Editing ThreeDimensionalConfigurations/ChallengesPt6 (section)

====Plot Off-Center, Slightly Flattened Ellipse====


<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\biggl[ \frac{y' - y'_0}{y'_\mathrm{radius}} \biggr]^2
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 - \biggl( \frac{x'}{x'_\mathrm{radius}} \biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ \frac{y' - y'_0}{y'_\mathrm{radius}} 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\pm \biggl[ 1 - \biggl( \frac{x'}{x'_\mathrm{radius}} \biggr)^2 \biggr]^{1 / 2}
</math>
  </td>
</tr>
</table>
This seems to work perfectly!  We used Excel to generate trajectories using this expression and the results matched earlier determinations of these trajectories to machine precision.  

Let's now examine the normal to the surface that is obtained from this compact trajectory expression.  Given that <math>y'_0</math>, <math>y'_\mathrm{radius}</math>, and <math>x'_\mathrm{radius}</math> are all constants, we have,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
G'
</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{y' - y'_0}{y'_\mathrm{radius}} \biggr]^2
+
\biggl( \frac{x'}{x'_\mathrm{radius}} \biggr)^2 - 1
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\frac{1}{2} \nabla G'
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl[ \frac{x'}{(x')^2_\mathrm{radius}} \biggr] 
+
\boldsymbol{\hat\jmath'} \biggl[ \frac{y' - y'_0}{(y')^2_\mathrm{radius}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl[ \frac{x'}{a^2} \biggr] 
\biggl[ \frac{(c^2  \cos^2\theta + b^2 \sin^2\theta )}{(c^2 - z_0^2) \cos^2\theta + b^2\sin^2\theta} \biggr]
+
\boldsymbol{\hat\jmath'} \biggl[ \frac{y' }{b^2 c^2} \biggr]
\biggl[ \frac{(c^2  \cos^2\theta + b^2 \sin^2\theta )^2}{(c^2 - z_0^2) \cos^2\theta + b^2\sin^2\theta} \biggr]
-
\boldsymbol{\hat\jmath'} \biggl[ \frac{y'_0}{b^2 c^2} \biggr]
\biggl[ \frac{(c^2  \cos^2\theta + b^2 \sin^2\theta )^2}{(c^2 - z_0^2) \cos^2\theta + b^2\sin^2\theta} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ \biggl[ \frac{ (c^2 - z_0^2) \cos^2\theta + b^2\sin^2\theta}{ (c^2  \cos^2\theta + b^2 \sin^2\theta ) } \biggr] \frac{1}{2} \nabla G'
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl[ \frac{x'}{a^2} \biggr] 
+
\boldsymbol{\hat\jmath'} \biggl[ \frac{y' }{b^2 c^2} \biggr]
\biggl[ (c^2  \cos^2\theta + b^2 \sin^2\theta ) \biggr]
-
\boldsymbol{\hat\jmath'} \biggl[ \frac{y'_0}{b^2 c^2} \biggr]
\biggl[ (c^2  \cos^2\theta + b^2 \sin^2\theta ) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl[ \frac{x'}{a^2} \biggr] 
+
\boldsymbol{\hat\jmath'} \biggl[ \frac{y' }{b^2 c^2} \biggr]
\biggl[ (c^2  \cos^2\theta + b^2 \sin^2\theta ) \biggr]
+
\boldsymbol{\hat\jmath'} \biggl[ \frac{1}{b^2 c^2} \biggr]
\biggl[ z_0 b^2 \sin\theta \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl[ \frac{x'}{a^2} \biggr] 
+
\boldsymbol{\hat\jmath'} \biggl[ \frac{1}{b^2 c^2} \biggr]\biggl\{
y'(c^2  \cos^2\theta + b^2 \sin^2\theta )
+
z_0 b^2 \sin\theta \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl[ \frac{x'}{a^2} \biggr] 
+
\boldsymbol{\hat\jmath'} \biggl\{
\frac{y' \cos^2\theta }{b^2}
+ \frac{(y'\sin\theta +z_0 )\sin\theta}{c^2}
\biggr\} \, .
</math>
  </td>
</tr>
</table>

As we have [[#gradP|already stated]] &#8212; but setting <math>z' = 0</math> and ignoring the <math>\mathbf{\hat{k}'}</math> component because there is no motion in that direction &#8212; 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{1}{2} \nabla P'(x', y', z')</math>
  </td>
  <td align="center">
<math>\rightarrow</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl(\frac{x'}{a^2}\biggr)
+ \boldsymbol{\hat\jmath'} \biggl\{ \biggl[\frac{( y'\cos\theta - \cancelto{0}{z'\sin\theta} )\cos\theta}{b^2}\biggr] 
+ \biggl[\frac{( z_0 + \cancelto{0}{z'\cos\theta} + y'\sin\theta) \sin\theta}{c^2}\biggr] \biggr\}

+ \mathbf{\hat{k}'} \cancelto{0}{\biggl\{- \biggl[\frac{( y'\cos\theta - z'\sin\theta) \sin\theta}{b^2}\biggr] 
+ \biggl[\frac{ ( z_0 + z'\cos\theta + y'\sin\theta) \cos\theta}{c^2}\biggr] \biggr\}} \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl(\frac{x'}{a^2}\biggr)
+ \boldsymbol{\hat\jmath'} \biggl\{ \biggl[\frac{( y'\cos\theta  )\cos\theta}{b^2}\biggr] 
+ \biggl[\frac{( z_0 + y'\sin\theta) \sin\theta}{c^2}\biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>

Since, properly normalized, <math>\nabla G'</math> is identical to <math>\nabla P'</math>, and since we have [[#Orthogonal|already shown]] that <math>\mathbf{u'}_\mathrm{EFE}</math> is everywhere orthogonal to <math>\nabla P'</math>, it must be true that <math>\nabla G'</math> is everywhere orthogonal to <math>\mathbf{u'}_\mathrm{EFE}</math>.

<b><font color="red">Hooray!!</font></b>