For $2N$ noninteracting electrons in a planar sample of area $A$, neglecting Zeeman splitting leaves two spin states for each orbital. Each Landau level therefore holds $D=eBA/(\pi \hslash )$ electrons. When $N\le D\le 2N$, fill the lowest level with $D$ electrons and the next with $2N-D$; summing their energies gives the displayed concave quadratic in $B$. The energies at the two integer-filling endpoints are equal.

In his 1883 paper, Osborne Reynolds investigated how the character of pipe flow depends jointly on velocity, pipe diameter and viscosity. His final dye apparatus drew water from a large glass tank through a horizontal glass tube with a smooth trumpet-shaped entrance. A separate reservoir supplied strongly colored water just upstream of the entrance; a valve below the tank controlled the outflow. The glass tube was immersed in the tank, so its internal dye band could be watched. Reynolds repeated the trials with different tube diameters and water temperatures. ^[ref]

The volumetric flow rate was measured independently of the appearance of the dye. Reynolds timed a one-inch fall of the tank's water level, corresponding to a calibrated discharged volume. In modern notation, if this volume is $V$ and the elapsed time is $\Delta t$, then $Q=V/\Delta t$ and $U=4V/(\pi D^2\Delta t)$. He determined the glass tubes' internal diameters by filling them with water. For a uniform bore, a known filled length $\ell$ and measured water volume $V_{\mathrm{tube}}$ determine $D=\sqrt{4V_{\mathrm{tube}}/(\pi \ell )}$; checking bore uniformity matters because the same volume would only give an average cross-section in a tapered tube. Thermometers at different tank levels checked both the temperature and its uniformity. Reynolds compared the viscosity at different temperatures using Poiseuille's results and checked his own water by separate experiments of that type. Thus the Reynolds number estimate did not come from judging the dye band or from assuming that transition occurred at a preset value. ^[ref]

A practical repetition can instead collect a known outflow volume in a measuring vessel and time its collection, or weigh the collected water and divide its mass by mass density. Measure the internal diameter, rather than the external diameter, and obtain kinematic viscosity at the measured water temperature from independently established properties or a separate viscosity measurement. Record the dye appearance and observation position alongside $Q$, $D$, temperature and the resulting Reynolds number. These are alternative measurement procedures following $Q=V/\Delta t$, not additional historical claims about Reynolds' dye apparatus.

Begin with settled water and a small dye supply, then open the outlet valve in small steps. Reynolds saw a thin straight band at sufficiently low speeds. With a tank that had not fully settled, the band could wander while retaining its coherence. At higher discharge it could abruptly spread into the surrounding water downstream of the entrance. Short illumination by an electric spark resolved curls within what otherwise looked like a colored cloud. This connected the visible breakup with internal irregular motion rather than merely the appearance of an opaque pipe. ^[ref]

The controls are part of the result. Reynolds used a smooth entrance and let the tank rest to reduce inlet disturbances. He found that a heated tank could retain slow circulation as it cooled, so measuring one temperature was insufficient: thermal uniformity and residual motion also mattered. His wire-coil trial broke up the band below the previous onset speed, but he explicitly cautioned that once the dye was mixed he could not tell from that dye alone whether the motion became orderly again farther along the tube. In a repetition, keep the dye supply small enough to act approximately as a passive scalar, avoid a strong injection jet, and record the inlet condition and downstream observation distance. If also testing the Hagen-Poiseuille equation, allow the profile to develop and measure pressure drop between positions within the developed region; do not treat the entrance and outlet losses as part of its uniform pressure gradient. The dye experiment by itself does not establish fully developed flow. ^[ref]

The experiment's usefulness is that it compares a visible change in the motion with independently measured scales. Repeating it with several diameters and temperatures, while controlling disturbances, tests the organizing role of the Reynolds number. Its limitations are equally instructive: dye is a tracer of transport, diffusion still acts in laminar flow, and a local dye observation cannot by itself prove that turbulence persists throughout an arbitrarily long pipe.

A regular Lagrangian $L:TQ\to \mathbb{R}$ defines a fibre derivative $\mathbb{F}L:TQ\to T^{\ast }Q$ and the Cartan one-form ${\theta }_L=(\mathbb{F}L{)}^{\ast }\theta$. Its two-form ${\omega }_L=-d{\theta }_L$ is nondegenerate when the velocity Hessian is invertible. The energy is $E_L=v^i{L}_{v^i}-L$. The equation ${\iota }_{{\Gamma }_L}{\omega }_L=d{E}_L$ determines the Euler–Lagrange vector field. A hyperregular Lagrangian makes the fibre derivative a global diffeomorphism and transports this dynamics to Hamiltonian phase space.

