Editing ThreeDimensionalConfigurations/FerrersPotential (section)

===The Case Where n = 1===

When <math>n = 1</math>, we have a specific heterogeneous density configuration, and the "interior" potential will be given by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{ \Phi_\mathrm{grav}(\mathbf{x})}{(-\pi G\rho_c)} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2} a_1 a_2 a_3  \int_0^\infty \frac{ du}{\Delta } \biggl[ 1 - \sum_{\ell = 1}^3 \frac{x_\ell^2}{ a_\ell^2 + u }  \biggr]^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2} a_1 a_2 a_3  \biggl\{  
\int_0^\infty \frac{ du}{\Delta } \biggl[ 1 - \sum_{\ell = 1}^3 \frac{x_\ell^2}{ a_\ell^2 + u }  \biggr]
- ~ x^2 \int_0^\infty \frac{ du}{\Delta (a_1^2 + u)} \biggl[ 1 - \sum_{\ell = 1}^3 \frac{x_\ell^2}{ a_\ell^2 + u }  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- ~y^2 \int_0^\infty \frac{ du}{\Delta (a_2^2 + u)}  \biggl[ 1 - \sum_{\ell = 1}^3 \frac{x_\ell^2}{ a_\ell^2 + u }  \biggr]
- ~z^2 \int_0^\infty \frac{ du}{\Delta (a_3^2 + u)} \biggl[ 1 - \sum_{\ell = 1}^3 \frac{x_\ell^2}{ a_\ell^2 + u }  \biggr]
\biggr\} \, .
</math>
  </td>
</tr>
</table>

The first definite-integral expression inside the curly braces is, to within a leading factor of <math>\tfrac{1}{2}</math>, identical to the entire expression for the normalized potential that was derived in the case where n = 0. That is, we can write,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{ \Phi_\mathrm{grav}(\mathbf{x})}{(-\pi G\rho_c)} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2} \biggl[ I_\mathrm{BT} a_1^2 - \biggl(A_1 x^2 + A_2 y^2 +A_3 z^2 \biggr) \biggr]
- \frac{1}{2} a_1 a_2 a_3  \biggl\{ ~ x^2 \int_0^\infty \frac{ du}{\Delta (a_1^2 + u)} \biggl[ 1 - \sum_{\ell = 1}^3 \frac{x_\ell^2}{ a_\ell^2 + u }  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+~y^2 \int_0^\infty \frac{ du}{\Delta (a_2^2 + u)}  \biggl[ 1 - \sum_{\ell = 1}^3 \frac{x_\ell^2}{ a_\ell^2 + u }  \biggr]
+ ~z^2 \int_0^\infty \frac{ du}{\Delta (a_3^2 + u)} \biggl[ 1 - \sum_{\ell = 1}^3 \frac{x_\ell^2}{ a_\ell^2 + u }  \biggr]
\biggr\} \, .
</math>
  </td>
</tr>
</table>
Then, from &sect;22, p. 56 of [[Appendix/References#EFE|EFE]], we see that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>a_1 a_2 a_3 \int_0^\infty \frac{ du}{\Delta (a_i^2 + u)}  \biggl[ 1 - \sum_{\ell = 1}^3 \frac{x_\ell^2}{ a_\ell^2 + u }  \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( A_i - \sum_{\ell=1}^3 A_{i\ell} x_\ell^2 \biggr) \, .
</math>
  </td>
</tr>
<tr><td align="center" colspan="3">
[ [[Appendix/References#EFE|EFE]], <font color="#00CC00">Chapter 3, &sect;22, p. 53, Eq. (125)</font> ]
</td></tr>
</table>

<span id="GravFor1">Applying this result to each of the other three definite integrals gives us,</span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{ \Phi_\mathrm{grav}(\mathbf{x})}{(-\pi G\rho_c)} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2} \biggl[ I_\mathrm{BT} a_1^2 - \biggl(A_1 x^2 + A_2 y^2 +A_3 z^2 \biggr) \biggr]
- \frac{x^2}{2}  \biggl( A_1 - \sum_{\ell=1}^3 A_{1\ell} x_\ell^2 \biggr)
- \frac{y^2}{2}  \biggl( A_2 - \sum_{\ell=1}^3 A_{2\ell} x_\ell^2 \biggr)
- \frac{z^2}{2}  \biggl( A_3 - \sum_{\ell=1}^3 A_{3\ell} x_\ell^2 \biggr) \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2} \biggl[ I_\mathrm{BT} a_1^2 - \biggl(A_1 x^2 + A_2 y^2 +A_3 z^2 \biggr) \biggr]
- \frac{x^2}{2}  \biggl[ A_1 - \biggl( A_{11}x^2 + A_{12}y^2 + A_{13}z^2 \biggr) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- \frac{y^2}{2}  \biggl[ A_2 - \biggl( A_{21}x^2 + A_{22}y^2 + A_{23}z^2 \biggr) \biggr]
- \frac{z^2}{2}  \biggl[ A_3 - \biggl( A_{31}x^2 + A_{32}y^2 + A_{33}z^2 \biggr) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2} I_\mathrm{BT} a_1^2 
- \biggl(A_1 x^2 + A_2 y^2 +A_3 z^2 \biggr) 
~+ \biggl( A_{12} x^2y^2 + A_{13} x^2z^2 + A_{23} y^2z^2\biggr)
~+ \frac{1}{2}  \biggl(A_{11}x^4 +  A_{22}y^4 + A_{33}z^4  \biggr) 
\, ,
</math>
  </td>
</tr>
</table>

where,
<table border="1" align="center" cellpadding="10" width="80%">
<tr>
  <td align="center" width="50%">

