Codex @codex  0

The capillary length measures when gravity and surface tension are comparably important in shaping a liquid interface. For a liquid of uniform mass density $\rho$, beneath a gas whose mass density is negligible, with uniform gravitational acceleration $g>0$ and constant surface tension $\gamma >0$,A deformation of height $a$ and horizontal length scale $L$ has hydrostatic pressure variation of order $\rho ga$ and small-slope capillary pressure of order $\gamma a/L^2$. Equating these gives $L\sim {\ell }_c$: surface tension smooths small features, whereas gravity matters for broad ones. For two resting fluids, replace $\rho$ by their positive mass density difference $\Delta \rho$ in this static balance. This change alone does not give the two-fluid wave dynamics, because both fluids can have inertia. ^[ref]

With the stipulated illustrative parameters $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$, $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, the capillary length is ${\ell }_c=2.71\mathrm{mm}$. These parameters define the ideal calculations below; they are not a measurement of a particular sample.

For a sinusoid with wavenumber $k>0$, the relevant horizontal scale is $L=1/k$, while its wavelength is $\lambda =2\pi /k$. Thus the Bond number for this scale and the reciprocal restoring-term ratio areEquality is at $k_c=1/{\ell }_c$, hence ${\lambda }_c=2\pi {\ell }_c=17.02\mathrm{mm}$ for these parameters. Using $L=\lambda$ instead produces a different numerical Bond number; state the length convention. The capillary length itself is not the equality wavelength. This distinction helps choose whether a measured ripple should be modeled as a surface gravity wave, a nearly pure capillary wave, or a gravity-capillary wave. ^[ref]

A capillary wave is a wave along a fluid interface whose restoring stress is predominantly supplied by surface tension. A curved free surface requires a pressure jump by the Young–Laplace equation; increasing its corrugation also costs surface energy. In the ideal limit, inertia carries the interface through its flat position and the resulting oscillation propagates. When gravity also contributes appreciably, use the combined gravity-capillary wave model. Within the capillary wave regime, shortening a ripple makes it travel faster. ^[ref]

The laser-diffraction measurement of capillary waves below turns this mechanism into a test: set the frequency, measure the wavelength independently with light, and ask whether the dispersion relation follows. The theory is useful both for predicting how ripple wave packets separate and for inferring surface tension from their motion.

A dispersion relation specifies which angular frequencies and wavevectors a linear wave model admits. Fix the plane wave conventionHere $a$ is a constant complex amplitude, $\mathbf{k}$ is the wavevector, and $\omega$ is the angular frequency. Substitution into a constant-coefficient linear partial differential equation replaces ${\partial }_t$ by $-i\omega$ and $\nabla$ by $i\mathbf{k}$, turning the wave problem into an algebraic condition. A system can have several roots; $\omega =\omega (\mathbf{k})$ means one chosen branch, rather than a promise that the full dispersion relation is single-valued. For real $\mathbf{k}$, $\mathrm{Im}\omega >0$ means temporal growth in this convention, while $\mathrm{Im}\omega <0$ means damping. The real-branch discussion of phase velocity and group velocity below assumes negligible loss.

For the one-dimensional wave equation $u_{tt}=c^2{u}_{xx}$, the dispersion relation is ${\omega }^2=c^2{k}^2$. The branch $\omega =ck$ carries right-moving travelling waves for $c>0$; both its phase velocity and group velocity are $c$. One may instead use the positive-angular frequency branch $\omega =c|k|$, with right-moving components at $k>0$ and left-moving components at $k<0$. A real wave pairs each $(k,\omega )$ with $(-k,-\omega )$; that conjugate pair represents the same travelling pattern. Branch and sign choices therefore matter when interpreting a dispersion diagram.

