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<quiz>
  <question type="description">
    <name>
      <text>28.3.4.10 Spherical polar coordinates</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h3>Spherical polar coordinates</h3>
<p>In spherical polar coordinates \((r,\theta,\phi)\), the 3 unit vectors are \(\underline{\hat{r}}\), \(\underline{\hat{\theta}}\) and \(\underline{\hat{\phi}}\) with scale factors \(h_{r}=1\), \(h_{\theta} = r\), \(h_{\phi}=r \sin \theta\). The quantities \(r\), \(\theta\) and \(\phi\) are related to \(x\), \(y\) and \(z\) by \(x=r \sin \theta \cos \phi\), \(y=r \sin \theta \sin \phi\) and \(z=r \cos \theta\). In spherical polar coordinates,</p>
<p>grad \(\displaystyle f = {\underline \nabla} f = \frac{\partial f}{\partial r} \underline{\hat{r}} + \frac{1}{r} \frac{\partial f}{\partial \theta} \underline{\hat{\theta}} + \frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi} \underline{\hat{\phi}}\)</p>
<p>If \(\underline{F} = F_{r} \underline{\hat{r}} + F_{\theta} \underline{\hat{\theta}} + F_{\phi} \underline{\hat{\phi}}\)<br> then</p>
<p>div \(\displaystyle \underline{F} = {\underline \nabla} \cdot \underline{F} = \frac{1}{r^2 \sin \theta} \left[ \frac{\partial}{\partial r}(r^2 \sin \theta F_{r}) + \frac{\partial}{\partial \theta}(r \sin \theta F_{\theta}) + \frac{\partial}{\partial \phi}(r F_{\phi}) \right]\)<br> curl \(\displaystyle \underline{F} = {\underline \nabla} \times \displaystyle \underline{F} = \frac{1}{r^2 \sin \theta} \left|\begin{array}{ccc} \underline{\hat{r}} &amp; r \underline{\hat{\theta}} &amp; r \sin \theta \underline{\hat{\phi}} \\ \\ \displaystyle \frac{\partial}{\partial r} &amp; \displaystyle \frac{\partial}{\partial \theta} &amp; \displaystyle \frac{\partial}{\partial \phi} \\ \\ F_{r} &amp; r F_{\theta} &amp; r \sin \theta F_{\phi} \end{array} \right|\) </p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>In spherical polar coordinates, find \({\underline \nabla} f\) for
<ol class="HELM_parts_inline">
<li> \(f = r\) </li><li> \(\displaystyle f = \frac{1}{r}\) </li>
<li> \(f = r^2 \sin(\phi + \theta)\)</li></ol></p>
<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
<li><p>
\[
\begin{aligned} {\underline \nabla} f &=
\frac{\partial f}{\partial r} \underline{\hat{r}} + \frac{1}{r}
\frac{\partial f}{\partial \theta} \underline{\hat{\theta}} +
\frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi}
\underline{\hat{\phi}}
\\
&=\frac{\partial (r)}{\partial r} \underline{\hat{r}} +
\frac{1}{r}  \frac{\partial (r)}{\partial
\theta} \underline{\hat{\theta}} + \frac{1}{r \sin \theta} 
\frac{\partial (r)}{\partial \phi} \underline{\hat{\phi}} \\
&= 1 \underline{\hat{r}} = \underline{\hat{r}}
\end{aligned}
\]</p></li>
<li><p>
\[\begin{aligned} {\underline \nabla} f &=
\frac{\partial f}{\partial r} \underline{\hat{r}} + \frac{1}{r}
\frac{\partial f}{\partial \theta} \underline{\hat{\theta}} +
\frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi}
\underline{\hat{\phi}}
\\
&=  \frac{\partial (\frac{1}{r})}{\partial r} \underline{\hat{r}} +
\frac{1}{r}  \frac{\partial (\frac{1}{r})}{\partial
