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Given a polar axis, the equatorial plane passes through the sphere's centre and is perpendicular to that axis. In spherical coordinates with polar angle $\theta$, it is the plane $\theta =\pi /2$. For motion in a spherically symmetric gravitational field, conservation of angular momentum permits rotation of axes so that a nonradial orbit lies in such a plane. This is a choice of orientation, rather than an additional physical restriction on a spherically symmetric problem.

The quoted expansion is nondegenerate perturbation theory for a known isolated eigenstate of a time-independent Hamiltonian. A useful small-mixing condition is $|⟨k|H_1|i⟩|\ll |E_i-E_k|$ for coupled states. First order is a diagonal expectation value; second order describes virtual mixing. A degenerate or nearly degenerate coupled subspace must first be treated by diagonalization there. Symmetry can eliminate terms, and convergence still needs checking. Ground-state second-order shifts are nonpositive because all denominators are negative.^[ref]

For the rotational Stark effect, $H_1=-pE\cos \theta$, where $\theta$ is the polar angle to the field. The source calls it azimuthal, which is inconsistent with this coupling. Rotor states have parity $(-1{)}^J$, while $\cos \theta$ is odd, so every first-order matrix element within one fixed-$J$ manifold vanishes. The $m$ degeneracy therefore causes no first-order splitting.

Only $J\pm 1$ with the same $m$ contribute in second order. Using their energy differences $2BJ$ and $-2B(J+1)$,For $J\ge 1$ this simplifies to the quadratic Stark shift of a polar linear rotor,For $J=0$, only $J^{\prime }=1,m=0$ contributes, with squared angular matrix element $1/3$ and denominator $-2B$:Here $B$ is the rotational energy constant, unlike the magnetic field in Question3.

The unperturbed line gives $B/h=6.08\mathrm{GHz}$. The $J=1$ shifts are $(pE{)}^2/(10B)$ for $m=0$ and $-(pE{)}^2/(20B)$ for $m=\pm 1$. Subtracting the common lower-state shift gives transition shifts $4(pE{)}^2/(15Bh)$ and $7(pE{)}^2/(60Bh)$. Their separation determines the dipole moment from a rotational Stark doublet:The $\mathrm{D}$ unit is the debye. Using external $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and $1\mathrm{D}=3.33564\times 10^{-30}\mathrm{C}\mathrm{m}$ gives $pE/B\simeq 0.0593$, consistent with weak-field perturbation theory. The two shifted frequencies are about $12.165707$ and $12.162497\mathrm{GHz}$; polarization selects $\Delta m=0$ or $\pm 1$. The inferred value is a magnitude; these shifts do not determine the dipole's sign.^[ref][ref]

There is no contradiction. The domain on which the field is smooth is the punctured plane, which is not simply connected. A loop enclosing the origin cannot be spanned by a planar region entirely inside that domain, so the smoothness hypothesis needed to apply Green's theorem to its full interior fails. A loop that bounds a region avoiding the origin does have zero circulation.

Locally the field is the gradient of the polar angle, but that angle cannot be made a single-valued continuous potential on the whole punctured plane. The nonzero period $2\pi$ around the hole measures this global obstruction, despite the field having zero local curl.

Far from the planet, the magnitude of angular momentum per unit mass is the magnitude of the cross product of position vector and velocity. The perpendicular distance from the incoming asymptote is the impact parameter $d$, soChoose $\theta =0$ at closest approach. The general solution of the differential equation in a, with $u^{\prime }(0)=0$, isThe specific orbital energy isAt infinity $\epsilon =v^2/2$, giving $e=\sqrt{1+h^2{v}^2/{\mu }^2}>1$. Thus the hyperbolic orbit hasA different zero of polar angle simply replaces $\theta$ by $\theta -{\theta }_0$.

The incoming and outgoing asymptotes have radial directions $-{\theta }_{\infty }$ and ${\theta }_{\infty }$, where $\cos {\theta }_{\infty }=-1/e$ and $\pi /2<{\theta }_{\infty }<\pi$. The incoming velocity points opposite its radial direction, so the deflection angle between incoming and outgoing velocities is $\delta =2{\theta }_{\infty }-\pi$. HencePutting $\delta =\pi /2$ gives

Divide Newton's second law by the mass and resolve along the radial and transverse unit vectors in polar coordinates. A central force has no transverse component, soMultiplication of the second equation by $r$ shows that $d(r^2\dot{\theta })/dt=0$. Thuswhere $h$ is the signed specific angular momentum perpendicular to the orbital plane.

