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Use spherical coordinates about the fixed pivot, with fixed rod length $l$, polar angle $\theta$ measured from the downward vertical, and azimuth $\varphi$. The bob's position iswhen the Cartesian $z$ axis points upward. Differentiation gives kinetic energy $T=m{l}^2({\dot{\theta }}^2+{\sin }^2\theta {\dot{\varphi }}^2)/2$. Taking zero gravitational potential energy at pivot height gives $V=-mgl\cos \theta$. The spherical pendulum's Lagrangian is thereforeThe rod is light, rigid and of fixed length; pivot friction and air drag are neglected. A rigid rod, unlike a string, may sustain either tensile or compressive constraint force.

The azimuth $\varphi$ is a cyclic coordinate, since $\partial \mathcal{L}/\partial \varphi =0$. Its Euler-Lagrange equation gives conservation of the vertical component of angular momentum:This $h$ is the actual angular momentum, not specific angular momentum. The Lagrangian also has no explicit time dependence. The conserved energy isAlong solutions, $\dot{E}=-\partial \mathcal{L}/\partial t=0$. These are the conserved quantities associated with rotational symmetry about the vertical and time-translation symmetry, respectively.

Choose the vertical $z$ axis downward, and measure the polar angle $\theta$ from it. With azimuth $\varphi$ and fixed rod length $r$, the Cartesian coordinates areThe massless rod has no kinetic energy. Differentiating these coordinates and adding the squared components of velocity yieldsThe gravitational potential energy is $V=-mgz=-mgr\cos \theta$, up to an arbitrary constant. The Lagrangian of this spherical pendulum is thereforeApplying the Euler-Lagrange equation to the two generalized coordinates givesEquivalently, away from the coordinate singularities at the poles,Since $L$ is independent of $\varphi$, the azimuth is a cyclic coordinate. Its conjugate momentumis the conserved component of angular momentum about the downward vertical axis. The polar angle is not a cyclic coordinate. Using an upward $z$ axis instead reverses the sign of the cosine term in $L$ and of the corresponding gravity term; it describes the same motion with the opposite polar angle convention.

For a regular Lagrangian, form the canonical momenta $p_i=L_{{\dot{q}}_i}$, invert for ${\dot{q}}_i(\mathbf{q},\mathbf{p},t)$, and take the Legendre transform $H={\sum }_i{p}_i{\dot{q}}_i-L$. On the upper cone let $s>0$ be distance along a generator from the apex and $\varphi$ the azimuth. Then the horizontal radius is $s\sin \theta$ and height is $s\cos \theta$, soConsequently

For nonzero $\chi$, the outer product $X=\chi {\chi }^{†}$ is rank-one Hermitian positive semidefinite, so $\det X=0$ and $ct={\chi }^{†}\chi /2>0$. It therefore represents a future-directed null vector. Conversely any nonzero future-null Hermitian $X$ has eigenvalues $2ct$ and zero, so its positive eigenvector multiplied by $\sqrt{2ct}$ supplies $\chi$. The phase $\chi \mapsto e^{i\zeta }\chi$ leaves $X$ unchanged. A past-directed null vector needs $X=-\chi {\chi }^{†}$, rather than the positive outer product alone.

For the printed normalized spinor, direct multiplication yieldsComparing with the printed $x+iy$ convention givesIts usual spatial azimuth is $-\varphi$, not $\varphi$. The source's asserted usual-angle identification has a sign inconsistency. Either conjugating the two spinor phases or replacing the upper-right matrix entry by $x-iy$ restores the usual azimuth $\varphi$. Neither alteration is made silently here. A further positive scale factor sets the magnitude of an arbitrary future-null vector; the displayed normalized spinor fixes $ct=1/2$ in the chosen coordinate units.

The spinor transformation is ${\chi }^{\prime }=S\chi$, with $S=\mathrm{diag}(e^{\kappa /2},e^{-\kappa /2})$. Its squared norm becomesDividing by $\sqrt{\mathcal{R}}$ isolates the new direction without discarding the physical scale factor in $X^{\prime }$. The lower-to-upper component ratio givesEquivalently,This is relativistic aberration for the active positive-$z$ boost specified in the question. At the poles, azimuth is undefined rather than a measurable transformed variable. Under the literal matrix convention from (c), the actual azimuth $-\varphi$ is likewise unchanged. The time coordinate is multiplied by $\mathcal{R}$; normalizing a spinor for its direction must not erase that boost-induced change of null-vector scale.

