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The sphere-plane intersection is governed by the orthogonal projection of the sphere's centre onto the plane. Since $\mathbf{n}$ has unit length, let $h=p-\mathbf{n}\cdot \mathbf{c}$, the signed distance from $\mathbf{c}$ to the plane along $\mathbf{n}$. The following parts determine the centre and radius and retain this sign; the radii alone cannot distinguish the two possible sides of the oriented plane.

Let $\Theta =\angle AOB$ and let the sphere have radius $R$. By rotational invariance, condition on $A$ and place it at the north pole. A band at polar angle $\theta$ has surface area $2\pi R^2\sin \theta d\theta$, while the sphere has area $4\pi R^2$. Thus the central angle between independent uniform sphere points has densitywith density zero outside that interval.

Condition now on $A,B$ with their angle fixed at $\theta$. For $\angle AOC$ and $\angle BOC$ to be acute, the unit position vector of $C$ must have positive dot products with both unit vectors of $A$ and $B$. Hence $C$ must lie in the intersection of the two corresponding open hemispheres. The great circles bounding them meet at the two antipodes perpendicular to the plane of $A,B$. Their common hemisphere region is a spherical lune of opening $\pi -\theta$: when $A=B$ it is a hemisphere of opening $\pi$, and as the two normals separate, that opening decreases by $\theta$. More explicitly, take the common boundary antipodes as poles. The longitude conditions each allow an interval of length $\pi$, with their centres separated by $\theta$, so the intersection has longitude width $\pi -\theta$.

Integrating the spherical area element $R^2\sin \phi d\phi d\lambda$ over that lune gives area $2R^2(\pi -\theta )$. Thus the conditional probability that the two angles involving $C$ are acute is $(\pi -\theta )/(2\pi )$. The remaining angle $\Theta$ must itself be acute, so the desired probability isIntegration by parts gives ${\int }_0^{\pi /2}\theta \sin \theta d\theta =1$, while ${\int }_0^{\pi /2}\sin \theta d\theta =1$. ThereforeThis calculates pairwise acute central angles of three uniform sphere points by conditional area, rather than incorrectly treating the three angle events as independent. Boundary right angles have probability zero, so choosing open rather than closed hemispheres does not change the result.

By orthonormality in the Generalized Bloch representation,A pure state has density operator $\rho =|\psi ⟩⟨\psi |$ with $⟨\psi |\psi ⟩=1$, so ${\rho }^2=\rho$ and its purity of a density operator is one. ConsequentlyEvery pure state therefore maps to the specified radius-$\sqrt{1-1/N}$ sphere in ${\mathbb{R}}^{N^2-1}$, of dimension $N^2-2$. The expansion is unique, so it distinguishes pure-state density operators, while vectors differing only by global phase describe the same point. This proves that pure states lie on the sphere, without asserting that they fill it.

Take any rank-one orthogonal projection $P$ and its generalized Bloch vector $s$. The antipodal vector $-s$ is on the same sphere, but its corresponding trace-one Hermitian matrix isOn the range of $P$, this matrix has eigenvalue $2/N-1$, which is negative for $N>2$. It is therefore not a positive semidefinite matrix and cannot be a density operator. Hence the pure states do not cover the sphere when $N>2$. This also shows directly why the purity bound for generalized Bloch vectors does not by itself guarantee a physical quantum state.

Use the nonnegative Laplacian convention $\tilde{\Delta }=-{\sum }_{j=1}^{d+1}{\partial }_j^2$ and $\Delta =-\mathrm{div}\mathrm{grad}$ on the unit sphere. In polar coordinates the Euclidean Riemannian metric is $d{r}^2+r^2{g}_{S^d}$, its volume form is $r^{d}drd{V}_{S^d}$, and therefore the Laplacian in polar coordinates isConsequently the required restriction identity isFor the opposite sign convention all eigenvalues below change sign, and the restriction identity changes accordingly.

Let ${\mathcal{P}}_n$ be the vector space of degree-$n$ homogeneous polynomials in $d+1$ variables, and let ${\mathcal{H}}_n$ be its subspace of harmonic polynomials. Write $H(r\omega )=r^{n}Y(\omega )$ for $H\in {\mathcal{H}}_n$. Substitution in the Laplacian identity givesThus each restricted harmonic polynomial is a spherical harmonic with eigenvalue ${\lambda }_n=n(n+d-1)$.

