Spherical Coordinates

A system of Curvilinear Coordinates which is natural for describing positions on a Sphere or Spheroid. Define $\theta$ to be the azimuthal Angle in the $xy$-Plane from the x-Axis with $0\leq\theta<2\pi$ (denoted $\lambda$ when referred to as the Longitude), $\phi$ to be the polar Angle from the z-Axis with $0\leq\phi\leq\pi$ (Colatitude, equal to $\phi=90^\circ-\delta$ where $\delta$ is the Latitude), and $r$ to be distance (Radius) from a point to the Origin.

Unfortunately, the convention in which the symbols $\theta$ and $\phi$ are reversed is frequently used, especially in physics, leading to unnecessary confusion. The symbol $\rho$ is sometimes also used in place of $r$. Arfken (1985) uses $(r, \phi, \theta)$, whereas Beyer (1987) uses $(\rho, \theta, \phi)$. Be very careful when consulting the literature.

In this work, the symbols for the azimuthal, polar, and radial coordinates are taken as $\theta$, $\phi$, and $r$, respectively. Note that this definition provides a logical extension of the usual Polar Coordinates notation, with $\theta$ remaining the Angle in the $xy$-Plane and $\phi$ becoming the Angle out of the Plane.

$$r = \sqrt{x^2+y^2+z^2}$$ (1)

$$\theta = \tan^{-1}\left({y\over x}\right)$$ (2)

$$\phi = \sin^{-1}\left({\sqrt{x^2+y^2}\over r}\right)= \cos^{-1}\left({z\over r}\right),$$ (3)

where $r \in [0, \infty)$, $\theta \in [0, 2\pi)$, and $\phi \in [0, \pi]$. In terms of Cartesian Coordinates,

$$x = r\cos\theta\sin\phi$$ (4)

$$y = r\sin\theta\sin\phi$$ (5)

$$z = r\cos\phi.$$ (6)

The Scale Factors are

$$h_r = 1$$ (7)

$$h_\theta = r\sin\phi$$ (8)

$$h_\phi = r,$$ (9)

so the Metric Coefficients are

$$g_{rr} = 1$$ (10)

$$g_{\theta\theta} = r^2\sin^2 \phi$$ (11)

$$g_{\phi\phi} = r^2.$$ (12)

The Line Element is

$$d{\bf s} = dr \hat{\bf r} + r\,d\phi\,\hat{\boldsymbol{\phi}} + r\sin \phi\,d\theta\,\hat {\boldsymbol{\theta}},$$ (13)

the Area element

$$d{\bf a} = r^2 \sin \phi\,d\theta\,d\phi\,\hat{\bf r},$$ (14)

and the Volume Element

$$dV = r^2 \sin \phi\,d\theta\,d\phi\,dr.$$ (15)

The Jacobian is

$$\left\vert{\partial (x,y,z)\over\partial(r,\theta,\phi)}\right\vert = r^2\vert\sin\phi\vert.$$ (16)

The Position Vector is

$${\bf r} \equiv\left[{\matrix{r\cos\theta\sin\phi\cr r\sin\theta\sin\phi\cr r\cos\phi\cr}}\right],$$ (17)

so the Unit Vectors are

$$\hat {\bf r} \equiv {{d{\bf r}\over dr}\over \left\vert{d{\bf r}\over dr}\right\vert}... ...array}{c}\cos\theta\sin\phi\\ \sin\theta\sin\phi\\ \cos\phi\end{array}\right]$$ (18)

$$\hat {\boldsymbol{\theta}} \equiv {{d{\bf r}\over d\theta}\over \left\vert{d{\bf r}\over d\theta}\r... ...\vert} = \left[\begin{array}{c}-\sin\theta\\ \cos\theta\\ 0\end{array}\right]$$ (19)

$$\hat {\boldsymbol{\phi}} \equiv {{d{\bf r}\over d\phi}\over \left\vert{d{\bf r}\over d\phi }\righ... ...ray}{c}\cos\theta\cos\phi\\ \sin\theta\cos\phi\\ -\sin\phi\end{array}\right].$$ (20)

