Editing Spherical coordinate system (section)

== Integration and differentiation in spherical coordinates ==
[[File:Kugelkoord-lokb-e.svg|thumb|Unit vectors in spherical coordinates]]

The following equations (Iyanaga 1977) assume that the colatitude {{mvar|θ}} is the inclination from the positive {{mvar|z}} axis, as in the ''physics convention'' discussed.

The [[line element]] for an infinitesimal displacement from {{math|(''r'', ''θ'', ''φ'')}} to {{math|(''r'' + d''r'', ''θ'' + d''θ'', ''φ'' + d''φ'')}} is
<math display="block">
\mathrm{d}\mathbf{r} = \mathrm{d}r\,\hat{\mathbf r} + r\,\mathrm{d}\theta \,\hat{\boldsymbol\theta } + r \sin{\theta} \, \mathrm{d}\varphi\,\mathbf{\hat{\boldsymbol\varphi}},</math>
where
<math display="block">\begin{align}
 \hat{\mathbf r} &= \sin \theta \cos \varphi \,\hat{\mathbf x} +
  \sin \theta \sin \varphi \,\hat{\mathbf y} + \cos \theta \,\hat{\mathbf z}, \\
 \hat{\boldsymbol\theta} &= \cos \theta \cos \varphi \,\hat{\mathbf x} +
  \cos \theta \sin \varphi \,\hat{\mathbf y} - \sin \theta \,\hat{\mathbf z}, \\
 \hat{\boldsymbol\varphi} &= - \sin \varphi \,\hat{\mathbf x} +
  \cos \varphi \,\hat{\mathbf y}
\end{align}</math>
are the local orthogonal [[unit vectors]] in the directions of increasing {{mvar|r}}, {{mvar|θ}}, and {{mvar|φ}}, respectively,
and {{math|'''x̂'''}}, {{math|'''ŷ'''}}, and {{math|'''ẑ'''}} are the unit vectors in Cartesian coordinates. The linear transformation to this right-handed coordinate triplet is a [[rotation matrix]],
<math display="block">R = \begin{pmatrix}
     \sin\theta\cos\varphi&\sin\theta\sin\varphi&\hphantom{-}\cos\theta\\
     \cos\theta\cos\varphi&\cos\theta\sin\varphi&-\sin\theta\\
    -\sin\varphi&\cos\varphi   &\hphantom{-}0
   \end{pmatrix}.
</math>

This gives the transformation from the Cartesian to the spherical, the other way around is given by its inverse.
Note: the matrix is an [[orthogonal matrix]], that is, its inverse is simply its [[transpose]].

The Cartesian unit vectors are thus related to the spherical unit vectors by:
<math display="block">\begin{bmatrix}\mathbf{\hat x} \\ \mathbf{\hat y}  \\ \mathbf{\hat z} \end{bmatrix}
  = \begin{bmatrix} \sin\theta\cos\varphi & \cos\theta\cos\varphi & -\sin\varphi \\
                    \sin\theta\sin\varphi & \cos\theta\sin\varphi & \hphantom{-}\cos\varphi \\
                    \cos\theta         & -\sin\theta        & \hphantom{-}0 \end{bmatrix}
    \begin{bmatrix} \boldsymbol{\hat{r}} \\ \boldsymbol{\hat\theta} \\ \boldsymbol{\hat\varphi} \end{bmatrix}</math>

The general form of the formula to prove the differential line element, is<ref name="q74503">{{cite web |title=Line element (dl) in spherical coordinates derivation/diagram |date=October 21, 2011 |work=[[Stack Exchange]] |url=https://math.stackexchange.com/q/74503}}</ref>
<math display="block">\mathrm{d}\mathbf{r} =
 \sum_i \frac{\partial \mathbf{r}}{\partial x_i} \,\mathrm{d}x_i =
 \sum_i \left|\frac{\partial \mathbf{r}}{\partial x_i}\right|
  \frac{\frac{\partial \mathbf{r}}{\partial x_i}}{\left|\frac{\partial \mathbf{r}}{\partial x_i}\right|} \, \mathrm{d}x_i =
 \sum_i \left|\frac{\partial \mathbf{r}}{\partial x_i}\right| \,\mathrm{d}x_i \, \hat{\boldsymbol{x}}_i,
</math>
that is, the change in <math>\mathbf r</math> is decomposed into individual changes corresponding to changes in the individual coordinates. 

