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For a smooth ambient function near the unit sphere, the chain rule along the great circles ${\gamma }_i(t)=p\cos t+e_i\sin t$ gives $(F\circ {\gamma }_i{)}^{\prime \prime }(0)=D^{2}F(e_i,e_i)-DF(p)$. Sum over an orthonormal basis of the tangent space and supplement it by the radial vector $p$. The geodesic trace formula for the Laplace-Beltrami operator then gives the displayed formula, evaluated at radius one. In particular a degree-$\ell$ harmonic polynomial restricts to a spherical harmonic with positive eigenvalue $\ell (\ell +d-1)$.

A planar circle other than the unit circle lifts to a great circle exactly when it meets the unit circle at two opposite points. A great circle other than the equator intersects the equator in an antipodal pair. Conversely the plane containing a lifted circle and an antipodal pair contains their midpoint, the sphere's centre. Algebraically, adding the equations $|z-c{|}^2=r^2$ and $|-z-c{|}^2=r^2$ with $|z|=1$ gives the displayed condition.

The arc length of a piecewise smooth curve $\gamma :[a,b]\to M$ on a Riemannian manifold is the integral of its speed:An odometer measures this accumulated travel rather than displacement between the endpoints. In Euclidean space, the Riemannian metric is the ordinary inner product, so the formula becomes ${\int }_a^b\sqrt{{\sum }_i{\dot{\gamma }}_i(t{)}^2}dt$.

For a circle of radius $R$ traversed through an angle $\alpha$, use $\gamma (t)=(R\cos (\alpha t),R\sin (\alpha t))$, $0\le t\le 1$. Its speed is $R|\alpha |$, hence its arc length is $R|\alpha |$. Traversing a full circle twice gives $4\pi R$, even though the starting and ending points coincide. The same formula measures the two great circle arcs pictured in geodesic.

The curve parameter controls how fast the journey happens, not its arc length. A regular increasing or decreasing reparametrization preserves the integral by change of variables; reversing direction leaves the speed norm unchanged. Retracing part of a route does add length, because it changes the actual journey.

The astronomical horizon is the great circle where the horizontal plane through the observer intersects the celestial sphere. It is perpendicular to the local vertical, so every direction on the astronomical horizon has altitude $0^{\circ }$ and zenith distance $90^{\circ }$. In the local horizontal coordinate system, it consists of the unit vectors satisfying $\mathbf{n}\cdot {\mathbf{e}}_U=0$. This geometric reference remains defined when buildings or mountains hide it; a visible skyline is not a calibration of zero altitude.

Three boundaries must be distinguished. The astronomical horizon fixes the angular reference. Terrain supplies a site-dependent boundary with apparent altitude $t(\eta )$ that changes with azimuth. Atmospheric refraction changes the apparent direction of incoming light, and can also change the apparent distant boundary. A point source clears an opaque terrain boundary only if ${\gamma }_{\mathrm{app}}>t(\eta )$ when both are expressed in the same apparent horizontal coordinate system; this clearance condition does not guarantee visibility through clouds or bright sky. In the synthetic diagram, a source with geometric altitude $3^{\circ }$ and an imposed illustrative upward correction $R=0.2^{\circ }$ remains behind a $4^{\circ }$ boundary. The chosen $R$ is an algebraic illustration, not a fitted atmospheric prediction.

Even an unobstructed sea boundary need not coincide with the astronomical horizon. On a spherical Earth of radius $r$, an observer at height $q$ sees the unrefracted tangent to the surface at a downward angle $D$ satisfying $\cos D=r/(r+q)$. The right-angle tangent triangle supplies this relation; the visible boundary then has altitude $-D$. With illustrative $r=6371\mathrm{km}$ and $q=10\mathrm{m}$, $D\simeq 0.1015^{\circ }$, before atmospheric refraction. Raising the eye therefore changes the visible boundary without redefining the horizontal plane.

The US Naval Observatory's rise/set definitions explicitly distinguish the geometric reference from apparent contact with a visible disk: the standard Sun calculation puts its centre $50^{\prime }$ below the horizontal plane, combining an average $16^{\prime }$ radius with $34^{\prime }$ of atmospheric refraction. An angular minute is $1^{\prime }=1/60^{\circ }$ in the minute and second of arc convention. This is an operational average under stated conditions, rather than a universal zero of altitude. The same source identifies weather, local topography and observer height as limits on predicted rise/set time. An observation log should therefore retain the level reference, site and eye height, terrain profile, observation time, and whether each reported altitude is geometric or apparent. ^[ref].

The celestial equator is the great circle perpendicular to the rotation axis of Earth; declination measures angular displacement from it.

An observer’s celestial meridian is the great circle containing the zenith and the celestial poles. Its upper half passes through the local north and south directions.

The celestial sphere is the sphere of viewing directions around an observer; distances to objects are suppressed. Great circles define coordinate reference planes.

For north-pole stereographic projection $z=(X+iY)/(1-Z)$ on the unit sphere, substitute $|z{|}^2=(1+Z)/(1-Z)$ into a planar circle equation. The displayed plane equation follows, and its intersection with the sphere is a circle. The plane contains the sphere's centre exactly when $r^2=1+|c{|}^2$, giving a great circle.