The generator acts only in the $ct,z$ plane, with its square equal to the identity on that two-dimensional subspace. Its exponential isFor the vector in (c), this giveswhile $x^{\prime }=x$ and $y^{\prime }=y$. Since the vector is future-null, its spatial magnitude is $c{t}^{\prime }$. ThereforeThe transverse azimuth does not change. These are exactly the direction and scale obtained from the spinor congruence. In particular $c{t}^{\prime }=\mathcal{R}ct$ and $\tan ({\theta }^{\prime }/2)=e^{-\kappa }\tan (\theta /2)$. The azimuth-sign qualification in (c) is compatible with this check because a boost along $z$ leaves both transverse coordinates unchanged.

In the usual physics convention for spherical coordinates, the polar angle is the angle between the positive $z$ axis and the position vector. For $r=\sqrt{x^2+y^2+z^2}>0$, it satisfies $\cos \theta =z/r$ with $0\le \theta \le \pi$. The north and south axes have $\theta =0$ and $\theta =\pi$, respectively; the plane $z=0$ has $\theta =\pi /2$. The angle is undefined at the origin. In plane polar coordinates, the same phrase usually denotes the oriented in-plane angle instead, so the reference axis and convention should always be stated. The azimuth supplies the other angular coordinate in three dimensions.

A point mass constrained to a fixed distance $r$ from a pivot and free to move in three dimensions under uniform gravitational acceleration. With the vertical axis downward and polar angle $\theta$ measured from it, the Lagrangian is the expression in the title. The azimuth $\varphi$ is a cyclic coordinate, so $m{r}^2{\sin }^2\theta \dot{\varphi }$ is conserved angular momentum about the vertical axis. A rigid massless rod enforces the constraint without the loss-of-tension limitation of a string.

Spherical trigonometry calculates the sides and interior angles of spherical triangles using trigonometric functions. Its sides must be shorter great circle arcs, which are geodesics on a round sphere. A curve at constant nonzero latitude does not qualify. The spherical law of sines and spherical law of cosines play the roles of the familiar flat-triangle rules, but account for the sphere's curvature.

Use uppercase $A,B,C$ both for the vertices and for their interior angles, and lowercase $a,b,c$ for the opposite angular sides. For example, $a$ is the central angle subtended by the arc from $B$ to $C$. On a sphere of radius $R$, its physical arc length is ${\ell }_a=Ra$ when $a$ is in radians; similarly ${\ell }_b=Rb$ and ${\ell }_c=Rc$. Only on the unit sphere do the angular and physical side lengths have the same numerical values. See J. B. Tatum, Celestial Mechanics, chapter 3, section 3.5: great-circle sides and angular units.

One useful application of spherical trigonometry is finding great-circle distance from spherical coordinates. Write a unit vector as $u(\varphi ,\lambda )=(\cos \varphi \cos \lambda ,\cos \varphi \sin \lambda ,\sin \varphi )$, where $\varphi$ is latitude and $\lambda$ is the equatorial azimuth, or longitude, measured from a chosen reference direction. The dot product of two such unit vectors givesHere $\delta \in [0,\pi ]$ is the central angle and $d$ is the great-circle distance. The same identity follows from the spherical law of cosines by adding the north pole as the third vertex, with angular sides $\pi /2-{\varphi }_1$ and $\pi /2-{\varphi }_2$. This connects spherical geometry to position calculations in astronomy and to navigation on a spherical model of Earth. Choosing the surface model matters: Paula Neeley and Anthony Narkawicz, Map Projection Induced Variations in Locations of Polygon Geofence Edges, NASA/TM-2017-219675 (2017), sections 2-3 distinguishes spherical and ellipsoidal geodesics and gives this coordinate identity.

