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<section id="unipolar-relaxation">
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<h1>Unipolar relaxation<a class="headerlink" href="#unipolar-relaxation" title="Permalink to this headline">¶</a></h1>
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<section id="problem-statement">
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<h2>Problem Statement<a class="headerlink" href="#problem-statement" title="Permalink to this headline">¶</a></h2>
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<p>Consider an initially uncharged conducting sphere placed in a uniform, cold, unipolar ion bath.
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After a while we bias the sphere to a polarity opposite that of the ions, and then immediately disconnect it, leaving the sphere electrically floating at a nonzero potential.
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The resulting unipolar current can be derived from the electric field</p>
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<div class="math notranslate nohighlight">
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\begin{eqnarray}
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E_r=\frac{Q}{4 \pi \varepsilon_0 r^2}
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\end{eqnarray}</div><p>and the definition of mobility</p>
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<div class="math notranslate nohighlight">
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\begin{eqnarray}
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\vec{v} = \mu \vec{E}
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\end{eqnarray}</div><p>to get</p>
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<div class="math notranslate nohighlight">
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\begin{eqnarray}
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I = n q v_r A = n q \cdot \mu E_r \cdot 4 \pi R^2
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\end{eqnarray}</div><p>where <span class="math notranslate nohighlight">\(n\)</span> is ion concentration, <span class="math notranslate nohighlight">\(q\)</span> is ion charge, and <span class="math notranslate nohighlight">\(R\)</span> is radius of the sphere. This yields the differential equation for charge on the sphere</p>
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<div class="math notranslate nohighlight">
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\begin{eqnarray}
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\dot{Q}=n q \mu \cdot \frac{Q}{4 \pi \varepsilon_0 R^2} \cdot 4 \pi R^2=Q \frac{n q \mu}{\varepsilon_0}
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\end{eqnarray}</div><p>Converting <span class="math notranslate nohighlight">\(Q\)</span> to <span class="math notranslate nohighlight">\(V\)</span> using the definition of capacitance</p>
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<div class="math notranslate nohighlight">
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\begin{eqnarray}
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C V = Q
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\end{eqnarray}</div><p>and noting that</p>
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<div class="math notranslate nohighlight">
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\begin{eqnarray}
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\sigma = n |q| \mu
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\end{eqnarray}</div><p>the solution is</p>
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<div class="math notranslate nohighlight">
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\begin{eqnarray}
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V(t) = V_0 \exp\left(-\frac{t}{\varepsilon_0/\sigma}\right)
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\end{eqnarray}</div><p>The negative sign in the exponential comes from the sign of <span class="math notranslate nohighlight">\(q\)</span> relative to <span class="math notranslate nohighlight">\(V_0\)</span>.</p>
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</section>
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<section id="numerical-validation">
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<h2>Numerical Validation<a class="headerlink" href="#numerical-validation" title="Permalink to this headline">¶</a></h2>
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<p>Runs simulating the relaxation phenomenon for a conducting sphere can be found in <code class="docutils literal notranslate"><span class="pre">dragion/runs/sphere</span></code>.</p>
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<p>Initially, positive unipolar ions are loaded uniformly into the domain, with concentration <span class="math notranslate nohighlight">\(n=\sqrt{\frac{q}{\alpha}}=2\times10^9 \: \mathrm{m}^{-3}\)</span> [Lorenz, MMRTG], and a sphere of radius <span class="math notranslate nohighlight">\(R=0.5 \mathrm{m}\)</span> is biased to <span class="math notranslate nohighlight">\(-10 \mathrm{V}\)</span> then left floating.
