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An antipodal simplicial map from a regular antipodal triangulation of a sphere $S^k$ to $F^n$ has an odd number of positive alternating $k$-simplices. For each $k$-simplex, the number of positive alternating facets is odd exactly when the simplex is positive or negative alternating; otherwise it is even, including collapsed images. Count these incidences in one hemisphere: interior facets contribute twice, boundary facets once. Antipodal pairing then gives $p_k\equiv p_{k-1}$, starting with one positive vertex on $S^0$. In particular, no such map exists when $k>n$.

When both endpoint bundles have the filled circular-cone geometry of etendue of a circular aperture and cone, conservation of etendue givesFor the same refractive index, quadrupling the area halves the sine of the cone half-angle. In the $10\mathrm{mm}$, $30^{\circ }$ example, an ideal output of diameter $20\mathrm{mm}$ must have $\sin {\theta }_2=1/4$, so ${\theta }_2\approx 14.4775^{\circ }$. The exact result differs from simply halving $30^{\circ }$. Conversely, reducing the diameter to $5\mathrm{mm}$ requires a complete forward hemisphere, ${\theta }_2=90^{\circ }$, as a limiting case.

Equal bounding area and solid angle do not alone prove equal etendue: an actual output can have unoccupied directions or position-direction correlations. The filled-cone equation is an exact model of the specified bundles and a capacity constraint on an optical instrument, not a guarantee of an aberration-free implementation.

Join three independent uniform points on a sphere when their central angle is acute. Given angle $\gamma$ between the first pair, the third point's two favorable hemispheres overlap in a spherical lune of area fraction $(\pi -\gamma )/(2\pi )$. If $\gamma <\pi /2$, connection requires membership in their union, of probability $1/2+\gamma /(2\pi )$. If $\gamma >\pi /2$, it requires membership in their intersection, of probability $1/2-\gamma /(2\pi )$. Integrating these against the central-angle density $\sin \gamma /2$ proves the displayed probability. Connectivity is weaker than all three pairwise angles being acute.

The probability that $N\ge 1$ independent uniformly oriented open hemispheres cover a sphere is the displayed quantity. Coverage fails exactly when their normals lie in a closed hemisphere. With probability one, no three normals are coplanar and no pair is antipodal; a closed hemisphere containing the normals can then be perturbed into an open one. Condition on the unoriented axes of the normals. Their $N$ orientation signs remain independent fair bits, and the regions of a general-position great-circle arrangement realize exactly $N(N-1)+2$ of the $2^N$ possible sign vectors. Each such vector corresponds to normals that fit into one open hemisphere, proving the complementary probability. The argument counts equiprobable orientation choices, not equal-area regions.

A diffuse thermal emitter has emitted radiance independent of outgoing direction over the outward hemisphere. Its emitted radiant intensity from a planar patch still varies with viewing angle through projected area, as described by Lambert's cosine law. Diffuse emission does not mean equal power per solid angle from the patch. This approximation does not by itself impose a grey body spectrum or specify how incident radiation is reflected.

Suppose an extended source fills every position of the entrance aperture with a cone of half-angle ${\theta }_{\mathrm{in}}$, and all these light rays must reach a one-sided planar receiver. If the receiver accepts at most a cone of half-angle ${\theta }_{\mathrm{out}}$, its available etendue bounds the admitted etendue:ConsequentlyThis original derivation combines conservation of etendue with the exact cone integral; the second inequality allows the complete output hemisphere. It bounds capacity and does not prove that a chosen shape attains the bound.

For the $30^{\circ }$ input cone with $n_{\mathrm{in}}=n_{\mathrm{out}}=1$, $C\le 4$: a $10\mathrm{mm}$ entrance cannot deliver the entire filled bundle to a circular receiver smaller than $5\mathrm{mm}$ in diameter. If the output half-angle is restricted to $60^{\circ }$, $C\le 3$ and the receiver diameter must be at least $10/\sqrt{3}\approx 5.7735\mathrm{mm}$. For a receiver medium with model refractive index $n_{\mathrm{out}}=1.5$, the full-hemisphere capacity bound instead gives $C\le 9$; that larger value also depends on transmitting the bundle into that medium.

