotonically with diameter: the single-floe reflection and the number of encounters must both be considered. Thickness increases areal inertia as ${\rho }_{i}h$ and bending stiffness as $h^3$, generally increasing wave mismatch and reflection, although detailed frequency-dependent resonances again prevent a universal monotonic law.

When $d\ll \lambda$, particularly for frazil ice and pancake ice, weak individual scattering leaves other processes dominant. Relative crystal and water motion causes viscous dissipation; an aggregate layer can behave as a viscous or viscoelastic material, and pancake ice collisions, rubbing and overwash remove energy. Their importance depends on concentration and wave amplitude.

For a uniform continuous sheet with horizontal dimensions much greater than the wavelength, there are no repeated floe edges in its interior. A perfectly elastic sheet over inviscid water supports undamped flexural-gravity waves, so internal scattering is not an explanation of decay there. Real attenuation can instead arise from internal ice anelasticity or viscoelasticity, a dissipative sub-ice viscous boundary layer, turbulence, cracks and brine-related processes. Small-floe mixtures and continuous sheets require dissipation models beyond the isolated-floe scattering picture.

For an axisymmetric thin elastic plate, define ${\nabla }_r^2=r^{-1}{\partial }_r(r{\partial }_r)$. Uniform excess pressure and bending stiffness $B$ give $B{\nabla }_r^{4}h=p$. The roof is attached to undeformed rock at the front, giving clamped boundary conditions $h(R)=h_r(R)=0$, and regularity excludes singular displacement or curvature at $r=0$. A regular particular solution is $p{r}^4/(64B)$; adding a constant and a multiple of $r^2$ and enforcing the two edge conditions gives the axisymmetric clamped-plate deflectionIntegrating the magma intrusion volume,therefore yieldsHere the axisymmetric phrase requires the radial plate biharmonic operator; a Cartesian one-dimensional beam equation would not give the same pressure coefficient. Hydrostatic contributions are excluded as instructed.

For the early front, let $U_p=\dot{R}$ be the leading propagation speed. The fluid and fracture radii differ only within a local region of length $l_p\ll R$. In lubrication theory, the elastic pressure gradient near the front is of order $B{h}_f/l_p^5$, and the volume flux through the gap is of order $B{h}_f^4/(\mu l_p^5)$. In a frame following the fluid front that flux is of order $U_p{h}_f$. The viscous peeling of an elastic plate law is thereforeMatch the peeling region to the interior curvature: $h_f/l_p^2\sim {\kappa }_i$. The stated vapour-gap relation $h_f\simeq p_T{l}_p^4/(24B)$, with $p_T>0$ interpreted as the relevant pressure magnitude, then givesSubstitution gives the vapour-tip peeling of a magma intrusion speedWith constant supplied volume flux $Q$ and negligible initial intrusion volume, $V=Qt$. Integrating $R^{14}\dot{R}\sim B^{3/2}{Q}^{7/2}{t}^{7/2}/(\mu p_T^{1/2})$ gives $R^{15}\sim B^{3/2}{Q}^{7/2}{t}^{9/2}/(\mu p_T^{1/2})$, orOrder-one coefficients require the full local peeling profile. This early regime is a front-controlled asymptotic regime, not an assertion that a continuum peeling zone remains thin at arbitrarily small $t$.

When fracture toughness instead imposes fixed ${\kappa }_f>0$, equating it to the outer edge curvature gives the toughness-controlled magma intrusion lawIts pressure can be written in either useful formThe pressure decreases as the intrusion grows on this advancing branch.

If $V_{r0}$ is the reservoir volume before any magma was transferred to the initially negligible intrusion, volume conservation and negligible conduit storage give $V_r=V_{r0}-V$. With the supplied pressure-volume law and conduit conductance, the reservoir-fed magma intrusion obeysIn this law $E$ is an effective reservoir pressure-volume stiffness, with dimensions of pressure per volume. If the stated initial reservoir volume refers instead to the start of the late regime, when the intrusion already has volume $V_i$, replace $V_{r0}$ by the total $V_T=V_{r0}+V_i$ throughout. An independent value of $V_i$ would then be needed.

