SSC/BipolytropeGeneralization/Pt4

Bipolytrope Generalization (Pt 4)

Part I:  Bipolytrope Generalization

Part II:  Derivations

Part III:  Examples

Part IV:  Best of the Best

Best of the Best

One Derivation of Free Energy

${\mathfrak{𝔊}}^{*}$	$=$	$-\frac{3\cdot {\mathfrak{𝔣}}_{WM}}{5}(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}+\frac{\nu s_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{({\gamma }_c-1)}[\left(\frac{3}{4\pi }\right)\frac{\nu }{q^3}{]}^{{\gamma }_c-1}[\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{-3({\gamma }_c-1)}$
		$+\frac{(1-\nu )s_{\mathrm{e}\mathrm{n}\mathrm{v}}}{({\gamma }_e-1)}\left(\frac{K_e}{K_c}\right)\left[\frac{K_c^3}{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{]}^{({\gamma }_c-{\gamma }_e)/(3{\gamma }_c-4)}\right[\left(\frac{3}{4\pi }\right)\frac{(1-\nu )}{(1-q^3)}{]}^{{\gamma }_e-1}[\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{-3({\gamma }_e-1)}.$

Another Derivation of Free Energy

Hence the renormalized gravitational potential energy becomes,

$\frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$

$=$

$-\left(\frac{3}{5}\right)\frac{{\nu }^2}{q}(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}\cdot f;$

and the two, renormalized contributions to the thermal energy become,

$\frac{U_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\frac{2}{3({\gamma }_c-1)}\left[\frac{S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]$	$=$	$\frac{4\pi q^3(1+\mathrm{\Lambda })}{3({\gamma }_c-1)}\left[\right(\frac{3}{4\pi }\left)\frac{\nu }{q^3}{]}^{{\gamma }_c}\right(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{3-3{\gamma }_c}$
	$=$	$\frac{\nu (1+\mathrm{\Lambda })}{({\gamma }_c-1)}\left[\right(\frac{3}{4\pi })\frac{\nu }{q^3}{]}^{{\gamma }_c-1}{\chi }^{3-3{\gamma }_c},$
$\frac{U_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\frac{2}{3({\gamma }_e-1)}\left[\frac{S_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]$	$=$	$\frac{2(2\pi )}{3({\gamma }_e-1)}\left[\frac{R^3{P}_{ie}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right][(1-q^3)+\frac{5}{2}\mathrm{\Lambda }(\frac{{\rho }_e}{{\rho }_0})(-2+3q-q^3)+\frac{3}{2q^2}\mathrm{\Lambda }(\frac{{\rho }_e}{{\rho }_0}{)}^2(-1+5q^2-5q^3+q^5)]$
	$=$	$\frac{2(2\pi )}{3({\gamma }_e-1)}\left[\frac{\mathrm{B}\mathrm{i}\mathrm{g}\mathrm{T}\mathrm{e}\mathrm{r}\mathrm{m}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]R^3{K}_e{\rho }_{ie}^{{\gamma }_e}$
	$=$	$\frac{2(2\pi )}{3({\gamma }_e-1)}\left[\frac{\mathrm{B}\mathrm{i}\mathrm{g}\mathrm{T}\mathrm{e}\mathrm{r}\mathrm{m}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]R^3{K}_e{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^{{\gamma }_e}\left(\frac{{\rho }_{ie}}{\overline{\rho }}{)}^{{\gamma }_e}\right(\frac{\overline{\rho }}{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{{\gamma }_e}$
