SSC/BipolytropeGeneralization/Pt3

Bipolytrope Generalization (Pt 3)

Part I:  Bipolytrope Generalization

Part II:  Derivations

Part III:  Examples

Part IV:  Best of the Best

Examples

(0, 0) Bipolytropes

Review

In an accompanying discussion we have derived analytic expressions describing the equilibrium structure and the stability of bipolytropes in which both the core and the envelope have uniform densities, that is, bipolytropes with $(n_c,n_e)=(0,0)$. From this work, we find that integrals over the mass and pressure distributions give:

$\frac{W}{R_{\mathrm{e}\mathrm{q}}^3{P}_i}=-\frac{A}{R_{\mathrm{e}\mathrm{q}}^4{P}_i}$	$=$	$-\frac{3}{5}\left[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R^4{P}_i}\right]\left(\frac{{\nu }^2}{q}\right)f$
	$=$	$-4\pi q^3\mathrm{\Lambda }f,$
$\frac{S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{R_{\mathrm{e}\mathrm{q}}^3{P}_i}=B_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$2\pi q^3(1+\mathrm{\Lambda }),$
$\frac{S_{\mathrm{e}\mathrm{n}\mathrm{v}}}{R_{\mathrm{e}\mathrm{q}}^3{P}_i}=B_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$2\pi [(1-q^3)+\frac{5}{2}\mathrm{\Lambda }(\frac{{\rho }_e}{{\rho }_0})(-2+3q-q^3)+\frac{3}{2q^2}\mathrm{\Lambda }(\frac{{\rho }_e}{{\rho }_0}{)}^2(-1+5q^2-5q^3+q^5)],$

where,

$\mathrm{\Lambda }$	$\equiv$	$\frac{3}{2^2\cdot 5\pi }\left(\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}^4{P}_i}\right)\frac{{\nu }^2}{q^4},$
$f(q,{\rho }_e/{\rho }_c)$	$\equiv$	$1+\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)(\frac{1}{q^2}-1)+\left(\frac{{\rho }_e}{{\rho }_c}{)}^2\right[(\frac{1}{q^5}-1)-\frac{5}{2}(\frac{1}{q^2}-1)]$
	$=$	$1+\frac{5}{2q^2}\left(\frac{{\rho }_e}{{\rho }_c}\right)(1-q^2)+\frac{1}{2q^5}(\frac{{\rho }_e}{{\rho }_c}{)}^2(2-5q^3+3q^5),$
$g^2(q,{\rho }_e/{\rho }_c)$	$\equiv$	$1+\left(\frac{{\rho }_e}{{\rho }_0}\right)\left[2\right(1-\frac{{\rho }_e}{{\rho }_0}\left)\right(1-q)+\frac{{\rho }_e}{{\rho }_0}(\frac{1}{q^2}-1\left)\right]$
	$\equiv$	$1+\left[2\right(\frac{{\rho }_e}{{\rho }_c})(1-q)+\frac{1}{q^2}(\frac{{\rho }_e}{{\rho }_c}{)}^2(1-3q^2+2q^3)],$

Renormalize

Let's renormalize these energy terms in order to more readily relate them to the generalized expressions derived above.

$R^3{P}_i$	$=$	$R^3{K}_c\left(\frac{{\rho }_{ic}}{\overline{\rho }}{)}^{{\gamma }_c}\right[\left(\frac{3}{4\pi }\right)\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{R^3}{]}^{{\gamma }_c}$
	$=$	$\left[\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{3-3{\gamma }_c}\right[\left(\frac{3}{4\pi }\right)\frac{{\rho }_{ic}}{\overline{\rho }}{]}^{{\gamma }_c}{K}_c{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{{\gamma }_c}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^{3-3{\gamma }_c}$
	$=$	$\left[\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{3-3{\gamma }_c}\right[\left(\frac{3}{4\pi }\right)\frac{{\rho }_{ic}}{\overline{\rho }}{]}^{{\gamma }_c}\left\{K_c^{3{\gamma }_c-4}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{{\gamma }_c(3{\gamma }_c-4)}\right[\left(\frac{K_c}{G}\right)M_{\mathrm{t}\mathrm{o}\mathrm{t}}^{{\gamma }_c-2}{]}^{3-3{\gamma }_c}{\}}^{1/(3{\gamma }_c-4)}$
	$=$	$\left[\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{3-3{\gamma }_c}\right[\left(\frac{3}{4\pi }\right)\frac{{\rho }_{ic}}{\overline{\rho }}{]}^{{\gamma }_c}\{\frac{G^{3{\gamma }_c-3}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{5{\gamma }_c-6}}{K_c}{\}}^{1/(3{\gamma }_c-4)}$
	$=$	$\left[\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{3-3{\gamma }_c}\right[\left(\frac{3}{4\pi }\right)\frac{{\rho }_{ic}}{\overline{\rho }}{]}^{{\gamma }_c}{E}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}.$