For an advancing solution, inflow ceases when the reservoir and intrusion pressures equalize. Its final volume is the larger positive root in $0<V<V_{r0}$ ofEquivalently, $y=\sqrt{V_{\infty }}$ solves $y^3-V_{r0}y+\mathcal{A}/E=0$. The larger root is essential: $\sqrt{V}(V_{r0}-V)$ rises from zero to a maximum $2V_{r0}^{3/2}/(3\sqrt{3})$ at $V=V_{r0}/3$ and then decreases. A nonempty interval of positive inflow exists only ifThere are then two positive equilibria $V_{-}<V_{r0}/3<V_{+}$. At equilibrium the derivative of the volume ordinary differential equation is $\beta E(V_{r0}-3V)/(2V)$, so $V_{-}$ is unstable and $V_{+}$ has asymptotic stability. An initial advancing intrusion with $V_{-}<V_i<V_{+}$ approaches $V_{\infty }=V_{+}$. The equality case has a double equilibrium and no positive-inflow interval; above the threshold there is no advancing equilibrium branch of this model.

The late formula $p\propto V^{-1/2}$ is singular at $V=0$ and cannot nucleate the intrusion: the earlier propagation regime must supply its finite seed. Nor does this fixed-edge-curvature propagation law prescribe how an irreversible fracture closes under negative inflow. The final-size answer is conditional on reaching the advancing late branch. In particular the reservoir is generally not emptied, and $\beta$ controls the approach rate rather than the equilibrium size.

To first order, vanishing tangential velocity on the elastic plate is simply $u(0)=0$. In the mode from part a this givesConsequentlyso $u(0)=0$, $w(0)=ikA{e}^{ikx}$, and $w_z(0)=0$. The dynamic pressure at the surface is

Evaluating the hydrostatic pressure at $z=\zeta$ contributes the restoring normal stress $\rho g\zeta$. Since ${\nabla }^4\zeta =k^4\zeta$ for this mode, the linearized normal-stress balance isIt follows thatThe two restoring effects are buoyancy and the plate's bending stiffness.

An Euler strut is a slender initially straight elastic column under axial compression. For a uniform bending stiffness $EI$ and a small transverse displacement $y(x)$, the second-order potential energy under a conservative compressive load $P$ isThe bending term is positive; shortening releases load potential. Stationarity yields $EI{y}^{(4)}+P{y}^{\prime \prime }=0$. Pinned ends impose $y(0)=y(L)=0$ and zero end bending moments, $y^{\prime \prime }(0)=y^{\prime \prime }(L)=0$. Substitution of $y=A\sin (n\pi x/L)$ givesThe straight configuration loses stability first in the longest admissible bending mode. The second moment of area $I=\int z^{2}dA$ is geometric, not a mass moment of inertia. Boundary conditions matter: a fixed-free column has effective length $2L$, while fixed-fixed ends have effective length $L/2$, changing the threshold by factors $1/4$ and $4$. Slenderness, elastic material response, alignment and the weak bending direction must be checked. Imperfections produce bending before the ideal threshold, and a short column can yield or crush first. Linear buckling theory does not predict an unlimited deflection or a complete postbuckling curve.

Use the Euler-Bernoulli beam equation for a uniform slender elastic beam, small deflection and small slope, neglecting axial tension, shear deformation, rotational inertia of a cross-section and damping. Let $E$ be Young's modulus, $I={\int }_A{z}^{2}dA$ the second moment of area about the neutral axis, and $\rho$ the mass per unit length. If density instead denotes mass per volume, replace this $\rho$ by ${\rho }_{\mathrm{bulk}}A$. The bending stiffness is $EI$; the small-slope curvature is $y_{xx}$. The bending and kinetic energies areTwice integrating the variation of $U$ by parts gives the restoring load $-EI{y}_{xxxx}$ per unit length. Newton's second law therefore givesAt a pivot, displacement is fixed but slope is free; zero end bending moment gives $EI{y}_{xx}=0$. This also follows from the natural boundary term $EI{y}_{xx}\delta y_x$ in the variation.

With separation of variables, $y=XT$ givesChoose the time normalization $T(0)=1$. Initial rest removes the sine time term, so $T(t)=\cos \omega t$. The spatial characteristic roots are $\pm ik,\pm k$, givingThe two conditions at $x=0$ give $B+D=0$ and $-B+D=0$, hence $B=D=0$. At $L$, addition and subtraction of $X(L)=X^{\prime \prime }(L)=0$ give $C\sinh kL=0$ and $A\sin kL=0$. Nontrivial positive-frequency modes therefore have $C=0$ and $k_n=n\pi /L$. Thus the simply supported beam normal modes areThe zero-frequency cubic solution also vanishes under these four endpoint conditions, so $n=0$ is not an additional mode.