Attempting $C=10$ in the air-to-air example forces rejection of some input. For the uniform radiance already specified, even filling the entire output hemisphere givesThe maximum mean irradiance gain remains $\eta C\le 4$. With nonuniform radiance, rejecting a given fraction of geometric etendue need not reject the same fraction of radiant flux.

These bounds use a filled extended-source bundle in passive reversible geometrical optics, comparing the same wavelength and the same radiation channels. An ideal optical point source does not fill that four-dimensional set. Diffraction, optical coherence, or coupling a Gaussian beam into an optical fiber require a physical optics calculation; the cone bound is not a universal guarantee for a single wave mode. Likewise, optical gain or fluorescence changes the propagation assumptions.

Consider a planar circular aperture of diameter $D$ in constant refractive index $n$. Suppose every position in its area $A=\pi D^2/4$ is filled with directions inside the same rotationally symmetric cone about the unit normal, with half-angle ${\theta }_{\max }$. Using spherical coordinates, the exact etendue iswhere $\mathrm{NA}=n\sin {\theta }_{\max }$ is the appropriate side's numerical aperture. The ordinary cone solid angle is $2\pi (1-\cos {\theta }_{\max })$; its cosine-weighted value is $\pi {\sin }^2{\theta }_{\max }$. A full forward hemisphere therefore gives $G=n^{2}A\pi$, rather than $2\pi n^{2}A$. ^[ref]

For an original model example, choose $n=1$, $D=10\mathrm{mm}$, and ${\theta }_{\max }=30^{\circ }$. ThenIf the radiance is uniform across this filled bundle and equals $L=100\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$, its radiant flux is $\Phi =LG/n^2\approx 6.1685\mathrm{mW}$. Neither the geometry nor the chosen radiance is an experimental measurement.

The Geiger-Marsden experiment probes the interior of a chemical atom by counting alpha particles deflected by metal foils. The surprising observation was a small population returning towards the incident side. Hans Geiger and Ernest Marsden performed the experiments under Ernest Rutherford's direction; Ernest Rutherford developed the Rutherford model in 1911. Observation and interpretation were distinct contributions.^[ref]

In 1909, a radon source and its radioactive decay products in a conical glass tube supplied alpha particles through a mica window towards a reflector. Lead shielded a fixed zinc sulfide screen from direct hits; an optical microscope resolved individual particle scintillations. Reflectors occupied the same position. Reflector-out counts tested air scattering; stacked gold foils and bare glass tested thickness and support contributions.^[ref]

Nearby source geometry blurred incidence angles. Extrapolating platinum counts uniformly over a hemisphere was an assumption, not a universal reflection probability.^[ref]

For a closed hemisphere of radius $a>0$, take a rectangular parallelepiped with a face on the flat base, base half-widths $x,y>0$ and height $z>0$. Its curved-boundary constraint is $x^2+y^2+z^2=a^2$ and its volume is $4xyz$. A Lagrange multiplier gives $x=y=z=a/\sqrt{3}$, yielding the displayed maximum. The full edge lengths are $2a/\sqrt{3},2a/\sqrt{3},a/\sqrt{3}$.

This bound remains valid when arbitrary orientations are allowed. Write a cuboid's half-edge lengths as $u_1,u_2,u_3$, its center components in its orthonormal edge basis as $v_i$, and the vertical unit-vector components as $n_i$. Containment above the base plane requires $\sum n_i{v}_i\ge \sum |n_i|u_i$. Hence $|v_j|\ge u_j$ for at least one $j$: otherwise every contributing term would be strictly smaller than its required bound. Its farthest vertex has squared distanceContainment in the sphere bounds the right side by $a^2$. The arithmetic-geometric mean inequality then gives $2u_1{u}_2{u}_3\le a^3/(3\sqrt{3})$, so the full volume $8u_1{u}_2{u}_3$ cannot exceed the displayed value. The constructed cuboid attains it. With strict containment and no boundary contact, the same value is a supremum.