Derivatives of the Unit Vectors are

$${\partial\hat {\bf r}\over\partial r} = {\bf0}$$ (21)

$${\partial\hat {\boldsymbol{\theta}}\over\partial r} = {\bf0}$$ (22)

$${\partial\hat {\boldsymbol{\phi}}\over\partial r} = {\bf0}$$ (23)

$${\partial\hat {\bf r}\over\partial\theta} = \left[\begin{array}{c}-\sin\theta\sin\phi\\ \cos\theta\sin\phi\\ 0\end{array}\right] =\sin\phi\,\hat{\boldsymbol{\theta}}$$ (24)

$${\partial\hat {\boldsymbol{\theta}}\over\partial\theta} = \left[\begin{array}{c}-\cos\theta\\ -\sin\theta\\ 0\end{array}\right]=-\cos\phi\,\hat{\boldsymbol{\phi}}-\sin\phi\,\hat{\bf r}$$ (25)

$${\partial\hat {\boldsymbol{\phi}}\over\partial\theta} = \left[\begin{array}{c}-\sin\theta\cos\phi\\ \cos\theta\cos\phi\\ 0\end{array}\right]=\cos\phi\,\hat{\boldsymbol{\theta}}$$ (26)

$${\partial\hat {\bf r}\over\partial\phi } = \left[\begin{array}{c}\cos\theta\\ \sin\theta\cos\phi\\ -\sin\phi\end{array}\right] =\hat {\boldsymbol{\phi}}$$ (27)

$${\partial\hat {\boldsymbol{\theta}}\over\partial\phi} = {\bf0}$$ (28)

$${\partial\hat {\boldsymbol{\phi}}\over\partial\phi} = \left[\begin{array}{c}-\cos\theta\sin\phi\\ -\sin\theta\sin\phi\\ -\cos\phi\end{array}\right]=- \hat {\bf r}.$$ (29)

The Gradient is

$$\nabla = \hat{\bf r}{\partial\over\partial r} + {1\over r}\h... ...in\phi}\hat{\boldsymbol{\theta}}{\partial\over\partial\theta},$$ (30)

so

$$\nabla_r \hat {\bf r} = {\bf0}$$ (31)

$$\nabla_r \hat {\boldsymbol{\theta}} = {\bf0}$$ (32)

$$\nabla_r \hat {\boldsymbol{\phi}} = \hat {\bf0}$$ (33)

$$\nabla_\theta \hat {\bf r} = {\sin\phi\,\hat {\boldsymbol{\theta}}\over r\sin\phi} ={1\over r}\hat {\boldsymbol{\theta}}$$ (34)

$$\nabla_\theta \hat {\boldsymbol{\theta}} = -{\cos\phi\,\hat {\boldsymbol{\phi}} +\sin\phi\,\hat {\bf r}\over r\sin\phi} = -{\cot\phi\over r}\hat {\boldsymbol{\phi}}-{1\over r}\hat {\bf r}$$ (35)

$$\nabla_\theta \hat {\boldsymbol{\phi}} = {\cos \phi \,\hat {\boldsymbol{\phi}}\over r\sin \phi}={1\over r}\cot\phi\,\hat {\boldsymbol{\theta}}.$$ (36)

Now, since the Connection Coefficients are given by $\Gamma_{jk}^i =\hat {\bf x}_i\cdot(\nabla_k \hat {\bf x}_j)$,

$$\Gamma^\theta = \left[\begin{array}{ccc}0 & {1\over r} & 0\\ 0 & 0 & 0\\ 0 & {\cot\phi\over r} & 0\end{array}\right]$$ (37)

$$\Gamma^\phi = \left[\begin{array}{ccc}0 & 0 & {1\over r}\\ 0 & -{\cot\phi\over r} & 0\\ 0 & 0 & 0\end{array}\right]$$ (38)

$$\Gamma^r = \left[\begin{array}{ccc}0 & 0 & 0\\ 0 & -{1\over r} & 0\\ 0 & 0 & -{1\over r}\end{array}\right].$$ (39)