For the specified release, small slope requires $|a|L\ll 1$. Expand the initial deflection in the Fourier sine series. Its coefficients areWriting ${\omega }_1={\pi }^2\sqrt{EI/\rho }/L^2$, the resulting motion isOnly odd normal modes occur because the initial profile is symmetric about the midpoint, and their amplitudes decrease as $n^{-3}$. The frequencies grow as $n^2$, so the motion is dispersive and is not a single shape multiplied by one cosine. Nevertheless the ideal undamped solution is exactly periodic with period $2\pi /{\omega }_1$. At $t=\pi /(2{\omega }_1)$ all odd-mode cosines vanish, so the bar is instantaneously straight; at $t=\pi /{\omega }_1$ it has the opposite initial deflection. The first mode dominates displacement, with third-mode amplitude only $1/27$ of the first.

There is initial curvature incompatibility at a moment-free beam end: the prescribed parabola has $y_{xx}(x,0)=2a$, whereas a moment-free pivot requires zero curvature at an endpoint. Therefore the data do not define a classical solution with curvature continuous through the two end-time corners. They do define a finite-energy weak solution after release of the forces or end moments that prepared the initial shape. The sine expansion represents that solution: its displacement reproduces the initial parabola, each mode satisfies the released endpoint conditions, and ${\sum }_n{n}^4|A_n{|}^2<\infty$ gives finite bending energy. One must not differentiate the initial series twice at its endpoints and assign the interior parabola's curvature there. This qualification leaves the modal displacement prediction intact.

An Euler strut is a slender elastic column under axial compression. Let its bending stiffness about the weaker section axis be $E{I}_A$ and its length be $L$. With small lateral deflection $y(x)$ and a dead compressive load $P$, the quadratic total potential energy isThe first term is bending energy; the negative term is work released by the load as bending shortens the end-to-end distance. Stationarity gives $E{I}_A{y}^{\prime \prime \prime \prime }+P{y}^{\prime \prime }=0$. For pinned ends, $y=y^{\prime \prime }=0$ at $0,L$, and $y\propto \sin (n\pi x/L)$ requires $P_n=n^2{\pi }^{2}E{I}_A/L^2$. The first instability is Euler's critical load,Below this load, the straight state has positive quadratic energy against these perturbations. At the threshold, the lowest bending mode has zero quadratic restoring energy; above it the straight state is unstable. For a rectangular section, $I_A=b{h}^3/12$ about the appropriate bending axis, so the smaller second moment of area controls the first buckling direction.

Endpoint constraints matter. Write $P_{\mathrm{cr}}={\pi }^{2}E{I}_A/(KL{)}^2$: $K=1$ for pinned-pinned, $2$ for fixed-free and $1/2$ for fixed-fixed conditions. This elastic instability differs from compressive material failure. Imperfections cause bending before the ideal threshold, and a short column may yield before Euler buckling occurs.

For a slender elastic beam undergoing small bending, assume plane cross-sections remain plane and normal to the neutral axis. Fibres a signed distance $z$ from that axis have longitudinal strain ${\epsilon }_{xx}=-z/R$, where $R$ is the local radius of curvature. Hooke's law gives stress ${\sigma }_{xx}=-Ez/R$, where $E$ is Young's modulus. Zero resultant axial force places the neutral axis through the section's centroid for a homogeneous beam.

The bending moment is the moment of this stress distribution. Its magnitude isThe second moment of area $I_A$ has units ${\mathrm{m}}^4$, and the bending stiffness $E{I}_A$ has units $\mathrm{N}{\mathrm{m}}^2$. A rectangular section of width $b$ and depth $h$ has $I_A=b{h}^3/12$, so placing more material far from the neutral axis strongly resists bending. This is the purpose of an I-section.

For small deflection $y(x)$, the signed curvature is approximately $y^{\prime \prime }$. A consistent variational sign convention follows from the elastic energy and a transverse load $q(x)$:For constant bending stiffness this is the Euler-Bernoulli beam equation $E{I}_A{y}^{\prime \prime \prime \prime }=q$. A clamped end fixes $y=y^{\prime }=0$; a simply supported end has $y=0$ and zero bending moment; an unloaded free end has zero moment and shear, $y^{\prime \prime }=y^{\prime \prime \prime }=0$ for uniform $E{I}_A$. Endpoint loads replace the corresponding zero-force boundary conditions.

For a cantilever of length $L$ carrying a tip force $F$, choose positive deflection along $F$. The internal bending gives $E{I}_A{y}^{\prime \prime }=F(L-x)$, with $y(0)=y^{\prime }(0)=0$. Two integrations giveThis approximation requires small slopes, elastic strains and a length large compared with section dimensions. Shear deformation, rotary inertia, plastic yielding and large rotations require extensions. Adding inertia gives flexural waves with ${\omega }^2=(E{I}_A/{\rho }_m{A}_s)k^4$ for a uniform beam, where ${\rho }_m{A}_s$ is mass per unit length.