To count these spherical harmonics, give polynomials the Fischer inner product $⟨x^{\alpha },x^{\beta }{⟩}_F=\alpha !{\delta }_{\alpha \beta }$. Multiplication by $x_j$ is adjoint to ${\partial }_j$, so multiplication by $r^2$ is adjoint to the ordinary signed Laplacian ${\sum }_j{\partial }_j^2$. In these finite-dimensional inner product spaces, the orthogonal complement of $r^2{\mathcal{P}}_{n-2}$ is exactly ${\mathcal{H}}_n$. Hence the Fischer decomposition isMultiplication by $r^2$ is injective, while restriction of a homogeneous polynomial to the sphere is injective: homogeneity recovers its value at every nonzero point. Counting monomials now gives the eigenvalue multiplicitywhere the second term is zero for $n<2$.

It remains to establish that these are all the eigenvalues and their full multiplicities. Iterating the Fischer decomposition expresses every restricted polynomial as a sum of the restricted harmonic polynomials. Restricted polynomials separate points and contain the constants, so the Stone-Weierstrass theorem makes their span dense in continuous functions on the sphere, hence dense in its L2 space. The Laplace-Beltrami operator is self-adjoint, and the distinct numbers ${\lambda }_n$ strictly increase for $d\ge 1$. Its different eigenspaces are therefore orthogonal. Any additional eigenfunction, or any additional vector in one of these eigenspaces orthogonal to its known spherical harmonics, would be orthogonal to a dense subspace and hence zero. The standard compact elliptic spectral theorem then shows that the spectrum of the round sphere is precisely ${\lambda }_n=n(n+d-1)$ with multiplicity $m_n$. For $S^1$ this gives multiplicity one at zero and two at $n^2$ for $n\ge 1$. If the degenerate case $d=0$ is included, $S^0$ has two points and its Laplacian is zero on a two-dimensional function space.

Work in the laboratory frame, with fluid at rest at infinity and the instantaneous sphere centre at the origin. In the Unscaled Papkovich–Neuber representation, take the harmonic potentialsDifferentiating the two terms gives the translating sphere in Stokes flow:Here $\mathbf{n}=\mathbf{x}/r$. The potentials are harmonic outside the sphere, so they satisfy the Stokes equations. At $r=a$, the two coefficients of $\mathbf{nn}$ cancel and the coefficient of $I$ is one, proving the no-slip boundary condition $\mathbf{u}=\mathbf{V}$; the field decays at infinity. Its leading part is a Stokeslet of force $6\pi \mu a\mathbf{V}$ exerted on the fluid.

Let $\hat{\mathbf{z}}$ point upwards, $\delta =V_1-V_2$, and $D=a_2{V}_2-a_1{V}_1$. Each isolated sphere moves with velocity $-V_i\hat{\mathbf{z}}$. To leading order its correction is the other's incident Stokeslet velocity:The forces generating these fields are the fixed sedimenting loads, so their leading strengths are set by the isolated speeds. Subtraction yields the leading interaction of two sedimenting spheres:Since $\hat{\mathbf{z}}=\cos \theta \hat{\mathbf{r}}-\sin \theta \hat{\mathbf{\theta }}$, this reduces toThe stipulated higher-order errors can be appended to both equations; they do not enter this leading calculation.

For the same-sign case, interchange labels if necessary to arrange $\delta >0,D>0$, and put $c=3D/4$. The vertical separation $z=r\cos \theta$ satisfiesIf sphere one is initially above sphere two, its vertical order reverses; in every case $z\to -\infty$, implying $r\to \infty$. For a noncollinear trajectory there is no collision in the point-force equations. Indeed, divide the two polar equations and integrate:At side-by-side passage, $\theta =\pi /2$, the closest separation is the positive root of $\delta r^2+2cr=C$. Before passage $\dot{r}<0$ and after passage $\dot{r}>0$, proving overtaking and unbounded subsequent separation. This passing invariant for unequal point-force spheres describes a physical far-field passage only when the closest separation also greatly exceeds the sphere radii. Exactly collinear approach instead reaches the near-contact regime, where this leading approximation cannot describe two finite spheres passing through one another.