	$=$	$\frac{2(2\pi )}{3({\gamma }_e-1)}\left[\frac{\mathrm{B}\mathrm{i}\mathrm{g}\mathrm{T}\mathrm{e}\mathrm{r}\mathrm{m}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]({\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3)K_e{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^{{\gamma }_e-1}[\frac{(1-\nu )}{(1-q^3)}{]}^{{\gamma }_e}{\chi }^{3-3{\gamma }_e}$
	$=$	$\frac{2(2\pi )}{3({\gamma }_e-1)}\left[\frac{(1-\nu )}{(1-q^3)}{]}^{{\gamma }_e}{\chi }^{3-3{\gamma }_e}\right[\mathrm{B}\mathrm{i}\mathrm{g}\mathrm{T}\mathrm{e}\mathrm{r}\mathrm{m}\left]\right(\frac{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4\pi }\left)\frac{K_e}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right(\frac{3}{4\pi }{)}^{{\gamma }_e-1}[\frac{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{K_c^3}{]}^{({\gamma }_e-1)/(3{\gamma }_c-4)}$
	$=$	$\frac{(1-\nu )}{(1-q^3)({\gamma }_e-1)}\left[\frac{3}{4\pi }\frac{(1-\nu )}{(1-q^3)}{]}^{{\gamma }_e-1}\right(\frac{K_e}{K_c}\left){\chi }^{3-3{\gamma }_e}\right[\mathrm{B}\mathrm{i}\mathrm{g}\mathrm{T}\mathrm{e}\mathrm{r}\mathrm{m}\left]\frac{K_c{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right[\frac{K_c^3}{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{]}^{(1-{\gamma }_e)/(3{\gamma }_c-4)}$
	$=$	$\frac{(1-\nu )}{(1-q^3)({\gamma }_e-1)}\left[\frac{3}{4\pi }\frac{(1-\nu )}{(1-q^3)}{]}^{{\gamma }_e-1}\right(\frac{K_e}{K_c}\left){\chi }^{3-3{\gamma }_e}\right[\mathrm{B}\mathrm{i}\mathrm{g}\mathrm{T}\mathrm{e}\mathrm{r}\mathrm{m}\left]\right[\frac{K_c^3}{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{]}^{(1-{\gamma }_e)/(3{\gamma }_c-4)}[\frac{K_c^{3{\gamma }_c-4}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{3{\gamma }_c-4}}{G^{3{\gamma }_c-3}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{5{\gamma }_c-6}{K}_c^{-1}}{]}^{1/(3{\gamma }_c-4)}$
	$=$	$\frac{(1-\nu )}{(1-q^3)({\gamma }_e-1)}\left[\frac{3}{4\pi }\frac{(1-\nu )}{(1-q^3)}{]}^{{\gamma }_e-1}\right(\frac{K_e}{K_c}\left){\chi }^{3-3{\gamma }_e}\right[\mathrm{B}\mathrm{i}\mathrm{g}\mathrm{T}\mathrm{e}\mathrm{r}\mathrm{m}\left]\right[\frac{K_c^3}{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{]}^{(1-{\gamma }_e)/(3{\gamma }_c-4)}[\frac{K_c^3}{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{]}^{({\gamma }_c-1)/(3{\gamma }_c-4)}$
	$=$	$\frac{(1-\nu )}{(1-q^3)({\gamma }_e-1)}\left[\frac{3}{4\pi }\frac{(1-\nu )}{(1-q^3)}{]}^{{\gamma }_e-1}\right(\frac{K_e}{K_c}\left)\right[\frac{K_c^3}{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{]}^{({\gamma }_c-{\gamma }_e)/(3{\gamma }_c-4)}\left[\mathrm{B}\mathrm{i}\mathrm{g}\mathrm{T}\mathrm{e}\mathrm{r}\mathrm{m}\right]{\chi }^{3-3{\gamma }_e}$

