Editing SSC/FreeEnergy/PolytropesEmbedded/Pt2 (section)

==Virial Equilibrium and Dynamical Stability==
The first (partial) derivative of <math>~\mathfrak{G}</math> with respect to <math>~R</math> is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial \mathfrak{G}}{\partial R}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{R} \biggl[
3\mathcal{A} GM^2 R^{-1} - 3\mathcal{B}KM^{(n+1)/n} R^{-3/n} + 4\pi P_e R^3
\biggr] \, ;
</math>
  </td>
</tr>
</table>
</div>
and the second (partial) derivative is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial^2 \mathfrak{G}}{\partial R^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{R^2} \biggl[
-6\mathcal{A} GM^2 R^{-1} + \biggl(\frac{n+3}{n}\biggr) 3\mathcal{B}KM^{(n+1)/n} R^{-3/n} + 8\pi P_e R^3
\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
The virial equilibrium radius is identified by setting the first derivative to zero. This means that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~3\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3\mathcal{A} GM^2 R_\mathrm{eq}^{-1}  + 4\pi P_e R_\mathrm{eq}^3 \, .
</math>
  </td>
</tr>
</table>
</div>

This expression can be usefully rewritten in the following forms:
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
  <td colspan="2" align="center">Virial Equilibrium Condition</td>
</tr>
<tr>
  <td align="center">Case 1:</td>
  <td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~3(n+3)\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3(n+3)\mathcal{A} GM^2 R_\mathrm{eq}^{-1}  + 4\pi (n+3) P_e R_\mathrm{eq}^3 
</math>
  </td>
</tr>
</table>
  </td>
</tr>

<tr>
  <td align="center">Case 2:</td>
  <td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
-6n\mathcal{A} GM^2 R_\mathrm{eq}^{-1}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 8\pi nP_e R_\mathrm{eq}^3 - 6n\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n}  </math>
  </td>
</tr>
</table>
  </td>
</tr>

<tr>
  <td align="center">Case 3:</td>
  <td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
8\pi nP_e R_\mathrm{eq}^3 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~6n\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} - 6n\mathcal{A} GM^2 R_\mathrm{eq}^{-1} </math>
  </td>
</tr>
</table>
  </td>
</tr>
</table>
</div>

Dynamical stability is determined by the sign of the second derivative expression ''evaluated at the equilibrium radius''; setting the second derivative to zero identifies the transition from stable to unstable configurations.  The criterion is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[
-6n\mathcal{A} GM^2 R^{-1} + 3(n+3) \mathcal{B}KM^{(n+1)/n} R^{-3/n} + 8\pi nP_e R^3\biggr]_{R_\mathrm{eq}}
</math>
  </td>
</tr>
</table>
</div>

===Case 1 Stability Criterion===
Using the "Case 1" virial expression to define the equilibrium radius means that the stability criterion is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-6n\mathcal{A} GM^2 R_\mathrm{eq}^{-1} + 3(n+3)\mathcal{A} GM^2 R_\mathrm{eq}^{-1}  + 4\pi (n+3) P_e R_\mathrm{eq}^3 + 8\pi nP_e R_\mathrm{eq}^3
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathcal{A} GM^2 R_\mathrm{eq}^{-1} [3(n+3)- 6n ] + 4\pi P_e R_\mathrm{eq}^3 [(n+3)   + 2n]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
4\pi P_e R_\mathrm{eq}^3 [3(n+1) ]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathcal{A} GM^2 R_\mathrm{eq}^{-1} [3(n-3)] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
4\pi P_e R_\mathrm{eq}^4 (n+1) 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathcal{A} GM^2 (n-3)
</math>
  </td>
</tr>
</table>
</div>

===Case 2 Stability Criterion===
Using the "Case 2" virial expression to define the equilibrium radius means that the stability criterion is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
 8\pi nP_e R_\mathrm{eq}^3 - 6n\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n}  + 3(n+3) \mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} + 8\pi nP_e R_\mathrm{eq}^3
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
16\pi nP_e R_\mathrm{eq}^3 - [3(n-3)]\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
16\pi nP_e R_\mathrm{eq}^3  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[3(n-3)]\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
16\pi nP_e R_\mathrm{eq}^{3(n+1)/n}  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[3(n-3)]\mathcal{B}KM^{(n+1)/n}  
</math>
  </td>
</tr>
</table>
</div>

===Case 3 Stability Criterion===
Using the "Case 3" virial expression to define the equilibrium radius means that the stability criterion is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-6n\mathcal{A} GM^2 R_\mathrm{eq}^{-1} + 3(n+3) \mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} + 6n\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} - 6n\mathcal{A} GM^2 R_\mathrm{eq}^{-1} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-12n\mathcal{A} GM^2 R_\mathrm{eq}^{-1}  
+ [6n +3(n+3)] \mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
9(n+1) \mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
12n\mathcal{A} GM^2 R_\mathrm{eq}^{-1}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
R_\mathrm{eq}^{n-3} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{4n\mathcal{A}}{3(n+1) \mathcal{B}} \biggr]^n \biggl(\frac{G}{K}\biggr)^n M^{n-1} 
</math>
  </td>
</tr>
</table>
</div>

