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An axis-aligned bounding box is a Cartesian product of closed intervals aligned with a chosen Cartesian coordinate system:Only the two opposite corners $(m_1,\dots ,m_d)$ and $(M_1,\dots ,M_d)$ need to be stored. In two dimensions it is a rectangle. Changing the Cartesian coordinate system can change how tightly the axis-aligned bounding box fits the same object.^[ref]

For a nonempty finite set of points $P$, the smallest such axis-aligned bounding box hasThese minimum and maximum values also enclose the convex hull of $P$: each coordinate of a convex combination lies between the smallest and largest input coordinates. In particular, three vertices of a polygon determine the axis-aligned bounding box of a filled triangle. For a general nonempty compact set, the same formulas use all its points; inspecting only an arbitrary finite sample is insufficient.

Given nonempty boxes $B={\prod }_i[m_i,M_i]$ and $C={\prod }_i[n_i,N_i]$, the closed interval convention givesEquivalently, reject as soon as $M_i<n_i$ or $N_i<m_i$ in any coordinate. The surviving set intersection has endpoints $\max (m_i,n_i)$ and $\min (M_i,N_i)$. This is an exact box test, while its conclusion about enclosed objects remains a conservative enclosure test.^[ref]

Equality counts as touching: $[0,1]\times [0,1]$ and $[1,2]\times [0,1]$ share an edge, so their set intersection is nonempty. Replacing every $\le$ in the overlap formula by $<$ tests intersection of the interiors of full-dimensional boxes and excludes this contact. A zero-width closed interval can represent a point or a lower-dimensional enclosure, and must not automatically be discarded when touching counts. A box with some $m_i>M_i$ is empty; test emptiness before applying rules intended for nonempty boxes. The broad phase must retain touching candidates whenever the narrow phase counts touching as a hit.

For nonempty boxes, the smallest enclosing axis-aligned bounding box of their set union isThis operation supplies the parent boxes in a bounding volume hierarchy. It is an enclosure of the set union, rather than generally the set union itself. In floating-point arithmetic, comparisons are conclusive only for endpoints that still enclose the intended exact objects; use certified bounds such as interval arithmetic with outward rounding when that guarantee is required.

// Target: analysis.bigb

For a smooth horizontal circular hoop of radius $a$, rotated at constant angular velocity $\omega$ about a vertical axis through a rim point, measure $\theta$ at that pivot from the diameter toward the bead. In rotating Cartesian coordinate system coordinates, $\mathbf{r}=a(1+\cos 2\theta ,\sin 2\theta ,0)$. The tangential component of the centrifugal acceleration gives the displayed equation; the Coriolis acceleration is normal to the hoop. The farthest point $\theta =0$ is stable for $\omega \ne 0$, with small oscillation angular frequency $|\omega |$. The pivot itself is unstable. This geometry differs from a vertical rotating hoop and has no tangential gravitational term.

A Cartesian basis is an ordered basis of unit vectors along the mutually perpendicular axes of a Cartesian coordinate system. It is an orthonormal basis of a real Euclidean space: $e_i\cdot e_j={\delta }_{ij}$, with ${\delta }_{ij}$ the Kronecker delta. In the plane, the familiar standard basis is $e_1=(1,0{)}^T$, $e_2=(0,1{)}^T$; the vector $p=(2,1{)}^T$ is $2e_1+e_2$. A rotated pair of perpendicular unit vectors is also a Cartesian basis, even though its basis vectors have different standard basis entries.

The advantage is that coordinates can be measured separately by dot products, and the squared norm is the sum of their squares:These are the real Euclidean space instances of the orthonormal basis expansion.^[ref]

An orthogonal basis only requires perpendicular basis vectors; their lengths need not be one. For example, $(2,0{)}^T,(0,3{)}^T$ are orthogonal, but $p=(2,1{)}^T$ has coordinates $(1,1/3{)}^T$, while its dot products with them are $4,3$. For an orthogonal basis $(a_i)$, the coordinate along $a_i$ is $(p\cdot a_i)/(a_i\cdot a_i)$. Dividing each nonzero basis vector by its norm makes it a unit vector without changing its direction. Unit length alone does not imply orthogonality: $(1,0{)}^T$ and $(1,1{)}^T/\sqrt{2}$ are both unit vectors but have dot product $1/\sqrt{2}$.