Finally, then, we can state that,

${\mathfrak{𝔣}}_{WM}$	$\equiv$	$\frac{{\nu }^2}{q}\cdot f,$
$s_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$\equiv$	$1+\mathrm{\Lambda },$
$(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$\equiv$	$(1-q^3)+\mathrm{\Lambda }\left[\frac{5}{2}\right(\frac{{\rho }_e}{{\rho }_0})(-2+3q-q^3)+\frac{3}{2q^2}(\frac{{\rho }_e}{{\rho }_0}{)}^2(-1+5q^2-5q^3+q^5)].$

Note,

$\mathrm{\Lambda }$

$\equiv$

$\frac{3}{2^2\cdot 5\pi }\frac{{\nu }^2}{q^4}\left[\right(\frac{3}{4\pi }\left)\frac{\nu }{q^3}{]}^{-{\gamma }_c}\right(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{3{\gamma }_c-4}=\frac{1}{5}\left(\frac{\nu }{q}\right)\left[\right(\frac{3}{4\pi })\frac{\nu }{q^3}{]}^{1-{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}.$

We also want to ensure that envelope pressure matches the core pressure at the interface. This means,

$K_e{\rho }_{ie}^{{\gamma }_e}$	$=$	$K_c{\rho }_{ic}^{{\gamma }_c}$
$\Rightarrow \frac{K_e}{K_c}$	$=$	${\rho }_{ic}^{{\gamma }_c}{\rho }_{ie}^{-{\gamma }_e}$
	$=$	$\left[\frac{{\rho }_{ic}}{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{{\gamma }_c}\right[\frac{{\rho }_{ie}}{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{-{\gamma }_e}{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^{{\gamma }_c-{\gamma }_e}$
	$=$	$\left[\frac{{\rho }_{ic}}{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{{\gamma }_c}\right[\frac{{\rho }_{ie}}{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{-{\gamma }_e}\left\{\frac{3}{4\pi }\right[\frac{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{K_c^3}{]}^{1/(3{\gamma }_c-4)}{\}}^{{\gamma }_c-{\gamma }_e}$
$\Rightarrow \frac{K_e}{K_c}\left[\frac{K_c^3}{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{]}^{({\gamma }_c-{\gamma }_e)/(3{\gamma }_c-4)}\right[\left(\frac{3}{4\pi }\right)\frac{{\rho }_{ie}}{\overline{\rho }}{]}^{{\gamma }_e-1}$	$=$	$\left[\right(\frac{3}{4\pi }\left)\frac{{\rho }_{ic}}{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{{\gamma }_c}\right[\left(\frac{3}{4\pi }\right)\frac{{\rho }_{ie}}{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{-{\gamma }_e}\left[\right(\frac{3}{4\pi })\frac{{\rho }_{ie}}{\overline{\rho }}{]}^{{\gamma }_e-1}$
	$=$	$\left[\right(\frac{3}{4\pi }\left)\frac{{\rho }_{ic}}{\overline{\rho }}{]}^{{\gamma }_c-1}\right(\frac{{\rho }_{ic}}{{\rho }_{ie}}\left)\right(\frac{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{\overline{\rho }}{)}^{{\gamma }_e-{\gamma }_c}$
	$=$	$\left[\right(\frac{3}{4\pi }\left)\frac{{\rho }_{ic}}{\overline{\rho }}{]}^{{\gamma }_c-1}\right(\frac{{\rho }_{ic}}{{\rho }_{ie}}\left)\right(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{3({\gamma }_e-{\gamma }_c)}$

Keep in mind that, if the envelope and core both have uniform (but different) densities, then ${\rho }_{ic}={\rho }_c$, ${\rho }_{ie}={\rho }_e$, and

$\frac{{\rho }_c}{\overline{\rho }}=\frac{\nu }{q^3};\frac{{\rho }_e}{\overline{\rho }}=\frac{1-\nu }{1-q^3};\frac{{\rho }_e}{{\rho }_c}=\frac{q^3(1-\nu )}{\nu (1-q^3)}.$

Summary

Understanding Free-Energy Behavior

Step 1: Pick values for the separate coefficients, ${𝒜},{ℬ},$ and ${𝒞},$ of the three terms in the normalized free-energy expression,

${\mathfrak{𝔊}}^{*}$

$=$

$-3{𝒜}{\chi }^{-1}-\frac{{ℬ}}{(1-{\gamma }_c)}{\chi }^{3-3{\gamma }_c}-\frac{{𝒞}}{(1-{\gamma }_e)}{\chi }^{3-3{\gamma }_e}$

then plot the function, ${\mathfrak{𝔊}}^{*}(\chi )$, and identify the value(s) of ${\chi }_{\mathrm{e}\mathrm{q}}$ at which the function has an extremum (or multiple extrema).

Step 2: Note that,

${𝒜}$	$\equiv$	$\frac{{\nu }^2}{5q}\{1+\frac{5}{2}(\frac{{\rho }_e}{{\rho }_c}\left)\right(\frac{1}{q^2}-1)+(\frac{{\rho }_e}{{\rho }_c}{)}^2\left[\right(\frac{1}{q^5}-1)-\frac{5}{2}(\frac{1}{q^2}-1\left)\right]\}$
	$=$	$\frac{1}{5}\left(\frac{\nu }{q^3}{)}^2\right[q^5+\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)(1-q^2)q^3+\left(\frac{{\rho }_e}{{\rho }_c}{)}^2\right(1-\frac{5}{2}{q}^3+\frac{3}{2}{q}^5\left)\right]$
${ℬ}$	$\equiv$	$\nu \left[\right(\frac{3}{4\pi }\left)\frac{\nu }{q^3}{]}^{{\gamma }_c-1}\right[1+{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}]$
${𝒞}$	$\equiv$	$(1-\nu )\left(\frac{K_e}{K_c}{)}^{*}\right[\left(\frac{3}{4\pi }\right)\frac{(1-\nu )}{(1-q^3)}{]}^{{\gamma }_e-1}\{1+\frac{{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}}{(1-q^3)}[\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)(-2+3q-q^3)+\frac{3}{2q^2}(\frac{{\rho }_e}{{\rho }_c}{)}^2(-1+5q^2-5q^3+q^5)]\}$
	$\equiv$	$\nu \left[\right(\frac{3}{4\pi }\left)\frac{\nu }{q^3}{]}^{{\gamma }_c-1}\right\{\frac{(1-q^3)}{q^3}+\frac{{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}}{q^3}\left[\frac{5}{2}\right(\frac{{\rho }_e}{{\rho }_c})(-2+3q-q^3)+\frac{3}{2q^2}(\frac{{\rho }_e}{{\rho }_c}{)}^2(-1+5q^2-5q^3+q^5)\left]\right\}{\chi }_{\mathrm{e}\mathrm{q}}^{3({\gamma }_e-{\gamma }_c)}$

where (see, for example, in the context of its original definition),

${\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}\equiv \frac{3}{2^2\pi \cdot 5}\left(\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}^4{P}_i}\right)\frac{{\nu }^2}{q^4}$

