Editing SSCpt1/Virial (section)

==Energy Extrema==
As is illustrated in [[#Visual_Representation|Figure 1]], the free energy surface generally will exhibit multiple local minima and local maxima, and may also possess one or more points of inflection. The locations along the energy surface where these special points arise identify equilibrium states, and the associated values of <math>(R/R_0)_\mathrm{eq}</math> give the radii of the equilibrium configurations.  

For a given choice of the set of physical parameters <math>M</math>, {{ Template:Math/MP_PolytropicConstant }}, <math>J</math>, <math>P_e</math>, and <math>\gamma_g</math>, extrema occur wherever,
<div align="center">
<math>
\frac{d\mathfrak{G^*}}{d\chi} = 0 \, .
</math>
</div>
For the free energy function identified above, 
<div align="center">
<math>
\frac{d\mathfrak{G^*}}{d\chi}  = 
3A\chi^{-2} -~ (1-\delta_{1\gamma_g})~3 B\chi^{2 -3\gamma_g} -~ \delta_{1\gamma_g} 3\chi^{-1} ~ -2C \chi^{-3} +~ 3D\chi^2  \, ,
</math>
</div>
<span id="GeneralVirial">so <math>\chi_\mathrm{eq} \equiv R_\mathrm{eq}/R_\mathrm{norm}</math> is obtained from the real root(s) of the equation,</span>
<div align="center">
<math>
2C \chi_\mathrm{eq}^{-2}  + ~ (1-\delta_{1\gamma_g})~3 B\chi_\mathrm{eq}^{3 -3\gamma_g} +~ \delta_{1\gamma_g} 3 ~
-~3A\chi_\mathrm{eq}^{-1}  -~ 3D\chi_\mathrm{eq}^3 = 0 \, .
</math>
</div>

<div id="Tohline85">
<table border="1" align="center" width="90%" cellpadding="20">
<tr><td align="left">
<b><font color="purple">ASIDE:</font></b> When we discuss the equilibrium of isothermal, rotating configurations that are immersed in an external medium, we will draw on the work of {{ Weber76full }} and the work of {{ Tohline85full }} which, in turn, draws upon {{ Tohline81full }}.  In preparation for that discussion, we will go ahead and show how the {{ Tohline85 }} statement of virial equilibrium &#8212; his equation (9) &#8212; is the same as the equation that defines free energy extrema that has been derived here; and we will show how the {{ Weber76 }} "energy integral" &#8212; his equation (B3) &#8212; relates to our dimensionless free-energy function.

----


<table border="1" cellpadding="10" align="center" width="75%">
<tr><td align="center">
<!-- [[Image:Tohline1985_Eq9.png|500px|center]] -->
<!--
The Astrophysical Journal, 292: 181-187, 1985 May 1<br />
STAR FORMATION: PHASE TRANSITION, NOT JEANS INSTABILITY<br />
Joel E. Tohline<br />
-->
{{ Tohline85figure }}<br />&nbsp;<br />
<math>\beta(\sin^{-1}e/e)^2 [\tfrac{1}{2} - \beta ] + kV^* - F_s^* = 0 \, .</math>&nbsp; &nbsp; &nbsp; &nbsp;(9)
</td></tr>
</table>