For a slender elastic beam, the Euler-Bernoulli beam equation assumes small strain, small deflection slopes and cross-sections that remain plane and normal to the deflected centreline. A fibre at signed distance $y$ from the neutral axis has longitudinal strain ${ϵ}_{xx}=-y\kappa$, where $\kappa$ is the centreline curvature. Hooke's law gives ${\sigma }_{xx}=-Ey\kappa$, with Young's modulus $E$.

Choose the internal bending-moment sign so that $M=-\int y{\sigma }_{xx}dA$. ThenHere $I_A$ is the second moment of area, with units ${\mathrm{m}}^4$, not the mass moment of inertia of a rotating body. For a uniform material under pure bending, zero axial resultant places the neutral axis through the section's centroid.

For small slopes, $\kappa \simeq w^{\prime \prime }(x)$. Combining the moment-curvature relation with transverse force and moment balance gives, with a consistent load sign convention,For constant stiffness this becomes $E{I}_A{w}^{(4)}=q$. The bending stiffness $E{I}_A$ has units $\mathrm{N}{\mathrm{m}}^2$. Boundary conditions specify displacement and slope at a clamped end, or bending moment and shear force at a free end.

For a cantilever of length $\ell$ carrying a transverse end load $F$, take deflection positive towards the load. The moment magnitude is $M(x)=F(\ell -x)$, and integration with $w(0)=w^{\prime }(0)=0$ givesThe free-end moment vanishes and the tip shear supplies the applied load. For a rectangular section of width $b$ and bending thickness $h$, $I_A=b{h}^3/12$, so increasing thickness strongly increases bending resistance. The model ceases to be reliable for large deflections, plastic strains or short thick beams where shear deformation matters.

Ideal and self-avoiding walks. An ideal chain has independent isotropic coarse-grained steps and permits self-intersections in its mathematical model. For $N$ steps of length $b$, the cross terms average to zero, so its root-mean-square end-to-end distance of a polymer chain obeysA self-avoiding walk excludes repeated occupancy or intersection at the segment scale. This represents positive excluded volume in a dilute good solvent. The nonlocal constraint correlates otherwise distant portions of the chain and swells it:The Flory approximation replaces this exponent by $3/5$, by balancing stretching free energy $R^2/(N{b}^2)$ with an interaction contribution $v{N}^2/R^3$. The result is a more extended and less internally dense coil than the ideal walk; self-avoidance changes the statistics, not merely the local step length.

Two ideal-chain regimes. At the theta temperature for a dilute polymer–solvent pair, leading effective segment attractions and repulsions cancel. Large-scale chain statistics are approximately ideal, even though microscopic atoms still have finite size. In a homogeneous polymer melt, surrounding chains screen long-range excluded volume interactions of a chosen chain, again making its equilibrium large-scale size approximately ideal. Local bond-angle correlations must be included in an effective statistical length such as the Kuhn length. Polymer entanglements can slow the dynamics without invalidating this leading equilibrium size law.

Semiflexible DNA. The nucleic acid double helix gives double-stranded DNA a bending stiffness and substantial local orientational persistence. Independent step orientations are therefore a poor description on short scales, while self-avoidance alone does not specify the energy cost of bending. The worm-like chain describes the molecule as a continuously bendable, locally inextensible curve $\mathbf{r}(s)$ parameterized by arc length $s$, with unit tangent $\mathbf{t}(s)=d\mathbf{r}/ds$. Its bending energy iswhere $A$ is bending stiffness. In the free three-dimensional model, the persistence length is ${\ell }_p=A/(k_{B}T)$ andThus ${\ell }_p$ is the contour separation over which tangent alignment falls by the factor $e^{-1}$; it quantifies stiffness relative to thermal bending. The model bridges locally rod-like and globally coil-like conformations. It omits long-range self-avoidance in its simplest form, so sufficiently long dilute DNA can require excluded-volume corrections.

Flexible and rigid limits. Integrating the tangent correlation over the two contour positions gives the end-to-end distance of a worm-like chain:For a very flexible chain on the scale of its total length, $L/{\ell }_p\gg 1$, the exponential is negligible andThis is the ideal chain result with Kuhn length $b_K=2{\ell }_p$ and effective step count $N_K=L/(2{\ell }_p)$. Many persistent sections have decorrelated orientations. The flexible limit means a large ratio $L/{\ell }_p$ within the continuum model's range, not an extrapolation of the persistence length below all molecular length scales.