\theta} \underline{\hat{\theta}} + \frac{1}{r \sin \theta}
\frac{\partial (\frac{1}{r})}{\partial \phi} \underline{\hat{\phi}} \\
&= -\frac{1}{r^2} \underline{\hat{r}}
\end{aligned}\]</p></li>
<li><p> \[\begin{aligned}\qquad\quad {\underline \nabla} f &=
\frac{\partial f}{\partial r} \underline{\hat{r}} + \frac{1}{r}
\frac{\partial f}{\partial \theta} \underline{\hat{\theta}} +
\frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi}
\underline{\hat{\phi}}
\\
&= \frac{\partial (r \sin(\phi
+ \theta)}{\partial r} \underline{\hat{r}}  + \frac{1}{r} 
\frac{\partial (r \sin(\phi + \theta)}{\partial \theta} \underline{\hat{\theta}} +
\frac{1}{r \sin \theta} 
\frac{\partial (r^2 \sin(\phi + \theta)}{\partial \phi} \underline{\hat{\phi}} \\
&=  2r \sin(\phi + \theta)\underline{\hat{r}} + \frac{1}{r}
 r^2 \cos(\phi + \theta)\underline{\hat{\theta}}  + \frac{1}{r
\sin \theta} 
 r^2 \cos(\phi + \theta) \underline{\hat{\phi}} \\
 &= 2r \sin(\phi + \theta) \underline{\hat{r}} + r \cos(\phi +
 \theta) \underline{\hat{\theta}} + \frac{r\cos (\phi + \theta)}{\sin
 \theta} \underline{\hat{\phi}} \end{aligned}\]
</p></li>
<hr>
<h4 class="HELM_example">Engineering Example</h4>
<p><b>Electric potential</b></p>
<p><strong>Introduction</strong></p>
<p>There is a scalar quantity \(V\), called the electric potential, which satisfies \(\qquad {\underline \nabla}V=-{\underline E}\) where \(\underline{E}\) is the electric field.</p>
<p>It is often easier to handle scalar fields rather than vector fields. It is therefore convenient to work with \(V\) and then derive \(\underline{E}\) from it.</p>
<p><strong>Problem in words</strong></p>
<p>Given the electric potential, find the electric field.</p>
<p><strong>Mathematical statement of problem</strong></p>
<p>For a point charge, \(Q\), the potential \(V\) is given by</p>
 <p>\(\qquad\displaystyle V=\dfrac{Q}{4\pi\epsilon_0 r}\)</p>
<p>Verify, using spherical polar coordinates, that \({\underline E}= -{\underline \nabla}V=\dfrac{Q}{4\pi\epsilon_0r^2}{\underline {\hat r}}\)</p>
<p><strong>Mathematical analysis</strong></p>
<p>In spherical polar coordinates:
\[\begin{aligned} {{\underline \nabla}}V&=\dfrac{\partial V}{\partial r}{\underline {\hat r}} +\dfrac{1}{r}\dfrac{\partial V}{\partial \theta} {\underline {\hat\theta}}+\dfrac{1}{r\sin \theta}\dfrac{\partial V}{\partial \phi}{\underline {\hat \phi}}\\
&=\dfrac{\partial V}{\partial r}{\underline {\hat r}}\quad\text{ as the other partial derivatives are zero}\\
&=\dfrac{\partial}{\partial r}\left[\dfrac{Q}{4\pi\epsilon_0r}\right]{\underline {\hat r}}\\
&=-\dfrac{Q}{4\pi\epsilon_0r^2}{\underline {\hat r}}\end{aligned}\]</p>
<p><strong>Interpretation</strong></p>
<p>So \({\underline E}=\dfrac{Q}{4\pi\epsilon_0r^2}{\underline {\hat r}}\) as required.
</p>
<p>This is a form of Coulomb’s Law. A positive charge will experience a positive repulsion radially <em>outwards</em> in the field of another positive charge. </p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Using spherical polar coordinates, find \({\underline \nabla} \cdot \underline{F}\) for the following vector functions.