Declination $\delta$ is the signed angle from the reference celestial equator to a direction, measured along its hour circle: $-90^{\circ }\le \delta \le +90^{\circ }$. Positive declination means celestial north, rather than necessarily north of the observer's zenith. A direction with $\delta =0$ lies on the celestial equator; $\delta =+90^{\circ }$ is the north reference celestial pole. Declination is an angle, not a height or a measure of Euclidean distance. ^[ref].

The connection to an observation follows from a dot product. In the elementary spherical-Earth, unrefracted model, an observer at geographical latitude $\varphi$ has zenith direction $\mathbf{z}=(\cos \varphi \cos \Theta ,\cos \varphi \sin \Theta ,\sin \varphi )$ in the same reference axes. Since $\mathbf{n}\cdot \mathbf{z}=\sin a$ for altitude $a$, the cosine addition formula givesAt upper meridian transit, $H=0$, so $a_{\mathrm{upper}}=90^{\circ }-|\varphi -\delta |$. For the synthetic site $\varphi =52^{\circ }$ and synthetic direction $\delta =20^{\circ }$, this yields $a_{\mathrm{upper}}=58^{\circ }$. A direction with $\delta =70^{\circ }$ instead reaches $72^{\circ }$, north of the zenith; positive declination does not imply a crossing on the southern half of the celestial meridian. A negative computed upper altitude means the direction never rises above the geometric astronomical horizon at that site.

For an angular readout, define signed zenith distance $z_s$ on the upper celestial meridian, positive toward the south and negative toward the north. Then $\delta =\varphi -z_s$. A synthetic southward readout $z_s=32^{\circ }$ at $\varphi =52^{\circ }$ recovers $\delta =20^{\circ }$. An unsigned distance $|z_s|$ alone would leave two possibilities, $\varphi \pm |z_s|$, when both are in range. A calibrated optical telescope must therefore record which side of the zenith it points toward. In this linear relation, a $1^{\prime }$ error in the assumed geographical latitude produces a $1^{\prime }$ error in inferred declination; averaging repeated readings cannot remove that shared offset. Near the astronomical horizon, uncorrected atmospheric refraction also invalidates the geometric readout.

The unit vector construction explains a subtle feature of sky maps. At fixed declination, changing right ascension traces a parallel whose radius on the unit sphere is $\cos \delta$. Its arc length is $\cos \delta |d\alpha |$, with angles in radians; at $\delta =60^{\circ }$, a $15^{\circ }$ right ascension increment spans a $7.5^{\circ }$ arc along that parallel. This arc is not generally the shortest great circle separation. At either celestial pole, all right ascension values represent the same direction: the coordinate singularity belongs to the coordinate grid. The unit vector remains regular there.

On the unit sphere with one point deleted, a distinct nonantipodal pair has a length minimizer exactly when its shorter great circle arc avoids the deleted point. Antipodal pairs admit an avoiding semicircle. If the unique shorter arc crosses the puncture, smooth detours approach its length without attaining it.

For two inertial frames with equal speed in a laboratory, a photon is measured at equal angular frequency by their comoving observers exactly when its laboratory propagation unit vector $\mathbf{n}$ obeys the displayed condition. Indeed, measured angular frequency is $-k\cdot U=\gamma {\omega }_{\mathrm{lab}}(1-\mathbf{n}\cdot \mathbf{v}/c)$, and equal laboratory Lorentz factors cancel. Thus the permitted directions form a great circle on the unit sphere. In the plane of the two laboratory velocities, only the internal angle bisector line remains. The velocity information alone fixes neither direction along that line nor coplanarity of the ray.

For a sphere with a chosen polar axis, its equator is the great circle in the plane through its center perpendicular to that axis. On the unit sphere in spherical coordinates it is $\theta =\pi /2$. Like every great circle, it is a geodesic.

A geodesic is a curve whose tangent vector is parallel along itself under the Levi-Civita connection. With an affine parameter it has zero covariant acceleration:This is the Riemannian manifold version of moving straight ahead with constant speed. Its coordinate form is the geodesic equation. In Euclidean space it becomes $\ddot{\gamma }=0$, giving straight lines. On an embedded surface with its induced metric, the ordinary acceleration of such a curve is normal to the surface: all acceleration is used to follow the surface, rather than to turn within it.

On a unit round sphere, the great circle $\gamma (t)=(\cos t,\sin t,0)$ has $\ddot{\gamma }(t)=-\gamma (t)$, which is normal to the sphere. Two points at $t=0$ and $t=\pi /3$ can be joined along that circle by the short arc of length $\pi /3$ or the long arc of length $5\pi /3$. Each admits a constant-speed geodesic parametrization, yet only the short arc realizes their great-circle distance.