The Divergence is

$\nabla\cdot {\bf F} = A^k_{,k}+\Gamma_{jk}^kA^j$
$= [A^r_{,r}+(\Gamma_{rr}^rA^r+\Gamma_{\theta r}^rA^\theta+\Gamma_{\phi r}^rA^\phi]$
$\quad +[A^\theta_{,\theta}+(\Gamma_{r\theta}^\theta A^r+\Gamma_{\theta\theta}^\theta A^\theta+\Gamma_{\phi\theta}^\theta A^\phi)]$
$\quad + [A^\phi_{,\phi}+(\Gamma_{r\phi}^\phi A^r+\Gamma_{\theta\phi}^\phi A^\theta +\Gamma_{\phi\phi}^\phi A^\phi)]$
$= {1\over g_r}{\partial A^r\over\partial r} +{1\over g_\theta}{\partial A^\theta\over\partial\theta} +{1\over g_\phi}{\partial A^\phi\over\partial\phi}+(0+0+0)$
$\quad +\left({{1\over r}A^r+0+{\cot\phi\over r}A^\phi}\right)+\left({{1\over r}A^r+0+0}\right)$
$= {\partial\over\partial r}A^r+{2\over r}A^r+{1\over r\sin\phi}{\partial\over\p... ...}A^\theta+{1\over r}{\partial\over\partial\phi}A^\phi +{\cot\phi\over r}A^\phi,$
	(40)

or, in Vector notation,

$$\nabla\cdot{\bf F} = \left({{2\over r}+{\partial\over\partial r}}\right)F_r + \left({{... ...\over r}}\right)F_\phi + {1\over\sin\phi}{\partial F_\theta\over\partial\theta}$$

$$= {1\over r^2}{\partial\over\partial r} (r^2 F_r) + {1\over r\sin\p... ...} (\sin\phi F_\phi) + {1\over r\sin\phi}{\partial F_\theta\over\partial\theta}.$$ (41)

The Covariant Derivatives are given by

$$A_{j;k}={1\over g_{kk}}{\partial A_j\over\partial x_k}-\Gamma_{jk}^i A_i,$$ (42)

so

$$A_{r;r} = {\partial A_r\over\partial r}-\Gamma_{rr}^iA_i ={\partial A_r\over\partial r}$$ (43)

$$A_{r;\theta} = {1\over r\sin\phi}{\partial A_r\over\partial\theta}-\Gamma_{r\the... ... = {1\over r\sin\phi}{\partial A_r\over\partial\theta}-\Gamma_{r\theta}A_\theta$$

$$= {1\over r\sin\phi}{\partial A_r\over\partial\phi}-{A_\theta\over r}$$ (44)

$$A_{r;\phi} = {1\over r}{\partial A_r\over\partial\phi}-\Gamma_{r\phi}^iA_i = {1\over r}{\partial A_r\over\partial\phi}-\Gamma_{r\phi}^\phi A_\phi$$

$$= {1\over r}\left({{\partial A_r\over\partial\phi}-A_\phi}\right)$$ (45)

$$A_{\theta;r} = {\partial A_\theta\over\partial r}-\Gamma_{\theta r}^iA_i ={\partial A_\theta \over\partial r}$$ (46)

$$A_{\theta;\theta} = {1\over r\sin\phi}{\partial A_\phi\over\partial\theta}-\Gamma_{\theta\theta}^iA_i$$

$$= {1\over r\sin\phi}{\partial A_\theta\partial\theta}-\Gamma_{\theta\theta}^\phi A_\phi -\Gamma_{\theta\theta}^rA_r$$