<ol class="HELM_parts"> 
<li>\(\displaystyle \underline{F} = r\underline{\hat{r}}\) </li>
<li> \(\underline{F} = r^2 \sin \theta \underline{\hat{r}}\) </li>
<li> \(\underline{F} = r \sin \theta \; \underline{\hat{r}} + r^2 \sin \phi \; \underline{\hat{\theta}} + r \cos \theta \; \underline{\hat{\phi}}\)</li></ol></p>
<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts"><li>
\[\begin{aligned}
{\underline \nabla} \cdot \underline{F} &= \frac{1}{r^2 \sin \theta} \left[
\frac{\partial}{\partial r}(r^2 \sin \theta F_{r}) +
\frac{\partial}{\partial \theta}(r \sin \theta F_{\theta}) +
\frac{\partial}{\partial \phi}(r  F_{\phi})  \right]\\
&= \frac{1}{r^2 \sin \theta} \left[ \frac{\partial}{\partial
r}(r^2 \sin \theta \times r) + \frac{\partial}{\partial \theta}(r
\sin \theta \times 0) + \frac{\partial}{\partial \phi}(r \times 0)
\right]\\
&= \frac{1}{r^2 \sin \theta} \left[ \frac{\partial}{\partial r}(r^3
\sin \theta) + \frac{\partial}{\partial \theta}(0) +
\frac{\partial}{\partial \phi}(0) \right]\\
&= \frac{1}{r^2 \sin \theta} \left[ 3 r^2 \sin \theta + 0 + 0
\right]= 3\end{aligned}\]
Note :- in Cartesian coordinates, the corresponding vector is
\(\underline{F} = x\underline{i} + y\underline{j}+z\underline{k}\)
with \({\underline \nabla} \cdot \underline{F} = 1+1+1 = 3\) (hence consistency). </li>

<li>\[\begin{aligned} {\underline \nabla} \cdot \underline{F} &= \frac{1}{r^2 \sin \theta} \left[
\frac{\partial}{\partial r}(r^2 \sin \theta \; F_{r}) +
\frac{\partial}{\partial \theta}(r \sin \theta \; F_{\theta}) +
\frac{\partial}{\partial \phi}(r  F_{\phi})  \right]\\
&= \frac{1}{r^2 \sin \theta} \left[ \frac{\partial}{\partial
r}(r^2 \sin \theta \; r^2 \sin \theta) + \frac{\partial}{\partial
\theta}(r \sin \theta \times 0) + \frac{\partial}{\partial \phi}(r
\times 0) \right]
\\ &= \frac{1}{r^2 \sin \theta} \left[ \frac{\partial}{\partial
r}(r^4 \sin^2 \theta) + \frac{\partial}{\partial \theta}(0) +
\frac{\partial}{\partial \phi}(0) \right]\\&= \frac{1}{r^2 \sin \theta} \left[ 4 r^3 \sin^2 \theta + 0 + 0
\right]= 4 r \sin \theta
\end{aligned}\]</li>

<li>\[\begin{aligned} {\underline \nabla} \cdot \underline{F} &= \frac{1}{r^2 \sin \theta} \left[
\frac{\partial}{\partial r}(r^2 \sin \theta \; F_{r}) +
\frac{\partial}{\partial \theta}(r \sin \theta \; F_{\theta}) +
\frac{\partial}{\partial \phi}(r  F_{\phi})  \right]\\
&= \frac{1}{r^2 \sin \theta} \left[ \frac{\partial}{\partial
r}(r^2 \sin \theta \; r \sin \theta) + \frac{\partial}{\partial
\theta}(r \sin \theta \times r^2 \sin \phi) +
\frac{\partial}{\partial \phi}(r \times \; r \cos \theta) \right]
\\ &= \frac{1}{r^2 \sin \theta} \left[ \frac{\partial}{\partial
r}(r^3 \sin^2 \theta) + \frac{\partial}{\partial \theta}(r^3 \sin
\theta \sin \phi ) +
\frac{\partial}{\partial \phi}(r^2 \cos \phi) \right]\\ &= \frac{1}{r^2 \sin \theta} \left[ 3 r^2 \sin^2 \theta + r^3
\cos \theta \; \sin \phi + 0
\right]=3 \sin \theta + r \cot \theta \; \sin \phi
\end{aligned}\]</li>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Using spherical polar coordinates, find \({\underline \nabla} \times \underline{F}\) for the following vector fields \(\underline{F}\).