Every geodesic minimizes arc length on sufficiently short subarcs. A long segment can cease to be globally minimizing, as the sphere example shows; the first loss from a starting point is described by the cut locus. Thus “locally straight” and “shortest route between these endpoints” are distinct assertions. On a circular cylinder, different winding gives the same distinction, made explicit by Riemannian distance on a circular cylinder.

This provides a useful division of work. The geodesic equation predicts a path from an initial point and tangent vector; comparing arc lengths decides which candidate solves a global shortest-route problem. Variations of energy of a curve give the equation as an Euler-Lagrange equation, offering a proof outline and a connection with classical mechanics. ^[ref].

The gnomonic projection intersects rays from a sphere’s centre with a tangent plane. If $\mathbf{b}$ is its unit normal and $\mathbf{e},\mathbf{n}$ are tangent basis vectors, a direction $\mathbf{s}$ maps to $f(\mathbf{s}\cdot \mathbf{e},\mathbf{s}\cdot \mathbf{n})/(\mathbf{s}\cdot \mathbf{b})$. It maps great circle arcs to straight lines and is singular at the tangent-plane horizon.

For points on a sphere of radius $R$, write their position vectors relative to the centre as $Ru,Rv$ with $u,v$ unit vectors. The great-circle distance is the shorter great circle arc length, $R$ times its central angle. Hence it equals $R\arccos (u\cdot v)$, including zero for identical points and $\pi R$ for antipodal points. At antipodal points the shortest arc is not unique, but its length is.

An hour circle is a great circle through both reference celestial poles and a chosen direction. Its two halves have right ascensions differing by $12^{\mathrm{h}}$; choose the half containing the direction when measuring declination. The observer's celestial meridian is the particular hour circle through the zenith. ^[ref].

A great circle passing through a chosen pair of antipodal poles, or the pole-to-pole half of that circle at fixed longitude. In spherical coordinates, a fixed azimuth selects such a half-circle. Meridians are special cases of a meridian of a surface of revolution and are geodesics of the sphere.

A finite set of distinct great circles not all passing through a common antipodal pair has a vertex lying on exactly two circles. Its cells are spherical polygons with at least three edges: a digon would have antipodal endpoints, and every circle's defining hemisphere would then have to contain both endpoints, forcing a common axis. With $V-E+F=2$ and $3F\le 2E$, the graph has a vertex of valence less than six. Valences are even and at least four, so one is exactly four. Central projection identifies this with an ordinary intersection in a real projective line arrangement. If affine line intersections are counted only at finite points, parallel-line configurations require a separate qualification.

Let $R=|\alpha |$, $S=|\beta |$ and $d=|\mathbf{b}-\mathbf{a}|$. For distinct centres, the two spherical surfaces intersect exactly whenStrict inequalities give a nondegenerate circle; equality gives a single tangency point. If $d=0$, equal radii give coincident spheres and unequal radii give no intersection, so the circle-centre formula below does not apply.

Subtracting the squared sphere equations gives the radical planeThe centre of the circle of intersection of two spheres is the projection of either sphere centre onto this plane. Write $\mathbf{c}=\mathbf{a}+\ell (\mathbf{b}-\mathbf{a})/d$; substitution gives $\ell =(d^2+{\alpha }^2-{\beta }^2)/(2d)$ and henceThe circle radius squared is $R^2-{\ell }^2$, providing another check of the intersection conditions.

Its centre equals $\mathbf{a}$ precisely when $\ell =0$, or $\boxed{{\beta }^2={\alpha }^2+d^2}$. For a positive first radius, the intersection is then a great circle of the first sphere. A triangle from its centre to an intersection point and then to the second centre is right-angled at $\mathbf{a}$, explaining the Pythagorean theorem relation geometrically.

For a spherical triangle on a sphere of radius $R$, bounded by shorter great circle arcs and with interior angles $\alpha ,\beta ,\gamma$, the spherical excess formula, the triangular form of the Gauss-Bonnet theorem, isHere is an area proof. On the unit sphere, a spherical lune of angle $\alpha$ has area $2\alpha$: rotating around its vertices shows it occupies a fraction $\alpha /(2\pi )$ of the sphere when one uses the full angular width, or directly integrates ${\int }_0^{\alpha }{\int }_0^{\pi }\sin \theta d\theta d\phi$. The three side great circles of the triangle divide the sphere into eight regions, including the triangle $T$ and its antipodal triangle $-T$. Choose, at each vertex, the lune containing $T$, and also its antipodal lune. Each of $T$ and $-T$ is covered three times by these six lunes; each other region is covered once. Thus, with $A$ the area of $T$,This proves $A=\alpha +\beta +\gamma -\pi$; scaling lengths by $R$ scales area by $R^2$.

Radially project each face of the regular dodecahedron from its centre onto the unit sphere. Each edge and the origin lie in a plane, so its image is a shorter great circle arc. The twelve faces give congruent convex regular spherical polygons partitioning the sphere. At each projected vertex three congruent face angles fill the tangent plane. Hence each spherical pentagon has angle $2\pi /3$ at each of its five vertices. Splitting a pentagon into three spherical triangles and summing the spherical excess formula givesThis also agrees with $4\pi /12$, as required by the spherical partition.