$$= {1\over r\sin\phi}{\partial A_\theta\over\partial\theta}+{\cot\phi\over r} A_\phi +{A_r\over r}$$ (47)

$$A_{\theta;\phi} = {1\over r}{\partial A_\theta\over\partial r}-\Gamma_{\phi r}^iA_i{\partial A_\theta\over\partial\phi}$$ (48)

$$A_{\phi;r} = {\partial A_\phi\over\partial r}-\Gamma_{\phi r}^iA_i={\partial A_\phi\over r}$$ (49)

$$A_{\phi;\theta} = {1\over r\sin\phi}{\partial A_\phi\over\partial\theta} -\Gamma_{\... ...\over r\sin\phi}{\partial A_\phi\over\partial\theta}-\Gamma_{\phi\theta}^\theta$$

$$= {1\over r\sin\phi}{\partial A_\phi\over\partial\theta} -{\cot\phi\over r}A_\theta$$ (50)

$$A_{\phi;\phi} = {1\over r}{\partial A_\phi\over\partial\phi}-\Gamma_{\phi\phi}^iA_i ={1\over r}{\partial A_\phi\over\partial\phi}-\Gamma_{\phi\phi}^rA_r$$

$$= {1\over r}{\partial A_\phi\over\partial\phi}+{A_r\over r}.$$ (51)

The Commutation Coefficients are given by

$$c_{\alpha\beta}^\mu\vec e_\mu = [\vec e_\alpha,\vec e_\beta] =\nabla_\alpha\vec e_\beta-\nabla_\beta\vec e_\alpha$$ (52)

$$[\hat{\bf r},\hat{\bf r}]= [{\hat{\boldsymbol{\theta}}},{\hat... ...[{\hat{\boldsymbol{\phi}}},{\hat{\boldsymbol{\phi}}}] ={\bf0},$$ (53)

so $c_{rr}^\alpha = c_{\theta\theta}^\alpha=c_{\phi\phi}^\alpha=0$, where $\alpha =r,\theta,\phi$.

$$[\hat{\bf r},\hat{\boldsymbol{\theta}}]= -[{\hat{\boldsymbol{... ...t{\boldsymbol{\theta}} = -{1\over r}\hat{\boldsymbol{\theta}},$$ (54)

so $c_{r\theta}^\theta =-c_{\theta r}^\theta =-{1\over r}$, $c_{r\theta}^r =c_{r\theta }^\phi = 0$.

$$[\hat{\bf r},\hat{\boldsymbol{\phi}}]=-[{\hat{\boldsymbol{\ph... ...}\hat{\boldsymbol{\phi}} = -{1\over r}\hat{\boldsymbol{\phi}},$$ (55)

so $c_{r\phi }^\phi =-c_{\phi r}^\phi ={1\over r}$.

$$[\hat{\boldsymbol{\theta}},\hat{\boldsymbol{\phi}}]= -[{\hat{... ...{\theta}}-{\bf0} ={1\over r}\cot\phi\hat{\boldsymbol{\theta}},$$ (56)

so

$$c_{\theta\phi }^\theta =-c_{\phi\theta }^\theta ={1\over r}\cot\phi.$$ (57)

Summarizing,

$$c^r = \left[\begin{array}{ccc}0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\end{array}\right]$$ (58)

$$c^\theta = \left[\begin{array}{ccc}0 & -{1\over r} & 0\\ {1\over r} & 0 &{1\over r}\cot\phi\\ 0 & -{1\over r}\cot\phi & 0\end{array}\right]$$ (59)

$$c^\phi = \left[\begin{array}{ccc}0 & 0 & -{1\over r}\\ 0 & 0 & 0\\ {1\over r} & 0 & 0\end{array}\right].$$ (60)