$=$

$\frac{1}{5}\left(\frac{\nu }{q}\right)\left[\right(\frac{3}{4\pi })\frac{\nu }{q^3}{]}^{1-{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}$

and, where,

$(\frac{K_e}{K_c}{)}^{*}\equiv \frac{K_e}{K_c}[\frac{K_c^3}{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{]}^{({\gamma }_c-{\gamma }_e)/(3{\gamma }_c-4)}$

$=$

$\left[\right(\frac{3}{4\pi }\left)\frac{1-\nu }{1-q^3}{]}^{-{\gamma }_e}\right[\left(\frac{3}{4\pi }\right)\frac{\nu }{q^3}{]}^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{3({\gamma }_e-{\gamma }_c)}.$

Also, keep in mind that, if the envelope and core both have uniform (but different) densities, then ${\rho }_{ic}={\rho }_c$, ${\rho }_{ie}={\rho }_e$, and

$\frac{{\rho }_c}{\overline{\rho }}=\frac{\nu }{q^3};\frac{{\rho }_e}{\overline{\rho }}=\frac{1-\nu }{1-q^3};\frac{{\rho }_e}{{\rho }_c}=\frac{q^3(1-\nu )}{\nu (1-q^3)}\Rightarrow \frac{q^3}{\nu }=\left(\frac{{\rho }_e}{{\rho }_c}\right)(1-q^3)+q^3.$

Step 3: An analytic evaluation tells us that the following should happen. Using the numerically derived value for ${\chi }_{\mathrm{e}\mathrm{q}}$, define,

${{𝒞}}^{\text{'}}\equiv {𝒞}{\chi }_{\mathrm{e}\mathrm{q}}^{3({\gamma }_c-{\gamma }_e)}.$

We should then discover that,

$\frac{{𝒜}}{{ℬ}+{{𝒞}}^{\text{'}}}={\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}=\frac{1}{{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}}\cdot \frac{1}{5}\left(\frac{\nu }{q}\right)\left[\right(\frac{3}{4\pi })\frac{\nu }{q^3}{]}^{1-{\gamma }_c}.$

Check It

${ℬ}+{{𝒞}}^{\text{'}}$

$=$

$\nu \left[\right(\frac{3}{4\pi }\left)\frac{\nu }{q^3}{]}^{{\gamma }_c-1}\right\{[1+{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}]+\frac{(1-q^3)}{q^3}+\frac{{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}}{q^3}\left[\frac{5}{2}\right(\frac{{\rho }_e}{{\rho }_c})(-2+3q-q^3)+\frac{3}{2q^2}(\frac{{\rho }_e}{{\rho }_c}{)}^2(-1+5q^2-5q^3+q^5)\left]\right\}$

$\Rightarrow {𝒜}[{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}\cdot 5(\frac{q}{{\nu }^2}\left)\right]$

$=$

$1+{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}+\frac{(1-q^3)}{q^3}+\frac{{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}}{q^3}\left[\frac{5}{2}\right(\frac{{\rho }_e}{{\rho }_c})(-2+3q-q^3)+\frac{3}{2q^2}(\frac{{\rho }_e}{{\rho }_c}{)}^2(-1+5q^2-5q^3+q^5)]$

$\Rightarrow {\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}\{1+\frac{5}{2}(\frac{{\rho }_e}{{\rho }_c}\left)\right(\frac{1}{q^2}-1)+(\frac{{\rho }_e}{{\rho }_c}{)}^2\left[\right(\frac{1}{q^5}-1)-\frac{5}{2}(\frac{1}{q^2}-1\left)\right]\}$

$=$

$1+{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}+\frac{(1-q^3)}{q^3}+\frac{{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}}{q^3}\left[\frac{5}{2}\right(\frac{{\rho }_e}{{\rho }_c})(-2+3q-q^3)+\frac{3}{2q^2}(\frac{{\rho }_e}{{\rho }_c}{)}^2(-1+5q^2-5q^3+q^5)]$

$\Rightarrow {\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}\left\{\frac{5}{2}\right(\frac{{\rho }_e}{{\rho }_c}\left)\right(\frac{1}{q^2}-1)+(\frac{{\rho }_e}{{\rho }_c}{)}^2\left[\right(\frac{1}{q^5}-1)-\frac{5}{2}(\frac{1}{q^2}-1\left)\right]\}$

$=$

$\frac{1}{q^3}+\frac{{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}}{q^3}\left[\frac{5}{2}\right(\frac{{\rho }_e}{{\rho }_c})(-2+3q-q^3)+\frac{3}{2q^2}(\frac{{\rho }_e}{{\rho }_c}{)}^2(-1+5q^2-5q^3+q^5)]$