First, in order to match sign conventions, we need to multiply our "free energy extrema" equation through by minus one; second, we should set <math>\delta_{1\gamma_g} = 1</math> because {{ Tohline85 }} was only concerned with isothermal systems; then, because {{ Tohline85 }} normalizes each energy term by
<div align="center">
<math>E^* \equiv \biggl( \frac{2^2 \cdot 3^2}{5^3} \biggr) \frac{G^2 M_\mathrm{tot}^5}{J^2} \, ,</math>
</div>
instead of by our <math>E_\mathrm{norm}</math>, we need to multiply our equation through by the ratio,
<div align="center">
<math>\frac{E_\mathrm{norm}}{E^*} = \biggl( \frac{5^3}{2^4 \cdot 3\pi} \biggr) \frac{J^2 c_\mathrm{norm}^2}{G^2 M_\mathrm{tot}^4} \, .</math>
</div>
With these three modifications, our "free energy extrema" relation becomes,
<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>\frac{3E_\mathrm{norm}}{E^*}\biggl[~A\chi_\mathrm{eq}^{-1} ~- \biggl( \frac{2C}{3}\biggr) \chi_\mathrm{eq}^{-2}    
~  +~ D\chi_\mathrm{eq}^3 - ~ B_I \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
Next, because {{ Tohline85 }} considered only uniform-density configurations, all of the dimensionless filling factors can be set to unity in the definitions of the leading coefficients of all of our energy terms; but, following {{ Tohline81 }}, the leading coefficients of two of our energy terms should be modified to include a factor involving the configuration's eccentricity,
<div align="center">
<math>e \equiv \biggl( 1 - \frac{Z_\mathrm{eq}^2}{R_\mathrm{eq}^2} \biggr)^{1/2} \, ,</math>
</div>
in order to account for rotational flattening.  Properly adjusted, the four coefficients are,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>A</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>\frac{1}{5} \biggl( \frac{\sin^{-1}e}{e} \biggr) \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>B_I</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
1 \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>C</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\frac{3\cdot 5}{2^4 \pi} \biggl[ \frac{J^2 c_\mathrm{norm}^2}{G^2 M_\mathrm{tot}^4} \biggr]  
= \biggl( \frac{3^2}{5^2} \biggr) \frac{E_\mathrm{norm}}{E^*}  \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>D</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{4\pi}{3} \biggr) \frac{P_e}{P_\mathrm{norm}} (1-e^2)^{1/2} \, .
</math>
  </td>
</tr>
</table>
</div>
Inserting these coefficient definitions, our "free energy extrema" relation becomes,
<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>\frac{3E_\mathrm{norm}}{E^*}
\biggl[~\frac{1}{5} \biggl( \frac{\sin^{-1}e}{e} \biggr) \chi_\mathrm{eq}^{-1} 
~- \frac{E_\mathrm{norm}}{E^*} \biggl( \frac{2\cdot 3}{5^2} \biggr)  \chi_\mathrm{eq}^{-2}    
~  +~ \biggl( \frac{4\pi}{3} \biggr) \frac{P_e}{P_\mathrm{norm}} (1-e^2)^{1/2} \chi_\mathrm{eq}^3 - ~ 1 \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
Next we need to appreciate that {{ Tohline85 }} adopted the dimensionless parameter, <math>\beta \equiv T_\mathrm{rot}/|W_\mathrm{grav}|</math>, instead of the normalized radius, <math>\chi</math>, as the order parameter that is varied when searching for extrema in the free-energy function.  So, in our equation that defines "free energy extrema" we need to replace <math>\chi_\mathrm{eq}</math> with <math>\beta_\mathrm{eq}</math>, using the relation,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\beta \equiv \frac{T_\mathrm{rot}}{|W_\mathrm{grav}|}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{C\chi^{-2}}{3A \chi^{-1}} = \biggl( \frac{3}{5} \biggr) \frac{E_\mathrm{norm}}{E^*} 