Time derivatives of the Position Vector are

$$\dot{\bf r} = \left[\begin{array}{c}\cos\theta\sin\phi\,\dot r-r\sin\theta\sin\... ...eta\cos\phi\,\dot\phi\\ \cos\phi\,\dot r-r\sin\phi\,\dot\phi\end{array}\right]$$

$$= \left[\begin{array}{c}\cos\theta\sin\phi\\ \sin\theta\sin\phi\\ ... ...\cos\theta\cos\phi\\ \sin\theta\cos\phi\\ -\sin\phi\end{array}\right]\dot\phi$$

$$= \dot r\,\hat{\bf r} + r\sin\phi\,\dot\theta\,\hat{\boldsymbol{\theta}} + r\,\dot\phi\,\hat{\boldsymbol{\phi}}.$$ (61)

The Speed is therefore given by

$$v \equiv \vert\dot {\bf r}\vert = \sqrt{\dot r^2 +r^2\sin^2\phi\dot\theta^2 + r^2\dot\phi^2}.$$ (62)

The Acceleration is

$$\ddot x = (-\sin\theta\sin\phi\dot\theta\dot r+\cos\theta\cos\phi\dot r\dot\phi+\cos\theta\sin\phi\ddot r)$$

$$\phantom{=} -(\sin\theta\sin\phi\dot r\dot\theta +r\cos\theta\sin\phi\dot\theta^2+r\sin\theta\cos\phi\dot\theta\dot\phi$$

$$\phantom{=} +r\sin\theta\sin\phi\ddot\theta)+(\cos\theta\cos\phi\dot r\dot\phi-r\sin\theta\cos\dot\theta\dot\phi$$

$$\phantom{=} -r\cos\theta\sin\phi\dot\phi^2+r\cos\theta\cos\phi\ddot\phi)$$

$$= -2\sin\theta\sin\phi\dot\theta\dot r+2\cos\theta\cos\phi\dot r\dot\phi-2r\sin\theta\cos\phi\dot\theta\dot\phi$$

$$\phantom{=} +\cos\theta\sin\phi\ddot r-r\sin\theta\sin\phi\ddot\theta+r\cos\theta\cos\phi\ddot\phi$$

$$\phantom{=} -r\cos\theta\sin\phi(\dot\theta^2+\dot\phi^2)$$ (63)

$$\ddot y = (\sin\theta\sin\phi\ddot r+r\cos\theta\sin\phi\dot\theta+r\cos\phi\sin\theta\dot\phi)$$

$$\phantom{=} {} +(\cos\theta\sin\phi\dot r\dot\theta-r\sin\theta\sin\phi\dot\theta^2+r\cos\theta\cos\phi\dot\theta\dot\phi$$

$$\phantom{=} +r\cos\theta\sin\phi\ddot\theta)+(\sin\theta\cos\phi\dot r\dot\phi+r\cos\theta\cos\phi\dot\theta\dot\phi$$

$$\phantom{=} -r\sin\theta\sin\phi\dot\phi^2+r\sin\theta\cos\phi\ddot\phi)$$

$$= 2\cos\theta\sin\phi\dot\theta\dot r+2\sin\theta\cos\phi\dot r\dot\phi +2r\cos\theta\cos\phi\dot\theta\dot\phi$$

$$\phantom{=} +\sin\theta\sin\phi\ddot r+r\cos\theta\sin\phi\ddot\theta+r\sin\theta\cos\phi\ddot\phi$$

$$\phantom{=} -r\sin\theta\sin\phi (\dot\theta^2 +\dot\phi^2)$$ (64)

$$\ddot z = (\cos\phi\ddot r-\sin\phi\dot r\dot\phi ) -(\dot r\sin\phi\dot\phi +r\cos\phi\dot\phi^2+r\sin\phi\ddot\phi )$$

$$= -r\cos\phi\dot\phi^2+\cos\phi\ddot r-2\sin\phi\dot\phi\dot r-r\sin\phi\ddot\phi.$$ (65)

