Spherical polar coordinates

Spherical polar coordinates

In mathematics and physics, spherical polar coordinates (also known as spherical coordinates) form a coordinate system for the three-dimensional real space $\scriptstyle \mathbb {R} ^{3}$. Three numbers, two angles and a length specify any point in $\scriptstyle \mathbb {R} ^{3}$. The two angles specify the position on the surface of a sphere and the length gives the radius of the sphere.

Spherical polar coordinates are useful in cases where there is (approximate) spherical symmetry, in interactions or in boundary conditions (or in both). In such cases spherical polar coordinates often allow the separation of variables simplifying the solution of partial differential equations and the evaluation of three-dimensional integrals.

Definition

Let x, y, z be Cartesian coordinates of a vector $\scriptstyle {\vec {\mathbf {r} }}$ in $\scriptstyle \mathbb {R} ^{3}$, that is,

${\vec {\mathbf {r} }}=({\vec {\mathbf {e} }}_{x},\,{\vec {\mathbf {e} }}_{y},\,{\vec {\mathbf {e} }}_{z}){\begin{pmatrix}x\\y\\z\\\end{pmatrix}}\equiv x\,{\vec {\mathbf {e} }}_{x}+y\,{\vec {\mathbf {e} }}_{y}+z\,{\vec {\mathbf {e} }}_{z},$

where $\scriptstyle {\vec {\mathbf {e} }}_{x},\,{\vec {\mathbf {e} }}_{y},\,{\vec {\mathbf {e} }}_{z}$ are unit vectors along the x, y, and z axis, respectively. The x, y, and z axes are orthogonal and so are the unit vectors along them.

The length r of the vector $\scriptstyle {\vec {\mathbf {r} }}$ is one of the three numbers necessary to give the position of the vector in three-dimensional space. By applying twice the theorem of Pythagoras we find that r² = x² + y² + z².

Let θ be the colatitude angle (see the figure) of the vector $\scriptstyle {\vec {\mathbf {r} }}$. In the usual system to describe a position on Earth, latitude has its zero at the equator, while the colatitude angle, introduced here, has its zero at the "North Pole". That is, the angle θ is zero when $\scriptstyle {\vec {\mathbf {r} }}$ is along the positive z-axis. The sum of latitude and colatitude of a point is 90⁰; these angles being complementary explains the name of the latter. The colatitude angle is also called polar or zenith angle in the literature.

The angle φ gives the angle with the x-axis of the projection $\scriptstyle {\vec {\mathbf {r} }}'$ of $\scriptstyle {\vec {\mathbf {r} }}$ on the x-y plane. The angle φ is the longitude angle (also known as the azimuth angle).

Note that the projection $\scriptstyle {\vec {\mathbf {r} }}'$ has length r sinθ. The length of the projection of $\scriptstyle {\vec {\mathbf {r} }}'$ on the x and y axis is therefore r sinθcosφ and r sinθsinφ, respectively. In summary, the spherical polar coordinates r, θ, and φ of $\scriptstyle {\vec {\mathbf {r} }}$ are related to its Cartesian coordinates by

${\begin{aligned}x&=r\sin \theta \cos \phi \\y&=r\sin \theta \sin \phi \\z&=r\cos \theta \end{aligned}}$

Given a spherical polar triplet (r, θ, φ) the corresponding Cartesian coordinates are readily obtained by application of these defining equations.

The figure makes clear that 0⁰ ≤ φ ≤ 360⁰, 0⁰ ≤ θ ≤ 180⁰, and r > 0. The coordinate surfaces are:

1. r constant, all θ and φ: surface of sphere.
2. θ constant, all r and φ: surface of a cone.
3. φ constant, all r and θ: plane.

The computation of spherical polar coordinates from Cartesian coordinates is somewhat more difficult than the converse, due to the fact that the spherical polar coordinate system has singularities, also known as points of indeterminacy. The first such point is immediately clear: if r = 0, we have a zero vector (a point in the origin). Then θ and φ are undetermined, that is to say, any values for these two parameters will give the correct result x = y = z = 0. Compare this to the case that one of the Cartesian coordinates is zero, say x = 0, then the other two coordinates are still determined (they fix a point in the yz-plane). Two other points of indeterminacy are the "North" and the "South Pole", θ = 0⁰ and θ = 180⁰, respectively (while r ≠ 0). On both poles the longitudinal angle φ is undetermined.

