Motional electromotive force is the contribution ${\oint }_C(\mathbf{v}\times \mathbf{B})\cdot d\mathbf{l}$ from an electrical conductor moving with local velocity $\mathbf{v}$ through a magnetic field. The cross product comes from the magnetic part of the Lorentz force. For a straight rod of length $\ell$, perpendicular to both its velocity and a uniform magnetic field, its magnitude is $B\ell v$. The sign depends on the direction of integration along the rod. ^[ref].

For a computed example, take $\ell =0.50\mathrm{m}$, velocity $\mathbf{v}=2.0\hat{\mathbf{x}}m/s$ to the right, and magnetic field $\mathbf{B}=0.40\hat{\mathbf{z}}\mathrm{T}$ out of the page. The rod lies along $y$. ThenPositive electric charge is driven downward. For an isolated rod after charge separation has equilibrated, $\mathbf{E}=-\mathbf{v}\times \mathbf{B}$ inside it, and the electric potential difference is $V_{\mathrm{bottom}}-V_{\mathrm{top}}=+0.40\mathrm{V}$; the bottom is positive. This open-circuit statement does not imply a sustained electric current.

Close the rod's path with stationary U-shaped rails and an electrical resistance $R>0$ at the left. With out-of-page unit normal, the positive loop direction is counterclockwise, hence upward along the rod. If the rod's horizontal position is $x(t)$, the enclosed area is $A=\ell x$ and the external magnetic flux is ${\Phi }_{\mathrm{ext}}=B\ell x$. ThusHere the resistive approximation neglects self-inductance. The negative electric current is clockwise, downward along the rod, consistent with the charge-separation calculation and Lenz's law.

The closed-loop result illustrates why the moving-circuit form of Faraday's law of induction includes motional electromotive force: the magnetic field is constant in time, so the electric field circulation alone would miss this drive. Its mechanical consequence is electromagnetic braking of a sliding rod.

The dual electromagnetic field tensor is obtained by applying the Hodge star operator to the electromagnetic field tensor. For metric signature $(+---)$ and ${ϵ}^{0123}=+1$, define ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$. With $F^{0i}=-E_i/c$ and $F^{ij}=-{ϵ}_{ijk}{B}_k$, it has components ${\tilde{F}}^{0i}=-B_i$ and ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$. The homogeneous Maxwell equations become ${\partial }_{\mu }{\tilde{F}}^{\mu \nu }=0$.

Assume an exact solution of finite penalty exists and select the unique $J$-minimizing exact solution $u^{†}$. Minimality gives $\Vert K{u}_{\alpha }^{\delta }-f^{\delta }{\Vert }^2/2+\alpha J(u_{\alpha }^{\delta })\le {\delta }^2/2+\alpha J(u^{†})$. If $\alpha \to 0$ and ${\delta }^2/\alpha \to 0$, boundedness, weak sequential compactness of bounded sequences in a reflexive Banach space and lower semicontinuity identify every weak limit with $u^{†}$. The separate lower semicontinuity of $\Vert u{\Vert }_1$ and $\Vert u{\Vert }_2^2$, together with the upper bound on $J$, forces the Hilbert norms to converge. The Radon-Riesz theorem then gives norm convergence.

For fixed $\alpha >0$ and nonzero bounded $K$, the reconstruction vanishes when $\Vert K^{\ast }f{\Vert }_{\infty }\le \alpha$, which includes a neighborhood of zero. If it were linear it would vanish everywhere. Choose a coordinate $e_j$ with $K{e}_j\ne 0$ and data $f=tK{e}_j$ with $t\Vert K{e}_j{\Vert }^2>\alpha$; the optimality inequality at zero fails, so its minimizer is nonzero. Thus the map is nonlinear. The qualification $K\ne 0$ is essential: $K=0$ gives the zero linear map.

There are no weight-four cusp forms at level two. A nonzero form would have a weight-twelve coset norm of a modular form with modular cusp order at least two, contradicting the simple modular cusp order and interior nonvanishing of the modular discriminant. The two independent displayed Eisenstein series span the weight-four space, since its cusp-constant map embeds it in a two-dimensional space.

The ramification polynomial records the normalised differences of conjugates of a uniformizer, including the zero difference from the identity automorphism. Its quotient by $X$ is the reduced ramification polynomial. In the usual coefficient-index convention, the finite slopes of its lower Newton polygon are negatives of the valuations of its nonzero roots; the coefficient at the zero root is treated as having infinite valuation.

