{title2=$\gamma$}
{wiki}

<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>:
$$
F_{\text{one interface}}=\gamma\ell,\qquad dE_{\mathrm s}=\gamma\,dA,\qquad
[\gamma]=\mathrm{N\,m^{-1}}=\mathrm{J\,m^{-2}}.
$$
The equality of units follows from $1\,\mathrm J=1\,\mathrm{N\,m}$; the equality of the mechanical coefficients follows from the moving-rim argument below. https://openstax.org/books/college-physics-2e/pages/11-8-cohesion-and-adhesion-in-liquids-surface-tension-and-capillary-action[OpenStax, College Physics 2e, §11.8: surface tension and the sliding wire]{ref} https://arxiv.org/html/2011.11403v2[Tomas Bohr and Bernhard Scheichl, Surface Tension and Energy Conservation in a Moving Fluid, equations (5) and (13)]{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 $\gamma=0.072\,\mathrm{N\,m^{-1}}$ 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.

\Image[/surface-tension-film-force.png]
{title=Two film faces double the force on a sliding rim}
{description=The plan view shows the outward pull and the inward <surface tension> <force> on a sliding rim; the two interfaces overlap in this view. The graph compares one and two interfaces at the same assumed constant $\gamma=0.072\,\mathrm{N\,m^{-1}}$, with rim length in millimetres and <force> in millinewtons.}

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. https://davidtong.org/pdfs/teaching/fluid-mechanics/fluids3.pdf[David Tong, Fluid Mechanics, §4.1.3, pp. 121–123]{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 $dA_{\mathrm{total}}=2\ell\,dx$. Slow pulling without <viscous dissipation> requires <work> $F\,dx=\gamma\,dA_{\mathrm{total}}$, 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.


 Surface tension (source code)