<ol class="HELM_parts">
<li>\(\underline{F} = r^k \underline{\hat{r}}\), where \(k\) is a constant</li>
<li>\(\underline{F} = r^2 \cos \theta \; \underline{\hat{r}} + \sin \theta \; \underline{\hat{\theta}} + \sin^2 \theta \; \underline{\hat{\phi}}\)</li>
</ol></p>
<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
<li>\[\begin{aligned}
{\underline \nabla}\! \times\! \displaystyle \underline{F} &= \frac{1}{r^2 \sin
\theta} \left|\begin{array}{ccc}
   \underline{\hat{r}} & r \underline{\hat{\theta}} & r^2 \sin \theta \; \underline{\hat{\phi}}
   \\
   \displaystyle \frac{\partial}{\partial r} & \displaystyle \frac{\partial}{\partial \theta} &
   \displaystyle \frac{\partial}{\partial \phi} \\ \\
  F_{r} & r F_{\theta} & r^2 \sin \theta F_{\phi}
\end{array} \right|\\ &=
\frac{1}{r^2 \sin \theta} \left|\begin{array}{ccc}
   \underline{\hat{r}} & r \underline{\hat{\theta}} & r^2 \sin \theta \; \underline{\hat{\phi}}
   \\
   \displaystyle \frac{\partial}{\partial r} & \displaystyle \frac{\partial}{\partial \theta} &
   \displaystyle \frac{\partial}{\partial \phi} \\
  r^k & r \times 0 & r^2 \sin \theta \times 0
\end{array} \right| \\
&= \frac{1}{r^2 \sin \theta} \left[ \left(
\frac{\partial}{\partial \theta}(0) - \frac{\partial}{\partial
\phi}(0) \right) \underline{\hat{r}} + \left(
\frac{\partial}{\partial \phi}(r^k) - \frac{\partial}{\partial
r}(0) \right) r \underline{\hat{\theta}} + \left( \frac{\partial}{\partial r}(0) - \frac{\partial}{\partial \theta}(r^k) \right)
r^2 \sin \theta \underline{\hat{\phi}} \right] \\
&= 0 \; \underline{\hat{r}} + 0 \; \underline{\hat{\theta}} + 0
\; \underline{\hat{\phi}} = \underline{0}
\end{aligned}\]</li>
<li>\[\begin{aligned} {\underline \nabla} \times \displaystyle \underline{F} &= \frac{1}{r^2 \sin \theta} \left|\begin{array}{ccc}
   \underline{\hat{r}} & r \underline{\hat{\theta}} & r \sin \theta \; \underline{\hat{\phi}}
   \\
   \displaystyle \frac{\partial}{\partial r} & \displaystyle \frac{\partial}{\partial \theta} &
   \displaystyle \frac{\partial}{\partial \phi} \\
  F_{r} & r F_{\theta} & r \sin \theta F_{\phi}
\end{array} \right| \\
&=\frac{1}{r^2 \sin \theta} \left|\begin{array}{ccc}
   \underline{\hat{r}} & r \underline{\hat{\theta}} & r \sin \theta \; \underline{\hat{\phi}}
   \\
   \displaystyle \frac{\partial}{\partial r} & \displaystyle \frac{\partial}{\partial \theta} &
   \displaystyle \frac{\partial}{\partial \phi} \\
  r^2 \cos \theta & r \times \sin \theta & r \sin \theta \times \sin^2
  \theta
\end{array} \right| \\
&= \frac{1}{r^2 \sin \theta} \left[ \left(
\frac{\partial}{\partial \theta}(r \sin^3 \theta) -
\frac{\partial}{\partial \phi}(r \sin \theta) \right)
\underline{\hat{r}} + \left( \frac{\partial}{\partial \phi}(r^2
\cos \theta) - \frac{\partial}{\partial r}(r \sin^3 \theta)
\right) r \underline{\hat{\theta}} + \left( \frac{\partial}{\partial r}(r \sin \theta) -
\frac{\partial}{\partial \theta}(r^2 \cos \theta) \right)
r \sin \theta \underline{\hat{\phi}} \right] \\
&= \frac{1}{r^2 \sin \theta} \left[ \left( 3 r \sin^2 \theta
\cos \theta +0 \right) \underline{\hat{r}} + \left( 0 -  \sin^3
\theta \right) r \underline{\hat{\theta}} \left( \sin \theta + r \sin \theta \right) r \sin \theta \underline{\hat{\phi}} \right] \\
&= \dfrac{3 \sin \theta \cos \theta}{r} \underline{\hat{r}} - \dfrac{ \sin^2\theta}{r}
\underline{\hat{\theta}} + \dfrac{(1 + r^2)}{r} \sin \theta
\underline{\hat{\phi}}
\end{aligned}\]</li>]]></text>
    </questiontext>
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    </generalfeedback>
    <defaultgrade>0</defaultgrade>
    <penalty>0</penalty>
    <hidden>0</hidden>
    <idnumber/>
  </question>
</quiz>