The examples below show why these distinctions matter: rotating a Cartesian basis changes the coordinate vector while keeping lengths easy to compute; using nonperpendicular basis vectors still gives unique coordinates, but requires the dual basis or a matrix inverse to extract them.

A conic section is obtained by intersecting a plane with a double circular cone. Cuts away from the vertex produce an ellipse, a parabola or a hyperbola; a perpendicular cut is the circle case of an ellipse. A cut through the vertex can instead produce a degenerate conic. One simple surface therefore connects several familiar plane curves.^[ref]

Consider the original familySubstituting the plane equation into the double circular cone equation givesThus $m=0$ gives a circle, $m=3/5$ an ellipse, $m=1$ a parabola, and $m=7/5$ a hyperbola. The last cut meets both sheets of the double circular cone, producing its two branches. These parameters specify a mathematical family, rather than measurements of a physical object.

The Cartesian coordinate system descriptions of these curves are second-degree equationsA rotation and a translation expose their principal directions and offsets. The conic type discriminant identifies the ordinary nondegenerate real type, while a separate degenerate conic check prevents confusing two crossing straight lines with a hyperbola.

This is useful because a difficult-looking equation can be understood through a small set of geometric features: foci of a conic, directrices of a conic, eccentricity of a conic and the semi-major and semi-minor axes. The same ellipse can be traced by a constant-length string or parametrized by $x=a\cos t$, $y=b\sin t$. These are different descriptions of one curve, with different advantages for construction and calculation.

One of the straight lines through the origin parallel to a standard basis vector in a Cartesian coordinate system. The positive direction corresponds to positive values of that coordinate.

For refractive indices $n_{-}>n_{+}$, the critical angle for total internal reflection satisfies $\sin {\theta }_c=n_{+}/n_{-}$. Above it, the transmitted normal wavenumber is imaginary and the transmitted field is an evanescent wave.

The critical angle for total internal reflection is the boundary between outgoing refraction and an evanescent wave, rather than an angle at which the external electric field abruptly disappears. Write $n_1=n_{-}$ and $n_2=n_{+}$. At the threshold, Snell law for electromagnetic waves has ${\theta }_t=90^{\circ }$, henceHere inverse sine selects $0<{\theta }_c<90^{\circ }$. For $n_1=1.5$ and $n_2=1$, $\sin {\theta }_c=2/3$ and ${\theta }_c\approx 41.8103149^{\circ }$. The threshold is independent of polarization of an electromagnetic wave under the stated assumptions.

A short proof using phase matching at a planar wave interface reveals all three cases at once. Choose a Cartesian coordinate system with boundary $z=0$, incident medium $z<0$, and transmitted medium $z>0$. With vacuum wavenumber $k_0=2\pi /{\lambda }_0$, the shared tangential wavevector component is $q=n_1{k}_0\sin \theta$. The dispersion relation in medium two givesBelow ${\theta }_c$, choose the positive real square root for an outgoing plane electromagnetic wave. At ${\theta }_c$, $k_{2z}=0$: there is a grazing wave, with zero normal average power but no exponential confinement in $z$. Above ${\theta }_c$, choose $k_{2z}=i\kappa$ with $\kappa >0$, so the wave amplitude decays into the second half-space.

If $n_1\le n_2$, no nongrazing angle of incidence produces this threshold. As $n_2/n_1\to 1$ from below, ${\theta }_c\to 90^{\circ }$; as that ratio tends to zero, ${\theta }_c\to 0$. These limits explain why a larger refractive index contrast permits confinement over a wider range of internal angles.

Euclidean geometry describes flat Euclidean space using lengths, angles, and rigid motions. A coordinate realization uses points $x,y\in {\mathbb{R}}^n$, Euclidean distanceand angles determined by the dot product. The coordinate axes are a convenience: translations and rotations change coordinates while preserving these measurements. This makes Euclidean geometry useful for comparing figures and converting measurements along different directions into a single distance.

In the Euclidean plane, through a point outside a line there is exactly one parallel line. This is a standard equivalent formulation of the parallel postulate. Euclid's fifth postulate in the Elements instead describes when two lines meet using the interior angles made by a crossing line.^[ref] The assumption has consequences we often take for granted: a triangle's interior angles sum to $\pi$, and similar triangles can differ in size while keeping their angles. Comparing with spherical geometry reveals that these are properties of a particular geometry.