===Case M===
Now, in our discussion of "Case M" sequence analyses, the configuration's radius is normalized to,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~R_\mathrm{norm}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~[ G^n K^{-n} M^{n-1} ]^{1/(n-3)} \, .</math>
  </td>
</tr>
</table>
</div>
Our "Case 3" stability criterion directly relates.  We conclude that the transition from stability to dynamical instability occurs when,

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

<tr>
  <td align="right">
<math>~
\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}}\biggr]_\mathrm{crit}^{n-3} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{4n\mathcal{A}}{3(n+1) \mathcal{B}} \biggr]^n 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}}\biggr]_\mathrm{crit}^{(n-3)/n} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{4n}{15(n+1) }  \biggl(\frac{4\pi}{3}\biggr)^{1/n}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_A \tilde{\mathfrak{f}}_M^{(n-1)/n}} 
</math>
  </td>
</tr>
</table>
</div>
Also in the "Case M" discussions, the external pressure is normalized to,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~P_\mathrm{norm}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~[ G^{-3(n+1)} K^{4n} M^{-2(n+1)} ]^{1/(n-3)} \, .</math>
  </td>
</tr>
</table>
</div>

If we raise the "Case 1" stability criterion expression to the <math>~(n-3)</math> power, then divide it by the "Case 3" stability criterion expression raised to the fourth power, we find,

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

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
[P_e]_\mathrm{crit}^{n-3} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{\mathcal{A} GM^2 (n-3)}{4\pi (n+1)}\biggr]^{n-3}\biggl\{ \biggl[\frac{4n\mathcal{A}}{3(n+1) \mathcal{B}} \biggr]^n \biggl(\frac{G}{K}\biggr)^n M^{n-1}  \biggr\}^{-4}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{\mathcal{A}  (n-3)}{4\pi (n+1)}\biggr]^{n-3} G^{n-3} M^{2(n-3)} \biggl[\frac{3(n+1) \mathcal{B}}{4n\mathcal{A}} \biggr]^{4n} \biggl(\frac{K}{G}\biggr)^{4n} M^{4(1-n)}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{  (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n} K^{4n} M^{-2(n+1)}  G^{-3(n+1)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
\biggl[\frac{P_e}{P_\mathrm{norm}} \biggr]_\mathrm{crit}^{n-3} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{  (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{  (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \biggl[ \frac{ 5\tilde{\mathfrak{f}}_M^2 }{ \tilde{\mathfrak{f}}_W } \biggr]^{3(n+1)} 
\biggl( \frac{3}{4\pi}\biggr)^4 \biggl[ \frac{ \tilde{\mathfrak{f}}_A }{ \tilde{\mathfrak{f}}_M^{(n+1)/n} } \biggr]^{4n} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{3}{4\pi}\biggr)^4 \biggl[\frac{  (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} 
\biggl[ \frac{ 5 }{ \tilde{\mathfrak{f}}_W } \biggr]^{3(n+1)} \tilde{\mathfrak{f}}_M^{2(n+1)} \tilde{\mathfrak{f}}_A^{4n} 
</math>
  </td>
</tr>
</table>
</div>

===Case P===
Flipping around this expression for <math>~[P_e]_\mathrm{crit}</math>, we also can write,

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

<tr>
  <td align="right">
<math>~
[M]_\mathrm{crit}^{2(n+1)}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{  (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n} K^{4n}   G^{-3(n+1)} P_e^{3-n} \, .
</math>
  </td>
</tr>
</table>
</div>

Now, in our "Case P" discussions we normalized the mass to
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M_\mathrm{SWS}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{n+1}{n}\biggr)^{3/2} G^{-3/2} K^{2n/(n+1)} P_e^{(3-n)/[2(n+1)]} \, .
</math>
  </td>
</tr>
</table>
</div>
Hence, we have,

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

<tr>
  <td align="right">
<math>~\biggl[\frac{M}{M_\mathrm{SWS}} \biggr]_\mathrm{crit}^{2(n+1)}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{  (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n} \biggl(\frac{n+1}{n}\biggr)^{-3(n+1)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{  (n-3)}{4\pi n}\biggr]^{n-3} \biggl(\frac{3 }{4} \biggr)^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n} \, ,
</math>
  </td>
</tr>
</table>
</div>

where the constants, 
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~\mathcal{A} \equiv \frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}</math>
  </td>
  <td align="center">
&nbsp; &nbsp;&nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;&nbsp;
  </td>
  <td align="left">
<math>\mathcal{B} \equiv \biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \, .</math>
  </td>
</tr>
</table>
</div>

