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For a sphere of radius $a$, the scattering efficiency is its effective scattering area divided by the projected geometric cross-section $\pi a^2$. It is a dimensionless quantity. For Rayleigh scattering by a small sphere with relative refractive index $m$ and incident wavenumber $k$,At fixed wavelength and fixed refractive index, the Rayleigh scattering cross-section scales as $a^6$, whereas the scattering efficiency scales as $a^4$. Doubling radius multiplies the former by $64$ and the latter by $16$. An efficiency here is an effective scattering area relative to a reference area, rather than a fraction of all light in an arbitrary beam.

For a worked toy example, take $a=10\mathrm{nm}$, $m=1.5$, and $\lambda =550\mathrm{nm}$. Then $ka=2\pi (10/550)\simeq 0.1142$ and $(m^2-1)/(m^2+2)=5/17$. Substitution gives Rayleigh scattering cross-section $\sigma \simeq 1.234\times 10^{-20}{\mathrm{m}}^2$ and scattering efficiency $Q_{\mathrm{sca}}\simeq 3.929\times 10^{-5}$. At incident light intensity $1\mathrm{W}{\mathrm{m}}^{-2}$, the scattered power is approximately $1.234\times 10^{-20}\mathrm{W}$. These are leading small-particle estimates, not a measured air-molecule cross-section.

For an oriented rank-$k$ bundle over $S^n$, with $k\ge 2$, normalize its clutching function to be based at the identity. A nowhere-zero section exists precisely when the displayed homotopy class vanishes. If it vanishes, extend $\gamma e_k$ across one hemisphere and use the constant section $e_k$ on the other. Conversely normalize any nowhere-zero section using the fiber metric. Its two hemisphere maps satisfy ${\sigma }_{-}=\gamma {\sigma }_{+}$ on the boundary. The first hemisphere makes ${\sigma }_{+}$ homotopic to a constant; the second makes $\gamma {\sigma }_{+}$ null-homotopic. Thus $\gamma e_k$ is null-homotopic. If $k>n$, the lower homotopy group of the sphere vanishes and a section always exists. Rank one is already trivial because $SO(1)$ is trivial.

The sedimentation coefficient $s$ is radial drift velocity divided by local centrifugal acceleration. It has units of time and combines particle properties with those of the surrounding liquid. In a dilute, deterministic mathematical model with negligible particle inertia, an effective displaced volume $V$, particle mass $m$, local liquid density ${\rho }_f$ and linear friction coefficient $f>0$ giveThe subtraction represents buoyancy. Equal mass does not imply equal $s$: changing shape changes drag force, and changing the medium changes both buoyancy and viscosity.^[ref]

For an isolated sphere of radius $b$ and uniform density ${\rho }_p$, Stokes's law supplies $f=6\pi \eta b$, where $\eta$ is dynamic viscosity, and $m=(4\pi /3)b^3{\rho }_p$. ThereforeThis original force-balance check predicts fourfold $s$ on doubling $b$ with everything else fixed, and zero drift when ${\rho }_p={\rho }_f$. It applies to dilute isolated spheres in Stokes flow, not arbitrary folded proteins or crowded suspensions. If $s$ and $\omega$ remain constant in a uniform medium, the ordinary differential equation $dr/dt=s{\omega }^{2}r$ has the exponential function solution $r(t)=r(0)e^{s{\omega }^{2}t}$. Arrival at an outer boundary therefore depends on starting position as well as $s$; a separation has no universal size cutoff.

A sphere of radius $R$ settling in a stationary cylindrical tube of radius $R+d$, with $d\ll R$ and zero net through-flow, has leading speedThe neck gap is $H(z)=d+z^2/(2R)$. In the sphere frame, leading bypass flux per circumferential length is $q\simeq UR/2$; wall-motion transport is smaller by $d/R$. The pressure-driven flux through a parabolic lubrication gap gives $\Delta p=9\pi \mu UR\sqrt{2R}/(4d^{5/2})$. Pressure dominates the sphere drag, $F\simeq \pi R^2\Delta p$, and balancing $F$ against $4\pi R^3\Delta \rho g/3$ proves the speed. Tube end conditions allowing appreciable net flow change the bypass balance and can give other drag regimes.

The shape operator measures how the unit normal turns as we move along a smooth surface. A plane has a constant unit normal, whereas the outward unit normal of a sphere turns in every direction in the tangent plane. Keeping the direction of this change, rather than just its magnitude, lets one recover the principal curvatures and their principal directions.