If $h\ne 0$ and $r>0$, the polar angle is locally a valid independent variable. For $u=1/r$, use primes for derivatives with respect to $\theta$. The chain rule givesSubstitution into the radial equation gives the Binet equationIf $h=0$, the orbit is radial and division by $h$ is unavailable; the first radial equation still applies with $\dot{\theta }=0$ away from the origin.

Using the physics convention for spherical coordinates, the polar angle $\theta$ is measured from positive $z$ and the azimuth $\varphi$ from positive $x$ toward positive $y$. The Cartesian coordinates are

Use the convention that $\theta$ is the polar angle from positive $z$, and $\varphi$ the azimuth from positive $x$. The radius isThen $\cos \theta =z/r=\sqrt{3}/2$, while $\tan \varphi =y/x=\sqrt{3}$ and both planar coordinates are positive. Thus the spherical coordinates are

Project the radius vector onto the $x$-$y$ plane. The polar angle $\theta =\pi /6$ lies between the positive $z$ axis and $OP$. The azimuth $\varphi =\pi /3$ lies in the $x$-$y$ plane between the positive $x$ axis and the projection of $OP$, measured toward positive $y$.

On the sphere, constant $\theta$ gives a latitude circle of radius $a\sin \theta$ and height $a\cos \theta$. Constant $\varphi$ gives a meridian semicircle in a vertical plane through the $z$ axis. The orthonormal basis of spherical coordinates isHere ${\mathbf{e}}_r$ points outward, ${\mathbf{e}}_{\theta }$ along increasing polar angle, and ${\mathbf{e}}_{\varphi }$ along increasing azimuth. At either pole the radial unit vector is unambiguous, but the two tangential coordinate vectors depend on the undefined longitude. The figure shows their directions away from the poles.

This coordinate ambiguity does not make a vector field singular if its tangential coefficient vanishes suitably. In Cartesian coordinates,The first is smooth at the north pole and zero there; the second is smooth and zero at the south pole. Their actual singular axes are the opposite ones. These are the regular patches of a two-patch vector potential for a radial monopole.

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.

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.

A circular orbit has constant $r=r_0>0$. Both its initial radial velocity and radial acceleration must vanish, givingEquivalently, the initial tangential speed and angular velocity obeyThe initial polar angle is arbitrary, and the energy is fixed at $E=V_{\mathrm{eff}}(r_0)$. Either orientation is possible. These conditions give an exact circular orbit even if it is unstable; stability requires an additional positive second derivative of the effective potential.

Use arc length $s$ as the affine parameter, and put a dot over derivatives with respect to $s$. The geodesic Lagrangian on the sphere isThe cyclic coordinate $\varphi$ gives $\partial L/\partial \dot{\varphi }=a^2{\sin }^2\theta \dot{\varphi }=\text{constant}$. With the constant normalized as in the paper,Here $h$ has units of inverse length, because $s$ is physical arc length. Unit-speed normalization givesFor $h\ne 0$, division by ${\dot{\varphi }}^2$ yields the spherical-geodesic first integralThe turning latitudes satisfy $\sin \theta =|ah|$. In the usual polar-coordinate range $0\le \theta \le \pi$,The maximum departure from the equator, often the geometric quantity intended by an angular amplitude, is insteadThus polar angle and signed latitude must not be interchanged. At $|ah|=1$ the geodesic is the equator. At $h=0$ it is a meridian through the poles; $\varphi$ cannot serve as a global parameter and the divided equation is inapplicable. The limiting polar range is then $[0,\pi ]$.

The original PDF prints an ambiguous derivative with the square on $\theta$ rather than a conventional derivative notation. If the intended equation was second order, differentiating the first integral, or eliminating $\dot{\theta }$ from the Euler-Lagrange equation, gives the equally useful formIt has the same normalized geodesics and the same turning angles. Giving both forms resolves the typographical ambiguity without pretending that a second derivative is the square of a first derivative. Geometrically these trajectories are great circles, with inclination $\arccos |ah|$ to the equator.