So we can furthermore conclude that,

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

<tr>
  <td align="right">
<math>~\biggl[\frac{M}{M_\mathrm{SWS}} \biggr]_\mathrm{crit}^{2(n+1)}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{  (n-3)}{4\pi n}\biggr]^{n-3} \biggl(\frac{3 }{4} \biggr)^{4n} \biggl\{ \frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr\}^{-3(n+1)} 
\biggl\{ \biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \biggr\}^{4n} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{3}{4\pi} \biggr)^{4}
\biggl[\frac{  (n-3)}{4\pi n}\biggr]^{n-3} \biggl(\frac{3\tilde{\mathfrak{f}}_A }{4} \biggr)^{4n} \biggl[ \frac{5^3\tilde{\mathfrak{f}}_M^2}{\tilde{\mathfrak{f}}_W^3} \biggr]^{(n+1)} \, .
</math>
  </td>
</tr>
</table>
</div>


Our expression for <math>~[M]_\mathrm{crit}^{2(n+1)}</math> can also be combined with the "Case 2 stability criterion" to eliminate the mass entirely, giving,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\biggl\{ 16\pi nP_e R_\mathrm{eq}^{3(n+1)/n}  \biggr\}^{2n}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ [3(n-3)]\mathcal{B}K\biggr\}^{2n} 
\biggl[\frac{  (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n} K^{4n}   G^{-3(n+1)} P_e^{3-n}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
R_\mathrm{eq}^{6(n+1)} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{ 3(n-3)}{16\pi n} \biggr]^{2n} 
\biggl[\frac{  (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{6n} K^{6n}   G^{-3(n+1)} P_e^{3(1-n)}   
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
R_\mathrm{eq}^{2(n+1)} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \biggl[ \frac{ (n-3)}{4\pi n} \biggr]^{2n} \biggl[\frac{  (n-3)}{4\pi n}\biggr]^{n-3} \biggl[\frac{(n+1) }{n} \biggr]^{4n+(3-n)} \biggl(\frac{3 }{4} \biggr)^{6n} \biggr\}^{1/3}
\mathcal{A}^{-(n+1)} \mathcal{B}^{2n} K^{2n}   G^{-(n+1)} P_e^{(1-n)}   
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{ (n-3)}{4\pi n} \biggr]^{(n-1)} \biggl[\frac{(n+1) }{n} \biggr]^{(n+1)} \biggl(\frac{3 }{4} \biggr)^{2n} 
\mathcal{A}^{-(n+1)} \mathcal{B}^{2n} K^{2n}   G^{-(n+1)} P_e^{(1-n)}   \, .
</math>
  </td>
</tr>
</table>
</div>

Finally, recognizing that in our "Case P" discussions we normalized the radius to
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~R_\mathrm{SWS}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{n+1}{n}\biggr)^{1/2} G^{-1/2} K^{n/(n+1)} P_e^{(1-n)/[2(n+1)]} \, ,
</math>
  </td>
</tr>
</table>
</div>
we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
[R_\mathrm{eq}]_\mathrm{crit}^{2(n+1)} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{ (n-3)}{4\pi n} \biggr]^{(n-1)} \biggl(\frac{n+1 }{n} \biggr)^{(n+1)} \biggl(\frac{3 }{4} \biggr)^{2n} 
\mathcal{A}^{-(n+1)} \mathcal{B}^{2n} \biggl\{ R_\mathrm{SWS}\biggl(\frac{n+1 }{n} \biggr)^{-1/2} \biggr\}^{2(n+1)}   
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\biggl[ \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]_\mathrm{crit}^{2(n+1)} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{ (n-3)}{4\pi n} \biggr]^{(n-1)} \biggl(\frac{3 }{4} \biggr)^{2n} 
\mathcal{A}^{-(n+1)} \mathcal{B}^{2n}    
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{n}{n-3} \biggr]^{(1-n)} (4\pi)^{1-n}\biggl(\frac{3 }{4} \biggr)^{2n} 
\biggl[ \frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr]^{-(n+1)} 
\biggl[ \biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \biggr]^{2n}    
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{n}{n-3} \biggr]^{(1-n)} (4\pi)^{1-n -2} 3^{2n+2} 4^{-2n}
\biggl[ \frac{5\tilde{\mathfrak{f}}_M^2}{\tilde{\mathfrak{f}}_W}\biggr]^{(n+1)} 
\biggl[ \frac{\tilde{\mathfrak{f}}_A^{2n}}{\tilde{\mathfrak{f}}_M^{2(n+1)}} \biggr]    
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{n}{n-3} \biggr]^{(1-n)} 
\biggl[ \frac{3^2\cdot 5}{4\pi \tilde{\mathfrak{f}}_W}\biggr]^{(n+1)} 
\biggl[\frac{ \tilde{\mathfrak{f}}_A}{4} \biggr]^{2n}    \, .
</math>
  </td>
</tr>
</table>
</div>