For a finite constrained optimum value $\varphi (b)=\inf \{f(x):h(x)=b,x\in X\}$ and optimization Lagrangian $L_b(x,\lambda )=f(x)-{\lambda }^T(h(x)-b)$, the strong Lagrangian property means that some finite Lagrange multiplier attains the dual lower bound exactly: ${\inf }_X{L}_b=\varphi (b)$. It includes dual attainment, but need not include primal attainment. It is equivalent to a non-vertical supporting hyperplane of a value function: a supporting slope gives $f(x)\ge \varphi (h(x))\ge \varphi (b)+{\lambda }^T(h(x)-b)$, hence ${\inf }_X{L}_b\ge \varphi (b)$; weak duality gives the opposite inequality. Conversely the infimum identity bounds every $L_b(x,\lambda )$ below by $\varphi (b)$; taking the infimum over $h(x)=u$ gives the supporting inequality. No convexity assumption is needed for this equivalence.

The Lagrangian perturbation compares a variable on the same moving fluid element before and after a displacement. Taylor expansion of the background value at the displaced position gives the displayed relation to an Eulerian perturbation. For an adiabatic ideal gas, $\Delta (p/{\rho }^{\gamma })=0$ even when the equilibrium entropy varies spatially.

The Eulerian perturbation compares field values at the same spatial location, rather than following a fluid element. To first order it satisfies $\delta q=\Delta q-\xi \cdot \nabla q$. Thus a field advected unchanged by individual elements can still have a nonzero Eulerian perturbation in an inhomogeneous background.

For real $a_i$ not all equal, let $S_1={\sum }_i{a}_i$, $S_2={\sum }_i{a}_i^2$ and $\Delta =n{S}_2-S_1^2>0$. The unique minimizer of ${\sum }_i{x}_i^2$ under ${\sum }_i{x}_i=1$, ${\sum }_i{a}_i{x}_i=0$ is $x_i^{\ast }=(S_2-S_1{a}_i)/\Delta$, with minimum $S_2/\Delta$. To prove sufficiency, the optimization Lagrangian with multipliers $\lambda =S_2/\Delta$, $\mu =-S_1/\Delta$ is a sum of squares in $x_i-\lambda -\mu a_i$ plus a constant. Equivalently every feasible $x$ satisfies ${\sum }_i{x}_i^2-{\sum }_i(x_i^{\ast }{)}^2={\sum }_i(x_i-x_i^{\ast }{)}^2$.

For a smooth constrained maximum $\varphi (b)=f(\bar{x}(b))$ subject to $g(\bar{x}(b))=b$, suppose smooth optimizers and multipliers satisfy $\nabla f(\bar{x})=Dg(\bar{x}{)}^T\lambda$. Differentiating the constraint gives $Dg(\bar{x})D_b\bar{x}=I$, and the chain rule gives $D_b\varphi ={\lambda }^T$. Thus a Lagrange multiplier measures the marginal value of relaxing its corresponding constraint. A global supporting Lagrangian duality representation implies the needed stationarity, but local stationarity and smoothness alone suffice for this derivative identity.

For radius $r>0$ and height $h>0$, volume is $\pi r^{2}h$ and area including both ends is $2\pi r^2+2\pi rh$. The Lagrange multiplier equations give $h=2r$. Substituting the constraint gives the one-variable function $2\pi r^2+2V/r$, which tends to infinity at both ends and has its only stationary point at $r^3=V/(2\pi )$. This proves the stationary cylinder is the unique global minimizer.

For an axis-aligned box, maximize $xyz$ subject to $x^2/a^2+y^2/b^2+z^2/c^2=1$. The Lagrange multiplier equations imply $x^2/a^2=y^2/b^2=z^2/c^2=1/3$. Boundary boxes have zero volume, so this gives the maximum. Arbitrary orientation cannot improve it: averaging the ellipsoid inequality over the eight vertices gives a sum of three squared transformed half-edge lengths at most one; the determinant bound and arithmetic-geometric mean bound their product by $1/(3\sqrt{3})$.

This nonlinear dispersive equation replaces the quadratic KdV nonlinearity by a cubic one. In the displayed sign convention, the Miura transformation maps its solutions to $u_t-6u{u}_x+u_{xxx}=0$. Its scaling symmetry is $(x,t,v)\mapsto (e^{s}x,e^{3s}t,e^{-s}v)$.

The cylindrical KdV equation in this normalization is $q_t+q_{xxx}+q{q}_x+q/(3t)=0$, for $t\ne 0$. The additional time-dependent term distinguishes it from the ordinary KdV equation. A scaling-invariant solution has $q=t^{-2/3}F(x{t}^{-1/3})$. Substitution gives $F^{(3)}+F{F}^{\prime }-(\xi F{)}^{\prime }/3=0$. Integration reduces the similarity problem to $F^{\prime \prime }+F^2/2-\xi F/3=C$. The integration constant depends on supplementary conditions and cannot be set to zero without them. Use real cube roots separately on either sign of $t$.

