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  <question type="description">
    <name>
      <text>29.1.3.13 Conservative vector fields</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h3>Conservative vector fields</h3>
<p>For some line integrals in the previous section, the value of the integral depended only on the vector field \(\underline{F}\) and the start and end points of the line but not on the actual path between the start and end points. However, for other line integrals, the result depended on the actual details of the path of the line.</p>
<p>Vector fields are classified according to whether the line integrals are path dependent or path independent. Those vector fields for which <em>all</em> line integrals between <em>all</em> pairs of points are path independent are called <strong>conservative vector fields</strong>.</p>
<p>There are five properties of a conservative vector field (1 to 5 below). It is impossible to check the value of every line integral over every path, but it <em>is</em> possible to use any one of these five properties (particularly property 3 below) to determine whether or not a vector field is conservative. These properties are also used to simplify calculations with conservative vector fields over non-closed paths.</p>
<ol>
<li>The line integral \(\displaystyle\displaystyle\int_A^B\underline{F}\cdot \underline{dr}\) depends only on the end points \(A\) and \(B\) and is independent of the actual path taken.</li>
<li>The line integral around any closed curve is zero. That is \(\displaystyle\displaystyle\oint_C\underline{F}\cdot \underline{dr}=0\) for all \(C\).</li>
<li>The curl of a conservative vector field \(\underline F\) is zero i.e. \({\underline \nabla} \times \underline{F}=\underline{0}\).</li>
<li>For any conservative vector field \(\underline{F}\), it is possible to find a scalar field \(\phi\) such that \({\underline \nabla} \phi=\underline{F}\). Then, \(\displaystyle\displaystyle\oint_C\underline{F}\cdot \underline{dr}=\phi(B)-\phi(A)\) where \(A\) and \(B\) are the start and end points of contour \(C\).</li>
<li>All gradient fields are conservative. That is, \(\underline{F} = {\underline \nabla} \phi\) is a conservative vector field for any scalar field \(\phi\).</li>
</ol>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Consider the following vector fields and line integral pairs:</p>
<ol class="HELM_parts">
<li>\(\underline{F}_1=y^2\underline{i}+xy\underline{j}\) and paths \(C_1\), \(C_2\) such that \(\displaystyle \int_{C_1} \underline{F}_1 \cdot \underline{\mathrm{d}r} = \dfrac{8}{3}\), \(\displaystyle \int_{C_2} \underline{F}_1 \cdot \underline{\mathrm{d}r} = \dfrac{12}{5}\), and \(C_1 - C_2\) forms a closed path.</li>
<li>\(\underline{F}_2=2x\underline{i}+2y\underline{j}\) and paths \(C_3\), \(C_4\) such that \(\displaystyle \int_{C_3} \underline{F}_2 \cdot \underline{\mathrm{d}r} = 20\), \(\displaystyle \int_{C_4} \underline{F}_2 \cdot \underline{\mathrm{d}r} = 20\) and \(C_3 - C_4\) forms a closed path.</li>
<li>\(\underline{F}_3=y^2z^3\underline{i}+2xyz^3\underline{j}+3xy^2z^2\underline{k}\) and paths \(C_5\), \(C_6\) such that \(\displaystyle \int_{C_5} \underline{F}_3 \cdot \underline{\mathrm{d}r} =1\), \(\displaystyle \int_{C_6} \underline{F}_3 \cdot \underline{\mathrm{d}r} = 1\) and \(C_5 - C_6\) forms a closed path.</li>
