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For a long axisymmetric bubble with slowly varying radius $a(z,t)$, the leading normal-stress balance and internal lubrication flux couple capillary pressure to conservation of the cross-sectional area.

A thin-film equation combines a lubrication flux with mass conservation to evolve a free-surface thickness. Gravity, capillary pressure, wall motion, and intermolecular forces may contribute to the flux.

If pressure approaches the same ambient value at both ends of a lubrication region, then ${\int }_{-\infty }^{\infty }{p}_{x}dx=0$, which determines the conserved flux.

Between surfaces of velocities $U_0,U_1$ separated by $h$, the local lubrication profile is

Near the closest approach of a circle of radius $a$ to a wall, the gap is $h(x)=h_0+x^2/(2a)+O(x^4/a^3)$ and has streamwise scale $\sqrt{a{h}_0}$.

For a film of thickness $h(x,t)$ flowing down a plane inclined by $\alpha$, lubrication theory gives the fluxwhen surface tension is neglected. Mass conservation gives the thin-film equation $h_t+q_x=0$.

For gap scale $H$, length $L$, lower-wall tangential speed $U$, upper-wall normal speed $V$, and induced horizontal scale $W=|U|+VL/H$, the lubrication limit requiresThe last condition includes both $\rho |U|H^2/(\mu L)\ll 1$ and $\rho VH/\mu \ll 1$.

The determinant-one Lorentz group in one space and one time dimension has two connected components. Its identity component consists of boostsand the other component is its negative.

Let $A$ be $k$ vertices of a bipartite graph, each of degree at least $k$. If the graph has no $4$-cycle, then $|N(A)|\ge 2k-1$. Indeed, any pair in $A$ has at most one common neighbour, soIf $|N(A)|\le 2k-2$, Cauchy--Schwarz and ${\sum }_y{d}_A(y)\ge k^2$ make the left side strictly larger than $\left(\genfrac{}{}{0ex}{}{k}{2}\right)$, a contradiction.

For treatment groups Control, LD, and SD and binary outcomes, equal LD and SD efficacy can be imposed while retaining separate group totals by using separate treatment main effects but one shared treated-versus-control interaction with outcome. Comparing this five-parameter model with the saturated six-parameter model gives a one-degree-of-freedom analysis of deviance for nested generalized linear models.

A saturated log-linear model has enough parameters to reproduce every observed contingency-table cell count and therefore has zero residual deviance.

Localization at a prime ideal $\mathfrak{p}$ means inverting $R\setminus \mathfrak{p}$. The resulting local ring $R_{\mathfrak{p}}$ has maximal ideal $\mathfrak{p}{R}_{\mathfrak{p}}$.

Every localization of a Noetherian ring is Noetherian. Each ideal $J\subseteq S^{-1}R$ is the extension of its contraction to $R$, whose finite generating set consequently generates $J$.

Extension and contraction give inverse inclusion-preserving bijections between prime ideals of $S^{-1}R$ and prime ideals of $R$ disjoint from $S$:

The localization map $\iota :R\to S^{-1}R$ sends every element of $S$ to a unit. If a ring homomorphism $f:R\to T$ also sends every element of $S$ to a unit, there is a unique homomorphism $\tilde{f}:S^{-1}R\to T$ with $\tilde{f}(r/s)=f(r)f(s{)}^{-1}$ and $\tilde{f}\iota =f$.

A valuation ring is an integral domain $R$ such that, for every nonzero element $x$ of its fraction field $K$, either $x\in R$ or $x^{-1}\in R$. Equivalently, its ideals are totally ordered by inclusion.

Quantum teleportation transfers an unknown qubit using one shared Bell pair, a Bell-basis measurement by the sender, two classical bits, and a Pauli correction by the receiver.

Two parties can distinguish $|{\Phi }^{+}⟩$ from $|{\Phi }^{-}⟩$ by both measuring in the Hadamard basis and comparing outcomes: equal outcomes identify $|{\Phi }^{+}⟩$, while unequal outcomes identify $|{\Phi }^{-}⟩$.

The maximal abelian extension $K^{\mathrm{ab}}$ of a field $K$ is the compositum in a fixed separable closure of all finite abelian extensions of $K$.

The roots of unity in a finite extension $K/{\mathbb{Q}}_p$ form a finite group. Each element of their prime-to-$p$ part is a Teichmuller representative, while any $p$-power part lies among the principal units.