Fix paired orthonormal bases and let $|\varphi {⟩}_A={\sum }_j{c}_j|j{⟩}_A$. Its conjugate index vector in the other Hilbert space uses the conjugated coefficients. For the unnormalized maximally entangled state $|\tilde{\Psi }⟩={\sum }_j|j{⟩}_A|j{⟩}_B$, the partial inner product is $(I\otimes ⟨{\varphi }_B^{\ast }|)|\tilde{\Psi }⟩=|\varphi {⟩}_A$. Both the conjugation and the lack of normalization of $\tilde{\Psi }$ matter. Applying a linear operator on the first subsystem gives the corresponding relative state of the operated vector.

The indicators $Y_i=1_{\{R_i=1\}}$ of successively smaller records are independent and have Bernoulli distributions of parameters $1/i$. Successively larger records use $R_i=i$ and have the same laws. For $K_n={\sum }_{i=1}^n{Y}_i$, expectation and variance additivity for independent random variables give $E{K}_n={\sum }_{i=1}^{n}1/i$ and $\mathrm{Var}{K}_n={\sum }_{i=1}^n(1/i-1/i^2)$.

Attach conditionally independent nonnegative marks to a Poisson point process, with a measurable mark law depending only on location. If a mark at $z$ has mean $m(z)$ and variance $v(z)$, and the displayed integrals are finite, their total $W$ is finite almost surely and has the displayed moments. The Campbell first-moment formula gives the first moment and finiteness. The Poisson factorial moment measure formula gives the square of that first moment for distinct pairs; the diagonal contributes $\int (v+m^2)d\mu$. Their difference is the variance. Signed marks require absolute first-moment integrability.

For three pairwise independent events of common probability $a$, write $x=\mathbb{P}(A_1\cap A_2\cap A_3)$. Inclusion-exclusion principle gives avoidance probability $1-3a+3a^2-x$, and $x\le a^2$ gives the displayed lower bound. It is strictly positive for $0<a<1/2$. For $1/2\le a<1$, a symmetric atomic law with no zero-success atom has mass $(1-a{)}^2$ on each singleton success pattern, $(1-a)(2a-1)$ on each doubleton, and $1-3a+3a^2$ on the tripleton. These masses sum to one, have marginal $a$, and have pair intersections $a^2$.

Suppose the recorded durations retain an exponential distribution of rate $\theta$, but an independent fraction $q$ are relabelled as failures and the rest as right-censored. Applying the ordinary survival likelihood gives $\hat{\theta }=d/{\sum }_i{y}_i\to q\theta$, rather than $\theta$. A censoring flag chosen independently of the recorded duration does not establish independent censoring of a latent failure time. Genuine simulation instead draws an event time and an independent censoring time and records their minimum.

A nonconstant smooth periodic roll obeys $\mu ⟨T_0^{\prime 2}⟩-⟨T_0^{\prime \prime 2}⟩-⟨T_0^{\prime 4}⟩=0$. The orthogonal-roll perturbation raises the increasing-gradient functional for poorly conducting convection by the displayed amount. The variance is strictly positive: a nonconstant periodic function has a derivative that vanishes somewhere and is nonzero somewhere. The Hessian of $V$ therefore has a positive direction. Its self-adjoint linearized evolution operator has a positive Rayleigh quotient and hence a positive eigenvalue, proving transverse linear instability wherever a nonconstant roll exists.

With exponentially decreasing injection, the advective approximation $\varphi h_t+u{h}_x=0$ and inlet condition $uh(0,t)=Q{e}^{-t/\tau }$ give $h_0=Q/u$ and the displayed solution along characteristic curves. It applies behind the advective nose $X=ut/\varphi$. The relative size of the discarded nonlinear spreading term for this profile is $2\varphi h\cot \theta /(u\tau )$. Smallness of that quantity, and matching through a narrow leading layer, are necessary qualifications: elapsed time alone does not make the approximation uniform at the nose. In a translating frame the full equation remains a porous medium equation with coefficient $u\cot \theta /\varphi$.

