Use lattice paths with $k$ north steps and $n-k$ east steps, beginning at $(0,0)$. Let $A(P)$ be the sum of the horizontal coordinates of its north steps, equivalently the number of unit cells to their left inside the $k$-by-$(n-k)$ rectangle. A north step in zero-based word position $j_i$ has $i-1$ earlier north steps, so its horizontal coordinate is $j_i-(i-1)$. The Gaussian binomial coefficient recurrence therefore counts these lattice paths with weight $t^{A(P)}$.

For $n=4,k=2$, the words NNEE, NENE, NEEN, ENNE, ENEN, EENN have respective areas $0,1,2,2,3,4$. The word NEEN contributes $t^2$ because the two north steps occur at horizontal coordinates zero and two. These paths never move south or west; allowing such extra steps would change the collection being counted.

For $n\ge 1$ and $1\le k\le n$, the displayed recurrence relation uses $G_{n,0}(t)=1$ and $G_{n,k}(t)=0$ when $k<0$ or $k>n$. At $t=1$ it becomes Pascal's identity.

An exact weighted-subset formula isThe exponent measures how far the selected positions lie beyond the earliest possible choices $0,\dots ,k-1$. Split the subsets according to whether they contain $n-1$. Those without it give $G_{n-1,k}(t)$; deleting it from those with it removes exactly $n-k$ from their exponent. This gives the displayed recurrence relation, explaining why it builds polynomials with nonnegative integer coefficients.

The Rayleigh range of a fundamental Gaussian beam in vacuum is the positive axial distance from the beam waist at which its radius grows from $w_0$ to $\sqrt{2}{w}_0$. Thus its contour area doubles and its peak irradiance halves. Within $|z|\le z_R$, the radius stays at most $\sqrt{2}{w}_0$; for $|z|\gg z_R$, beam divergence dominates. At fixed vacuum wavelength, the Rayleigh range scales quadratically with the beam waist radius. The formula here uses the vacuum wavelength, not an unlabelled wavelength in a material.

The beam waist is the narrowest cross-section of a freely propagating Gaussian beam. Its radius $w_0$ uses the wave amplitude $1/e$, or light intensity $1/e^2$, convention; its diameter is $2w_0$. Specifying a “spot size” without saying radius or diameter can introduce a factor-of-two error. Within the scalar paraxial approximation, the wavefront at the beam waist is planar. The minimum radius sets both the Rayleigh range and the far-field beam divergence.

For uniform charge density ${\rho }_e=3Q/(4\pi R^3)$, Gauss's law gives the interior electric field $E={\rho }_{e}r/(3{\epsilon }_0)$. Integrate its vertical force density ${\rho }_e^{2}z/(3{\epsilon }_0)$ over the upper hemisphere: $\int zdV=\pi R^4/4$. Internal pair forces in that hemisphere cancel, so its net force is exactly the force from the lower hemisphere. The result is upward and has magnitude $3Q^2/(64\pi {\epsilon }_0{R}^2)$.

In static time times round three-dimensional hyperbolic space of curvature radius $R$, a geodesic sphere has area $4\pi R^2{\sinh }^2(\xi /R)$. The Gauss law therefore gives the displayed radial field. It approaches the flat-space inverse-square field near the charge and falls exponentially at large geodesic distance. The curvature changes the area of a sphere, not the conserved flux.

Place the same positive electric charge $Q$ uniformly on the sphere of radius $R$, with no external field. Every concentric interior Gaussian surface encloses zero charge; spherical symmetry then makes the interior electric field zero. Outside, it is again $Q/(4\pi {\epsilon }_0{r}^2)$. At $R$, the inner and outer limits differ by $Q/(4\pi {\epsilon }_0{R}^2)=\sigma /{\epsilon }_0$, with surface charge density $\sigma =Q/(4\pi R^2)$. The ideal surface charge makes the field discontinuous, so the plot displays one-sided limits rather than assigning a single field value on the charged sheet. An isolated charged conducting sphere in electrostatic equilibrium realizes this exterior/interior pattern.

Distribute a positive electric charge $Q$ uniformly through a closed ball of radius $R$ in vacuum, with no external field. Its volume charge density is $3Q/(4\pi R^3)$. For $0<r<R$, the enclosed fraction is $(r/R{)}^3$. Applying Gauss's law on a concentric Gaussian surface givesThe electric field grows linearly inside and decays as an inverse square outside. With $E_R=Q/(4\pi {\epsilon }_0{R}^2)$, its values at $r/R=1/2,1,2$ are $E_R/2,E_R,E_R/4$. The field vanishes at the centre by spherical symmetry and is continuous at the surface because this ideal volume distribution has no singular surface layer. Its interior behavior differs from a conductor in electrostatic equilibrium, where mobile charge redistributes.

For a minimal surface, the shape operator has eigenvalues $k,-k$, so $⟨dN(v),dN(w)⟩=k^2⟨v,w⟩$. Conversely this metric identity only implies that the two principal curvatures have equal squares. A sphere has equal nonzero principal curvatures and satisfies the identity while having nonzero mean curvature. The distinction is between opposite and equal signs, not equality of magnitudes.

Radially project a convex polyhedron from an interior point onto the unit sphere and triangulate each face without adding vertices. There are $T=2E-2F$ spherical triangles. The Gauss-Bonnet theorem gives each area as angle sum minus $\pi$. Summing angles around the projected vertices gives $2\pi V$, while the total area is $4\pi$. Thus $4\pi =2\pi V-\pi T$, proving the formula. Convexity guarantees that the projected triangles cover the sphere without overlaps of interiors.