One normalization of the first Hamiltonian structure uses the constant differential Poisson operator $J=-2{\partial }_x$. For functionals with variational derivatives $f,g$, its Poisson bracket is $\int fJgdx=-2\int f{g}^{\prime }dx$ under periodic or vanishing-boundary conventions. Thus the field bracket is $\{u(x),u(y)\}=-2{\partial }_x\delta (x-y)$. The factor and sign change with normalization; the constant first-order operator is the invariant structural feature.

Dehn surgery removes tubular neighborhoods of a link in a three-dimensional manifold and performs Dehn filling on their boundary tori with specified primitive meridian slopes. Dehn surgery on a framed link is the integral-slope special case.

For otherwise outbred opposite-sex first cousins, the probability that the mother's randomly transmitted X copy matches the father's X by descent depends on which of their parents are siblings. It is zero when the male cousin is related through his father, $1/8$ when the female cousin is related through her father and the male through his mother, and $3/16$ when their mothers are sisters. Daughters can be X-autozygous; sons have one X and a Y and are not homozygous for a paired X chromosome. Recombination and pseudoautosomal regions require a locus-based formulation instead of whole-chromosome blocks.

For $N$ individuals in monogamous families of exactly two offspring, let ${\Phi }_G$ be mean pairwise kinship coefficient, $F_G$ mean inbreeding and ${\Phi }_{m,G}$ mean kinship of mates. A random distinct offspring pair are siblings with probability $1/(N-1)$. Accounting separately for within-family and cross-family pairs gives the displayed recurrence, while $F_{G+1}={\Phi }_{m,G}$. Random mating makes ${\Phi }_{m,G}={\Phi }_G$. Sibling avoidance changes the mating-pair mean; it does not create unrelated founders or stop eventual drift to ancestral identity in a closed finite biological population.

The Vlasov equation describes collisionless evolution of a phase-space distribution function in a self-consistent force field: ${\partial }_{t}f+\mathbf{v}\cdot {\nabla }_{x}f+\mathbf{F}\cdot {\nabla }_{p}f=0$. The force can be electromagnetic or gravitational; the Collisionless dark-matter Vlasov equation is the cosmological gravitational specialization.

Thermal transpiration is gas transport driven by an imposed temperature difference in a rarefied connection. Gas can move from cold to hot even when the two end pressures start equal. If the reservoirs are closed, redistribution builds a counteracting pressure difference until the net transfer stops. This final stationary state is not global thermal equilibrium: the two temperatures remain different. It explains why a heated pressure sensor can read a different pressure from the vessel to which it is connected. The correction depends on gas, connection geometry and molecular interaction with its surfaces. ^[ref]

The simple molecular limit is beautiful: equal opposing particle rates need not mean equal pressures. Consider effusion through a thin hole connecting two locally equilibrated reservoirs of the same gas, at $T_h>T_c$. Equating their one-way rates givesAt equal pressure, the hotter reservoir has lower number density; its faster molecules do not fully compensate for that reduction in outgoing particle rate. With $T_c=300\mathrm{K}$ and $T_h=351\mathrm{K}$, the ideal zero-transfer ratio is $\sqrt{351/300}\simeq 1.0817$. The hydrodynamic, zero-flow limit instead has equal pressures when other forces are absent. These are limiting models, not an interpolation formula for a glass capillary. ^[ref] In a real tube, incomplete exchange of molecular energy with the wall can alter the limiting ratio; a Knudsen number does not specify that exchange. ^[ref]

Effusion is molecular passage through a small opening whose dimensions, including its thickness, are much smaller than the intermolecular mean free path. For an equilibrium ideal gas, the one-way particle rate through a thin hole of area $A$ is $\dot{N}=Ap/\sqrt{2\pi m{k}_{B}T}$. This weights molecules by their wall-normal velocity, rather than just counting all molecules in the reservoir. ^[ref]

For a Killing vector field $K$, its metric-dual differential one-form produces the electromagnetic field $F_{ab}=2{\nabla }_{[a}{K}_{b]}=2{\nabla }_a{K}_b$. The exterior derivative gives $dF=0$, while the second covariant derivative of a Killing vector gives ${\nabla }^a{F}_{ab}=-2R_{bd}{K}^d$. Thus this construction solves the source-free Maxwell equations in a vacuum spacetime with zero cosmological constant. The field is a test field: adding its stress-energy tensor would generally change the original metric tensor.