For a toy example, put two points at latitude $30^{\circ }$, with equatorial azimuths $-60^{\circ }$ and $60^{\circ }$. The great-circle distance and the arc length along constant latitude areThe great circle takes a shorter route by bending towards the pole in these spherical coordinates. To locate its highest latitude, the endpoint unit vectors are $p=(\sqrt{3}/4,-3/4,1/2)$ and $q=(\sqrt{3}/4,3/4,1/2)$. Their great circle lies in the plane through the origin with $z=2x/\sqrt{3}$. On its shorter arc, the maximum latitude occurs at $y=0$ and is $\arctan (2/\sqrt{3})\approx 49.106605^{\circ }$. Thus a geodesic can change latitude substantially even when its endpoints have equal latitude. These are lengths on an ideal sphere; the example illustrates the geometry rather than specifying an actual flight route.

Historically, spherical trigonometry was prominent in teaching as well as in astronomy. John Conway and Alex Ryba note its presence in early twentieth-century school textbooks in Remembering spherical trigonometry, The Mathematical Gazette 100(547), March 2016, pages 1-8, DOI 10.1017/mag.2016.3. Their introductory observation documents that teaching history; it does not identify a single inventor of the spherical law of sines or spherical law of cosines.

With $X_{12}=x+iy$, a spinor whose components have phases $e^{-i\varphi /2}$ and $e^{i\varphi /2}$ has $x+iy\propto e^{-i\varphi }$, so its physical azimuth is $-\varphi$. The usual positive azimuth for those phases uses $X_{12}=x-iy$. Both conventions are valid if used consistently.

The radial vector field $B=e_r/r^2$ on punctured three-dimensional Euclidean space has flux $4\pi$ through every centered sphere. It has no vector potential smooth on an entire enclosing sphere: the Stokes theorem on a closed surface would force zero flux. Two overlapping patches suffice. In spherical coordinates take $A_{+}=(1-\cos \theta )e_{\varphi }/(r\sin \theta )$, regular away from the negative $z$ axis, and $A_{-}=-(1+\cos \theta )e_{\varphi }/(r\sin \theta )$, regular away from the positive $z$ axis. Their curls both equal $B$.

The Cartesian forms $A_{+}=(-y,x,0)/[r(r+z)]$ and $A_{-}=(y,-x,0)/[r(r-z)]$ show that the first is smooth at the north pole and the second at the south pole. On the overlap their difference is $A_{+}-A_{-}=2\nabla \varphi$, where the azimuth $\varphi$ is defined locally. Its multivaluedness around the axis explains the circulation difference $4\pi$. Splitting the sphere along a latitude, applying the Stokes theorem with $A_{+}$ on the north and $A_{-}$ on the south, and using their opposite boundary orientations recovers the total flux. A coordinate singularity in the tangential spherical basis does not by itself make a vanishing smooth vector field singular at a pole.

The zenith is the point on the celestial sphere in the direction of the observer's local upward vertical. In the horizontal coordinate system, its unit vector is ${\mathbf{e}}_U$ and its altitude is $90^{\circ }$. The zenith belongs to the observer's site; it is not generally the north celestial pole, whose altitude depends on geographical latitude. A direction's zenith distance is its nonnegative angle from the zenith, so $z=90^{\circ }-\gamma$. A direction of altitude $30^{\circ }$ therefore has zenith distance $60^{\circ }$. ^[ref].

At the zenith, every value of azimuth produces $(E,N,U)=(0,0,1)$. There is no horizontal component to identify a bearing: $atan2(0,0)$ cannot provide a geometrically meaningful azimuth, even if software returns a number. This coordinate singularity is a defect of the angular description; the viewing direction itself remains regular. The same ambiguity occurs at the opposite downward vertical. Near the zenith, two directions at fixed altitude separated by a small azimuth increment $d\eta$ have separation $\cos \gamma |d\eta |$ to first order, with angles in radians. At altitude $89^{\circ }$, a $1^{\circ }$ azimuth increment spans only about $0.01745^{\circ }$ on that small arc. An optical telescope can consequently require a large bearing change for a small change of pointing near the zenith; working with unit vectors avoids inventing a preferred bearing at the singular point.

Zenith distance is the angle from the zenith to a viewing direction: $z=\arccos (\mathbf{n}\cdot {\mathbf{e}}_U)=90^{\circ }-\gamma$, where $\gamma$ is altitude and $0\le z\le 180^{\circ }$. The astronomical horizon has $z=90^{\circ }$, while the zenith has $z=0$. This unsigned separation supplies no azimuth and is distinct from the signed north/south angular readout used under declination.