$\Rightarrow \frac{1}{{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}}$	$=$	$q^3\left\{\frac{5}{2}\right(\frac{{\rho }_e}{{\rho }_c}\left)\right(\frac{1}{q^2}-1)+(\frac{{\rho }_e}{{\rho }_c}{)}^2\left[\right(\frac{1}{q^5}-1)-\frac{5}{2}(\frac{1}{q^2}-1\left)\right]\}-[\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)(-2+3q-q^3)+\frac{3}{2q^2}(\frac{{\rho }_e}{{\rho }_c}{)}^2(-1+5q^2-5q^3+q^5)]$
$\Rightarrow \frac{2}{{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}}$	$=$	$5\left(\frac{{\rho }_e}{{\rho }_c}\right)(q-q^3)+\frac{2}{q^2}\left(\frac{{\rho }_e}{{\rho }_c}{)}^2\right[(1-q^5)-\frac{5}{2}(q^3-q^5)]-5(\frac{{\rho }_e}{{\rho }_c})(-2+3q-q^3)-\frac{3}{q^2}(\frac{{\rho }_e}{{\rho }_c}{)}^2(-1+5q^2-5q^3+q^5)$
	$=$	$5\left(\frac{{\rho }_e}{{\rho }_c}\right)[(q-q^3)-(-2+3q-q^3)]+\frac{1}{q^2}\left(\frac{{\rho }_e}{{\rho }_c}{)}^2\right[2(1-q^5)-5(q^3-q^5)-3(-1+5q^2-5q^3+q^5)]$
	$=$	$10\left(\frac{{\rho }_e}{{\rho }_c}\right)[1-q]+\frac{5}{q^2}\left(\frac{{\rho }_e}{{\rho }_c}{)}^2\right[1-3q^2+2q^3]$
$\Rightarrow \frac{1}{{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}}\left[\frac{2q^2}{5}\right(\frac{{\rho }_e}{{\rho }_c}{)}^{-1}]$	$=$	$2q^2(1-q)+\left(\frac{{\rho }_e}{{\rho }_c}\right)(1-3q^2+2q^3)$

Fortunately, this precisely matches our earlier derivation, which states that,

$\frac{1}{\mathrm{\Lambda }}$

$=$

$\frac{5}{2}(g^2-1)=\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_0}\right)\left[2\right(1-\frac{{\rho }_e}{{\rho }_0}\left)\right(1-q)+\frac{{\rho }_e}{{\rho }_0}(\frac{1}{q^2}-1\left)\right].$

Playing With One Example

By setting,

${\gamma }_c=6/5;{\gamma }_e=2$
${𝒜}$	${ℬ}$	${𝒞}$
2.5	1.0	2.0

a plot of ${\mathfrak{𝔊}}^{*}$ versus $\chi$ exhibits the following, two extrema:

extremum	${\chi }_{\mathrm{e}\mathrm{q}}$	${\mathfrak{𝔊}}^{*}$		${\chi }_{\mathrm{e}\mathrm{q}}^{3({\gamma }_c-{\gamma }_e)}$	${{𝒞}}^{\text{'}}$	${\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}$	$\frac{{𝒜}}{{ℬ}+{{𝒞}}^{\text{'}}}$
MIN	$1.1824$	$-0.611367$	$\Rightarrow$	$0.66891$	$1.3378$	$1.0693$	$1.0694$
MAX	$9.6722$	$+0.508104$	$\Rightarrow$	$0.004313$	$0.008625$	$2.4786$	$2.4786$

The last two columns of this table confirm the internal consistency of the relationships presented in Step 3, above. But what does this mean in terms of the values of $\nu$, $q$, and the related ratio of densities at the interface, ${\rho }_e/{\rho }_c$?

Let's assume that what we're trying to display and examine is the behavior of the free-energy surface for a fixed value of the ratio of densities at the interface. Once the value of ${\rho }_e/{\rho }_c$ has been specified, it is clear that the value of $q$ (and, hence, also $\nu$) is set because ${𝒜}$ has also been specified. But our specification of ${ℬ}$ along with ${\rho }_e/{\rho }_c$ also forces a particular value of $q$. It is unlikely that these two values of $q$ will be the same. In reality, once ${𝒜}$ and ${ℬ}$ have both been specified, they force a particular $(\nu ,q)$ pair. How do we (easily) figure out what this pair is?