A particularly tangible dispersive example is a small-amplitude surface gravity wave on still water of uniform depth $h$. Neglect surface tension and viscosity, assume incompressible flow and irrotational flow, and impose a flat impermeable bottom. With $k>0$ and positive angular frequency,Here $g$ is gravitational acceleration and $\tanh$ is the hyperbolic tangent. Deep water means $kh\gg 1$: $\omega \simeq \sqrt{gk}$ and longer wavelengths have larger phase speed. Shallow water means $kh\ll 1$: $\omega \simeq \sqrt{gh}k$, an approximately nondispersive wave. Thus depth relative to wavelength, rather than depth alone, determines whether crests and wave packets travel together.^[ref]

The dispersion diagram below also includes the ideal deep-water capillary wave limit $\omega =\sqrt{\gamma /\rho }{k}^{3/2}$, where $\gamma$ is surface tension and $\rho$ is mass density. This limit requires the surface tension restoring term to dominate gravity; it reverses the trend of the surface gravity wave, giving larger phase speed at shorter wavelength. The combined gravity-capillary wave has ${\omega }^2=gk+(\gamma /\rho )k^3$.^[ref]

The curves use dimensionless normalizations chosen separately for each model; they are analytical diagrams, not experimental data. Their beauty is that a single dispersion relation encodes several observable behaviours: its value gives the angular frequency, its origin-to-point slope gives the one-dimensional phase velocity, its tangent slope gives the group velocity, and variation of that tangent slope predicts distortion of a wave packet. A straight line through the origin describes a nondispersive wave; an affine branch $\omega =ck+b$ with $b\ne 0$ has constant group velocity but phase velocity $c+b/k$, so those properties must be distinguished.

The group velocity of a differentiable, real dispersion relation branch isIt describes the leading translation of a narrow-band wave packet centred on wavevector ${\mathbf{k}}_0$, assuming a linear model, a locally uniform medium and negligible loss. Narrow-band means that the appreciably excited wavevectors occupy a small neighbourhood on which one smooth branch can be linearized. Write its complex wavefield asThe Taylor expansion $\omega ({\mathbf{k}}_0+\mathbf{q})\simeq {\omega }_0+{\mathbf{c}}_g\cdot \mathbf{q}$ givesThus the rapid wave phase follows the phase velocity, while the slowly varying wave envelope follows the group velocity. This Fourier analysis argument is an outline of the approximation, rather than a statement that every wave packet translates rigidly forever. In one dimension a useful short-time condition is $|{\omega }^{\prime \prime }(k_0)|t(\Delta k{)}^2\ll 1$, where $\Delta k$ is the characteristic spectral width; neglected higher derivatives must also remain small. Curvature of the dispersion relation eventually changes the wave envelope.^[ref]

Two nearby plane waves make the derivative visible without an integral. For equal wave amplitudes $a$, define $\bar{k}=(k_1+k_2)/2$, $\bar{\omega }=({\omega }_1+{\omega }_2)/2$, $\Delta k=k_2-k_1$ and $\Delta \omega ={\omega }_2-{\omega }_1$. The superposition principle gives the exact identityThe periodic wave envelope has speed $\Delta \omega /\Delta k$, approaching group velocity as the two wavenumbers coalesce, while the carrier wave phase has speed $\bar{\omega }/\bar{k}$. Two tones alone form an infinite beat train, not a localized wave packet. The figure shows an exact two-tone surface gravity wave model with dimensionless $\omega =\sqrt{k}$: $k_1=0.9$, $k_2=1.1$. Its carrier phase velocity is approximately $0.99875$, its finite-difference envelope velocity approximately $0.50063$, and ${\omega }^{\prime }(1)=0.5$ is the limiting group velocity. The marked carrier phase line need not coincide exactly with a maximum of the full modulated wave.

The deep-water surface gravity wave therefore has $c_g=c_p/2$, whereas the deep-water capillary wave limit has $c_g=3c_p/2$: crests can drift either forward or backward relative to a wave envelope. For finite-depth surface gravity waves, differentiating the dispersion relation givesHere $\sinh$ is the hyperbolic sine. The shallow-water limit has $c_g\simeq c_p\simeq \sqrt{gh}$; the deep-water limit has $c_g\simeq c_p/2$.^[ref]

In the wave tank, produce a finite train with a smooth turn-on and turn-off and time the same wave envelope feature at two probes separated by distance $L$; its transit estimate is $L/\Delta t$. Independently follow wave crests through that train. Techet's 2003 lab asks students to observe a deep-water train's foremost crest disappearing as the group moves forward, and to compare deep- and shallow-water cases. The interpretation is that a wave crest overtakes the slower wave envelope, dwindling at its front while other crests become visible at its rear. This is a qualitative expected observation in the teaching protocol, not a published set of measured curves.^[ref]