For three independent uniform points on a sphere, condition on the central angle $\theta$ between the first two. The third point makes acute angles with both exactly when it lies in the intersection of their positive hemispheres. That intersection is a spherical lune of opening $\pi -\theta$, hence has probability $(\pi -\theta )/(2\pi )$. The first two points also need $\theta <\pi /2$. Integrating against the central angle between independent uniform sphere points density gives $(4\pi {)}^{-1}{\int }_0^{\pi /2}(\pi -\theta )\sin \theta d\theta =(\pi -1)/(4\pi )$.

By rotational invariance, fix the first landing point at the north pole of a sphere of radius $R$. A band of central angle $\theta$ and width $d\theta$ has area $2\pi R^2\sin \theta d\theta$. Dividing by total area gives the central angle between independent uniform sphere points densityand zero outside this range. The radius cancels.

Given the first pair's central angle $\gamma$, the third point can communicate with each of them exactly when it lies in that point's positive hemisphere. Choose their intersection line as a polar axis. Each hemisphere permits an azimuth interval of length $\pi$, and the intervals' centres are separated by $\gamma$. Their overlap is a spherical lune of opening $\pi -\gamma$, with area $2R^2(\pi -\gamma )$. ThereforeThe union probability follows from inclusion-exclusion for two hemispheres:Thus the requested conditional answers are“Either” here is inclusive: a third ship linked to both is also linked to at least one.

For network connectivity, if $\gamma <\pi /2$ the first pair is already linked, and the third must link to at least one of them. If $\gamma >\pi /2$, the first pair is not linked, and both must link through the third. Consequently the connectivity of three acute-angle sphere points isUsing ${\int }_0^{\pi /2}\gamma \sin \gamma d\gamma =1$ and ${\int }_{\pi /2}^{\pi }\gamma \sin \gamma d\gamma =\pi -1$ givesAngles exactly equal to the threshold have probability zero. Connectivity allows relaying and is weaker than requiring direct links between every pair.

Because the events are pairwise disjoint events, their count can only be zero or one. The event that the count is zero is the complement of an event of their union. Finite additivity therefore gives

Now use $N\ge 1$ for the fixed number of independently located ships. Normalize the radius to one and write their positions as unit vectors $a_1,\dots ,a_N$. A point $x$ on the sphere is reached by ship $i$ precisely when $a_i\cdot x>0$, so each ship reaches an open hemisphere. Coverage fails exactly when there is a unit vector $x$ with $a_i\cdot x\le 0$ for every $i$. Equivalently, all ship positions lie in some closed hemisphere.

With probability one no pair of ship vectors is antipodal and no three are linearly dependent: the exceptional choices lie on sets of zero surface area. For such a configuration, containment in a closed hemisphere implies containment in an open hemisphere. To see this, choose its inward normal $u$, so $u\cdot a_i\ge 0$. At most two vectors lie on its boundary. If there are two, their sum has positive dot product with each, since they are not antipodal; if there is one, use that vector itself. Perturb $u$ slightly in this direction. The boundary dot products become positive while all already positive dot products remain positive. Thus the strict and non-strict containment conditions have the same probability here.

Condition on the unoriented axes $\{a_i,-a_i\}$. Choose representatives $b_i$ by any fixed sign convention. By uniform sampling and independence, the actual positions are $a_i={\epsilon }_i{b}_i$ with $N$ independent fair signs ${\epsilon }_i\in \{-1,1\}$. Conditional on these axes, each of the $2^N$ sign choices has probability $2^{-N}$.