So, when going from Cartesian coordinates to spherical polar coordinates, one has to watch for the singularities, especially when the transformation is performed by a computer program. Given x, y and z, the consecutive steps are

${\begin{aligned}r&={\sqrt {x^{2}+y^{2}+z^{2}}}\\\theta &=\arccos(z/r),\quad r\neq 0\\r'&=r\sin \theta ={\sqrt {x^{2}+y^{2}}}\\\phi &={\begin{cases}\arccos(x/r')&\quad {\hbox{if}}\quad y\geq 0,\quad r'\neq 0\\360^{0}-\arccos(x/r')&\quad {\hbox{if}}\quad y<0,\quad r'\neq 0\\\end{cases}}\end{aligned}}$

Other convention

The convention introduced above (θ for the colatitude angle, φ for the azimuth angle) is used universally in physics. In mathematics—especially in the older and the European literature—it is very widespread, too. To quote a few prestiguous mathematical books that apply it: Abramowitz and Stegun (p. 332), Whittaker and Watson (p. 391), Courant and Hilbert (p.195), and Kline (p. 527). Until the 1960s this convention was used universally, also in mathematical textbooks, see e.g. the 1959 edition of Spiegel (p. 138).

Somewhere in the 1960s it became custom in American mathematical textbooks to use a convention in which φ and θ are interchanged, see e.g. Kay (p. 24) and Apostol (p. 419). This was done in order to not confuse students by changing the meaning of the Greek letter θ in the transition form 2D to 3D polar coordinates, as can be gathered from the following quotation of Eric Weisstein:

In this work, following the mathematics convention, the symbols for the radial, azimuth, and zenith coordinates are taken as r, θ, and φ, respectively. Note that this definition provides a logical extension of the usual polar coordinates notation, with θ remaining the angle in the xy-plane and φ becoming the angle out of that plane. The sole exception to this convention in this work is in spherical harmonics, where the convention used in the physics literature is retained (resulting, it is hoped, in a bit less confusion than a foolish rigorous consistency might engender).

In more advanced treatises—also American—on spherical functions the old convention remains in use, see e.g. Miller (p. 164). The swapping of θ and φ can only be called unfortunate, because it meant a break with the huge existing mathematics and physics literature covering more than a century, and since there exists an obvious pedagogical alternative, namely, call the angle, which appears in the 2D polar coordinates, φ instead of θ.

The convention, which calls the angle between the vector, whose coordinates are described, and the z-axis φ, is followed by the Maple algebraic program package and also by the numerical package Matlab. (Matlab also redefines the zero of the colatitude angle to be on the equator). The Mathematica package follows the convention that has θ as the angle between the vector and the z-axis.

Unit vectors

Unit vectors. $\scriptstyle {\vec {\mathbf {e} }}_{r}$ is perpendicular to the surface of the sphere, while $\scriptstyle {\vec {\mathbf {e} }}_{\theta }$ and $\scriptstyle {\vec {\mathbf {e} }}_{\phi }$ are tangent to the surface.

We will define algebraically the orthogonal set (a coordinate frame) of spherical polar unit vectors depicted on the right. In order to do this, we note that the spherical polar angles are two of the three Euler angles that describe any rotation of $\scriptstyle \mathbb {R} ^{3}$.

Indeed, start with a vector along the z-axis, rotate it around the z-axis over an angle φ. Perform the same rotation on the x, y, z coordinate frame. This rotates the x and y axis over a positive angle φ. The y axis goes to the y'-axis. Rotate then the vector and the new frame over an angle θ around the y'-axis. The vector that was initially on the z-axis is now a vector with spherical polar angles θ and φ with respect to the original (unrotated) frame. In formula