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Note: this simulation does not include a uniform background source or collisional sink.</p>
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<p>The resulting time-progression is illustrated by a sequence of 3D screenshots:</p>
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<ul class="simple">
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<li><p><span class="math notranslate nohighlight">\(t=0 \mathrm{s}\)</span></p></li>
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</ul>
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<img alt="../_images/relaxation-sphere-3d_0.png" src="../_images/relaxation-sphere-3d_0.png" />
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<ul class="simple">
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<li><p><span class="math notranslate nohighlight">\(t=500 \mathrm{s}\)</span></p></li>
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</ul>
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<img alt="../_images/relaxation-sphere-3d_1.png" src="../_images/relaxation-sphere-3d_1.png" />
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<ul class="simple">
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<li><p><span class="math notranslate nohighlight">\(t=1000 \mathrm{s}\)</span></p></li>
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</ul>
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<img alt="../_images/relaxation-sphere-3d_2.png" src="../_images/relaxation-sphere-3d_2.png" />
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<p>As the sphere collects nearby ions its (negative) charge level and electric field strength diminish.
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The voltage drop is exponential, and its slope (in log-voltage) can be used to back out an estimate of atmospheric conductivity, <span class="math notranslate nohighlight">\(\sigma = \mathrm{slope} \cdot \varepsilon_0\)</span>.</p>
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<p>Here is a plot of (log10) voltage over time:</p>
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<img alt="../_images/relaxation-sphere.png" src="../_images/relaxation-sphere.png" />
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<p>Notice that overall the <strong>dragion</strong> sphere relaxes more slowly than predicted by theory.
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However, there is a slight inflection in the simulated voltage at <span class="math notranslate nohighlight">\(t=1200 \mathrm{s}\)</span>.
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A comparison of slopes before and after this knee, and the full end-to-end slope provide estimates for conductivity, using <span class="math notranslate nohighlight">\(\sigma = \mathrm{slope} \cdot \varepsilon_0\)</span>:</p>
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<table class="docutils align-default">
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<colgroup>
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<col style="width: 27%" />
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<col style="width: 37%" />
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<col style="width: 36%" />
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</colgroup>
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<thead>
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<tr class="row-odd"><th class="head"><p>Fit</p></th>
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<th class="head"><p>Slope (relaxation rate) <span class="math notranslate nohighlight">\([\mathrm{s}^{-1}]\)</span></p></th>
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<th class="head"><p><span class="math notranslate nohighlight">\(\sigma [\mathrm{S}/\mathrm{m}]\)</span></p></th>
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</tr>
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</thead>
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<tbody>
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<tr class="row-even"><td><p>End-to-end</p>
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<p>Before <span class="math notranslate nohighlight">\(t=1200 \: \mathrm{s}\)</span></p>
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<p>After <span class="math notranslate nohighlight">\(t=1200 \:\mathrm{s}\)</span></p>
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</td>
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<td><p>-0.0011</p>
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<p>-0.0009</p>
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<p>-0.0013</p>
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</td>
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<td><p><span class="math notranslate nohighlight">\(0.95\times10^{-14}\)</span></p>
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<p><span class="math notranslate nohighlight">\(0.79\times10^{-14}\)</span></p>
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<p><span class="math notranslate nohighlight">\(1.2\times10^{-14}\)</span></p>
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</td>
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</tr>
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<tr class="row-odd"><td><p>Theoretical</p></td>
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<td><p>-0.0016</p></td>
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<td><p><span class="math notranslate nohighlight">\(1.4\times10^{-14}\)</span></p></td>
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</tr>
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</tbody>
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</table>
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<p><em>More discussion is needed. Why is there an inflection point? It is probably related to domain shape and size. Why is the late-time estimate the best?</em></p>
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<p>At very late times the near-surface ion concentration will be depleted, so without a background source charging will eventually halt. In this regime the assumption of steady near-surface concentration would be violated and the theoretical model breaks down.</p>
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<li class="toctree-l1"><a class="reference internal" href="../getstarted.html">Getting started</a></li>
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<li class="toctree-l1 current"><a class="reference internal" href="../physvalid.html">Physics validation</a><ul class="current">
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<li class="toctree-l2"><a class="reference internal" href="unipolar_source_loss.html">Unipolar ions: uniform source and loss</a></li>
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<li class="toctree-l2 current"><a class="current reference internal" href="#">Unipolar relaxation</a></li>
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<li>Previous: <a href="unipolar_source_loss.html" title="previous chapter">Unipolar ions: uniform source and loss</a></li>
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