Let's begin by rewriting the expressions for ${𝒜}$ and ${ℬ}$ in terms of just $q$ and the ratio, ${\rho }_e/{\rho }_c$, keeping in mind that, for the case of a uniform-density core (of density, ${\rho }_c$) and a uniform-density envelope (of density, ${\rho }_e$),

$\frac{{\rho }_e}{{\rho }_c}$

$=$

$\frac{q^3(1-\nu )}{\nu (1-q^3)},$

hence,

$\nu$

$=$

$[1+(\frac{{\rho }_e}{{\rho }_c})\frac{(1-q^3)}{q^3}{]}^{-1}$

and

$\frac{q^3}{\nu }$

$=$

$\left[\right(\frac{{\rho }_e}{{\rho }_c})(1-q^3)+q^3].$

Putting the expression for ${𝒜}$ in the desired form is simple because $\nu$ only appears as a leading factor. Specifically, we have,

${𝒜}$	$=$	$\frac{\pi q^5}{5}\left[\right(\frac{{\rho }_e}{{\rho }_c})(1-q^3)+q^3{]}^{-2}\{1+\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)(\frac{1}{q^2}-1)+\left(\frac{{\rho }_e}{{\rho }_c}{)}^2\right[(\frac{1}{q^5}-1)-\frac{5}{2}(\frac{1}{q^2}-1)\left]\right\}$
	$=$	$\frac{\pi }{5}\left[\right(\frac{{\rho }_e}{{\rho }_c})(1-q^3)+q^3{]}^{-2}\{q^5+\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)(q^3-q^5)+\left(\frac{{\rho }_e}{{\rho }_c}{)}^2\right[1-\frac{5}{2}{q}^3+\frac{3}{2}{q}^5\left]\right\}.$

The expression for ${ℬ}$ can be written in the form,

${ℬ}$	$=$	$\nu \left[\right(\frac{3}{4\pi }\left)\frac{\nu }{q^3}{]}^{{\gamma }_c-1}\right\{1+\frac{\pi }{5}\left(\frac{\nu }{q}\right)\left[\right(\frac{3}{4\pi }\left)\frac{\nu }{q^3}{]}^{1-{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}\right\}$
	$=$	$\nu \left[\right(\frac{4\pi }{3})\frac{q^3}{\nu }{]}^{1-{\gamma }_c}+\frac{\pi q^5}{5}(\frac{{\nu }^2}{q^6}){\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}$
	$=$	$q^3\left(\frac{4\pi }{3}{)}^{1-{\gamma }_c}\right[\frac{q^3}{\nu }{]}^{-{\gamma }_c}+\frac{\pi q^5}{5}(\frac{q^3}{\nu }{)}^{-2}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}$
	$=$	$q^3\left(\frac{4\pi }{3}{)}^{1-{\gamma }_c}\right[\left(\frac{{\rho }_e}{{\rho }_c}\right)(1-q^3)+q^3{]}^{-{\gamma }_c}+\frac{\pi q^5}{5}\left[\right(\frac{{\rho }_e}{{\rho }_c})(1-q^3)+q^3{]}^{-2}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}.$

Generally speaking, the equilibrium radius, ${\chi }_{\mathrm{e}\mathrm{q}}$, which appears in the expression for ${ℬ}$, is not known ahead of time. Indeed, as is illustrated in our simple example immediately above, the normal path is to pick values for the coefficients, ${𝒜}$, ${ℬ}$, and ${𝒞}$, and determine the equilibrium radius by looking for extrema in the free-energy function. And because ${\chi }_{\mathrm{e}\mathrm{q}}$ is not known ahead of time, it isn't clear how to (easily) figure out what pair of physical parameter values, $(\nu ,q)$, give self-consistent values for the coefficient pair, $({𝒜},{ℬ})$.

Because we are using a uniform density core and uniform density envelope as our base model, however, we do know the analytic solution for ${\chi }_{\mathrm{e}\mathrm{q}}$. As stated above, it is,

${\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}$	$=$	$\frac{1}{{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}}\cdot \frac{\pi }{5}\left(\frac{\nu }{q}\right)\left[\right(\frac{3}{4\pi })\frac{\nu }{q^3}{]}^{1-{\gamma }_c}$
	$=$	$\frac{\pi q^2}{2}\left(\frac{3}{4\pi }{)}^{1-{\gamma }_c}\right(\frac{\nu }{q^3}\left)\right(\frac{{\rho }_e}{{\rho }_c}\left)\right[2(1-\frac{{\rho }_e}{{\rho }_c})(1-q)+\frac{{\rho }_e}{{\rho }_c}(\frac{1}{q^2}-1)]\cdot [\frac{q^3}{\nu }{]}^{{\gamma }_c-1}$
	$=$	$\frac{\pi }{2}\left(\frac{3}{4\pi }{)}^{1-{\gamma }_c}\right(\frac{{\rho }_e}{{\rho }_c}\left)\right[2(1-\frac{{\rho }_e}{{\rho }_c})(q^2-q^3)+\frac{{\rho }_e}{{\rho }_c}(1-q^2)]\cdot [\left(\frac{{\rho }_e}{{\rho }_c}\right)(1-q^3)+q^3{]}^{{\gamma }_c-2}$

Combining this expression with the one for ${ℬ}$ gives us the desired result — although, strictly speaking, it is cheating! We can now methodically choose $(\nu ,q)$ pairs and map them into the corresponding values of ${𝒜}$ and ${ℬ}$. And, via an analogous "cheat," the choice of $(\nu ,q)$ also gives us the self-consistent value of ${𝒞}$. In this manner, we should be able to map out the free-energy surface for any desired set of physical parameters.