As a computed example, the diagonal from $(0,0)$ to $(4,3)$ in a rectangle has length $\sqrt{4^2+3^2}=5$. The Pythagorean theorem explains why perpendicular contributions add as squares. If the directions cease to meet at a right angle, the law of cosines supplies the missing angle-dependent term. The beauty is that a single distance formula unites geometric constructions, coordinate calculations, and orthogonality in linear algebra.

Make the assumptions visible: in a Cartesian coordinate system with orthogonal axes and the same length unit on each axis, take $A=(0,0)$, $B=(4,0)$ and $C=(0,3)$. The vectors $B-A=(4,0)$ and $C-A=(0,3)$ have dot product zero, establishing a right angle at $A$ before using the Pythagorean theorem. A sketch that merely looks like a right triangle supplies no such guarantee. Likewise, an image stretched horizontally and vertically by different amounts can distort both lengths and angles; a displayed shape and the assumptions defining it must be kept separate.

For a less origin-dependent measurement, take $P=(1,2)$ and $Q=(5,5)$. Their displacement vector is $Q-P=(4,3)$, so their Euclidean distance is again $5$. Its Euclidean norm records the direct separation, whereas travelling first horizontally and then vertically covers length $4+3=7$. This connects measuring a diagonal to locating two positions from a Cartesian coordinate system. A translation changes where the positions are recorded without changing their separation; a rotation changes the component values without changing the Euclidean norm.

A Euclidean vector is a vector in a finite-dimensional real inner product space. Choosing an orthonormal basis identifies it with its real vector components in ${\mathbb{R}}^n$. A geometric arrow represents a Euclidean vector by a length and a direction; translating the arrow without changing either represents the same Euclidean vector. The zero vector has length zero and no distinguished direction.^[ref]

For points $P=(1,1)$ and $Q=(4,5)$ in a Cartesian coordinate system, the displacement is $Q-P=(3,4)$. The arrow from $(-2,0)$ to $(1,4)$ represents the same Euclidean vector, even though its endpoints differ. An endpoint is a position; their difference is a vector. In an orthonormal basis $({\mathbf{e}}_1,{\mathbf{e}}_2)$,The Euclidean norm comes from the inner product: $\Vert \mathbf{v}{\Vert }_2^2=⟨\mathbf{v},\mathbf{v}⟩$. The sum of squared vector components is valid in an orthonormal basis; an arbitrary basis requires the appropriate Gram matrix. This distinction lets the geometric vector stay the same while its coordinate vector changes.^[ref]

Vector addition is componentwise. With $\mathbf{u}=(2,1)$ and $\mathbf{w}=(1,3)$,Geometrically, translate the second arrow so its tail starts at the first arrow's head. The vector sum joins the starting point to the final endpoint. Reversing the order gives the other route around the same parallelogram, so vector addition is commutative. This is the useful bookkeeping behind adding successive displacements, velocities, or forces: direction information survives the calculation. For a force, the Euclidean vector alone does not specify its point of application, which also affects the torque.^[ref]

The magnitude of a vector sum is generally different from the sum of the magnitudes. Here $\Vert \mathbf{u}+\mathbf{w}{\Vert }_2=5$, whereas $\Vert \mathbf{u}{\Vert }_2+\Vert \mathbf{w}{\Vert }_2=\sqrt{5}+\sqrt{10}>5$. The triangle inequality explains why the direct displacement cannot exceed the length of this two-segment route. A return displacement $(-3,-4)$ makes the net vector sum zero, although the lengths of the three travelled segments sum to a positive number.

The dot product exposes direction quantitatively: for nonzero Euclidean vectors, $⟨\mathbf{u},\mathbf{w}⟩=\Vert \mathbf{u}{\Vert }_2\Vert \mathbf{w}{\Vert }_2\cos \theta$. For the displayed pair, the dot product is $2\cdot 1+1\cdot 3=5$, so $\cos \theta =1/\sqrt{2}$ and the angle is $45^{\circ }$. For orthogonal vectors, that dot product is zero. Dividing by magnitudes separates direction from size, which is the purpose of vector normalization.