For opposite signs, choose $\delta >0,D<0$ and consider the invariant configuration with sphere one vertically above sphere two, $\theta =0$. The radial equation isIf $r>r_{\ast }$, separation decreases; if $r<r_{\ast }$, it increases. Thus it approaches this equilibrium along the vertical line. Linearization gives $\dot{\delta r}=-(\delta /r_{\ast })\delta r$. But a small tilt obeys $\dot{\theta }=(\delta /(2r_{\ast }))\theta$, so the vertical bound pair in the point-force sedimentation model is stable only against vertical perturbations and is a saddle in the full separation dynamics. The opposite vertical orientation has the reversed stability. Reversing gravity reverses both $\delta$ and $D$, leaves $r_{\ast }$ unchanged, and reverses every trajectory; radial attraction becomes repulsion. This is fully consistent with kinematic reversibility of Stokes flow.

There is an important restriction in the printed data. Its hierarchy $|D|\ll (a_1+a_2)|\delta |$ implies $r_{\ast }\ll a_1+a_2$, so the proposed equilibrium would overlap the physical spheres and lie outside the well-separated approximation. With that hierarchy there is no justified physical far-field bound pair. The intended far-field equilibrium instead requires the first inequality reversed, $|D|\gg (a_1+a_2)|\delta |$. The formal equilibrium and its vertical stability above show the intended mechanism, while also identifying the missing validity and transverse-stability qualifications.

The leading disturbance in sphere Stokes flow has size $u\sim Ua/r$. Consequently viscous diffusion has size $\nu u/r^2$, background advection has size $Uu/r$, and their ratio is $Ur/\nu$. Even if the sphere Reynolds number $Ua/\nu$ is small, this ratio becomes order one at $r\sim \nu /U$. Thus the Stokes approximation is nonuniform at distances of order $\nu /U$. At that outer distance the disturbance itself is small relative to $U$, so self-convection is smaller than background advection and the Oseen approximation is appropriate:Write $\mathbf{u}=\nabla \varphi +\nu \nabla \chi -\mathbf{U}\chi$. It is sufficient outside the origin to impose $\Delta \varphi =0$ and $(\nu \Delta -\mathbf{U}\cdot \nabla )\chi =0$: these give zero divergence, and substitution into momentum gives $p=-\rho \mathbf{U}\cdot \nabla \varphi$. With $\chi =h{e}^{\mathbf{U}\cdot \mathbf{x}/(2\nu )}$, differentiation yieldsA decaying radial choice is $h=C{e}^{-Ur/(2\nu )}/r$. ThereforeThe exponent is nonpositive, so this scalar decays at infinity, including as $1/r$ directly downstream. In the overlap $a\ll r\ll \nu /U$,Choosing $\varphi =-\nu C/r$ cancels the first term's gradient in $\nabla \varphi +\nu \nabla \chi$. The remaining leading velocity isMatching fixes $C=3a/2$. Thus the potential-source and wake decomposition of sphere Oseen flow isIts pressure is $p=-3\mu a(\mathbf{U}\cdot \mathbf{x})/(2r^3)$, the sign appropriate to a fixed sphere in an incident flow. The velocity decays in every direction and matches the requested leading inner disturbance; the sphere's potential-dipole term enters a higher matching order.

The scalar potential is a point source: $\nabla \varphi =(3a\nu /2)\mathbf{x}/r^3$, so integration over a centred sphere givesTo see the fluid wake, take $z$ downstream along $\mathbf{U}$ and transverse radius $R_{\perp }$. For $z\gg \nu /U$, $r-z\simeq R_{\perp }^2/(2z)$, givingThe fluid wake is concentrated within $R_{\perp }=O(\sqrt{\nu z/U})$. Its integrated mass deficit is $\rho U\int \chi dA\simeq 6\pi \mu a$, balancing the outward potential-source mass flux. The corresponding momentum deficit iswhich equals the leading sphere drag. The Gaussian integral used here is ${\int }_0^{\infty }{e}^{-U{R}_{\perp }^2/(4\nu z)}2\pi R_{\perp }d{R}_{\perp }=4\pi \nu z/U$.

The PDF labels $6\pi \mu aU$ as the mass-source strength, but this has dimensions of force. The actual mass-source strength is $6\pi \mu a$; $6\pi \mu aU$ is the momentum deficit or drag. The decomposition above supplies both quantities and identifies the printed normalization error.