For the conformal metric with $g_{0i}=-a^2{\partial }_{i}B$ and trace-free scalar spatial shear $E_{ij}=({\partial }_i{\partial }_j-{\delta }_{ij}{\nabla }^2/3)E$, the passive coordinate change $\tilde{\eta }=\eta +T$, ${\tilde{x}}^i=x^i+{\partial }^{i}L$ gives the displayed rules and $\tilde{\varphi }=\varphi +\mathcal{H}T+{\nabla }^{2}L/3$. The combination $C=\varphi +{\nabla }^{2}E/3$ changes by $\mathcal{H}T$, while $B+v$ changes by $T$. Hence $\mathcal{R}=-C+\mathcal{H}(B+v)$ is the comoving curvature perturbation in this sign convention.

For a nonzero Fourier mode in synchronous gauge, write $h_{ij}=h{\delta }_{ij}/3+h_s({\hat{k}}_i{\hat{k}}_j-{\delta }_{ij}/3)$. Under the passive displacement ${\xi }^{\mu }=(T,i{\hat{k}}_{i}L)e^{i\mathbf{k}\cdot \mathbf{x}}$, cancellation of spatial scalar shear requires $h_s+2kL=0$, and cancellation of the shift requires $kT-L^{\prime }=0$. The resulting cosmological Newtonian potential is $\Phi =(h_s^{\prime \prime }+\mathcal{H}{h}_s^{\prime })/(2k^2)$ and the spatial potential is $\Psi =-(h-h_s)/6-\mathcal{H}{h}_s^{\prime }/(2k^2)$. Both remain unchanged under residual synchronous-gauge freedom, for which $T^{\prime }+\mathcal{H}T=0$ and $L^{\prime }=kT$.

A smooth finite-energy monopole in an $SU(2)$ Yang-Mills theory with an adjoint Higgs field uses the normalized Higgs direction at infinity to define a map $S_{\infty }^2\to S^2$. Its nonzero degree supplies the magnetic topological sector. Unlike a singular Abelian Dirac monopole, the non-Abelian gauge and Higgs fields can have a regular core. In the zero-potential limit the Bogomolny-Prasad-Sommerfield monopole equations are $B_i=\pm D_i\Phi$.

Localizing a parameter of a global symmetry produces its Noether current times a derivative of that parameter. Introduce a gauge field with an inhomogeneous transformation and couple it to the Noether current to cancel this variation. Recompute the remaining variation and add the required interaction and transformation terms. Closure of the local symmetry algebra constrains the completion. For supersymmetry, its supercurrent couples to the gravitino, and the translation in the supersymmetry algebra requires a vierbein.

A holomorphic gauge coupling is the coefficient of the gauge kinetic F-term in a local Wilsonian effective action written in holomorphic variables. In a common normalization its real part is $1/g_h^2$. The continuous perturbative shift of its topological angle, together with holomorphy, restricts its renormalization-group beta function to one-loop order. Wave-function renormalization and anomalous canonical field rescalings distinguish it from the physical gauge coupling.

// Target: supersymmetry.bigb

With the reversed convention in curvature-sign convention in Killing derivative identities, choose the representative $h_{00}=2\Phi$, $h_{0i}=V_i$, $h_{ij}=2\Psi {\delta }_{ij}+h_{ij}^{\mathrm{TT}}$. The linearized Einstein components are $G_{00}^{(1)}=-2\Delta \Psi$, $G_{0i}^{(1)}=-2{\partial }_i\dot{\Psi }-\Delta V_i/2$ and $G_{ij}^{(1)}=({\partial }_i{\partial }_j-{\delta }_{ij}\Delta )(\Phi -\Psi )-2{\delta }_{ij}\ddot{\Psi }-({\partial }_i{\dot{V}}_j+{\partial }_j{\dot{V}}_i)/2+\square h_{ij}^{\mathrm{TT}}/2$. On all space with regular decaying fields, or for nonzero spatial Fourier modes, the vacuum constraints remove the scalar and vector parts, leaving two radiative gravitational wave polarizations. Singular Coulomb fields, spatial boundaries and homogeneous modes are not covered by that removal argument.

For a matrix-valued gauge field $A$, let $F=dA+A\wedge A$ be its gauge field strength; products below combine matrix multiplication with the wedge product of differential forms. The five-form can be written $Q_5=\mathrm{Tr}(A(dA{)}^2+\frac{3}{2}{A}^{3}dA+\frac{3}{5}{A}^5)$. Applying the exterior derivative, using its graded product rule and the graded cyclic property of the matrix trace, gives $d{Q}_5=\mathrm{Tr}(F^3)$. The six-form is an invariant polynomial in the gauge curvature, while its local potential depends on the connection.

The Euler--Mascheroni constant is

On $(0,\infty )$, the Gamma function is the unique positive function $f$ satisfying $f(1)=1$, $f(x+1)=xf(x)$, and the condition that $\log f$ is a convex function. This last condition concerns the natural logarithm of $f$, rather than $f$ itself, and removes the interpolation freedom left by the Gamma function recurrence.^[ref]

Rotating the integral defining the Gamma function toward the positive imaginary axis introduces the factor $i{e}^{-\pi n/2}$. After scaling by $n$, the remaining phase is $\log s-s$, whose unique stationary point is $s=1$ with second derivative $-1$. The stationary phase method gives $e^{-in-i\pi /4}\sqrt{2\pi /n}$ for this scaled integral, producing the displayed asymptotic. The boundary contour integral must be Abel-regularized; its off-saddle tails are smaller by integration by parts.