The great circles $b_i\cdot u=0$ partition the sphere into $R_N=N(N-1)+2$ open regions. In each region the sign vector $\sigma (u)=(\mathrm{sgn}(b_i\cdot u){)}_i$ is constant. No sign vector labels two different regions: its strict inequalities define a convex cone, and normalized line segments within that cone connect any two points of its spherical section. The regions of a general-position great-circle arrangement therefore yield exactly $R_N$ realized sign vectors.

The chosen positions lie in an open hemisphere if and only if some $u$ has $u\cdot a_i>0$ for every $i$, that is, ${\epsilon }_i=\mathrm{sgn}(b_i\cdot u)$ for every $i$. Failure of coverage thus corresponds exactly to one of the $R_N$ realized sign choices. These choices are mutually exclusive events, each of conditional probability $2^{-N}$, so the first part applies:The result is independent of the axes. Averaging gives the coverage of a sphere by random open hemispheres:For $N=1,2,3$ this is zero; for $N=4$ it is $1/8$. If no ship lands, coverage has probability zero separately. The counting uses equiprobable orientation signs, not an assumption that the spherical regions have equal area.

The standard simplicial decomposition of a sphere $F^n$ is the boundary of the cross-polytope with vertices $\pm e_1,\dots ,\pm e_{n+1}$, identified radially with $S^n$. A face chooses distinct coordinate indices and one sign for each; it never contains an antipodal pair.

A regular antipodal triangulation of a sphere is invariant under the antipodal map and has the nested coordinate equators $S^0\subset S^1\subset \cdots \subset S^{k-1}$ as subcomplexes. In particular each equator separates its sphere into two triangulated hemispheres. These conditions are preserved by barycentric subdivision.

For a simplicial map $f:F\to F^n$, a positive alternating simplex of dimension $j$ has distinct image labelsThe opposite alternating pattern is negative; all other simplices, including collapsed images, are neutral.

To prove the antipodal alternating-simplex parity lemma, let $p_j$ count positive $j$-simplices in the equatorial $S^j$. There is exactly one positive vertex in $S^0$, so $p_0=1$. A $j$-simplex has an odd number of positive alternating facets precisely when it is positive or negative alternating. Indeed an alternating sequence has exactly one positive alternating deletion; a nonalternating sequence has either zero or two. For a collapsed image, a positive facet can occur only when exactly one label is repeated; deleting either copy gives two such facets.

Count incidences of positive $(j-1)$-faces with $j$-simplices in one closed hemisphere. Interior faces contribute twice and equatorial faces once. The count modulo two is therefore $p_{j-1}$. It is also the number of positive or negative $j$-simplices in that hemisphere modulo two. Because $f$ is antipodal, each alternating pair $\sigma ,-\sigma$ has one positive and one negative member, with one member in each hemisphere. ThusInduction gives odd $p_k$, proving the existence of a positive $k$-simplex. In particular, an antipodal simplicial map $F\to F^{k-1}$ is impossible: a positive $k$-simplex would require $k+1$ distinct coordinate labels but only $k$ are available.

The Borsuk-Ulam theorem states that every continuous map $g:S^k\to {\mathbb{R}}^k$ has some $x$ with $g(x)=g(-x)$. First rule out a continuous antipodal map $h:S^k\to S^{k-1}$ for $k\ge 1$. By uniform continuity, take a sufficiently fine regular barycentric subdivision so that endpoints of each edge have images less than $2/\sqrt{k}$ apart. Label a vertex by $\mathrm{sgn}(h_i)e_i$, where $i$ is the smallest index maximizing $|h_i|$. Since $h$ is unit length, this maximum is at least $1/\sqrt{k}$; two adjacent vertices cannot have opposite labels. The labels therefore define an antipodal simplicial map to $F^{k-1}$, contradicting the preceding result.

If the stated Borsuk-Ulam theorem failed, the mapwould be just such a continuous antipodal map. This proves the required version; $k=0$ is immediate.