${\begin{aligned}{\vec {\mathbf {r} }}&=({\vec {\mathbf {e} }}_{x},\,{\vec {\mathbf {e} }}_{y},\,{\vec {\mathbf {e} }}_{z}){\begin{pmatrix}r\cos \phi \sin \theta \\r\sin \phi \sin \theta \\r\cos \theta \\\end{pmatrix}}\\&=({\vec {\mathbf {e} }}_{x},\,{\vec {\mathbf {e} }}_{y},\,{\vec {\mathbf {e} }}_{z})\mathbb {R} _{z}(\phi )\mathbb {R} _{y}(\theta ){\begin{pmatrix}0\\0\\r\\\end{pmatrix}},\end{aligned}}$

where the rotation matrices are defined by

$\mathbb {R} _{z}(\phi )\equiv {\begin{pmatrix}\cos \phi &-\sin \phi &0\\\sin \phi &\cos \phi &0\\0&0&1\\\end{pmatrix}},\qquad \mathbb {R} _{y}(\theta )\equiv {\begin{pmatrix}\cos \theta &0&\sin \theta \\0&1&0\\-\sin \theta &0&\cos \theta \\\end{pmatrix}}.$

By direct matrix multiplication the matrix expression for the spherical polar coordinates of $\scriptstyle {\vec {\mathbf {r} }}$ is easily verified—it could have been postulated without reference to Euler rotations and proved by verification.

We now introduce the coordinate frame depicted in the figure on the right:

${\begin{aligned}{\vec {\mathbf {r} }}&=({\vec {\mathbf {e} }}_{x},\,{\vec {\mathbf {e} }}_{y},\,{\vec {\mathbf {e} }}_{z})\mathbb {R} _{z}(\phi )\mathbb {R} _{y}(\theta ){\begin{pmatrix}0\\0\\r\\\end{pmatrix}}\equiv ({\vec {\mathbf {e} }}_{\theta },\,{\vec {\mathbf {e} }}_{\phi },\,{\vec {\mathbf {e} }}_{r}){\begin{pmatrix}0\\0\\r\\\end{pmatrix}}.\end{aligned}}$

That is, the new frame, depicted in the figure, is related to the old frame along the x, y, and z axes by rotation,

$({\vec {\mathbf {e} }}_{\theta },\,{\vec {\mathbf {e} }}_{\phi },\,{\vec {\mathbf {e} }}_{r})\equiv ({\vec {\mathbf {e} }}_{x},\,{\vec {\mathbf {e} }}_{y},\,{\vec {\mathbf {e} }}_{z})\mathbb {R} _{z}(\phi )\mathbb {R} _{y}(\theta ).$

Written out:

${\begin{aligned}{\vec {\mathbf {e} }}_{\theta }&={\vec {\mathbf {e} }}_{x}\cos \phi \cos \theta +{\vec {\mathbf {e} }}_{y}\sin \phi \cos \theta -\sin \theta {\vec {\mathbf {e} }}_{z}\\{\vec {\mathbf {e} }}_{\phi }&=-{\vec {\mathbf {e} }}_{x}\sin \phi +{\vec {\mathbf {e} }}_{y}\cos \phi \\{\vec {\mathbf {e} }}_{r}&={\vec {\mathbf {e} }}_{x}\cos \phi \sin \theta +{\vec {\mathbf {e} }}_{y}\sin \phi \sin \theta +\cos \theta {\vec {\mathbf {e} }}_{z}.\\\end{aligned}}$

Inverting this set of equations is very easy, since rotation matrices are orthogonal, that is, their inverse is equal to their transpose.

Apparently $\scriptstyle {\vec {\mathbf {r} }}$ is along $\scriptstyle {\vec {\mathbf {e} }}_{r}$. Since the two rotation matrices are orthogonal (have orthonormal rows and columns), the new frame is orthogonal. Since the two rotation matrices have unit determinant (are proper rotations), the new frame is right-handed.

Recall, parenthetically, that free parallel vectors have the same coordinate triplet without respect to a given coordinate frame. Or, equivalently, coordinate frames may be freely translated in a parallel manner. That is, the frame in the figure could have been drawn equally well with its origin in the crossing of the x, y, and z axes, which, however, would have obscured the fact that $\scriptstyle {\vec {\mathbf {e} }}_{\phi }$ and $\scriptstyle {\vec {\mathbf {e} }}_{\theta }$ are tangent to the surface of the sphere.