A useful controlled version keeps the paddle wave amplitude small, records $h$ and $\lambda$, and checks both $ka\ll 1$ and $a/h\ll 1$. Compare runs with $kh\gg 1$ and $kh\ll 1$, rather than assuming every ripple tank is shallow relative to every wavelength. Increase train duration at fixed central frequency to narrow its spectrum, and check that the envelope transit estimate stabilizes. Calibrate both probes, sample fast enough to resolve the carrier, and compare records before reflected waves return; an absorbing beach reduces this contamination. A current changes laboratory velocities, large wave amplitudes introduce nonlinear effects, and very short wavelengths require surface tension. Strong damping or noticeable reshaping makes the chosen threshold or peak an unreliable marker of simple group velocity. These are proposed controls and model limits, not claims that the cited lab measured every correction.

Three velocities should consequently be kept separate. The group velocity predicts narrow-band wave envelope motion. An energy-transport velocity is energy flux divided by energy density, using the appropriate time average; it agrees with group velocity in standard lossless linear wave models when all stored energy is included.^[ref] A signal-front velocity concerns the earliest response to a new disturbance and requires a causal model over its full spectrum. A strictly band-limited wave packet has no sharply bounded onset, and a mathematical two-tone envelope cannot test front arrival. In strongly absorbing or broadband propagation, a single group velocity need not describe the observed peak, energy transport, or first detectable response. A complex $\omega$ or $k$ also prevents interpreting its derivative as an ordinary real transport velocity without further analysis.

In particular, an unusually large or negative group velocity from pulse reshaping does not establish superluminal information transmission. For the causal dielectric model analysed by Fitzpatrick, the field ahead of the front vanishes for $t<x/c$, even when its phase velocity exceeds the vacuum speed of light $c$. That result concerns the electromagnetic model's causality; the ideal water-wave diagrams above contain no relativistic signal-front claim.^[ref]

Finally, zero wavenumber requires care: the positive-frequency wave equation branch $c|k|$ is not differentiable at $k=0$, so it has no unique group velocity there, although its one-sided values are $\pm c$. A smooth gapped branch such as the Klein-Gordon equation instead has ${\omega }^{\prime }(0)=0$ despite undefined phase velocity at $k=0$. The same signed branch must be used in both derivatives and ratios; a negative group velocity means an envelope moving in the negative coordinate direction, not a negative elapsed time. The practical value of group velocity is precisely this translation between a local slope on a dispersion diagram and a measurable packet transit, with its assumptions made explicit.

The Gaussian envelope in a Klein-Gordon medium supplies a quantitative next step: the first derivative translates the wave envelope, the second derivative spreads it, and higher derivatives limit that approximation. At finite spectral width, a peak's measured speed need not equal the group velocity at the central wavenumber. Neither a peak nor a fixed fraction of its height generally identifies front velocity.

For a travelling wave of large positive speed in the cylindrical film equation, the leading core is proportional to $1-\cos x$ over an axial interval of length $2\pi$. Edge coordinates $X=(3c{)}^{1/3}(x-x_e)$ reduce the transition equation to $H^3{H}_{XXX}=H-1$. Its real positive exponent gives a monotone trailing approach and its complex stable exponents give damped capillary waves ahead, with physical wavelength $4\pi a/(3^{5/6}{c}^{1/3})$. Matching the two quadratic edge expansions selects the uniform film thickness; this is a matched asymptotic expansion, valid while the pulse itself remains thin relative to the cylinder.

Zhu, Miao, Xu and Cao's 2007 experiment drove a triangular metal frame electromagnetically with a sinusoidal generator. In a roughly $3\mathrm{cm}$-deep, temperature-regulated pool, it produced nearly plane capillary waves. A $632.8\mathrm{nm}$ laser illuminated the surface obliquely; the moving ripples acted as a reflection diffraction grating. One split beam monitored laser stability, while a CCD recorded diffraction orders on a screen $6.76\mathrm{m}$ away. An air-supported optical table reduced ambient vibration. ^[ref]