To apply this to the present case, one needs to calculate how <math>\mathbf r</math> changes with each of the coordinates. In the conventions used, 
<math display="block">\mathbf{r} = \begin{bmatrix}
 r \sin\theta \, \cos\varphi \\
 r \sin\theta \, \sin\varphi \\
 r \cos\theta
\end{bmatrix},
x_1=r, x_2=\theta, x_3=\varphi.</math>

Thus,
<math display="block">
\frac{\partial\mathbf r}{\partial r} = \begin{bmatrix}
 \sin\theta \, \cos\varphi \\
 \sin\theta \, \sin\varphi \\
 \cos\theta
\end{bmatrix}=\mathbf{\hat r}, \quad
\frac{\partial\mathbf r}{\partial \theta} = \begin{bmatrix}
 r \cos\theta \, \cos\varphi \\
 r \cos\theta \, \sin\varphi \\
 -r \sin\theta
\end{bmatrix}=r\,\hat{\boldsymbol\theta }, \quad
\frac{\partial\mathbf r}{\partial \varphi} = \begin{bmatrix}
 -r \sin\theta \, \sin\varphi \\
 \hphantom{-}r \sin\theta \, \cos\varphi \\
 0
\end{bmatrix}
=
r \sin\theta\,\mathbf{\hat{\boldsymbol\varphi}}  .
</math>

The desired coefficients are the magnitudes of these vectors:<ref name="q74503" />
<math display="block">
\left|\frac{\partial\mathbf r}{\partial r}\right| = 1, \quad
\left|\frac{\partial\mathbf r}{\partial \theta}\right| = r, \quad
\left|\frac{\partial\mathbf r}{\partial \varphi}\right| = r \sin\theta.
</math>

The [[Surface integral|surface element]] spanning from {{mvar|θ}} to {{math|''θ'' + d''θ''}} and {{mvar|φ}} to {{math|''φ'' + d''φ''}} on a spherical surface at (constant) radius {{mvar|r}} is then
<math display="block">
\mathrm{d}S_r =
 \left\|\frac{\partial {\mathbf r}}{\partial \theta} \times \frac{\partial {\mathbf r}}{\partial \varphi}\right\| \mathrm{d}\theta \,\mathrm{d}\varphi =
\left|r {\hat \boldsymbol\theta} \times r \sin \theta {\boldsymbol\hat \varphi} \right|\mathrm{d}\theta \,\mathrm{d}\varphi= 
 r^2 \sin\theta \,\mathrm{d}\theta \,\mathrm{d}\varphi  ~.
</math>

Thus the differential [[solid angle]] is
<math display="block">\mathrm{d}\Omega = \frac{\mathrm{d}S_r}{r^2} = \sin\theta \,\mathrm{d}\theta \,\mathrm{d}\varphi.</math>

The surface element in a surface of polar angle {{mvar|θ}} constant (a cone with vertex at the origin) is
<math display="block">\mathrm{d}S_\theta = r \sin\theta \,\mathrm{d}\varphi \,\mathrm{d}r.</math>

The surface element in a surface of azimuth {{mvar|φ}} constant (a vertical half-plane) is
<math display="block">\mathrm{d}S_\varphi = r \,\mathrm{d}r \,\mathrm{d}\theta.</math>

The [[volume element]] spanning from {{mvar|r}} to {{math|''r'' + d''r''}}, {{mvar|θ}} to {{math|''θ'' + d''θ''}}, and {{mvar|φ}} to {{math|''φ'' + d''φ''}} is specified by the [[determinant]] of the [[Jacobian matrix]] of [[partial derivative]]s, 
<math display="block">
J =\frac{\partial(x,y,z)}{\partial(r,\theta,\varphi)}
  =\begin{pmatrix}
     \sin\theta\cos\varphi & r\cos\theta\cos\varphi & -r\sin\theta\sin\varphi\\
     \sin\theta\sin\varphi & r\cos\theta\sin\varphi & \hphantom{-}r\sin\theta\cos\varphi\\
     \cos\theta & -r\sin\theta & \hphantom{-}0
   \end{pmatrix},
</math>
namely 
<math display="block">
\mathrm{d}V = \left|\frac{\partial(x, y, z)}{\partial(r, \theta, \varphi)}\right| \,\mathrm{d}r \,\mathrm{d}\theta \,\mathrm{d}\varphi=
 r^2 \sin\theta \,\mathrm{d}r \,\mathrm{d}\theta \,\mathrm{d}\varphi =
 r^2 \,\mathrm{d}r \,\mathrm{d}\Omega ~.
</math>