Plugging these in gives

$$\ddot {\bf r} = (\ddot r-r\dot\phi^2)\left[\begin{array}{c}\cos\theta\sin\phi\\ \sin\theta\sin\phi\\ \cos\phi\end{array}\right]$$

$$\phantom{=} +(2r\cos\phi\dot\theta\dot\phi +r\sin\phi\ddot\theta)\left[\begin{array}{c}-\sin\theta\\ \cos\theta\\ 0\end{array}\right]$$

$$\phantom{=} \mathop{+}(2\dot r\dot\phi +r\ddot\phi)\left[\begin{array}{c}\cos... ...\left[\begin{array}{c}\cos\theta\\ \sin\theta\\ 0\end{array}\right],\nonumber$$ (66)

but

$$\sin\phi\hat{\bf r}+\cos\phi\hat{\boldsymbol{\phi}} = \left[\begin{array}{c}\cos\theta\sin^2\phi +\cos\theta\cos^2\phi\\ \sin\theta\sin^2\phi +\sin\theta\cos^2\phi\\ 0\end{array}\right]$$

$$= \left[\begin{array}{c}\cos\theta\\ \sin\theta\\ 0\end{array}\right],$$ (67)

so

$$\ddot {\bf r} = (\ddot r-r\dot\phi^2)\hat{\bf r}+(2r\cos\phi\dot\theta\dot\phi +2\sin\phi\dot\theta\dot r+r\sin\phi\ddot\theta )\hat{\boldsymbol{\theta}}$$

$$\phantom{=} \mathop{+}(2\dot r\dot\phi +r\ddot\phi )\hat{\boldsymbol{\phi}}-r\sin\phi\dot\theta^2 (\sin\phi\hat{\bf r}+\cos\phi\hat{\boldsymbol{\phi}})$$

$$= (\ddot r-r\dot\phi^2-r\sin^2\phi\dot\theta^2)\hat{\bf r}$$

$$\phantom{=} + (2\sin\phi\dot\theta\dot r+2r\cos\phi\dot\theta\dot\phi +r\sin\phi\ddot\theta )\hat{\boldsymbol{\theta}}$$

$$\phantom{=} \mathop{+}(2\dot r\dot\phi +r\ddot\phi -r\sin\phi\cos\phi\dot\theta^2)\hat{\boldsymbol{\phi}}.$$ (68)

Time Derivatives of the Unit Vectors are

$$\dot{\hat{\bf r}} = \left[\begin{array}{c}-\sin\theta\sin\phi\,\dot\theta+\cos\theta\... ...sin\phi\,\dot\theta\hat {\boldsymbol{\theta}} +\dot\phi\hat {\boldsymbol{\phi}}$$ (69)

$$\dot{\hat{\boldsymbol{\theta}}} = \left[\begin{array}{c}-\cos\theta\dot\theta\\ -\sin\theta\dot\th... ...nd{array}\right]=-\dot\theta(\sin\phi\hat{\bf r}+\cos\phi\hat\boldsymbol{\phi})$$ (70)

$$\dot{\hat{\boldsymbol{\phi}}} = \left[\begin{array}{c}-\sin\theta\cos\phi\,\dot\theta-\cos\theta\... ...}\right] = -\dot\phi\hat {\bf r} +\cos\phi\dot\theta\hat {\boldsymbol{\theta}}.$$ (71)

The Curl is

$\nabla\times {\bf F} = {1\over r\sin\phi}\left[{{\partial\over\partial\phi}(\sin\phi F_\theta)-{\partial F_\phi\over\partial\theta}}\right]\hat{\bf r}$
$+ {1\over r}\left[{{1\over\sin\phi}{\partial F_r\over\partial\theta} - {\parti... ...}(rF_\phi) - {\partial F_r\over\partial\phi}}\right]\hat {\boldsymbol{\theta}}.$
	(72)