\biggl( \frac{\sin^{-1}e}{e} \biggr)^{-1} \chi^{-1}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~~\chi_\mathrm{eq}^{-1} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{5}{3} \biggr) \frac{E^*}{E_\mathrm{norm}} 
\biggl( \frac{\sin^{-1}e}{e} \biggr)\beta_\mathrm{eq} \, .
</math>
  </td>
</tr>
</table>
</div>
Hence, our expression for the "free energy extrema" becomes,
<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( \frac{\sin^{-1}e}{e} \biggr)^2 \beta_\mathrm{eq} ~- 2\biggl( \frac{\sin^{-1}e}{e} \biggr)^2 \beta_\mathrm{eq}^{2}    
~  +~ \frac{4\pi P_e}{P_\mathrm{norm}} (1-e^2)^{1/2} 
\biggl[ \biggl( \frac{3^3}{5^3} \biggr) \biggl( \frac{E_\mathrm{norm}}{E^*} \biggr)^4
\biggl( \frac{\sin^{-1}e}{e} \biggr)^{-3}\biggr] \beta_\mathrm{eq}^{-3}
- ~ \frac{3E_\mathrm{norm}}{E^*}  </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
2 \biggl\{
\beta_\mathrm{eq} \biggl( \frac{\sin^{-1}e}{e} \biggr)^2 \biggl( \frac{1}{2} - \beta_\mathrm{eq} \biggr)    
~  + \frac{2\pi \cdot 3^3}{5^3} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \biggl( \frac{E_\mathrm{norm}}{E^*} \biggr)^4
\biggl[ 
\beta_\mathrm{eq}^{-3}\biggl( \frac{\sin^{-1}e}{e} \biggr)^{-3} (1-e^2)^{1/2}\biggr] 
- ~ \biggl( \frac{3}{2} \biggr) \frac{5^3}{2^4 \cdot 3 \pi} \biggl[ \frac{J^2 c_\mathrm{norm}^2}{G^2 M_\mathrm{tot}^4} \biggr]
\biggr\} \, .</math>
  </td>
</tr>
</table>
</div>
Now,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{2\pi \cdot 3^3}{5^3} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \biggl( \frac{E_\mathrm{norm}}{E^*} \biggr)^4</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{2\pi \cdot 3^3}{5^3} \biggl[ \frac{P_e}{(E^*)^4} \biggr] ( GM_\mathrm{tot}^2)^3</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{2\pi \cdot 3^3}{5^3} \biggl[\biggl( \frac{2^2 \cdot 3^2}{5^3} \biggr) \frac{G^2 M_\mathrm{tot}^5}{J^2} \biggr]^{-4} ( P_e G^3 M_\mathrm{tot}^6)</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\pi\biggl( \frac{5^{9}}{2^7 \cdot 3^5} \biggr) \frac{J^8 P_e }{G^5 M_\mathrm{tot}^{14}}  
= \frac{10 \pi}{3} \biggl( \frac{5^{2}}{2^2 \cdot 3} \biggr)^4 \frac{J^8 P_e }{G^5 M_\mathrm{tot}^{14}} \, ,
</math>
  </td>
</tr>
</table>
</div>
which is the definition of the coefficient "<math>k</math>" that is provided as equation (7) of {{ Tohline85 }}.  Hence, dropping the factor of two out front, our expression for "free energy extrema" becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\beta_\mathrm{eq} \biggl( \frac{\sin^{-1}e}{e} \biggr)^2 \biggl( \frac{1}{2} - \beta_\mathrm{eq} \biggr)    
~  + \frac{10 \pi}{3} \biggl( \frac{5^{2}}{2^2 \cdot 3} \biggr)^4 \frac{J^8 P_e }{G^5 M_\mathrm{tot}^{14}} 
\biggl[ 
\beta_\mathrm{eq}^{-3}\biggl( \frac{\sin^{-1}e}{e} \biggr)^{-3} (1-e^2)^{1/2}\biggr] 
- ~ \frac{3}{4\pi} \biggr( \frac{5^3}{2^3 \cdot 3} \biggr) \biggl[ \frac{J^2 c_\mathrm{norm}^2}{G^2 M_\mathrm{tot}^4} \biggr]
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0 \, .</math>
  </td>
</tr>
</table>
</div>
Finally, realizing that the square of the sound speed is related to our <math>c_\mathrm{norm}^2</math> via the relation [note that {{ Tohline85 }} uses <math>a^2</math> in place of <math>c_s^2</math>],
<div align="center">
<math>c_s^2 = \biggl( \frac{3}{4\pi} \biggr) c_\mathrm{norm}^2 \, ,</math>
</div>
it is clear that this last form of our "free energy extrema" expression is identical to the {{ Tohline85 }} virial equilibrium equation (9), which appears in print in a simpler but also more cryptic form as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\beta_\mathrm{eq} \biggl( \frac{\sin^{-1}e}{e} \biggr)^2 \biggl( \frac{1}{2} - \beta_\mathrm{eq} \biggr)    + kV^* - F_s^*
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0 \, .</math>
  </td>
</tr>
</table>
</div>


