Revision as of 19:51, 19 August 2016

Supporting Derivations for Free-Energy PowerPoint Presentation

The derivations presented here are an extension of our accompanying free-energy synopsis. These additional details proved to be helpful while developing an overarching PowerPoint presentation.

General Free-Energy Expression

We're considering a free-energy function of the following form:

<math>~\mathfrak{G}^*_\mathrm{type}</math>

<math>~-ax^{-1} + b x^{-3/n} + c x^{-3/j} + \mathfrak{G}_0 \, ,</math>

where,

<math>~x \equiv \frac{R}{R_0} \, .</math>

As we have shown, setting,

<math>~\frac{\partial \mathfrak{G}^*_\mathrm{type}}{\partial x}</math>

generates a mathematical statement of virial equilibrium, namely,

<math>~\frac{ b}{nc}\cdot x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3c} + \frac{1}{j}\cdot x^{(j-3)/j}_\mathrm{eq} </math>

And equilibrium configurations for which the second (as well as first) derivative of the free energy is zero are found at "critical" radii given by the expression,

<math>~[x_\mathrm{eq}^{(j-3)/j}]_\mathrm{crit} </math>

<math>~ \frac{a}{3^2c}\biggl[ \frac{j^2(n-3)}{n-j} \biggr] \, . </math>

Pressure-Truncated Polytropes

For pressure-truncated polytropes, set <math>~j=-1</math> and let <math>~n</math> be the chosen polytropic index. In this case, the statement of virial equilibrium is,

<math>~\frac{ b}{nc}\cdot x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3c} - x^{4}_\mathrm{eq} </math>

and the critical equilibrium configuration has,

<math>~[x_\mathrm{eq}]_\mathrm{crit} </math>

<math>~ \biggl[ \frac{a(n-3)}{3^2c (n+1)}\biggr]^{1/4} \, . </math>

Case M

Set <math>~K</math> and <math>~M_\mathrm{tot}</math> constant and examine how the free-energy behaves as a function of the coordinates, <math>~(R,P_e)</math>. In this case (see, for example, here),

<math>~a</math>	<math>~\equiv</math>	<math>~\frac{3}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}\, , </math>
<math>~b</math>	<math>~\equiv</math>	<math>~n\biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \, , </math>
<math>~c</math>	<math>~\equiv</math>	<math>~\frac{4\pi}{3}\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \, , </math>

where the structural form factors for pressure-truncated polytropes are precisely defined here. And (see, for example, here),

<math>~R_0 = R_\mathrm{norm}</math>	<math>~\equiv</math>	<math>~\biggl[ \biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{n-1} \biggr]^{1/(n-3)} \, ,</math>
<math>~P_\mathrm{norm}</math>	<math>~\equiv</math>	<math>~\biggl[ \frac{K^{4n}}{G^{3(n+1)} M_\mathrm{tot}^{2(n+1)}} \biggr]^{1/(n-3)} \, .</math>

If we set all three structural form-factors to unity, we have,

<math>~\frac{a}{3c}</math>	<math>~=</math>	<math>~\frac{3}{2^2\cdot 5\pi}\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)^{-1} \, ,</math>
<math>~\frac{b}{nc}</math>	<math>~=</math>	<math>~\biggl(\frac{3}{4\pi} \biggr)^{(n+1)/n} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)^{-1} \, .</math>

Virial Equilibrium

So the statement of virial equilibrium becomes,

<math>~ x^{4}_\mathrm{eq} </math>	<math>~=</math>	<math>~\biggl[ \biggl(\frac{3}{4\pi} \biggr)^{(n+1)/n} x^{(n-3)/n }_\mathrm{eq} - \frac{3}{2^2\cdot 5\pi}\biggr]\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)^{-1} </math>
	<math>~=</math>	<math>~\frac{3}{2^2\cdot 5\pi}\biggl[ 5\biggl(\frac{3}{4\pi} \biggr)^{1/n} x^{(n-3)/n }_\mathrm{eq} - 1\biggr]\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)^{-1} </math>
<math>~ \Rightarrow ~~~ \frac{P_e}{P_\mathrm{norm}}</math>	<math>~=</math>	<math>~\frac{3}{2^2\cdot 5\pi x^{4}_\mathrm{eq} }\biggl[ 5\biggl(\frac{3}{4\pi} \biggr)^{1/n} x^{(n-3)/n }_\mathrm{eq} - 1\biggr] \, . </math>

Dynamical Instability

And, along the equilibrium sequence that is generated by this virial expression, the transition from stable to unstable configurations occurs at,