For the Kneser conjecture, equivalently Lovász theorem on Kneser graphs, the claim isThe Kneser graph joins disjoint $r$-subsets of $[N]$. Let $d=N-2r+1$. Colour each $r$-set by its least element in $[d]$ when it has one, and otherwise by one final colour. A colour class indexed by $i$ is an intersecting family because all its members contain $i$; the final class lies in a $(2r-1)$-set and is also intersecting. This proves the upper bound $d+1$.

For the lower bound, first deduce the open-cover Lusternik-Schnirelmann-Borsuk theorem. An open cover $U_0,\dots ,U_d$ of $S^d$ admits a continuous partition of unity ${\varphi }_0,\dots ,{\varphi }_d$ subordinate to it. Apply the Borsuk-Ulam theorem to $({\varphi }_1,\dots ,{\varphi }_d)$. Equal values at $x,-x$ and ${\sum }_i{\varphi }_i=1$ make all values equal. At least one common value is positive, so some $U_i$ contains an antipodal pair.

Now suppose the $r$-sets could be coloured with only $d$ colours, each an intersecting family. Choose $N$ points $v_1,\dots ,v_N\in S^d$ such that every $d+1$ of their vectors are linearly independent. For example, normalize $(1,t_i,\dots ,t_i^d)$ for distinct real $t_i$; the Vandermonde determinant proves this property. The general-position hemisphere count says that at most $d$ points lie on any equator. Since $N-d=2r-1$, at least one of the opposite open hemispheres determined by $x,-x$ contains an $r$-set of the labelled points.

Let $U_i$ be the open set of directions whose positive open hemisphere contains an $r$-set of colour $i$. It contains no antipodal pair, as such a pair would give two disjoint sets of that colour. The compact set $K=S^d\setminus {\bigcup }_{i=1}^d{U}_i$ also contains no antipodal pair: otherwise each of the two opposite hemispheres would have at most $r-1$ points, contradicting the count $2r-1$.

If $K$ is nonempty, compactness gives a positive set distance between $K$ and $-K$; an open neighbourhood $U_0$ of radius less than half that set distance still has no antipodal pair. If $K$ is empty, take $U_0=\varnothing$. In either case $U_0,U_1,\dots ,U_d$ is an open cover with no member containing an antipodal pair, contradicting the Lusternik-Schnirelmann-Borsuk theorem. Hence at least $d+1$ colours are required.

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.

Normalize the surface area of the spherical cap by writingTwo elementary descriptions of a spherical cap are useful. The spherical polar coordinates formula givesProjecting orthogonally onto the last $n-1$ coordinates givesIndeed, the upper hemisphere is the graph $x_1=\sqrt{1-\Vert x^{\prime }{\Vert }^2}$, whose surface area factor is $(1-\Vert x^{\prime }{\Vert }^2{)}^{-1/2}\le 1/\cos \rho$ on $\Vert x^{\prime }\Vert \le \sin \rho$. This establishes the upper projection bound directly.

Set $F(\rho )={\sin }^n\rho -q(\rho )$. The supplied ratio bound implies $c_n<\sqrt{(n+1)/(2\pi )}/n\le 1/2$ for $n\ge 2$. Consequently, at $\rho =\pi /4$,Also,On $[\pi /4,\pi /2]$, the expression in parentheses is strictly decreasing; it starts positive and ends negative. Thus $F$ first increases and then decreases, so its minimum on this interval is at an endpoint. Since $F(\pi /2)=1/2$, both endpoint values are nonnegative. The spherical cap area upper bound follows:

Let $c>0$ denote the stated unit cost of cylindrical wall, and assume $V>0$. The circular cylinder and hemisphere give the volume constraint and total costOnly the walls and cap have specified costs, so no separate base cost is included. Eliminating $h$ giveswith $0<r\le (3V/(2\pi ){)}^{1/3}$ from $h\ge 0$. Now $C^{\prime }(r)=-2cV/r^2+(16/3)\pi cr$ vanishes at $r^3=3V/(8\pi )$. Moreover $C^{\prime \prime }(r)=4cV/r^3+(16/3)\pi c>0$ and the stationary point is feasible, so it is the unique global minimum. The optimal dimensions are

Use cylindrical coordinates. The real square root restricts the projection to $0\le r\le 1$, and there $r-1\le 0\le \sqrt{1-r^2}$. Since $\nabla x={\mathbf{e}}_x$, the volume integral isFor an independent boundary calculation, split $\partial V$ into an upper hemisphere $S_{+}$ and a lower right circular conical surface $S_{-}$. They meet in the circle $r=1,z=0$ and there is no additional planar face.

On $S_{+}$, use the parametrizationThe outward unit normal vector is $\mathbf{R}$, and the surface area element is $\sin \alpha d\alpha d\theta$. The $y$ and $z$ components of ${\iint }_{S_{+}}x\mathbf{n}dS$ vanish by the angular integrals of $\sin \theta \cos \theta$ and $\cos \theta$. The $x$ component isOn $S_{-}$, the parametrization $(r\cos \theta ,r\sin \theta ,r-1)$ has outward vector area elementIts negative vertical component points out below the solid. Multiplying by $x=r\cos \theta$ and using the same angular symmetry in integration givesTherefore the independently evaluated surface integral isin agreement with the volume integral. The isolated apex of the cone does not contribute to the surface integral.

Use source spherical coordinates with radius $s$, polar angle $\alpha$ measured from the positive $z$-axis, and azimuth $\theta$. The lower half of the spherical shell has $a\le s\le 2a$, $\pi /2\le \alpha \le \pi$, and $0\le \theta \le 2\pi$. Its volume element is $s^2\sin \alpha dsd\alpha d\theta$. At the observation point $(0,0,h)$, $h>0$, the separation is $\sqrt{s^2+h^2-2sh\cos \alpha }$, soThe angular integral is elementary: put $q=\cos \alpha$ to obtainWe used $\sqrt{(s+h{)}^2}=s+h$ since both $s,h$ are positive. HenceThus the axial potential of a uniform hemispherical shell, at every point of the positive $z$-axis, isThere is no need to split into $h<a$, $a<h<2a$, and $h>2a$: the observation point is outside the lower hemisphere in all three cases. As checks, $\varphi (h)\to -3\pi G\rho a^2$ as $h\downarrow 0$, and at large $h$,The last factor is exactly the total mass of the lower half of the spherical shell.

On the unit circle, the positively oriented unit tangent vector is $\mathbf{u}=(-y,x,0)$, and $|\mathbf{u}|=1$. Hence $d\mathbf{r}=\mathbf{u}ds$ andSince the scalar field $\psi$ has no $z$ dependence, the curl product rule for a scalar field givesOn the unit upper hemisphere $S_1$, the outward unit normal vector is $(x,y,z)$, so its vertical component is $z=\cos \theta$. Applying the Stokes theorem to the vector field $\psi \mathbf{u}$ therefore provesThe smoothness needed here is that $\psi$ extend as a continuously differentiable scalar field to a neighbourhood of the projected unit disk. The left side is a scalar line integral with respect to arc length; the positive boundary orientation is used when identifying it with the vector line integral.

Choose the outward unit normal vector on the upper hemisphere. The curl of $\mathbf{F}=(4y,x,2z)$ isIn spherical polar coordinates, putHere $n_z=\cos \phi$ and $dS=a^2\sin \phi d\phi d\theta$. The surface integral is thereforeThe compatible boundary orientation follows increasing $t$ in $\mathbf{r}(t)=(a\cos t,a\sin t,0)$, $0\le t\le 2\pi$. On this circle,Thus the line integral iswhich agrees with the surface integral as required by the Stokes theorem.