For porosity $P{z}^{\alpha -1}$ and permeability of a porous medium $K{z}^{\beta -1}$ above an inclined impermeable plane, stored liquid volume per area is $P{h}^{\alpha }/\alpha$. The depth-integrated Darcy flux is $\rho gK{h}^{\beta }(\sin \theta -\cos \theta h_x,-\cos \theta h_y)/(\beta \mu )$. Conservation of mass yields the displayed equation with $X=x\tan \theta$, $Y=y\tan \theta$ and $T=t\alpha \rho gK\sin \theta \tan \theta /(\beta \mu P)$.

With constant porosity $P$ and $X=x\tan \theta$, $Y=y\tan \theta$, a translating frame $\xi =X-T$, $\eta =Y$ has area Jacobian determinant $dxdy={\cot }^2\theta d\xi d\eta$. Actual fluid volume is $P\iint hdxdy$, proving the displayed normalization. Replacing the cotangent factor by a tangent factor is incompatible with these coordinates except at a slope of $\pi /4$.

Integrating the closed-collection balance in momentum conservation and applying the fundamental theorem of calculus givesFor one fixed-mass point mass, this becomes ${\mathbf{J}}_{\mathrm{net}}=\Delta \mathbf{p}=m\Delta \mathbf{v}$. Zero external impulse implies equal initial and final total momenta; momentum conservation throughout the interval requires zero net external force throughout it.

For stages $k_1=f(y)$ and $k_2=f(y+(1-a)h{k}_1+ah{k}_2)$ with update $y+h(k_1+k_2)/2$, the method has order two for every fixed real $a$. The third-order coefficient of $f^{\prime \prime }{f}^2$ is always $1/4$, not $1/6$, so no parameter gives third order on general nonlinear equations. The displayed stability function is bounded throughout the left half-plane only when $a=1/2$, when it is the A-stable trapezoidal function $(1+z/2)/(1-z/2)$. It is not L-stable.

A population jumps from $n$ to $n+2$ at a constant immigration rate $\lambda$ and from $n$ to $n-1$ at rate $\beta n$. Applying its Markov jump-process generator to $n$ and $n^2$ gives the displayed mean and variance equations. For $\beta >0$ and finite initial second moment, the stationary mean and variance are $2\lambda /\beta$ and $3\lambda /\beta$. Interpreting the pair-arrival rate as proportional to population instead would produce a different branching process.

The image of a charge $q$ at height $z$ is $-q$ at height $-z$. Their separation is $2z$, giving force $-q^2/(16\pi {\epsilon }_0{z}^2)$. Starting at rest at height $d$, energy conservation gives ${\dot{z}}^2=q^2(1/z-1/d)/(8\pi {\epsilon }_{0}m)$. Integrating the reciprocal speed to the plane yields the stated finite time in the electrostatic Newtonian model. The induced-conductor potential energy is $-q^2/(16\pi {\epsilon }_{0}z)$; using the full image-pair energy would double the force incorrectly.

Reduced magnetohydrodynamics is the leading ideal-fluid description of small, anisotropic fluctuations about a strong uniform guide field, with $k_{\parallel }/k_{\perp }\sim \delta B_{\perp }/B_0\sim \delta u_{\perp }/v_A\ll 1$. Its solenoidal perpendicular Elsässer variables obeyThe parallel magnetic derivative includes bending of field lines: ${\nabla }_{\parallel }={\partial }_z+(\delta {\mathbf{B}}_{\perp }/B_0)\cdot {\nabla }_{\perp }$. Slow compressive and entropy mode fluctuations are passively transported at this order. The system excludes the rapidly propagating fast magnetosonic wave sector rather than describing all possible MHD perturbations.

The Grad-Shafranov equation governs the poloidal magnetic flux of an axisymmetric magnetohydrostatic equilibrium. Pressure and toroidal-field contributions depend on the flux function. The Grad-Shafranov equation for a force-free magnetic field is its zero-pressure-gradient reduction.

