= Soap film pressure balance
{title2=$\Delta p=-2\tau H$}
For an ideal soap film with equal constant <surface tension> $\gamma$ on both faces, the effective tension is $\tau=2\gamma$. Choose a <unit normal> $N$ from the minus side to the plus side, and $\Delta p=p_- -p_+$. Translating the <Young–Laplace equation> to $S=-dN$ gives
$$
\Delta p=-\tau(k_1+k_2)=-2\tau H.
$$
A single interface uses $\tau=\gamma$. An outward-oriented spherical soap bubble has $H=-1/R$, giving $\Delta p=4\gamma/R$. The positive-convex curvature sum in the <Young–Laplace equation> article is $-2H$. <18.357 Interfacial Phenomena lecture notes (Fall 2010)>[18.357 Interfacial Phenomena, §2.6, pp. 9–10: drop pressure and the two-interface soap-bubble factor]{ref}
An open film with the same resting <gas> on both sides has zero <pressure> jump. Neglecting film weight, <soap film pressure balance> gives $H=0$: a <minimal surface>. A pressurized bubble instead has nonzero <mean curvature>. These statements assume <equilibrium>, constant <surface tension>, and negligible normal <viscous stress>. <Surfactant> gradients or motion require the fuller <normal stress> balance. <18.357 Interfacial Phenomena lecture notes (Fall 2010)>[§2.4, p. 7, equations (2.3)–(2.4): normal and tangential interfacial stress balances]{ref}
The variational outline is short. With fixed boundary, a normal displacement $fN$ gives $\delta A=-2\int_M Hf\,dA$ by the <first variation of area formula>. For a closed interface oriented outward, $\delta V=\int_M f\,dA$, so stationarity of $\tau A-\Delta p\,V$ gives $-2\tau H-\Delta p=0$. This is a <first variation> calculation; testing for a <stable minimal surface> needs the <second variation>, and deciding a <global minimum> needs a comparison of competitors.
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