<li>\(\underline{F}_4=x\underline{i}+\left(4x-y\right)\underline{j}\) and paths \(C_7\), \(C_8\) such that \(\displaystyle \int_{C_7} \underline{F}_4 \cdot \underline{\mathrm{d}r}= 2\), \(\displaystyle \int_{C_8} \underline{F}_4 \cdot \underline{\mathrm{d}r} = \dfrac{4}{3}\) and \(C_7 - C_8\) forms a closed path.</li>
</ol>
<p>Determine which of these vector fields are conservative where possible by referring to the given line integrals. For those that are conservative find a scalar field \(\phi\) such that \(\underline{F}={\underline \nabla} \phi\) and use property 4 to verify the values of the line integrals.</p>
<p></p>
<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
<li>
<p>Two different values were obtained for line integrals over the paths \(C_1\) and \(C_2\). Hence, by property 1, \(\underline{F}_1\) is not conservative. [It is also possible to reach this conclusion from P3 by finding that \({\underline \nabla} \times \underline{F}= -y\underline{k} \ne \underline{0}\)].</p>
</li>
<li>
<p>For the closed path consisting of \(C_3\) and \(C_4\) we obtain the value \(20+(-20)=0\). This alone does not mean that \(\underline{F}_2\) is conservative as there could be other paths giving different values. So by using property 3</p>
<p>\(\qquad\displaystyle {\underline \nabla} \times \underline{F}_2= \left|\begin{array} {ccc}\underline{i}&amp;\underline{j}&amp;\underline{k}\\ \\ \displaystyle\frac{\partial}{\partial x}&amp;\displaystyle\frac{\partial}{\partial y}&amp;\displaystyle\frac{\partial}{\partial z}\\ \\ 2x&amp;2y&amp;0 \end{array}\right|=\underline{i}(0-0)-\underline{j}(0-0)+\underline{k}(0-0)=\underline{0} \)</p>
<p>As \({\underline \nabla} \times \underline{F}_2 = \underline{0}\), property 3 gives that \(\underline{F}_2\) is a conservative vector field. <br>Now, find a \(\phi\) such that \(\underline{F}_2={\underline \nabla} \phi\). Then \(\displaystyle\frac{\partial\phi}{\partial x}\underline{i}+\frac{\partial\phi}{\partial y}\underline{j}=2x\underline{i}+2y\underline{j}.\)<br>Thus \(\qquad\displaystyle \left.\begin{array}{ccc} \displaystyle\frac{\partial\phi}{\partial x}=2x&amp;\Rightarrow&amp;\phi=x^2+f(y)\\ &amp;&amp;\\ \displaystyle\frac{\partial\phi}{\partial y}=2y&amp;\Rightarrow&amp;\phi=y^2+g(x)\end{array}\right\}\Rightarrow\phi=x^2+y^2(+\mbox{ constant})\)</p>
<p>Using property 4:</p>
<p>\(\displaystyle\displaystyle\int_{(0,0)}^{(4,2)}\underline{F}_2\cdot \underline{dr}=\displaystyle\int_{(0,0)}^ {(4,2)}({\underline \nabla} \phi)\cdot \underline{dr}=\phi(4,2)-\phi(0,0)=(4^2+2^2)-(0^2+0^2)=20\).</p>
</li>
<li>
<p>The fact that line integrals along two different paths between the same start and end points have the same value is consistent with \(\underline{F}_3\) being a conservative field according to property 1. So too is the fact that the integral around a closed path is zero according to property 2. However, neither fact can be used to <i>conclude</i> that \(\underline{F}_3\) is a conservative field. This can be done by showing that \({\underline \nabla} \times \underline{F}_3 = \underline{0}\).</p>