The dipole coupling $-pE\cos \theta$ connects only $J\pm 1$ at fixed $m$. For $J\ge 1$, second order gives $(pE{)}^2[J(J+1)-3m^2]/[2BJ(J+1)(2J-1)(2J+3)]$; the ground-state result is separately $-(pE{)}^2/(6B)$. Here $B$ has energy units and $\theta$ is the polar angle to the field.

A planar orbit whose radius oscillates as its polar angle increases, producing successive lobes around the force center. For nonconstant $r$ with fundamental angular radial period $2\pi /\nu$ and nonzero specific angular momentum, the full position and velocity repeat precisely when $\nu$ is a rational number, provided the radius is bounded above and away from zero. If $\nu =p/q$ in lowest terms, $p$ radial periods advance the angle by $2\pi q$; $dt=r^{2}d\theta /h$ then gives a finite time period. This proves closure rather than merely a spatial self-intersection.

A family of algorithms that exploit monotonically changing geometric directions while traversing sorted points or the boundary of a convex polygon. For example, after sorting rays by polar angle around a point, a second index can advance monotonically to identify all rays lying in each open half-plane. Collinear rays require consistent boundary handling.

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.

A stress concentration is a localized increase of stress caused by geometry or another disturbance of the load path. Holes, notches and abrupt section changes can make the largest local normal stress much greater than a stress calculated from the total force divided by a reference cross-sectional area. A stress concentration factor $K_t={\sigma }_{\max }/{\sigma }_{\mathrm{ref}}$ is meaningful only after both the stress component and its reference have been specified.

The Kirsch equations give an instructive exact result for a circular hole of radius $a$ in an infinite, isotropic linear elastic elastic plate under remote tensile stress ${\sigma }_{\infty }$ in one direction, using a plane stress model. Measure the polar angle $\theta$ from the loading direction. At the hole boundary,At $\theta =\pi /2,3\pi /2$, the tangential normal stress reaches $3{\sigma }_{\infty }$, giving $K_t=3$ relative to remote stress. At $\theta =0,\pi$, it is $-{\sigma }_{\infty }$. A traction-free hole therefore does not require every component of its stress tensor to vanish: it requires zero traction on the radial boundary.^[ref]

For a finite plate, distinguish gross and net reference areas. A plate of width $W$, thickness $t$ and central hole diameter $d$ transmits force $P$ through gross area $A_g=tW$ and minimum net area $A_n=t(W-d)$. Defining ${\sigma }_g=P/A_g$ and ${\sigma }_n=P/A_n$ gives the exact conversionThis is a change of normalization, not a solution for the finite plate's stress field. For an original numerical example, take $W=40\mathrm{mm}$, $t=2\mathrm{mm}$, $d=10\mathrm{mm}$ and $P=800\mathrm{N}$. The reference stresses are $10$ and $13.333\mathrm{MPa}$. If the local maximum is stipulated as $32\mathrm{MPa}$, the same physical maximum gives $K_{tg}=3.2$ and $K_{tn}=2.4$. Neither number changes the local stress by choosing a different denominator.

The distinction appears in real historical measurements. In 1939, C. Dumont loaded a perforated aluminum-alloy plate in out-of-plane bending and measured strain and deflection at two loads. The reported peak tangential stress was 1.59 times the computed net-section stress, but 1.85 times the computed stress for an unperforated plate under the same bending forces. These are bending results for that geometry, rather than the infinite-plate tensile factor three.^[ref]

Modern demonstrations can combine a metal plate instrumented with strain gauges and a transparent polymer viewed in a polariscope. The latter uses stress-induced birefringence; its fringes respond to differences of principal stresses. The MIT laboratory compares local and remote strain measurements with qualitative photoelasticity observations around holes and a slot.^[ref]

The elastic stress concentration factor is dimensionless. It differs from the stress intensity factor used to describe a sharp crack's singular field. Local yielding or a crack changes the mechanical problem; multiplying a remote stress by the elastic $K_t$ does not, by itself, establish a failure load.

With polar angle $0\le \theta \le \pi$ and $0<|ah|\le 1$, the spherical-geodesic first integral gives ${\theta }_{\min }=\arcsin |ah|$ and ${\theta }_{\max }=\pi -\arcsin |ah|$. The maximum absolute latitude is the different angle $\arccos |ah|$. At $|ah|=1$ the great circle is the equator; at $h=0$ it is a meridian reaching both poles.