For a nearly rigid chain, $L/{\ell }_p\ll 1$, the constant and linear terms in the exponential cancel those outside it. The Taylor series givessoThis is the expected rod limit: all tangents are nearly aligned, and the end-to-end distance approaches the arc length.

Helix–coil curves. A helix-coil transition model allows ordered helical runs and disordered coil regions. Use $s$ and $\sigma$ with exactly the free-energy conventions defined in the paper. A negative $\Delta F_{\mathrm{hc}}$ gives $s>1$ and favors the helical state; a positive junction cost gives $0<\sigma <1$ and penalizes starting new domains. The supplied helical fraction is the long-chain form associated with the Zimm-Bragg model.

Its main features follow directly without assuming a sharp transition. Put $Q=(s-1{)}^2+4s\sigma$. For every $\sigma >0$,and differentiating givesThus all curves rise monotonically and cross at $(1,1/2)$. Increasing the boundary free-energy cost reduces $\sigma$, suppresses frequent coil–helix boundaries and makes longer runs favorable. The transition region becomes narrower and steeper: balancing $(s-1{)}^2$ against $4s\sigma$ gives a width of order $\sqrt{\sigma }$ near $s=1$.

For $\sigma =1$, the fraction simplifies to $\boxed{f_h=s/(1+s)}$, the noncooperative two-state result. For $0<\sigma \ll 1$, the steep curve describes cooperative helix formation. In the singular limit $\sigma \downarrow 0$, it approaches zero for $s<1$ and one for $s>1$; its midpoint remains $1/2$ when approached through positive $\sigma$. Setting $\sigma =0$ directly makes the formula indeterminate at $s=1$. For any fixed positive $\sigma$, the given long-chain curve is smooth: it is a cooperative crossover, not a true discontinuity. The plot varies $s$ at fixed $\sigma$; in an actual temperature scan both parameters may change, and a finite chain can further round the crossover.

Interpret $\varphi$ as the small transverse displacement of the elastic beam. The first term is the positive kinetic energy density $\rho {\varphi }_t^2/2$, with $\rho$ the mass per unit length. The Lagrangian density is kinetic minus potential energy, so the potential density encoded by the other two terms isFor a compressed rod, transverse deflection releases compression energy: the small-slope geometric change in length is proportional to ${\varphi }_x^2/2$, so the applied compressive load lowers the potential by $\alpha {\varphi }_x^2/2$. This explains the positive sign of the $\alpha$ term in the Lagrangian. Tension would give the opposite sign and a restoring gradient term.

Bending costs elastic energy. At small slopes the curvature is ${\varphi }_{xx}$, so positive bending stiffness $\beta$ gives potential $\beta {\varphi }_{xx}^2/2$, and hence a negative term in the Lagrangian. For a uniform slender beam $\beta$ is its flexural rigidity. The competition between destabilizing compression and stabilizing bending is responsible for Euler buckling of an elastic filament.

The sheet elastocapillary energy ratio compares capillary and bending energy scales for an elastic sheet:Here $L$ is a lateral size, $\gamma$ is surface tension, $B$ is plate bending stiffness and ${\ell }_{\mathrm{ec}}$ is the elastocapillary length. The ratio is a dimensionless quantity: $\gamma L^2$ and $B$ both have units of energy.

The estimate assumes bending through an angle of order one over size $L$. Then curvature scales as $1/L$ and integrating the bending energy density over area $L^2$ gives a scale $B$. A comparable change in interfacial area gives a surface energy scale $\gamma L^2$. The same balance underlies the capillary origami study's definition of elastocapillary length.^[ref]

For an original calculation, take $B=2\times 10^{-8}\mathrm{N}\mathrm{m}$, $\gamma =0.05N/m$ and $L=1\mathrm{mm}$. Then $\eta =2.5$. Doubling $L$ alone raises it to 10. Halving thickness at fixed $L$, Young's modulus and Poisson's ratio divides $B$ by eight and raises $\eta$ to 20.

Changing all dimensions gives another trend. Write $s=L/L_0$ and keep material properties and surface tension fixed. At fixed thickness, $\eta /{\eta }_0=s^2$; at fixed aspect ratio $h/L$, $B/B_0=s^3$ and $\eta /{\eta }_0=s^{-1}$. Thus halving $L$ alone gives one quarter of the initial ratio, while halving both $L$ and $h$ doubles it.

A larger ratio makes capillary energy more competitive with bending, but does not establish complete closure. Actual wetting, contact angle, droplet volume, contact-line pinning and accessible deformation paths also matter. The ratio compares scales rather than computing the full change in energy along a folding path.