Second Example

Explain Logic

File:FreeEnergyExample.jpg

The figure presented here, on the right, shows a plot of the free energy, as a function of the dimensionless radius,

${\mathfrak{𝔊}}^{*}(\chi )$

, where,

${\mathfrak{𝔊}}^{*}$

$=$

$-3{𝒜}{\chi }^{-1}-\frac{{ℬ}}{(1-{\gamma }_c)}{\chi }^{3-3{\gamma }_c}-\frac{{𝒞}}{(1-{\gamma }_e)}{\chi }^{3-3{\gamma }_e},$

and, where we have used the parameter values,

${\gamma }_c=6/5;{\gamma }_e=2$
${𝒜}$	${ℬ}$	${𝒞}$
0.201707	0.0896	0.002484

Directly from this plot we deduce that this free-energy function exhibits a minimum at ${\chi }_{\mathrm{e}\mathrm{q}}=0.1235$ and that, at this equilibrium radius, the configuration has a free-energy value, ${\mathfrak{𝔊}}^{*}({\chi }_{\mathrm{e}\mathrm{q}})=-2.0097$. Via the steps described below, we demonstrate that this identified equilibrium radius is appropriate for an $(n_c,n_e)=(0,0)$ bipolytrope (with the just-specified core and envelope adiabatic indexes) that has the following physical properties:

• Fractional core mass, $\nu =0.1$;
• Core-envelope interface located at $r_i/R=q=0.435$;
• Density jump at the core-envelope interface, ${\rho }_e/{\rho }_c=0.8$.

Step 1: Because the ratio, $q^3/\nu$, is a linear function of the density ratio, ${\rho }_e/{\rho }_c$, the full definition of the free-energy coefficient, ${𝒜}$, can be restructured into a quadratic equation that gives the density ratio for any choice of the parameter pair, $(q,{𝒜})$. Specifically,

$5(\frac{q^3}{\nu }{)}^2{𝒜}$	$=$	$q^5+\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)(1-q^2)q^3+\left(\frac{{\rho }_e}{{\rho }_c}{)}^2\right(1-\frac{5}{2}{q}^3+\frac{3}{2}{q}^5)$
$\Rightarrow 5{𝒜}\left[\right(\frac{{\rho }_e}{{\rho }_c})(1-q^3)+q^3{]}^2$	$=$	$q^5+\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)(1-q^2)q^3+\left(\frac{{\rho }_e}{{\rho }_c}{)}^2\right(1-\frac{5}{2}{q}^3+\frac{3}{2}{q}^5),$

and this can be written in the form,

$(\frac{{\rho }_e}{{\rho }_c}{)}^{2}a+(\frac{{\rho }_e}{{\rho }_c})b+c$

$=$

$0,$

where,

$a$	$\equiv$	$5{𝒜}(1-q^3{)}^2-1+\frac{5}{2}{q}^3-\frac{3}{2}{q}^5,$
$b$	$\equiv$	$10{𝒜}{q}^3(1-q^3)-\frac{5}{2}{q}^3(1-q^2),$
$c$	$\equiv$	$5{𝒜}{q}^6-q^5.$

Hence,

$\frac{{\rho }_e}{{\rho }_c}$

$=$

$\frac{1}{2a}[\pm (b^2-4ac{)}^{1/2}-b].$

(For our physical problem it appears as though only the positive root is relevant.) For the purposes of this example, we set ${𝒜}=0.2017$ and examined a range of values of $q$ to find a physically interesting value for the density ratio. We picked:

${𝒜}$	$q$		$a$	$b$	$c$		$\frac{{\rho }_e}{{\rho }_c}$		$\nu$
$0.2017$	$0.435$	$\Rightarrow$	$0.03173$	$-0.01448$	$-0.008743$	$\Rightarrow$	$0.80068$	$\Rightarrow$	$0.10074$

Step 2: Next, we chose the parameter pair,

$(q,\frac{{\rho }_e}{{\rho }_c})=(0.43500,0.80000)$

and determined the following parameter values from the known analytic solution:

$\nu$	$f(q,\frac{{\rho }_e}{{\rho }_c})$	$g^2(q,\frac{{\rho }_e}{{\rho }_c})$	${\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}$	${\chi }_{\mathrm{e}\mathrm{q}}$	${𝒜}$	${ℬ}$	${𝒞}$
$0.100816$	$43.16365$	$3.923017$	$0.13684$	$0.12349$	$0.201707$	$0.089625$	$0.002484$