<math>~[x_\mathrm{eq}]^4_\mathrm{crit} </math>	<math>~=</math>	<math>~ \biggl[ \frac{(n-3)}{3(n+1)}\biggr] \frac{3}{2^2\cdot 5\pi}\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)^{-1} </math>
<math>~\Rightarrow ~~~ \frac{2^2\cdot 5\pi}{3}\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)[x_\mathrm{eq}]^4_\mathrm{crit} </math>	<math>~=</math>	<math>~ \biggl[ \frac{(n-3)}{3(n+1)}\biggr] </math>

which, in combination with the virial equilibrium condition gives,

<math>~5\biggl(\frac{3}{4\pi} \biggr)^{1/n} [x_\mathrm{eq}]^{(n-3)/n }_\mathrm{crit} -1</math>	<math>~=</math>	<math>~ \biggl[ \frac{(n-3)}{3(n+1)}\biggr] </math>
<math>~\Rightarrow~~~ [x_\mathrm{eq}]_\mathrm{crit} </math>	<math>~=</math>	<math>~\biggl[ \frac{4n}{3\cdot 5(n+1)} \biggl(\frac{4\pi}{3} \biggr)^{1/n}\biggr]^{n/(n-3)} \, . </math>

Turning Point

Let's examine the curvature of the equilibrium sequence.

<math>~ \frac{d}{dx}\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)</math>	<math>~=</math>	<math>~ - \frac{3}{ 5\pi x^{5} }\biggl[ 5\biggl(\frac{3}{4\pi} \biggr)^{1/n} x^{(n-3)/n } - 1\biggr] + \frac{3(n-3)}{2^2n \pi x^{4} }\biggl(\frac{3}{4\pi} \biggr)^{1/n} x^{-3/n } </math>
	<math>~=</math>	<math>~\frac{3}{ 5\pi x^{5} } + \frac{3}{4\pi}\biggl(\frac{3}{4\pi} \biggr)^{1/n} \biggl[ \frac{(n-3)}{n } - 4\biggr] \frac{x^{(n-3)/n } }{x^5} </math>
	<math>~=</math>	<math>~\frac{3}{ 5\pi x^{5} } - 3\biggl(\frac{3}{4\pi} \biggr)^{(n+1)/n} \biggl[ \frac{n+1}{n } \biggr] \frac{x^{(n-3)/n } }{x^5} \, . </math>

Setting this derivative to zero let's us identify the location of the turning point that identifies <math>~P_\mathrm{max}.</math>

<math>~ [ x_\mathrm{eq}^{(n-3)/n } ]_\mathrm{turn} </math>	<math>~=</math>	<math>~ \frac{1}{ 5\pi }\biggl[ \frac{n }{n+1} \biggr] \biggl(\frac{4\pi}{3} \biggr)^{(n+1)/n} </math>
<math>~\Rightarrow~~~ [ x_\mathrm{eq} ]_\mathrm{turn} </math>	<math>~=</math>	<math>~ \biggl[ \frac{4n}{ 15(n+1)}\biggl(\frac{4\pi}{3} \biggr)^{1/n} \biggr]^{n/(n-3)} \, . </math>

And, returning to the virial equilibrium expression, we find that, associated with this equilibrium radius,

<math>~ \frac{P_\mathrm{max}}{P_\mathrm{norm}}</math>	<math>~=</math>	<math>~\frac{3}{2^2\cdot 5\pi x^{4}_\mathrm{turn} }\biggl[ 5\biggl(\frac{3}{4\pi} \biggr)^{1/n} x^{(n-3)/n }_\mathrm{turn} - 1\biggr] </math>
<math>~ \Rightarrow ~~~2^2\cdot 5\pi x^{4}_\mathrm{turn} \biggl(\frac{P_\mathrm{max}}{P_\mathrm{norm}}\biggr)</math>	<math>~=</math>	<math>~15\biggl(\frac{3}{4\pi} \biggr)^{1/n} \biggl[ \frac{4n}{ 15(n+1)}\biggl(\frac{4\pi}{3} \biggr)^{1/n} \biggr] - 3 </math>
	<math>~=</math>	<math>~\biggl(\frac{n-3}{ n+1} \biggr) </math>
<math>~ \Rightarrow ~~~\frac{P_\mathrm{max}}{P_\mathrm{norm}}</math>	<math>~=</math>	<math>~\frac{1}{20\pi}\biggl(\frac{n-3}{ n+1} \biggr) \biggl[ \frac{ 15(n+1)}{4n}\biggl(\frac{3}{4\pi} \biggr)^{1/n} \biggr]^{4n/(n-3)} \, .</math>

Notice that, under the assumption that all three structural filling-factors are unity, <math>~[x_\mathrm{eq}]_\mathrm{turn} = [x_\mathrm{eq}]_\mathrm{crit}</math>, that is, the location of the turning point coincides precisely with the point along the sequence where the transition from stable to unstable equilibrium configurations occurs.

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+Notice that, under the assumption that all three structural filling-factors are unity, <math>~[x_\mathrm{eq}]_\mathrm{turn} = [x_\mathrm{eq}]_\mathrm{crit}</math>, that is, the location of the turning point coincides precisely with the point along the sequence where the transition from stable to unstable equilibrium configurations occurs.
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