<table border="0" cellpadding="5" align="center">
<tr><td align="center" colspan="3">for <math>i \ne j</math></td></tr>
<tr>
  <td align="right">
<math>A_{ij}</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>-\frac{A_i-A_j}{(a_i^2 - a_j^2)} </math>
  </td>
</tr>
<tr><td align="center" colspan="3">[ [[Appendix/References#EFE|EFE]], <font color="#00CC00">&sect;21, Eq. (107)</font> ]</td></tr>
</table>

  </td>
  <td align="center" width="50%">

<table border="0" cellpadding="5" align="center">
<tr><td align="center" colspan="3">for <math>i = j</math></td></tr>
<tr>
  <td align="right">
<math>2A_{ii} + \sum_{\ell = 1}^3 A_{i\ell}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{2}{a_i^2} </math>
  </td>
</tr>
<tr><td align="center" colspan="3">[ [[Appendix/References#EFE|EFE]], <font color="#00CC00">&sect;21, Eq. (109)</font> ]</td></tr>
</table>

  </td>
</tr>
</table>


and we have made use of the symmetry relation, <math>A_{ij} = A_{ji}</math>.  Again, as a check, let's see if this scalar potential satisfies the differential form of the
<div align="center">
<span id="PGE:Poisson"><font color="#770000">'''Poisson Equation'''</font></span><br />

{{Math/EQ_Poisson01}}
</div>
We find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\nabla^2 \biggl[ \frac{\Phi_\mathrm{grav}}{-2\pi G \rho_c} \biggr] </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2}\biggl[\frac{\partial^2}{\partial x^2}  + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\biggr] 
\biggl[- \biggl(A_1 x^2 + A_2 y^2 +A_3 z^2 \biggr) 
+ \biggl( A_{12} x^2y^2 + A_{13} x^2z^2 + A_{23} y^2z^2\biggr)
+ \frac{1}{2}  \biggl(A_{11}x^4 +  A_{22}y^4 + A_{33}z^4  \biggr) 
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\partial}{\partial x}   \biggl[- A_1 x  ~+ A_{12} x y^2 + A_{13} x z^2 ~+ A_{11}x^3  \biggr]
+\frac{\partial}{\partial y}  \biggl[- A_2 y   ~+  A_{12} x^2y  + A_{23} y z^2~+ A_{22}y^3  \biggr]
+ \frac{\partial}{\partial z}\biggl[- A_3 z  ~+ A_{13} x^2z + A_{23} y^2z~+ A_{33}z^3  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[- A_1  + A_{12} y^2 + A_{13} z^2 ~+ 3A_{11}x^2  \biggr]
+ \biggl[- A_2 +  A_{12} x^2  + A_{23} z^2~+ 3A_{22}y^2  \biggr]
+ \biggl[- A_3  + A_{13} x^2 + A_{23} y^2~+ 3A_{33}z^2  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- (A_1 + A_2 + A_3) + x^2(3A_{11} + A_{12} + A_{13}) + y^2( 3A_{22} + A_{12} + A_{23}) + z^2( 3A_{33} + A_{13} + A_{23})\, .
</math>
  </td>
</tr>
</table>

In addition to recognizing, as [[#SumTo2|stated above]], that <math>(A_1 + A_2 + A_3) = 2</math>, and making explicit use of the relation,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>2A_{ii} + \sum_{\ell = 1}^3 A_{i\ell}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{2}{a_i^2} \, ,</math>
  </td>
</tr>
</table>
this last expression can be simplified to discover that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\nabla^2 \biggl[ \frac{\Phi_\mathrm{grav}}{-2\pi G \rho_c} \biggr] </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- (2) + \frac{2x^2}{a_1^2} + \frac{2y^2}{a_2^2} + \frac{2z^2}{a_3^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \nabla^2 \Phi_\mathrm{grav}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>4\pi G \rho_c \biggl[ 1 -  \biggl( \frac{x^2}{a_1^2} + \frac{y^2}{a_2^2} + \frac{z^2}{a_3^2}\biggr) \biggr] \, .
</math>
  </td>
</tr>
</table>
This does indeed demonstrate that the derived gravitational potential is consistent with [[#DensitySpecification|our selected mass distribution in the case where n = 1]], namely,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>\rho</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\rho_c \biggl[ 1 -  \biggl( \frac{x^2}{a_1^2} + \frac{y^2}{a_2^2} + \frac{z^2}{a_3^2}\biggr) \biggr] \, .
</math>
  </td>
</tr>
</table>

Q.E.D.