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A synthetic fluid membrane is a thin self-assembled sheet of synthetic amphiphiles, fluid under static in-plane shear while resisting area changes and bending. It provides a nonbiological realization of a two-dimensional fluid interface with curvature elasticity. A Helfrich energy describes long-wavelength shape costs; its material parameters and any spontaneous curvature depend on composition and conditions. Lateral fluidity does not eliminate membrane bending rigidity or membrane tension. The membrane fluctuation spectrum under tension separates bending-dominated and tension-dominated length scales.

For a twice continuously differentiable regular curve describing motion, let $v=\Vert r^{\prime }(t)\Vert >0$, $T=r^{\prime }/v$, and let $N$ be the principal normal vector where the curvature $\kappa$ is positive. Differentiating the velocity vector $r^{\prime }=vT$ givesThe tangential component changes the speed; the normal component changes the direction of the velocity vector. A particle can therefore have nonzero acceleration at constant speed, as in circular motion with constant speed. The formula follows from $dT/dt=(dT/ds)(ds/dt)=\kappa vN$ and the product rule. When $\kappa =0$, the normal contribution vanishes even though the principal normal vector is undefined there. ^[ref]

In dimensionless coordinates, take $r(t)=(t,t^2)$. At $t=1$,The resulting vector components areBoth increasing speed and turning contribute to the same acceleration.

A multilinear operation on vector fields is tensorial when it is linear over smooth functions in every argument. Its value at a point then depends only on the argument vectors at that point. To see this, use a smooth cutoff function to reduce to local fields, expand them in a local frame, and apply linearity over smooth functions to their coefficients. A smoothly valued tensorial operation therefore defines a tensor field. The curvature of an affine connection is tensorial, whereas a covariant derivative differentiates a scalar coefficient in its second argument and is not tensorial there.

Antisymmetrization over selected tensor indices averages their permutations with the sign of a permutation. Square brackets denote it, for example $T_{[ab]}=(T_{ab}-T_{ba})/2$. It projects onto the alternating part. Antisymmetrized second derivatives of a scalar vanish for a torsion-free connection, but antisymmetrized later derivatives act on a tensor and can produce curvature.

With the axisymmetric clamped-plate deflection and fixed front curvature ${\kappa }_f>0$, a magma intrusion hasThese relations describe the advancing fracture branch, with roof bending rather than hydrostatic pressure controlling the interior load.

If a regular plane curve has positive curvature $\kappa$ and a particle moves with speed $v=1/\kappa$ in the chosen time units, its elapsed time is $\int \kappa ds$, the total curvature traversed. On a logarithmic spiral $r=a{e}^{b\theta }$, both $ds/d\theta$ and $1/\kappa$ equal $r\sqrt{1+b^2}$. Hence $dt=d\theta$ and every complete turn takes time $2\pi$, regardless of its radius.

For a regular curve $\mathbf{r}(t)$ with nonzero velocity vector, its unit tangent vector isThe Euclidean norm in the denominator removes the traversal speed, leaving a unit vector that records direction. A positive change of traversal speed leaves this direction unchanged; reversing the direction of travel reverses $T$. In arc length coordinates, $T=d\mathbf{r}/ds$.

For the parabola $\mathbf{r}(t)=(t,t^2)$, the velocity vector is $(1,2t)$ and the speed is $\sqrt{1+4t^2}$. At $t=1$, $T=(1,2{)}^T/\sqrt{5}$. The tangent vector $(1,2{)}^T$ and the unit tangent vector point in the same direction, but only the latter has Euclidean norm one. The principal normal vector is a further choice based on how this direction changes, where the curvature is nonzero. ^[ref]

If a positive vapour-tip pressure magnitude $p_T$ sets $h_f\sim p_T{l}_p^4/B$, matching $h_f/l_p^2$ to the interior curvature ${\kappa }_i$ gives $l_p\sim (B{\kappa }_i/p_T{)}^{1/2}$. The viscous peeling of an elastic plate law then yields $\dot{R}\sim B^{3/2}{\kappa }_i^{7/2}/(\mu p_T^{1/2})$. For constant injected volume flux $Q$, $V=Qt$ and ${\kappa }_i\sim V/R^4$, henceOrder-one coefficients require solving the local peeling profile. The scaling assumes a front region much shorter than the intrusion radius.

The Young–Laplace equation relates the pressure jump across a stationary interface to its curvature and surface tension. Choose a unit normal vector $\mathbf{n}$ pointing from the inside region to the outside region, and define $\kappa ={\nabla }_s\cdot \mathbf{n}$ to be positive for a convex sphere. In mechanical equilibrium, with constant surface tension and no additional normal viscous stresses,Here $k_1,k_2$ are the signed principal curvatures in this convention. Their sum is twice the corresponding signed mean curvature; stating the convention prevents a factor-of-two ambiguity. The Young–Laplace equation is a local normal stress balance for one interface. ^[ref]

A spherical liquid droplet of radius $R$ in a gas has one interface, with $k_1=k_2=1/R$. An ideal thin soap bubble has two liquid–gas interfaces. Adding their pressure jumps givesFor the soap bubble, $\Delta p$ now means the inner gas pressure minus the outer gas pressure. The second formula assumes equal surface tensions on both faces and film thickness negligible relative to $R$, so their radii can be identified. Retaining distinct radii gives $2\gamma /R_{\mathrm{inner}}+2\gamma /R_{\mathrm{outer}}$. A gas bubble inside bulk liquid has only one interface and therefore uses $2\gamma /R$. ^[ref]

With the assumed constant toy surface tension $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$, a droplet with $R=1\mathrm{mm}$ has excess pressure $144\mathrm{Pa}$; at $R=0.5\mathrm{mm}$ it has $288\mathrm{Pa}$. A thin soap bubble assigned the same $\gamma$ and $R=1\mathrm{mm}$ has excess pressure $288\mathrm{Pa}$. These are differences from the surrounding pressure, rather than absolute pressures. Halving the radius doubles the capillary pressure; doubling the interface count also doubles it at fixed radius and surface tension.

The spherical case has a short work argument. A virtual increase in radius gives $dV=4\pi R^{2}dR$ and $dA=8\pi RdR$. At mechanical equilibrium, the net pressure work balances the increase in surface energy:For a thin soap bubble, both interfaces gain approximately the same surface area, doubling the right-hand side and giving $4\gamma /R$. This is a virtual work balance between neighboring spherical configurations, rather than an assumption that a fixed-volume droplet actually expands. The general Young–Laplace equation replaces the spherical area-to-volume ratio by local curvature, making geometry directly measurable through pressure.