Thus, for example, a function {{math|''f''(''r'', ''θ'', ''φ'')}} can be integrated over every point in {{math|'''R'''<sup>3</sup>}} by the [[Multiple integral#Spherical coordinates|triple integral]]
<math display="block">\int\limits_0^{2\pi} \int\limits_0^\pi \int\limits_0^\infty f(r, \theta, \varphi) r^2 \sin\theta \,\mathrm{d}r \,\mathrm{d}\theta \,\mathrm{d}\varphi ~.</math>

The [[del]] operator in this system leads to the following expressions for the [[gradient]] and [[Laplacian]] for scalar fields,
<math display="block">\begin{align}
\nabla f &= {\partial f \over \partial r}\hat{\mathbf r}
  + {1 \over r}{\partial f \over \partial \theta}\hat{\boldsymbol\theta}
  + {1 \over r\sin\theta}{\partial f \over \partial \varphi}\hat{\boldsymbol\varphi},  \\[8pt]
\nabla^2 f &= {1 \over r^2}{\partial \over \partial r} \left(r^2 {\partial f \over \partial r}\right) + {1 \over r^2 \sin\theta}{\partial \over \partial \theta} \left(\sin\theta {\partial f \over \partial \theta}\right)
+ {1 \over r^2 \sin^2\theta}{\partial^2 f \over \partial \varphi^2} \\[8pt]
& = \left(\frac{\partial^2}{\partial r^2} + \frac{2}{r} \frac{\partial}{\partial r}\right) f + {1 \over r^2 \sin\theta}{\partial \over \partial \theta} \left(\sin\theta \frac{\partial}{\partial \theta}\right) f + \frac{1}{r^2 \sin^2\theta}\frac{\partial^2}{\partial \varphi^2}f ~, \\[8pt]
\end{align}</math>And it leads to the following expressions for the [[divergence]] and [[curl (mathematics)|curl]] of [[Vector field|vector fields]],

<math display="block">\nabla \cdot \mathbf{A}
  = \frac{1}{r^2}{\partial \over \partial r}\left( r^2 A_r \right)
  + \frac{1}{r \sin\theta}{\partial \over \partial\theta} \left( \sin\theta A_\theta \right)
  + \frac{1}{r \sin \theta} {\partial A_\varphi \over \partial \varphi},</math><math display="block">\begin{align}
\nabla \times \mathbf{A} = {}
&      \frac{1}{r\sin\theta} \left[{\partial \over \partial \theta} \left( A_\varphi\sin\theta \right)
    - {\partial A_\theta \over \partial \varphi}\right] \hat{\mathbf r} \\[4pt]
& {} + \frac 1 r \left[{1 \over \sin\theta}{\partial A_r \over \partial \varphi}
    - {\partial \over \partial r} \left( r A_\varphi \right) \right] \hat{\boldsymbol\theta} \\[4pt]
& {} + \frac 1 r \left[{\partial \over \partial r} \left( r A_\theta \right)
    - {\partial A_r \over \partial \theta}\right] \hat{\boldsymbol\varphi},
\end{align}</math>

Further, the inverse Jacobian in Cartesian coordinates is
<math display="block">J^{-1} = \begin{pmatrix}
  \dfrac{x}{r}&\dfrac{y}{r}&\dfrac{z}{r}\\\\
  \dfrac{xz}{r^2\sqrt{x^2+y^2}}&\dfrac{yz}{r^2\sqrt{x^2+y^2}}&\dfrac{-\left(x^2 + y^2\right)}{r^2\sqrt{x^2+y^2}}\\\\
  \dfrac{-y}{x^2+y^2}&\dfrac{x}{x^2+y^2}&0
\end{pmatrix}.</math>
The [[metric tensor]] in the spherical coordinate system is <math>g = J^T J </math>.