Metric tensor

In curvilinear coordinates q ⁱ the metric tensor (with elements g _ij ) defines the square of an infinitesimal distance,

$ds^{2}\equiv \sum _{i=1}^{3}g_{ij}dq^{i}dq^{j}.$

The Cartesian metric tensor is the identity matrix and hence in Cartesian coordinates,

$ds^{2}=dx^{2}+dy^{2}+dz^{2}=(dx\,dy\,dz){\begin{pmatrix}dx\\dy\\dz\end{pmatrix}}.$

Consider the following expression between differentials,

${\begin{pmatrix}dx\\dy\\dz\\\end{pmatrix}}=\mathbb {J} {\begin{pmatrix}d\theta \\d\phi \\dr\\\end{pmatrix}},$

with the Jacobi matrix, which is obtained by application of the chain rule, having the following form,

$\mathbb {J} \equiv {\begin{pmatrix}{\frac {\partial x}{\partial \theta }}&{\frac {\partial x}{\partial \phi }}&{\frac {\partial x}{\partial r}}\\{\frac {\partial y}{\partial \theta }}&{\frac {\partial y}{\partial \phi }}&{\frac {\partial y}{\partial r}}\\{\frac {\partial z}{\partial \theta }}&{\frac {\partial z}{\partial \phi }}&{\frac {\partial z}{\partial r}}\\\end{pmatrix}}={\begin{pmatrix}r\cos \phi \cos \theta &-r\sin \phi \sin \theta &\cos \phi \sin \theta \\r\sin \phi \cos \theta &r\cos \phi \sin \theta &\sin \phi \sin \theta \\-r\sin \theta &0&\cos \theta \\\end{pmatrix}}.$

By inspection it follows that

$\mathbb {J} =\mathbb {R} _{z}(\phi )\mathbb {R} _{y}(\theta ){\begin{pmatrix}r&0&0\\0&r\sin \theta &0\\0&0&1\\\end{pmatrix}}.$

The columns of $\scriptstyle \mathbb {R} _{z}(\phi )\mathbb {R} _{y}(\theta )$ are orthogonal vectors that are normalized to unity. Hence the columns of the Jacobi matrix, which are proportional to the columns of $\scriptstyle \mathbb {R} _{z}(\phi )\mathbb {R} _{y}(\theta )$, are orthogonal, but not normalized. The inverses of the normalization factors are on the diagonal of the matrix on the right of the expression. These inverse normalization factors are known as scale factors or Lamé factors. Usually they are denoted by h. Hence the spherical polar scale factors are

$h_{\theta }=r,\qquad h_{\phi }=r\sin \theta ,\qquad h_{r}=1.$

The infinitesimal distance can be written as follows

$ds^{2}=(dx\,dy\,dz){\begin{pmatrix}dx\\dy\\dz\end{pmatrix}}=(d\theta \,d\phi \,dr){\begin{pmatrix}r^{2}&0&0\\0&r^{2}\sin ^{2}\theta &0\\0&0&1\\\end{pmatrix}}{\begin{pmatrix}d\theta \\d\phi \\dr\end{pmatrix}}$

where we used that the rotation matrices are orthogonal (matrix times its transpose gives the identity matrix), so that

$\left[\mathbb {R} _{z}(\phi )\mathbb {R} _{y}(\theta ){\begin{pmatrix}r&0&0\\0&r\sin \theta &0\\0&0&1\\\end{pmatrix}}\right]^{T}\mathbb {R} _{z}(\phi )\mathbb {R} _{y}(\theta ){\begin{pmatrix}r&0&0\\0&r\sin \theta &0\\0&0&1\\\end{pmatrix}}={\begin{pmatrix}r^{2}&0&0\\0&r^{2}\sin ^{2}\theta &0\\0&0&1\\\end{pmatrix}}.$

The rightmost matrix being the metric tensor associated with spherical polar coordinates, we find

$ds^{2}=rd\theta ^{2}+r^{2}\sin ^{2}\theta d\phi ^{2}+dr^{2}\,.$

The fact that the metric tensor is diagonal is expressed by stating that the spherical polar coordinate system is orthogonal. We see that the metric tensor has the squares of the respective scale factors on the diagonal.