The Faraday effect is magnetic field-induced circular birefringence: opposite circular polarizations acquire different wave phases, rotating transmitted linear polarization. For a transparent medium with isotropy before applying the magnetic field, in the linear magnetic-response regime at fixed vacuum wavelength ${\lambda }_0$ and temperature $T$,Here $V$ is the Verdet constant, $L$ is the material path length, and $B_{\parallel }$ is the applied magnetic flux density component along the ray. This $\mathbf{B}$ is the external bias, distinct from the oscillating magnetic field of the light itself. Michael Faraday discovered the effect in 1845.^[ref]

Fix the sign before using the formula. On the forward path choose a right-handed Cartesian coordinate system, propagation along $+z$, and measure $\theta$ from $+x$ toward $+y$ for the electric field's oscillation line. Positive $\theta$ is counterclockwise in the $x$-right, $y$-up end view seen from $z>0$, looking toward the source. Then $B_{\parallel }=B_z$ and positive $V$ means positive rotation for positive $B_z$. This convention differs from the clockwise-positive convention in optical rotation. A linear polarization line is unchanged by $180^{\circ }$; a measured orientation determines accumulated rotation only modulo $\pi$, unless continuity selects a branch.

For uniform $B_{\parallel }$, the formula reduces to $\theta =VBL$. Its appeal is that a weak change of refractive index becomes a visible change of orientation. Magnetic reversal measurement of Faraday rotation turns this into an inference of $V$ or of the magnetic field path integral. Faraday rotation on a round trip explains the directional behavior used by an optical isolator. In an astronomical cold plasma, the high-frequency limit instead leads to the rotation measure law $\theta =\mathrm{RM}{\lambda }_0^2$; that wavelength dependence is not a universal law for transparent solids.

For $\dot{r}=r(a-g(r))$, $\dot{\theta }=1$ and $g(r)=1+[1+(r-1{)}^2{]}^{-1}$, the origin loses stability at $a=3/2$ and a stable small circular orbit appears for larger $a$. Its radius is $1-\sqrt{(2-a)/(a-1)}\sim 2(a-3/2)$. The Cartesian vector field is continuously differentiable but not generally twice differentiable at the origin, so the usual smooth Hopf normal form and its square-root amplitude law do not apply directly. The explicit radial equation proves the transition.

The Laplacian takes the divergence of a scalar field's gradient. For a real-valued function $f$ on an open region of Euclidean space, with continuous partial derivatives through order two, its expression in a Cartesian coordinate system isThus the Laplacian, also called the Laplace operator, adds the coordinate second derivatives; equivalently it is the matrix trace of the Hessian matrix. Its output is another scalar field.^[ref]

Three original examples distinguish shape from this sum. The bowl $f(x,y)=x^2+y^2$ has $\Delta f=2+2=4$. The saddle $g(x,y)=x^2-y^2$ instead has $\Delta g=2-2=0$: opposite second derivatives cancel. It is a harmonic function, although its graph of a function is not a plane. For the Gaussian function $h(x,y)=e^{-r^2}$, with $r^2=x^2+y^2$, differentiation givesIts Laplacian is negative inside the circle of radius one, zero on that circle, and positive outside it. The Gaussian function itself is positive throughout: its value and its Laplacian describe different properties.

The sum does not favor a particular orientation. Under a change of Cartesian coordinate system represented by an orthogonal matrix $Q$, the Hessian matrix becomes $Q^{\mathsf{T}}(D^{2}f)Q$. The cyclic property of the trace and $Q{Q}^{\mathsf{T}}=I$ preserve its matrix trace. This makes the Laplacian useful when the behavior depends on spatial differences equally in every direction, rather than on the arbitrary choice of axes.

For a vector field, the Euclidean Laplacian acts separately on components expressed in a fixed Cartesian coordinate system. Components in a position-dependent basis require differentiating that basis too; applying a scalar formula separately to polar components omits those terms.^[ref] The scalar formula in Laplacian in polar coordinates is valid away from $r=0$; smoothness at the origin must be handled in coordinates that remain regular there.

The point assigned all-zero coordinates in a chosen Cartesian coordinate system. Choosing an origin identifies position displacements with vectors; a bare affine space has no distinguished origin.

Two vectors are orthogonal when their inner product is zero.