<p>Now, \(\left|\begin{array}{ccc} \underline{i} &amp; \underline{j} &amp; \underline{k} \\ &amp; &amp; \\ \dfrac{\partial}{\partial x} &amp; \dfrac{\partial}{\partial y} &amp; \dfrac{\partial}{\partial z} \\ &amp; &amp; \\ y^2 z^3 &amp; 2 x y z^3 &amp; 3 x y^2 z^2 \end{array}\right| = (6xyz^2 - 6xyz^2)\underline{i} - (3 y^2 z^2 - 3 y^2 z^2 ) \underline{j} + (2yz^3 - 2yz^3)\underline{k} = \underline{0}.\)</p>
<p>As \({\underline \nabla} \times \underline{F}_3 = \underline{0}\), property 3 gives that \(\underline{F}_3\) is a conservative field.</p>
<p>To find \(\phi\) that satisfies \({\underline \nabla} \phi = \underline{F}_3\), it is necessary to satisfy \(\qquad\left. \begin{array}{lcc} \dfrac{\partial \phi}{\partial x} = y^2 z^3 &amp; \rightarrow &amp; \phi = x y^2 z^3 + f(y,z) \\ \\ \dfrac{\partial \phi}{\partial y} = 2 x y z^3 &amp; \rightarrow &amp; \phi = x y^2 z^3 + g(x,z) \\ \\ \dfrac{\partial \phi}{\partial z} =3 x y^2 z^2 &amp; \rightarrow &amp; \phi = x y^2 z^3 + h(x,y) \end{array}\right\} \rightarrow \phi = x y^2 z^3\)</p>
<p>Using property 4: \(\displaystyle\displaystyle\int_{(0,0,0)}^{(1,1,1)}\underline{F}_3\cdot \underline{dr}=\phi(1,1,1)-\phi(0,0,0) = 1-0 = 1\).</p>
</li>
<li>
<p>As the integral along \(C_7\) is \(2\) and the integral along \(C_8\) (same start and end points but different intermediate points) is \(\frac{4}{3}\), \(F_4\) is <strong>not</strong> a conservative field using property 1.</p>
<p>Note that \({\underline \nabla} \times \underline{F}_4 = 4\underline{k} \ne \underline{0}\) so, using property 3, this is an independent conclusion that \(\underline{F}_4\) is <strong> not</strong> conservative.</p>
</li>
</ol>
<hr>
<h4 class="HELM_example">Engineering Example</h4>
<p><strong>Work done moving a charge in an electric field</strong></p>
<p><strong>Introduction</strong></p>
<p>If a charge, \(q\), is moved through an electric field, \(\underline{E}\), from \(A\) to \(B\), then the work required is given by the line integral</p>
<p>\(\qquad\displaystyle W_{AB} = -q \int_A^B \underline{E} \cdot \underline{dr}\)</p>
<p><strong>Problem in words</strong></p>
<p>Compare the work done in moving a charge through the electric field around a point charge in a vacuum via two different paths.</p>
<p><strong>Mathematical statement of problem</strong></p>
<p>An electric field \(\underline{E}\) is given by \[\begin{aligned} \underline{E} &amp;= \frac{Q}{4\pi \varepsilon_0 r^2} \underline{\hat{r}}\\ &amp;= \frac{Q}{4\pi \varepsilon_0 (x^2 + y^2 + z^2)}\times \frac{x\underline{i} + y\underline{j} + z\underline{k}}{\sqrt{x^2+y^2+z^2}}\\ &amp;= \frac{Q(x\underline{i} + y\underline{j} + z\underline{k})}{4\pi\varepsilon_0(x^2+y^2+z^2)^{\frac{3}{2}}}\end{aligned}\] where \(\underline r\) is the position vector with magnitude \(r\) and unit vector \(\hat{\underline r}\), and \(\dfrac{1}{4\pi\epsilon_0}\) is a combination of constants of proportionality, where \(\epsilon_0=10^{-9}/36\pi \ {\rm F\ m}^{-1}\).</p>
<p>Given that \(Q\) = \(10^{-8}\)C, find the work done in bringing a charge of \(q\) = \(10^{-10}\)C from the point \(A=(10,10,0)\) to the point \(B=(1,1,0)\) (where the dimensions are in metres)</p>
<ol class="HELM_parts">
<li>by the direct straight line \(y=x\), \(z=0\)</li>
<li>by the straight line pair via \(C=(10,1,0)\)</li>
</ol>
<p></p>