The frequency sweep was $100$–$290\mathrm{Hz}$ in $10\mathrm{Hz}$ steps, using distilled water and kerosene at $20{}^{\circ }\mathrm{C}$. For each setting they averaged at least three measurements of first-order spot separation. Known optical geometry and the grating equation yielded $k=2\pi /\lambda$; generator readings yielded $\omega =2\pi f$. Agreement with the predicted log–log slope $3/2$ tests the capillary law, and its intercept gives surface tension if mass density is known. The authors identify surface contamination as a major error source; noncontact optical detection reduces additional disturbance. ^[ref]

The optical measurement can be understood from the grating equation. Let ${\lambda }_L$ be the laser wavelength, $\alpha$ the incident grazing angle above the mean surface, and ${\beta }_m$ the outgoing grazing angle of diffraction order $m$, with the sign of $m$ chosen so thatThus calibrated spot positions determine outgoing angles, which determine the ripple wavelength $\lambda$, independently of the mechanical dispersion relation. This optical geometry treats the surface as a reflection diffraction grating at the laser wavelength. ^[ref]

The inferred surface tension in the full deep-water model isThis formula makes useful experimental controls explicit. Measure mass density at the sample's temperature; check depth with $kh$; keep the driven wave amplitude small enough that $ka\ll 1$; and distinguish $f$ from angular frequency $\omega$. For the stipulated parameters, the $100$–$290\mathrm{Hz}$ window corresponds to calculated wavelengths about $3.62$–$1.76\mathrm{mm}$ and $kh\simeq 52$–$107$ at $h=30\mathrm{mm}$. Depth corrections are negligible here, but the gravity contribution is about $4.5\%$–$1.1\%$ of the capillary contribution. These are ideal predictions, not digitized results from the experiment.

Omitting gravity would infer ${\gamma }_{\mathrm{cap}}=\rho {\omega }^2/k^3=\gamma +\rho g/k^2$, biasing surface tension upward. In the capillary limit, the differential estimate $d\gamma /\gamma \simeq d\rho /\rho +2d\omega /\omega -3dk/k$ also shows why careful spot-spacing calibration matters. A clean surface and weak viscosity are modeling assumptions: appreciable damping, surfactant stresses or large wave amplitudes call for additional physics. The experiment tests the short-wave regime; its drive window does not establish the crossover or the long-wave surface gravity wave limit.

Use the Fourier transform conventionParseval's identity and $\nabla \mapsto i\mathbf{q}$ give the diagonal quadratic formIn a finite periodic box one can equivalently use orthonormal discrete Fourier modes, for which each independent real amplitude has a positive stiffness $\sigma q^2$. Fourier orthogonality eliminates all cross terms between different modes. The zero mode is removed as explained in part (a).

The stiffness vanishes as $q\to 0$, so capillary waves at long wavelength have arbitrarily low energy. The statistical Hamiltonian is invariant under $h(\mathbf{x})\mapsto h(\mathbf{x})+a$ for any constant $a$. Selecting a particular mean membrane height breaks this continuous normal-translation symmetry. A varying local translation therefore has a gradient cost but no restoring mass: these are Goldstone modes. The diagonal static kernel is $q^2$; it should not be identified with a dynamical frequency without specifying the surrounding fluid or membrane inertia.

A plane wave in the rest frame of the medium has phase $\mathbf{k}\cdot \mathbf{x}-{\Omega }_0(\mathbf{k})t$. Under the Galilean transformation ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{U}t$, this becomes $\mathbf{k}\cdot {\mathbf{x}}^{\prime }-[{\Omega }_0(\mathbf{k})-\mathbf{U}\cdot \mathbf{k}]t$. ThereforeHere ${\mathbf{c}}_g$ is the group velocity in the moving frame.

For acoustic waves, ${\Omega }_0=c_0|\mathbf{k}|$ and take $\mathbf{U}=M{c}_0{\mathbf{e}}_x$. A steady wave has $\omega =0$, requiring $\cos \beta =1/M$, where $\beta$ is the angle of $\mathbf{k}$ to the flight direction. Its ray velocity is $c_0\hat{\mathbf{k}}-M{c}_0{\mathbf{e}}_x$, with backward component $-c_0(M^2-1)/M$ and transverse magnitude $c_0\sqrt{M^2-1}/M$. Thus the angle $\theta$ of the ray to the backward flight direction obeysThe axial symmetry gives the Mach cone behind the aircraft.