For real Euclidean vectors, the inner product is the dot product, so $(3,4)$ and $(4,-3)$ are orthogonal: $3(4)+4(-3)=0$. For nonzero Euclidean vectors, this means their angle is $90^{\circ }$. The zero vector is orthogonal to every vector, although it has no direction or defined angle.^[ref]

In an abstract inner product space, orthogonality always refers to its chosen inner product. For example, the weighted inner product $⟨u,v{⟩}_w=4u_1{v}_1+u_2{v}_2$ makes $(1,1)$ and $(1,-4)$ orthogonal, since $4-4=0$. Their ordinary dot product is $-3$, so their arrows are not perpendicular in the usual Cartesian coordinate system. The algebraic definition also applies to functions, where arrows need not be available.

The original PDF measures $\theta$ at the rim point $O$, from the positive $x^{\prime }$-axis to $OP$; it is not the angle at the centre of the circle. The converted TeX omits this entire subpart. In rotating Cartesian coordinate system, the centre is $(a,0)$, and the nonzero intersection of the ray $\mathbf{r}=\rho (\cos \theta ,\sin \theta )$ with $(x^{\prime }-a{)}^2+y^{\prime 2}=a^2$ has $\rho =2a\cos \theta$. ThereforeThe second form stays regular even at the rim point $O$, where the polar radius is zero. Define the outward unit normal from the centre and the increasing-angle unit tangent vector byThe first two derivatives in the rotating reference frame are ${\mathbf{v}}^{\prime }=2a\dot{\theta }\mathbf{\tau }$ and ${\mathbf{a}}^{\prime }=2a\ddot{\theta }\mathbf{\tau }-4a{\dot{\theta }}^2\mathbf{n}$. These describe a bead on a horizontal hoop rotating about its rim.

Let the free surface be at $r=R_1$ and take the Newtonian gravitational potential to vanish at infinity. Mass conservation fixes $R_1=(3M_1/(4\pi {\rho }_1){)}^{1/3}$. The enclosed mass is $m(r)=4\pi {\rho }_1{r}^3/3$, and hydrostatic equilibrium requires $d{p}_0/dr=-{\rho }_{1}Gm(r)/r^2$. With vacuum outside, $p_0(R_1)=0$, givingThe pressure is zero outside. A prescribed constant external pressure would simply add that constant to $p_0$.

The bulk Eulerian and Lagrangian fluid perturbations obey $\delta \rho =0$ because the mass density is uniform and the Lagrangian displacement has zero divergence. The linearized Euler momentum equation in the interior isTaking the curl gives ${\omega }^2\nabla \times \mathbf{\xi }=0$. Thus every nonzero-frequency normal mode has an irrotational flow displacement; the simply connected space formed by the interior admits $\mathbf{\xi }=\nabla U$ and incompressibility gives the Laplace equation ${\nabla }^{2}U=0$. For a regular center, one spherical harmonic component is $U=A_l{r}^l{Y}_l^m$, with $|m|\le l$.

It is essential to retain the surface gravity perturbation of a uniform-density sphere. Although the bulk mass density perturbation is zero, the displaced density discontinuity givesThe Poisson equation implies that $\delta \Phi$ is continuous and its outward radial derivative has the jumpUsing regularity at the center and decay at infinity, writeThe derivative jump is $-(2l+1)C_l{R}_1^{l-1}{Y}_l^m$, so $C_l=-4\pi G{\rho }_{1}l{A}_l/(2l+1)$.

Integrating the bulk Euler momentum equation gives $\delta p/{\rho }_1={\omega }^{2}U-\delta \Phi$ for $l\ge 1$. At the free surface the Lagrangian pressure perturbation vanishes:Combining these relations yields the Kelvin stellar mode frequencyThe $2l+1$ choices of $m$ are degenerate because of spherical symmetry. For $l=1$, $r{Y}_1^m$ is a linear combination of coordinates in a Cartesian coordinate system, so $\nabla U$ is a constant displacement: the zero frequency is rigid translation of the whole isolated body. Such a translation has no restoring force. The constant $l=0$ potential generates no displacement; a radial breathing motion is excluded by incompressibility and a regular center.

The printed irrotational assertion needs the nonzero-frequency qualification, or restriction to the potential incompressible stellar surface modes. It is false for all neutral displacements: $\mathbf{\xi }=\mathbf{\Omega }\times \mathbf{r}$ has zero divergence, zero normal displacement at the surface, and $\delta p=\delta \Phi =0$, so it is a zero-frequency displacement, but $\nabla \times \mathbf{\xi }=2\mathbf{\Omega }\ne 0$. This does not change the Kelvin stellar mode spectrum just derived.