<figure class="HELM_figure"><img class="img-fluid" src="@@PLUGINFILE@@/twopath-1.svg" alt="A triangle ABC on which are depicted 2 routes (a and b) along which a charge can move through an electric field." width="300" height="243">
<figcaption>Two routes (\({a}\) and \({b}\)) along which a charge can move through an electric field.</figcaption>
</figure>
<p></p>
<p>The path comprises two straight lines from \(A=(10,10,0)\) to \(B=(1,1,0)\) via \(C=(10,1,0)\) (see Figure 4).</p>
<p><strong>Mathematical analysis</strong></p>
<ol class="HELM_parts">
<li>Here \(Q/(4\pi\varepsilon_0)\) = 90 so \(\displaystyle \underline{E} = \frac{90[x\underline{i} + y\underline{j}]}{(x^2+y^2)^{\frac{3}{2}}}\) as \(z=0\) over the region of interest. The work done \[\begin{aligned} W_{AB} &amp;=-q \int_A^B \underline{E} \cdot \underline{dr} \\ &amp;=-10^{-10} \int_A^B \frac{90}{(x^2+y^2)^{\frac{3}{2}}}\;[x\underline{i}+y\underline{j}] \cdot [\mathrm{d}x\underline{i}+\mathrm{d}y\underline{j}] \end{aligned}\] Using \(\;\;y=x, \;\; \mathrm{d}y=\mathrm{d}x\) \[\begin{aligned} W_{AB}&amp;=-10^{-10} \int_{x=10}^1 \frac{90}{(2x^2)^\frac{3}{2}} \;\;\{x\;\mathrm{d}x + x\;\mathrm{d}x\} \\ &amp;= -10^{-10} \int_{10}^1 \frac{90}{(2\sqrt{2})} \;\; x^{-3} \; 2x \; \mathrm{d}x\\ &amp;=\frac{90 \times -10^{-10}}{\sqrt{2}} \int_{10}^1 x^{-2} \; \mathrm{d}x \\ &amp;= \frac{9 \times -10^{-9}}{\sqrt{2}} \left[\phantom{\dfrac{1}{1}}\!\!\! -x^{-1} \right]_{10}^1 \\ &amp;=\frac{9 \times 10^{-9}}{\sqrt{2}} \left[\phantom{\dfrac{1}{1}}\!\!\! x^{-1} \right]_{10}^1 \\ &amp;=\frac{9 \times 10^{-9}}{\sqrt{2}} \left[ 1 - 0.1 \right] \\ &amp;=5.73 \times 10^{-9} \mbox{ J} \end{aligned}\]</li>
<li>The first part of the path is \(A\) to \(C\) where \(x=10\), \(\mathrm{d}x=0\) and \(y\) goes from 10 to 1. \[\begin{aligned} W_{AC} &amp;= -q \int_A^C \underline{E} \cdot \underline{dr} \\ &amp; = -10^{-10} \int_{y=10}^1\frac{90}{(100+y^2)^\frac{3}{2}} \;\; [x\underline{i} + y\underline{j}] \cdot [0\underline{i}+\mathrm{d}y\underline{j}]\\ &amp;=-10^{-10} \int_{10}^1\frac{90 y \; \mathrm{d}y}{(100+y^2)^\frac{3}{2}}\\ &amp;= -10^{-10} \int_{u=200}^{101}\frac{45 \; du}{u^\frac{3}{2}} \;\; \mbox{ (substituting } \;\; u=100+y^2, \; du = 2y\;\mathrm{d}y)\\ &amp;= -45 \times 10^{-10} \int_{200}^{101} u^{-\frac{3}{2}} \; du\\ &amp;=-45 \times 10^{-10} \left[ -2u^{-\frac{1}{2}} \right]_{200}^{101} \\ &amp;=45 \times 10^{-10} \left( \frac{2}{\sqrt{101}} - \frac{2}{\sqrt{200}} \right) = 2.59 \times 10^{-10} \mbox{J} \end{aligned}\] The second part is \(C\) to \(B\), where \(y=1\), \(\mathrm{d}y=0\) and \(x\) goes from 10 to 1. \[\begin{aligned} W_{CB} &amp;= -10^{-10} \int_{x=10}^1\frac{90}{(x^2+1)^\frac{3}{2}} \;\; [x\underline{i} + y\underline{j}] \cdot [\mathrm{d}x\underline{i}+0\underline{j}]\\ &amp;=-10^{-10} \int_{10}^1\frac{90x \; \mathrm{d}x}{(x^2+1)^\frac{3}{2}}\\ &amp;= -10^{-10} \int_{u=101}^{2}\frac{45 \; du}{u^\frac{3}{2}} \;\; \mbox{ (substituting }\;\; u=x^2+1, \; du = 2x\;\mathrm{d}x)\qquad\qquad\\ &amp;= -45 \times 10^{-10} \int_{101}^{2} u^{-\frac{3}{2}} \; du\\ &amp;=-45 \times 10^{-10} \left[ -2u^{-\frac{1}{2}} \right]_{101}^{2} \\ &amp;=45 \times 10^{-10} \left(\frac{2}{\sqrt{2}} - \frac{2}{\sqrt{101}} \right) = 5.468 \times 10^{-9} \mbox{J} \end{aligned}\] The sum of the two components \(W_{AC}\) and \(W_{CB}\) is \(5.73 \times 10^{-9}\)J. <br>Therefore the work done over the two paths (a) and (b) is identical.</li>