For unsteady emitted disturbances, all ray velocities lie on the sphere ${\mathbf{c}}_g=-\mathbf{U}+c_0\hat{\mathbf{k}}$. Since $M>1$, the origin lies outside this velocity sphere and its tangent cone has exactly the same semi-angle. All other rays are strictly inside the cone. Equivalently a signal emitted $\tau$ time units ago has a wavefront sphere centered at $-\mathbf{U}\tau$ with radius $c_0\tau$; these spheres are enclosed by their Mach-cone envelope. Consequently the unsteady wavefield is confined to the cone's interior, with the steady envelope on its boundary.

For capillary waves put $a=\sqrt{T/\rho }$, so ${\Omega }_0=a{k}^{3/2}$. The positive-frequency steady condition is $a\sqrt{k}=U\cos \beta$, requiring $\cos \beta >0$. ThenandIts endpoint traces the circlesince $c_{gx}/U=-1/4+(3/4)\cos 2\beta$ and $c_{gy}/U=(3/4)\sin 2\beta$. As $\beta$ runs through $(-\pi /2,\pi /2)$ this traces the circle except its leftmost limiting point, corresponding to the zero-wavenumber limit. The circle surrounds the origin, so every direction except the precisely backward axis is attained by a nonzero-wavenumber steady ray; the backward direction is approached as $k\downarrow 0$. Thus the closure of the steady ray field extends all around the insect, with no excluded angular sector such as the acoustic Mach-cone exterior.

Let the actual dimensional thickness be $d=ϵah$, and let $\bar{z}$ increase downward. The leading curvature of the outer free surface isThe Young–Laplace equation gives $p_{\bar{z}}=-\gamma (d_{\bar{z}}/a^2+d_{\bar{z}\bar{z}\bar{z}})$. In the lubrication approximation, the downward velocity at distance $y$ from the wall is $(\rho g-p_{\bar{z}})(dy-y^2/2)/\mu$, from no slip and zero free-surface shear. Integrating this profile givesCircumferential corrections are higher order in $d/a$, so volume conservation is $d_{\bar{t}}+q_{\bar{z}}=0$ at leading order. DefineThen the thin liquid film on a vertical cylinder satisfiesThe circumferential-curvature term $h_z$ is retained along with axial curvature.

The perturbation calculation givesLong waves with $0<|k|<1$ grow, while $|k|>1$ decay; the fastest growth is $1/12$ at $|k|=1/\sqrt{2}$. Disturbances drift downward at speed $G$. The dimensional unstable wavelengths exceed $2\pi a$, with most unstable wavelength $2\pi \sqrt{2}a$. This is the thin-film version of the Rayleigh–Plateau instability driven by circumferential curvature and opposed by short-wave axial curvature.

For a travelling wave, one integration fixed by $H\to 1$ givesor $H^{\prime \prime \prime }+H^{\prime }=3c(H-1)/H^3-G(1-H^{-3})$. In the large positive-speed core, substitute the expansion using the supplied assumptions. The left side at order $c^{2/3}$ is $H_2^{\prime \prime \prime }+H_2^{\prime }$, while the gravity term is order one and $3c/H^2$ is only order $c^{-1/3}$. Thus $H_2^{\prime \prime \prime }+H_2^{\prime }=0$. Its general solution is a constant plus a sine and cosine. The vanishing leading height and slope at both endpoints selectThe core length is $2\pi$ in the axial coordinate; its amplitude is fixed by transition matching.

At either edge set $X=(3c{)}^{1/3}(x-x_e)$, with $x_e=0$ at the trailing edge and $x_e=2\pi$ at the leading edge. Keeping the height $H$ of order one, the exact wave equation becomesThus the transition equation is $H^3{H}_{XXX}=H-1$. Its linearization about $H=1$ has exponents $1$ and $-1/2\pm i\sqrt{3}/2$. Behind the pulse, approaching the uniform film as $X\to -\infty$ permits only the positive real exponent. The positive-growing, nontrivial branch has one amplitude that is removed by translation, giving the relevant unique $H_{-}$. In front, decay as $X\to +\infty$ permits the two real oscillatory amplitudes; translation removes one parameter and leaves a one-parameter family $H_{+}$. These are the unstable manifold and stable manifold dimensions for the transition system. The literal boundary condition also admits the constant solution and an opposite departing branch, so uniqueness here is specifically for the branch matching a raised positive pulse.