Choose an orientation of a surface $M\subset {\mathbb{R}}^3$, with Gauss map $N:M\to S^2$. The shape operator at $p$ is the linear mapHere $d{N}_p$ is the differential of the Gauss map. Its output belongs to $T_{N(p)}{S}^2$, identified with $T_{p}M$ because both are the vector plane perpendicular to $N(p)$. The minus sign is our convention throughout this branch. ^[ref]

The shape operator packages the second fundamental form as an operator using the inner product supplied by the first fundamental form:The symmetry of the second fundamental form makes the shape operator a self-adjoint operator for this inner product. Thus an orthonormal basis of principal directions diagonalizes it. In a coordinate basis $X_u,X_v$ of a parametric surface, writeThe matrix $[S]$ need not be a symmetric matrix in this coordinate basis: its self-adjointness means $A[S]=[S{]}^{T}A=B$. Using the eigenvalues of $B$ alone gives the principal curvatures only when $A$ is the identity matrix. ^[ref]

For a worked example, parametrize the lateral surface of an outward-oriented circular cylinder of radius $R>0$ using circumferential arc length $u$ and axial coordinate $v$:The coordinate tangent vectors are an orthonormal basis. Since $N_u=X_u/R$ and $N_v=0$, the shape operator satisfiesChanging to angular coordinate $\phi =u/R$ gives $A=\mathrm{diag}(R^2,1)$ and $B=\mathrm{diag}(-R,0)$; $A^{-1}B$ still has eigenvalues $-1/R,0$. This is why the first fundamental form must participate in the calculation.

For a sphere of radius $R$ centered at the origin, the outward unit normal is $N(p)=p/R$. Hence $S_{p}v=-v/R$, so both principal curvatures are $-1/R$. For a plane, $dN=0$ and the shape operator vanishes. Reversing the unit normal negates the shape operator and both principal curvatures; the Gaussian curvature remains unchanged while the mean curvature changes sign, as in orientation reversal of surface curvature.

For standard unknotted spheres contained in four-balls in $X$ and $Y$, the standard fiber-reversing gluing gives the displayed connected sum. The complement of an unknotted $S^2\times D^2$ in $S^4$ is $S^1\times D^3$, and its double is $S^1\times S^3$. The portions of $X$ and $Y$ outside the four-balls remain as connected summands. Taking both ambient manifolds to be the Complex projective plane gives a smooth fiber sum with no symplectic form, by the symplectic connected-sum obstruction.

A smooth complex degree-two projective plane curve is a projective linear image of the degree-two Veronese map and is diffeomorphic to a sphere. Its class in ${\mathbb{CP}}^2$ is twice the class of a projective line. Its self-intersection number is four, so its oriented normal bundle has Euler class evaluating to four.

For the positive Laplace-Beltrami operator on the unit sphere $S^d$, with $d\ge 1$, the eigenvalues are $n(n+d-1)$ for integers $n\ge 0$. Their eigenspaces consist of restrictions of degree-$n$ harmonic polynomials and have dimensions $\left(\genfrac{}{}{0ex}{}{n+d}{d}\right)-\left(\genfrac{}{}{0ex}{}{n+d-2}{d}\right)$, with the second term zero for $n<2$. The Laplacian in polar coordinates gives the eigenvalue; Fischer decomposition and the Stone-Weierstrass theorem prove completeness.

For a positively oriented basis $\mathbf{a},\mathbf{b},\mathbf{c}$, put $\tau =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. The center of an interior sphere at distance $r$ from each of the three tetrahedron face planes through the origin has the displayed expression. The three inward unit vectors are the normalized cross products $\mathbf{b}\times \mathbf{c}$, $\mathbf{c}\times \mathbf{a}$, $\mathbf{a}\times \mathbf{b}$. Taking inner products with these vectors solves three independent signed-distance equations. Reversing some signs instead produces centers on exterior sides. Existence of a sphere contained in the full tetrahedron also requires clearance from the fourth face.

For a sphere falling with speed $U$ through a tube of radius $a$, let $Q$ be the downward laboratory volume flux. In the sphere frame the wall moves upwards and the narrow annulus carries upward flux per unit circumference $q=(\pi a^{2}U-Q)/(2\pi a)$. Couette-Poiseuille flow in a thin gap gives $q=Uh/2-h^3{p}_z/(12\mu )$. Thus the pressure below minus above the sphere is $\Delta P=6\mu (2q{I}_3-U{I}_2)$, using lubrication resistance integrals for a nearly occluding sphere. A nearly piston-like laboratory flux can coexist with a nonzero relative gap flux.

In a normed vector space, this set consists of all points at a fixed positive norm distance $r$ from $x_0$. It also makes sense in an infinite-dimensional Hilbert space, unlike a specifically Euclidean sphere. In a real Hilbert space, the sphere in a normed vector space $\Vert u{\Vert }^2=2E_0>0$ has tangent variations $v$ satisfying $(u,v)=0$; this follows by differentiating the fixed squared norm. This is the geometric constraint used in fixed-energy initial-condition optimality.

Arrangement of spheres without interior overlap. The occupied volume fraction is its packing fraction; it depends on the spatial arrangement and on the radii.

Suppose disjoint separating spheres cut a 3-manifold into pieces none of which is a punctured three-sphere. Compress one sphere along a disk disjoint from the others. The old sphere and the two new spheres enclose a three-times punctured sphere. At least one of the two pieces on the compressed side is not a punctured sphere; retain that new sphere and discard the other. The piece on its opposite side cannot become a punctured sphere, since capping its boundaries and applying the Three-dimensional Schoenflies theorem would make the original opposite piece a punctured sphere as well. Hence the number of spheres and the nontriviality condition can be preserved.