Construction Multiple Curves to Define a Free-Energy Surface

Okay. Now that we have the hang of this, let's construct a sequence of curves that represent physical evolution at a fixed interface-density ratio, ${\rho }_e/{\rho }_c$, but for steadily increasing core-to-total mass ratio, $\nu$. Specifically, we choose,

$\frac{{\rho }_e}{{\rho }_c}=\frac{1}{2}.$

From the known analytic solution, here are parameters defining several different equilibrium models:

Identification of Local Minimum in Free Energy
$\nu$	$q$	$f(q,\frac{{\rho }_e}{{\rho }_c})$	$g^2(q,\frac{{\rho }_e}{{\rho }_c})$	${\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}$	${\chi }_{\mathrm{e}\mathrm{q}}$	${𝒜}$	${ℬ}$	${𝒞}$
$0.2$	$9^{-1/3}=0.48075$	$12.5644$	$2.091312$	$0.366531$	$0.037453$	$0.2090801$	$0.2308269$	$2.06252\times 10^{-4}$
$0.4$	$4^{-1/3}=0.62996$	$4.21974$	$1.56498$	$0.707989$	$0.0220475$	$0.2143496$	$0.5635746$	$4.4626\times 10^{-5}$
$0.5$	$3^{-1/3}=0.693361$	$2.985115$	$1.42334$	$0.9448663$	$0.0152116$	$0.2152641$	$0.791882$	$1.5464\times 10^{-5}$
Here we are examining the behavior of the free-energy function for bipolytropic models having $(n_c,n_e)=(0,0)$, $({\gamma }_c,{\gamma }_e)=(6/5,2)$, and a density ratio at the core-envelope interface of ${\rho }_e/{\rho }_c=1/2$. The figure shown here, on the right, displays the three separate free-energy curves, ${\mathfrak{𝔊}}^{*}(\chi )$ — where, $\chi \equiv R/R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$ is the normalized configuration radius — that correspond to the three values of $\nu \equiv M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}/M_{\mathrm{t}\mathrm{o}\mathrm{t}}$ given in the first column of the above table. Along each curve, the local free-energy minimum corresponds to the the equilibrium radius, ${\chi }_{\mathrm{e}\mathrm{q}}$, recorded in column 6 of the above table.					File:ThreeFreeEnergyCurves.png

Each of the free-energy curves shown above has been entirely defined by our specification of the three coefficients in the free-energy function, ${𝒜},{ℬ}$, and ${𝒞}$. In each case, the values of these three coefficients was judiciously chosen to produce a curve with a local minimum at the correct value of ${\chi }_{\mathrm{e}\mathrm{q}}$ corresponding to an equilibrium configuration having the desired $(\nu ,{\rho }_e/{\rho }_c)$ model parameters. Upon plotting these three curves, we noticed that two of the curves — curves for $\nu =0.4$ and $\nu =0.5$ — also display a local maximum. Presumably, these maxima also identify equilibrium configurations, albeit unstable ones. From a careful inspection of the plotted curves, we have identified the value of ${\chi }_{\mathrm{e}\mathrm{q}}$ that corresponds to the two newly discovered (unstable) equilibrium models; these values are recorded in the table that immediately follows this paragraph. By construction, we also know what values of ${𝒜},{ℬ}$, and ${𝒞}$ are associated with these two identified equilibria; these values also have been recorded in the table. But it is not immediately obvious what the values are of the $(\nu ,{\rho }_e/{\rho }_c)$ model parameters that correspond to these two equilibrium models.

Subsequently Identified Local Energy Maxima
${\chi }_{\mathrm{e}\mathrm{q}}$	${\mathfrak{𝔊}}^{*}$	${\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}$	$\therefore$	${𝒜}$	${ℬ}$	${𝒞}$	${{𝒞}}^{\text{'}}={𝒜}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}-{ℬ}$	$(\frac{{𝒞}}{{{𝒞}}^{\text{'}}}{)}^{1/(3{\gamma }_e-3{\gamma }_c)}$
$0.08255$	$+4.87562$	$0.368715$	$0$	$0.2143496$	$0.5635746$	$4.4626\times 10^{-5}$	$1.7768\times 10^{-2}$	$0.08254$
$0.032196$	$+11.5187$	$0.25300$	$0$	$0.2152641$	$0.791882$	$1.5464\times 10^{-5}$	$5.8964\times 10^{-2}$	$0.032196$

Related Discussions

• Newer, Version2 of Bipolytrope Generalization derivations.
• Analytic solution with $n_c=0$ and $n_e=0$.
• Analytic solution with $n_c=5$ and $n_e=1$.