A polynomial may fail the Eisenstein polynomial test while a translate $f(T+a)$ passes it. Translation does not change the generated extension: if $f(\alpha )=0$, then $f(T+a)$ vanishes at $\alpha -a$. For $f(T)=T^3-m$, the translate by $m$ is Eisenstein at $3$ whenever $3\nmid m$ and $m\not\equiv \pm 1(\mathrm{mod}9)$.

Let a real symmetric matrix $M$ have ordered eigenvalues ${\lambda }_1\ge {\lambda }_2\ge \cdots \ge {\lambda }_n$, with $n\ge 2$. If $A$ is obtained by deleting one matching row and column, its largest eigenvalue ${\mu }_1$ satisfiesThe Rayleigh quotient of $A$ is that of $M$ restricted to vectors whose deleted coordinate is zero. This gives the upper bound. Some nonzero linear combination of the top two orthonormal eigenvectors has that coordinate zero; normalizing it gives a quotient at least ${\lambda }_2$. The argument does not require distinct eigenvalues.

Geometric multiplicity is the dimension of an eigenspace; algebraic multiplicity is its multiplicity as a root of the characteristic polynomial. For a real symmetric matrix they coincide because it has an orthonormal eigenbasis.

The spectral abscissa is the largest real part of a matrix's eigenvalues. It describes asymptotic exponential rates of its matrix exponential, but a non-normal matrix can also have transient amplification. It is generally distinct from the Euclidean logarithmic norm, which gives an immediate Euclidean energy-growth bound.

For countably many unary predicates $P_i$, let $M$ contain infinitely many copies of each finite membership pattern and let $N$ add an element satisfying all predicates. Every finite vocabulary reduct has infinitely many elements in every Boolean cell in both structures, so $M\equiv N$. The Spoiler nevertheless wins the one-round full-language game by selecting the new element, whose complete atomic pattern has no counterpart in $M$.

A proper subvariety locally defined by one non-zero-divisor is an effective Cartier divisor. On an integral variety a nonzero local equation is automatically a non-zero-divisor. Its conormal sheaf is a line bundle, since $A/(f)\cong (f)/(f^2)$ via multiplication by $f$. Allowing the zero equation would include the whole variety and would not give this conclusion.

The terrestrial atmosphere has regions with opposite signs of the atmospheric lapse rate: the troposphere usually cools upward, the stratosphere warms upward through ultraviolet heating, the mesosphere cools, and the thermosphere warms. A monotone atmospheric pressure-temperature profile can approximate one such interval, but cannot describe all their transitions with one fixed branch.

For unwrapped grain and parent azimuths, define a signed lag $\Delta \theta ={\theta }_d-{\theta }_p$ and a positive trailing distance $s=-\Delta \theta$. A grain released without a velocity kick from a circular parent orbit has $\Delta \theta =2\beta (\sin {\theta }_d-{\theta }_d)+O({\beta }^2{\theta }_d)$. Keeping track of this sign avoids mistaking trailing motion for leading motion.

For a single-stream dust tail with injection rate $\dot{N}$, monotone angular distance $s(t)$, and survival probability $P(t)$, the number per unit angle is $n(s)=\dot{N}P(t(s))/\dot{s}$. A vanishing angular drift creates an idealized density enhancement; velocity dispersion rounds it.

For conjugate exponents $p,q\in (1,\infty )$, every bounded linear functional on an Lp space is uniquely $h\mapsto \int hgd\mu$ for $g\in L^q$, with functional norm $\Vert g{\Vert }_q$. This holds on arbitrary measure spaces. Apply the Radon-Nikodym theorem on finite-measure pieces, then use support localization of an Lp functional to obtain one global density without assuming sigma-finiteness of the entire measure.

For a dual pair in a monoidal category, the contraction $e:X\otimes Y\to I$. In a left closed monoidal category, evaluation also denotes the adjunction map $[M,N]\otimes M\to N$.

The duality map $n:I\to Y\otimes X$ from the monoidal unit object in a dual pair in a monoidal category, constrained together with evaluation by both snake identities.

The two triangular identities $(e\otimes 1_X)(1_X\otimes n)=1_X$ and $(1_Y\otimes e)(n\otimes 1_Y)=1_Y$, with canonical associators and unitors inserted, for a dual pair in a monoidal category.

For a linear map between finite-dimensional vector spaces, the image of its dual map is the annihilator of a vector subspace given by its kernel. Inclusion follows by composition. Conversely a linear functional vanishing on the kernel descends to the image and extends to the target by basis extension. The identity and the rank-nullity theorem show that a linear map and its dual have equal rank.