From tensor analysis it is known that an infinitesimal surface element spanned by two coordinates is equal to

$dA^{(ik)}={\sqrt {g_{ii}g_{kk}-g_{ik}^{2}}}\,dq^{i}dq^{k}.$

For spherical polar coordinates it follows that

${\begin{aligned}dA^{(\theta \phi )}&=r^{2}\sin \theta \,d\theta d\phi \\dA^{(r\theta )}&=r\,drd\theta \\dA^{(\phi r)}&=r\sin \theta d\phi dr\\\end{aligned}}$

As an example we compute the area of the surface of a sphere with radius R,

$A=\int _{0}^{\pi }\int _{0}^{2\pi }R^{2}\sin \theta \,d\theta d\phi =4\pi R^{2}.$

The weight appearing in the infinitesimal volume element is the determinant of the Jacobi matrix,

$\det {\big [}\mathbb {J} {\big ]}=\det {\big [}\mathbb {R} _{z}(\phi ){\big ]}\,\det {\big [}\mathbb {R} _{y}(\theta ){\big ]}\,r^{2}\sin \theta =r^{2}\sin \theta ,$

where we used that the determinant of a diagonal matrix is the product of its diagonal elements and the fact that the determinants of proper rotation matrices are unity. Hence, the volume V of a sphere with radius R is,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V= \int_{0}^R \int_{0}^\pi \int_{0} ^{2\pi} r^2\sin\theta\, dr d\theta d\phi =\frac{4}{3}\pi R^3. }

Differential operators

In vector analysis a number of differential operators expressed in curvilinear coordinates play an important role. They are the gradient, the divergence, the curl, and the Laplace operator. It is possible to derive general expressions for these operators that are valid in any coordinate system and are based on the metric tensor associated with the coordinate system. In the case of orthogonal systems (diagonal metric tensors) only the square roots of the diagonal elements (the scale factors) appear in the expressions. Since these general relations exist, we will not give derivations for the special case of spherical polar coordinates, but depart from the general expressions.

Above we derived the following scale factors for the spherical polar coordinates,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h_r = 1, \qquad h_\theta = r, \qquad h_\phi = r\sin\theta }

and we showed that the unit vectors Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle \vec\mathbf{e}_\theta, \, \vec\mathbf{e}_\phi, \, \vec\mathbf{e}_r} are obtained by two rotations of a Cartesian system.

The gradient of a scalar function Φ is,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla \Phi = \frac{\partial \Phi}{\partial r}\vec\mathbf{e}_r +\frac{1}{r} \frac{\partial \Phi}{\partial \theta}\vec\mathbf{e}_\theta +\frac{1}{r\sin\theta} \frac{\partial \Phi}{\partial \phi}\vec\mathbf{e}_\phi. }

If the vector function A is,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{A} = A_r \vec\mathbf{e}_r + A_\theta\vec\mathbf{e}_\theta +A_\phi\vec\mathbf{e}_\phi. }

then its divergence is,

${\begin{aligned}\nabla \cdot \mathbf {A} &={\frac {1}{r^{2}\sin \theta }}\left[{\frac {\partial r^{2}\sin \theta A_{r}}{\partial r}}+{\frac {\partial r\sin \theta A_{\theta }}{\partial \theta }}+{\frac {\partial rA_{\phi }}{\partial \phi }}\right]\\&={\frac {2}{r}}A_{r}+{\frac {\partial A_{r}}{\partial r}}+{\frac {\cos \theta }{r\sin \theta }}A_{\theta }+{\frac {1}{r}}{\frac {A_{\theta }}{\partial \theta }}+{\frac {1}{r\sin \theta }}{\frac {\partial A_{\phi }}{\partial \phi }},\end{aligned}}$

and its curl is given by the determinant,

$\nabla \times \mathbf {A} ={\frac {1}{r^{2}\sin \theta }}{\begin{vmatrix}{\vec {\mathbf {e} }}_{r}&r{\vec {\mathbf {e} }}_{\theta }&r\sin \theta {\vec {\mathbf {e} }}_{\phi }\\{\frac {\partial }{\partial r}}&{\frac {\partial }{\partial \theta }}&{\frac {\partial }{\partial \phi }}\\A_{r}&rA_{\theta }&r\sin \theta A_{\phi }\\\end{vmatrix}}.$

The Laplace operator of the scalar function Φ is,

$\nabla ^{2}\Phi ={\frac {1}{r^{2}}}\left[{\frac {\partial }{\partial r}}r^{2}{\frac {\partial \Phi }{\partial r}}+{\frac {1}{\sin \theta }}{\frac {\partial }{\partial \theta }}\sin \theta {\frac {\partial \Phi }{\partial \theta }}+{\frac {1}{\sin ^{2}\theta }}{\frac {\partial ^{2}\Phi }{\partial \phi ^{2}}}\right].$