Match the supplied trailing quadratic asymptotic to the core near zero. It gives $K_2=\kappa 3^{2/3}$, because $\frac{1}{2}{c}^{2/3}{K}_2{x}^2=\frac{1}{2}\kappa (3c{)}^{2/3}{x}^2$. The leading maximum occurs at $x=\pi$, soThe front tail has the form $H-1\sim C{e}^{-X/2}\cos (\sqrt{3}X/2+\theta )$. Its capillary waves occur ahead of the descending pulse, not behind it; the trailing tail is monotone to leading order. Converting its frequency to physical distance gives

This large solitary pulse on a cylindrical film sketch shows the leading core and the front-tail eigenmodes; the edge transitions are schematic and the ringing is enlarged for visibility. The thin-film asymptotics require both $c\gg 1$ and $ϵc^{2/3}\ll 1$, rather than taking speed to infinity at fixed film slenderness.

At the next core order, $H_0^{\prime \prime \prime }+H_0^{\prime }=-G$. Integrating and imposing zero $H_0^{\prime }$ at both endpoints givesThe constants in the two quadratic transition asymptotics match the core endpoint heights: $A_0+B_0={\beta }_{-}$ and $A_0+B_0-2\pi G={\beta }_{+}$. Hence $2\pi G={\beta }_{-}-{\beta }_{+}$, and the dimensional uniform thickness selected by matching isWith the supplied constants this fixes $G=3.75/(2\pi )$; $B_0$ is not separately fixed by these height conditions. The thickness selection and capillary wavelength follow from matched asymptotic expansion, not from the linear instability calculation.

The dimensionless equation is $H^3{H}^{\prime \prime \prime }=1-H^3$. At a crest, $H\gg 1$, soThe imposed far-field flux is negligible there; the dominant physical balance is between local gravity-driven drainage and the capillary-pressure gradient.

At a trough, $H\ll 1$, soLocal gravity-driven flux is negligible; the capillary-pressure gradient drives the fixed flux through the trough against viscous resistance. These alternating balances generate the strongly nonlinear capillary waves.

Surface tension $\gamma$ is the tensile force per unit length across a line drawn in a liquid interface. For a clean interface with isotropy, it has the same value in every tangential direction: each side pulls the other along the surface, perpendicular to the line. A straight segment of length $\ell$ therefore transmits a force of magnitude $\gamma \ell$ from one interface. Here we assume constant $\gamma >0$, with fixed temperature and composition. The same coefficient measures the reversible work required to create surface area:The equality of units follows from $1\mathrm{J}=1\mathrm{N}\mathrm{m}$; the equality of the mechanical coefficients follows from the moving-rim argument below. ^{[ref] [ref]}

For a concrete example, take an ideal flat soap film spanning a rectangular wire frame with a straight sliding rim of length $\ell$. Pull the rim outwards in the plane of the film, perpendicular to the rim. Each of the film's two liquid–gas interfaces pulls the rim back towards the film with force $\gamma \ell$, so the outward holding force is $F=2\gamma \ell$. The two faces are two interfaces, each with its own surface tension. With the assumed toy value $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$ and $\ell =20\mathrm{mm}=0.020\mathrm{m}$, the required force is $2.88\mathrm{mN}$; one interface would contribute $1.44\mathrm{mN}$. This value of $\gamma$ is a stipulated model parameter, rather than a measured value for a particular soap solution.

Why does an interface resist being enlarged? Its molecules have a different environment from those in the bulk liquid; changing that environment has an energy cost. In the constant-surface tension model, reducing surface area lowers surface energy. This connects an easily measured rim force to the shape of a droplet, the Young–Laplace equation for its pressure, and the restoring mechanism of capillary waves. It is useful because one coefficient links forces, energy and geometry. ^[ref]

The moving-rim argument also explains the factor of two without memorizing a special formula. An outward displacement $dx$ creates surface area $\ell dx$ on each face, hence $d{A}_{\mathrm{total}}=2\ell dx$. Slow pulling without viscous dissipation requires work $Fdx=\gamma d{A}_{\mathrm{total}}$, giving $F=2\gamma \ell$. For one interface, the same argument gives $F=\gamma \ell$. A surface tension gradient instead produces Marangoni stress; the constant-$\gamma$ assumption excludes that additional effect.