----


<table border="1" cellpadding="10" align="center" width="75%">
<tr><td align="center">
<!-- [[Image:Weber1976_EqB3.png|500px|center]] -->
<!--
The Astrophysical Journal, 208: 113-126, 1976 August 15<br />
OSCILLATION AND COLLAPSE OF INTERSTELLAR CLOUDS<br />
Stephen V. Weber<br />
-->
{{ Weber76figure }}<br />&nbsp;<br />
<math>E = \dot{a}^2 + \dot\gamma^2/2 + J^2/\alpha^2 
- \tfrac{4}{3} \log \alpha^2 \gamma + \tfrac{1}{3}\alpha^2 \gamma P_\mathrm{ext} -
\begin{cases}\biggl(\frac{3}{\alpha e}\biggr) \sin^{-1}e, & e > 0, \\ \biggl( \frac{3}{2\alpha e} \biggr) \log Q , & e<0 \, . \end{cases}
</math>&nbsp; &nbsp; &nbsp; &nbsp;(B3)
</td></tr>
</table>


Plugging the same set of modified leading coefficients into our derived expression for the free energy gives,
<div align="center">
<math>
\mathfrak{G}^* = \frac{3\cdot 5}{2^4 \pi} \biggl[ \frac{J^2 c_\mathrm{norm}^2}{G^2 M_\mathrm{tot}^4} \biggr] \chi^{-2} 
 -~ 3 \ln \chi +~ \biggl( \frac{4\pi}{3} \biggr) \frac{P_e}{P_\mathrm{norm}} (1-e^2)^{1/2} \chi^3 
- \frac{3}{5} \biggl( \frac{\sin^{-1}e}{e} \biggr) \chi^{-1}\, .
</math>
</div>
Now, recognize that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\chi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\alpha \biggl( \frac{R_0}{R_\mathrm{norm}} \biggr) = \biggl( \frac{2^2}{3\cdot 5} \biggr) \alpha \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>(1 - e^2)^{1/2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{Z}{R} = \frac{\gamma}{\alpha} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{P_e}{P_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{P_e}{P_0} \cdot \frac{P_0}{P_\mathrm{norm}} = \frac{3^4 \cdot 5^3}{2^{10} \pi} \biggl[ P_\mathrm{ext} \biggr]_\mathrm{Weber} 
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{3\cdot 5}{2^4 \pi} \biggl[ \frac{J^2 c_\mathrm{norm}^2}{G^2 M_\mathrm{tot}^4} \biggr]</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{3} \biggl( \frac{2}{5} J_\mathrm{Weber} \biggr)^2
\, ,</math>
  </td>
</tr>
</table>
</div>
where, for axisymmetric configurations (set <math>\beta=\alpha</math> in equation 12 of {{ Weber76 }}),
<div align="center">
<math>J_\mathrm{Weber} \equiv \alpha^2 \Omega = \biggl( \frac{R}{R_0} \biggr)^2 (\dot\varphi_c t_0)^2 \, .</math>
</div>
Hence, our expression for the free energy may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\mathfrak{G}^*</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{3} \biggl( \frac{2}{5} J_\mathrm{Weber}\biggr)^2 \biggl( \frac{3\cdot 5}{2^2} \biggr)^2 \alpha^{-2} 
 -~ 3 \ln \chi +~ \biggl( \frac{4\pi}{3} \biggr) \frac{3^4 \cdot 5^3}{2^{10} \pi} \biggl[ P_\mathrm{ext} \biggr]_\mathrm{Weber} \biggl( \frac{2^2}{3\cdot 5} \biggr)^3 \alpha^2 \gamma
- \frac{3}{5} \biggl( \frac{\sin^{-1}e}{e} \biggr) \biggl( \frac{3\cdot 5}{2^2} \biggr) \alpha^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{3}{2^2} \biggr)  J^2_\mathrm{Weber} \alpha^{-2} 
 -~ \ln \chi^3 +~ \frac{1}{2^{2} } \alpha^2 \gamma \biggl[ P_\mathrm{ext} \biggr]_\mathrm{Weber}  
- \frac{3^2}{2^2} \biggl( \frac{\sin^{-1}e}{e} \biggr) \alpha^{-1} \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~~\frac{4}{3} \mathfrak{G}^*</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
J^2_\mathrm{Weber} \alpha^{-2} 
 -~ \frac{4}{3} \ln \chi^3 +~ \frac{1}{3} \alpha^2 \gamma \biggl[ P_\mathrm{ext} \biggr]_\mathrm{Weber}  
- 3 \biggl( \frac{\sin^{-1}e}{e} \biggr) \alpha^{-1} \, .
</math>
  </td>
</tr>
</table>
</div>
The right-hand-side of this expression exactly matches the {{ Weber76 }} "energy integral" for oblate-spheroidal configurations &#8212; see his equation (B3) for the case, <math>e > 0</math> &#8212; except that Weber's energy integral includes an additional pair of terms (<math>{\dot\alpha}^2 + {\dot\gamma}^2/2</math>) to account for kinetic energy associated with the overall collapse or expansion of the configuration.  [NOTE:  The logarithmic term ultimately needs to be <math>\ln \alpha^2\gamma</math> instead of <math>\ln\chi^3</math> in order to reflect an oblate-spheroidal, rather than spherical, volume.  This term also needs to be fixed in the above discussion of Tohline's work.]
</td></tr>
</table>
</div>