</ol>
<p><strong>Interpretation</strong></p>
<p>In fact, the work done is independent of the route taken as the electric field \(\underline{E}\) around a point charge in a vacuum is a <strong>conservative</strong> field.</p>
<hr>
<h4 class="HELM_example">Example</h4>
<ol class="HELM_parts">
<li>Show that \(I = \displaystyle\int_{(0,0)}^{(2,1)} \left\{ (2xy+1)\mathrm{d}x + (x^2-2y)\mathrm{d}y\right\}\) is independent of the path taken.</li>
<li>Find \(I\) using property property 1.</li>
<li>Find \(I\) using property property 4.</li>
<li>Find \(I = \displaystyle\oint_C \left\{ (2xy+1)\mathrm{d}x + (x^2-2y)\mathrm{d}y\right\}\) where \(C\) is
<ol class="HELM_parts">
<li>the circle \(x^2 + y^2 = 1\)</li>
<li>the square with vertices \((0,0)\), \((1,0)\), \((1,1)\), \((0,1)\).</li>
</ol>
</li>
</ol>
<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
<li>
<p>The integral \(I = \displaystyle\int_{(0,0)}^{(2,1)} \left\{ (2xy+1)\mathrm{d}x + (x^2-2y)\mathrm{d}y\right\}\) may be re-written \(\displaystyle\int_C \underline{F} \cdot \underline{dr}\) where</p>
<p>\(\underline{F} = (2xy+1)\underline{i} + (x^2 - 2y) \underline{j}\).</p>
<p>Now \({\underline \nabla} \times \underline{F} = \left|\begin{array}{ccc} \underline{i} &amp; \underline{j} &amp; \underline{k} \\ &amp; &amp; \\ \dfrac{\partial}{\partial x} &amp; \dfrac{\partial}{\partial y} &amp; \dfrac{\partial}{\partial z} \\ &amp; &amp; \\ 2xy+1 &amp; x^2-2y &amp; 0 \end{array}\right| = 0\underline{i} + 0\underline{j} + 0\underline{k} = \underline{0}\)</p>
<p>As \({\underline \nabla} \times \underline{F} = \underline{0}\), \(\underline{F}\) is a conservative field and \(I\) is independent of the path taken between \((0,0)\) and \((2,1)\).</p>
</li>
<li>
<p>As \(I\) is independent of the path taken from \((0,0)\) to \((2,1)\), it can be evaluated along <strong> any</strong> such path. One possibility is the straight line \(y = \frac{1}{2}x\). On this line, \(\mathrm{d}y = \frac{1}{2}\mathrm{d}x\). The integral \(I\) becomes</p>
\[\begin{eqnarray*} I &amp;=&amp; \displaystyle\int_{(0,0)}^{(2,1)} \left\{ (2xy+1)\mathrm{d}x + (x^2-2y)\mathrm{d}y \right\}\\ &amp;=&amp; \displaystyle\int_{x=0}^2 \left\{ (2 x \times \frac{1}{2}x + 1)\mathrm{d}x + (x^2 - x) \frac{1}{2}\mathrm{d}x \right\}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \\ &amp;=&amp; \displaystyle\int_0^2 (\frac{3}{2}x^2 -\frac{1}{2}x + 1)\mathrm{d}x \\ &amp;=&amp; \left[ \frac{1}{2}x^3 - \frac{1}{4}x^2 + x\right]_0^2 = 4 - 1 + 2 - 0 = 5 \end{eqnarray*}\]</li>
<li>
<p>If \(\underline{F} = {\underline \nabla} \phi\) then\(\left. \begin{array}{ccc} \dfrac{\partial \phi}{\partial x} = 2xy+1 &amp; \rightarrow &amp; \phi = x^2 y + x + f(y) \\ \\ \dfrac{\partial \phi}{\partial y} = x^2-2y &amp; \rightarrow &amp; \phi = x^2 y - y^2 + g(x) \end{array}\right\} \rightarrow \phi = x^2 y+x-y^2+C.\)</p>
<p>These are consistent if \(\phi = x^2 y + x - y^2\) (plus a constant which may be omitted since it cancels).</p>
<p>So \(I = \phi(2,1) - \phi(0,0) = (4 + 2 - 1) - 0 = 5\)</p>
</li>
<li>
<p>As \(\underline{F}\) is a conservative field, all integrals around a closed contour are zero.</p>
</li>
</ol>]]></text>
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