Use the gnomonic projection associated with an undistorted focal plane. Write $\Delta =\alpha -A$ and rotate the celestial Cartesian coordinate system so that the pointing meridian has longitude zero. The stellar direction and an orthonormal basis adapted to the optical axis areHere $\mathbf{e}$ points east and $\mathbf{n}$ points north. Intersect the ray $t\mathbf{s}$ with the plane $\mathbf{x}\cdot \mathbf{b}=f$. It gives $t=f/(\mathbf{s}\cdot \mathbf{b})$, so the detector coordinates areThese equations also avoid spurious singularities caused by writing individual tangents or cotangents.

To put this gnomonic projection in the desired form, let $\rho =({\cos }^2\delta {\cos }^2\Delta +{\sin }^2\delta {)}^{1/2}$ and choose $q$ locally byThus $\cot q=\cot \delta \cos \Delta$. The denominator becomes $\rho \cos (q-D)$, the numerator for $\eta$ becomes $\rho \sin (q-D)$, and $\cos \delta \sin \Delta =\rho \cos q\tan \Delta$. ConsequentlyUse the local meridian chart $\cos \Delta >0$, with $q\in [-\pi /2,\pi /2]$ chosen continuously near $D$ and $q=D$ at the image centre. A visible gnomonic projection requires $\mathbf{s}\cdot \mathbf{b}>0$, but this front-hemisphere condition alone does not select that meridian chart. A narrow field near a celestial pole can cross the opposite meridian; for such fields use the Cartesian expressions above and distinguish the oriented great circle parameter from ordinary declination. In particular, the small-field limit is $\xi \simeq f\cos D(\alpha -A)$ and $\eta \simeq f(\delta -D)$, with angles in radians. The factor $\cos D$ is the shrinking angular distance per unit right ascension near the pole.

The magnetic field is $B=\nabla \times A$. The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$, so the curl of the curl identity givesA gauge transformation $A\mapsto A+\nabla \chi$ does not change $B$. We may therefore choose the Coulomb gauge $\nabla \cdot A=0$ by solving the Poisson equation ${\nabla }^2\chi =-\nabla \cdot A$. It follows component by component in a Cartesian coordinate system thatThe free-space Green function of the Laplacian then givesFor a localized steady current density, conservation of electric charge gives ${\nabla }^{\prime }\cdot J=0$; integration by parts confirms directly that this integral has zero divergence and hence obeys the Coulomb gauge.

Write $\times$ for the paper's $\wedge$. In a right-handed Cartesian coordinate system, the cross product isIt is perpendicular to both vectors, has norm $|\mathbf{a}||\mathbf{b}|\sin \varphi$, where $\varphi$ is their angle, and has direction given by the right-hand rule when nonzero. Its norm is the area of the parallelogram spanned by the two vectors.

If $\mathbf{b}\times \mathbf{c}\ne 0$, use that parallelogram as the base of the parallelepiped. Its perpendicular height is $|\mathbf{a}\cdot \mathbf{n}|$, where $\mathbf{n}=(\mathbf{b}\times \mathbf{c})/|\mathbf{b}\times \mathbf{c}|$. Thus its volume isIf $\mathbf{b}\times \mathbf{c}=0$, both the volume and this scalar triple product vanish. If the cross product is nonzero, the scalar triple product vanishes exactly when $\mathbf{a}$ lies in the plane spanned by $\mathbf{b},\mathbf{c}$. If it is zero, $\mathbf{b},\mathbf{c}$ are linearly dependent, and adjoining $\mathbf{a}$ still spans at most a plane. Consequently the three vectors are coplanar if and only if $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=0$, including cases involving a zero vector.

For $r>0$, differentiate $r=\sqrt{x^2+y^2}$ while keeping the other coordinate fixed in the Cartesian coordinate system:The origin is excluded from this polar coordinates chart.

The volume element is the absolute Jacobian determinant times the coordinate differentials. In the Cartesian coordinate system, cylindrical coordinates $(r,\theta ,z)$ and spherical coordinates $(\rho ,\vartheta ,\phi )$, respectively,Here $\rho$ is distance from the origin, $\vartheta$ is the polar angle from the positive $z$ axis, and $\phi$ is the azimuthal angle. The spherical coordinates convention fixes which angle enters the sine factor.