For a stationary, $z$-independent ideal magnetohydrodynamics flow with $E_z=0$, nonzero horizontal magnetic field and nonzero magnetohydrodynamic mass loading $k$, the displayed decomposition has $\mathbf{B}\cdot \nabla k=\mathbf{B}\cdot \nabla U=0$. The continuity equation gives the first invariant, and the ideal magnetohydrodynamic induction equation gives the second. Entropy and $V=u_z-B_z/({\mu }_{0}k)$ are also constant on each magnetic field line. Stationarity alone permits constant nonzero $E_z$, so it does not imply this decomposition.

Magnetic relaxation converts magnetic energy into material motion through the Lorentz force density, and dissipates the resulting kinetic energy through viscosity. In a perfect conductor, magnetic flux freezing preserves field-line connectivity throughout every smooth stage. A regular limiting state has zero velocity and is a magnetostatic equilibrium, but its magnetic field can contain current sheets. Conservation of one global magnetic helicity is weaker than preservation of the complete frozen field-line structure.

Consider a particle of mass $m$ in a rigid rectangular box of length $L$ along $x$ and wall surface area $A$, so $V=AL$ is the volume. Choose $+x$ toward the right wall. For an incoming normal velocity $v_x=v>0$, an elastic collision reverses the normal momentum:The impulse delivered to the wall points outward, toward $+x$, by Newton's third law. The particle's opposite momentum change is not the wall's impulse. A tangential velocity component is unchanged by the ideal smooth-wall collision.

Ignoring intervening particle-particle collisions in this counting model, the particle travels a normal distance $2L$ before hitting the same wall again. Its hit rate is $|v_x|/(2L)$; multiplying by the impulse per hit gives the time-averaged normal force and its contribution to pressure:For example, stipulate $m=4\times 10^{-26}\mathrm{kg}$, $|v_x|=500\mathrm{m}{\mathrm{s}}^{-1}$, $L=0.10\mathrm{m}$, and $A=0.010{\mathrm{m}}^2$. Then $V=0.0010{\mathrm{m}}^3$, the impulse per hit is $4\times 10^{-23}\mathrm{N}\mathrm{s}$, the hit rate is $2500{\mathrm{s}}^{-1}$, and $P_1=10^{-17}\mathrm{Pa}$. This is one particle's tiny time-averaged contribution to one wall's pressure, not an isotropic equilibrium gas by itself.

For $N$ particles of equal mass, add their force contributions and define $⟨v_x^2⟩=N^{-1}{\sum }_{i=1}^N{v}_{x,i}^2$. In the gas's rest frame, isotropy gives $⟨v_x^2⟩=⟨v_y^2⟩=⟨v_z^2⟩=⟨v^2⟩/3$, where $v^2=v_x^2+v_y^2+v_z^2$. HenceThe final equality uses the classical equipartition theorem. The kinetic pressure of an isotropic gas therefore recovers the ideal gas law from momentum transfers. Doubling the normal speed would double both the impulse per hit and the hit rate, giving four times the pressure contribution; pressure measures a mean squared speed. Interparticle elastic collisions redistribute velocities and help establish isotropy; the individual round-trip times need not remain $2L/|v_x|$, but the equilibrium momentum flux result survives. ^[ref]

Parity and valuations at 3 and 7 make the ideals $(y+\sqrt{-21})$ and $(y-\sqrt{-21})$ coprime in any hypothetical solution. Their product is a cube, so the first is an ideal cube. The ideal class group has exponent two, forcing its cube root ideal to be principal. Cubing $a+b\sqrt{-21}$ would then give coefficient $3b(a^2-7b^2)$ on $\sqrt{-21}$, which cannot equal one. This proves the nonexistence of integer solutions without assuming unique element factorization.

The ideal consisting only of zero. It is the principal ideal generated by zero.