The Laplacian is

$$\nabla^2 \equiv {1\over r^2} {\partial\over\partial r}\left({r^2{\partial\over\pa... ... {\partial\over\partial\phi}\left({\sin\phi{\partial \over\partial\phi}}\right)$$

$$= {1\over r^2}\left({r^2{\partial^2\over\partial r^2}+2r{\partial\o... ...\phi{\partial\over\partial\phi}+\sin\phi{\partial^2\over\partial\phi^2}}\right)$$

$$= {\partial^2\over\partial r^2} + {2\over r} {\partial\over\partial... ...hi} {\partial\over\partial\phi} + {1\over r^2} {\partial^2\over\partial\phi^2}.$$ (73)

The vector Laplacian is

$$\nabla^2 {\bf v}=\left[\begin{array}{l} {1\over r}{\partial^... ...al\phi} -{v_\phi\over r^2\sin^2\theta}}\\ \end{array}\right].$$ (74)

To express Partial Derivatives with respect to Cartesian axes in terms of Partial Derivatives of the spherical coordinates,

$$\left[\begin{array}{c}x\\ y\\ z\end{array}\right] = \left[\begin{array}{c}r\cos\theta\sin\phi\\ r\sin\theta\sin\phi\\ r\cos\phi\end{array}\right]$$ (75)

$$\left[\begin{array}{c}dx\\ dy\\ dz\end{array}\right] = \left[\begin{array}{c}\cos\theta\sin\phi\,dr-r\sin\theta\sin\phi\... ...a+r\sin\theta\cos\phi\,d\phi\\ \cos\phi\,dr-r\sin\phi\,d\phi\end{array}\right]$$

$$= \left[\begin{array}{ccc}\cos\theta\sin\phi & -r\sin\theta\sin\phi... ...nd{array}\right]\left[\begin{array}{c}dr\\ d\theta\\ d\phi\end{array}\right].$$ (76)

Upon inversion, the result is

$$\left[{\matrix{dr\cr d\theta\cr d\phi\cr}}\right] =\left[{\... ...trix{dx\cr dy\cr dz\cr}}\right]. \hrule width 0pt height 6.1pt$$ (77)

The Cartesian Partial Derivatives in spherical coordinates are therefore

$${\partial\over\partial x} = {\partial r\over\partial x}{\partial\over\partial r} +{\partial\t... ...\over\partial\theta} +{\partial\phi\over\partial x} {\partial\over\partial\phi}$$

$$= \cos\theta\sin\phi {\partial\over\partial r} - {\sin\theta\over r... ...l\over\partial\theta} + {\cos\theta\cos\phi\over r} {\partial\over\partial\phi}$$ (78)

$${\partial\over\partial y} = {\partial r\over\partial y} {\partial\over\partial r} + {\partial... ...over\partial\theta} + {\partial\phi\over\partial y} {\partial\over\partial\phi}$$

$$= \sin\theta\sin\phi {\partial\over\partial r} + {\cos\theta\over r... ...l\over\partial\theta} + {\sin\theta\cos\phi\over r} {\partial\over\partial\phi}$$ (79)

$${\partial\over\partial z} = {\partial r\over\partial z} {\partial\over\partial r} + {\partial... ...over\partial\theta} + {\partial\phi\over\partial z} {\partial\over\partial\phi}$$

$$= \cos\phi {\partial\over\partial r} - {\sin\phi\over r} {\partial\over\partial\phi}$$ (80)

(Gasiorowicz 1974, pp. 167-168).

The Helmholtz Differential Equation is separable in spherical coordinates.

See also Colatitude, Great Circle, Helmholtz Differential Equation--Spherical Coordinates, Latitude, Longitude, Oblate Spheroidal Coordinates, Prolate Spheroidal Coordinates

References

Arfken, G. ``Spherical Polar Coordinates.'' §2.5 in Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 102-111, 1985.

Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, p. 212, 1987.

Gasiorowicz, S. Quantum Physics. New York: Wiley, 1974.

Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, p. 658, 1953.