Also,

$[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R}{]}^{3{\gamma }_c-4}$	$=$	$\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-(3{\gamma }_c-4)}{G}^{3{\gamma }_c-4}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{6{\gamma }_c-8}\right(\frac{G}{K_c})M_{\mathrm{t}\mathrm{o}\mathrm{t}}^{2-{\gamma }_c}$
	$=$	$\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-(3{\gamma }_c-4)}\right[\frac{G^{3{\gamma }_c-3}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{5{\gamma }_c-6}}{K_c}]$
	$=$	$(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-(3{\gamma }_c-4)}{E}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^{3{\gamma }_c-4}.$
$\Rightarrow \frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R^4{P}_i}$	$=$	$\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}\right[\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{3{\gamma }_c-3}\left[\right(\frac{3}{4\pi })\frac{{\rho }_{ic}}{\overline{\rho }}{]}^{-{\gamma }_c}$
	$=$	$\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{3{\gamma }_c-4}\right[\left(\frac{3}{4\pi }\right)\frac{{\rho }_{ic}}{\overline{\rho }}{]}^{-{\gamma }_c}.$

Hence,

$\mathrm{\Lambda }$

$\equiv$

$\frac{3}{2^2\cdot 5\pi }\frac{{\nu }^2}{q^4}\left[\right(\frac{3}{4\pi }\left)\frac{{\rho }_{ic}}{\overline{\rho }}{]}^{-{\gamma }_c}\right(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{3{\gamma }_c-4}.$

Given that ${\rho }_{ic}/\overline{\rho }=\nu /q^3$ for the $(n_c,n_e)=(0,0)$ bipolytrope, we can finally write,

$\frac{R^3{P}_i}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$

$=$

$\left[\right(\frac{3}{4\pi }\left)\frac{\nu }{q^3}{]}^{{\gamma }_c}\right(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{3-3{\gamma }_c},$

and,

$\mathrm{\Lambda }$

$\equiv$

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

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,

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

Finally, then, we can state that,

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

Virial Equilibrium and Stability Evaluation

With these expressions in hand, we can deduce the equilibrium radius and relativity stability of $(n_c,n_e)=(0,0)$ bipolytropes using the generalized expressions provided above. For example, from the statement of virial equilibrium $(2S_{\mathrm{t}\mathrm{o}\mathrm{t}}=-W)$ we obtain,

$q^3(1+\mathrm{\Lambda })+(1-q^3)+\frac{5}{2}\mathrm{\Lambda }\left(\frac{{\rho }_e}{{\rho }_0}\right)(-2+3q-q^3)+\frac{3}{2q^2}\mathrm{\Lambda }(\frac{{\rho }_e}{{\rho }_0}{)}^2(-1+5q^2-5q^3+q^5)$

$=$

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

$\Rightarrow \frac{1}{\mathrm{\Lambda }}$	$=$	$\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)(q-q^3)+\frac{1}{2q^2}(\frac{{\rho }_e}{{\rho }_c}{)}^2(2-5q^3+3q^5)-[\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_0}\right)(-2+3q-q^3)+\frac{3}{2q^2}(\frac{{\rho }_e}{{\rho }_0}{)}^2(-1+5q^2-5q^3+q^5)]$
	$=$	$\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)(q-q^3+2-3q+q^3)+\frac{1}{2q^2}(\frac{{\rho }_e}{{\rho }_c}{)}^2(2-5q^3+3q^5+3-15q^2+15q^3-3q^5)$
	$=$	$\frac{5}{2}\left[2\right(\frac{{\rho }_e}{{\rho }_c})(1-q)+\frac{1}{q^2}(\frac{{\rho }_e}{{\rho }_c}{)}^2(1-3q^2+2q^3)]$
	$=$	$\frac{5}{2}(g^2-1)$
$\Rightarrow \left[\frac{P_i}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}\right]R_{\mathrm{e}\mathrm{q}}^4$	$=$	$\left(\frac{3}{2^3\pi }\right)\frac{{\nu }^2}{q^4}(g^2-1).$

Or, given the above renormalization, this expression can be written as,

$\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{4-3{\gamma }_c}\right[\left(\frac{3}{4\pi }\right)\frac{{\rho }_{ic}}{\overline{\rho }}{]}^{{\gamma }_c}$	$=$	$\left(\frac{3}{2^3\pi }\right)\frac{{\nu }^2}{q^4}(g^2-1)$
$\Rightarrow \frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$\left\{\right(\frac{3}{2^3\pi })\frac{{\nu }^2}{q^4}(g^2-1)[\left(\frac{3}{4\pi }\right)\frac{{\rho }_{ic}}{\overline{\rho }}{]}^{{\gamma }_c}{\}}^{1/(4-3{\gamma }_c)}.$

And the condition for dynamical stability is,

$-\frac{W}{2}({\gamma }_e-\frac{4}{3})-({\gamma }_e-{\gamma }_c)S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$>$	$0.$
$\Rightarrow 2\pi q^3\mathrm{\Lambda }\left[\right({\gamma }_e-\frac{4}{3})f-({\gamma }_e-{\gamma }_c)(1+\frac{1}{\mathrm{\Lambda }}\left)\right]$	$>$	$0.$
$({\gamma }_e-\frac{4}{3})f-({\gamma }_e-{\gamma }_c)[1+\frac{5}{2}(g^2-1)]$	$>$	$0.$

(5, 1) Bipolytropes

In another accompanying discussion we have derived analytic expressions describing the equilibrium structure of bipolytropes with $(n_c,n_e)=(5,1)$. Can we perform a similar stability analysis of these configurations? Work in progress!

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 $n_e=1$.