SSC/Structure/BiiPolytropes/FreeEnergy51

Free Energy of BiPolytrope with (n_c, n_e) = (5, 1)

Part I:  Mass Profile

Part II:  Gravitational Potential Energy

Part III:  Thermal Energy Reservoir

Free Energy of Bipolytropes (nc, ne) = (5, 1)

Here we present a specific example of the equilibrium structure of a bipolytrope as determined from a free-energy analysis. The example is a bipolytrope whose core has a polytropic index, $n_c=5$, and whose envelope has a polytropic index, $n_e=1$. The details presented here build upon an overview of the free energy of bipolytropes that has been presented elsewhere.  
 
 
 
 

Mass Profile

The Core

The core has $n_c=5\Rightarrow {\gamma }_c=1+1/n_c=6/5$. Referring to the general relation as established in our accompanying overview, and using ${\rho }_0$ to represent the central density, we can write,

$(\mathrm{F}\mathrm{o}\mathrm{r}0\le x\le q)$       $M_r$

$=$

$M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left(\frac{\nu }{q^3}\right)\left(\frac{{\rho }_0}{{\overline{\rho }}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{)}_{\mathrm{e}\mathrm{q}}\underset{0}{\overset{x}{\int }}3\right[\frac{\rho (x)}{{\rho }_0}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{x}^{2}dx.$

Drawing on the derivation of detailed force-balance models of $(n_c,n_e)=(5,1)$ bipolytropes, the density profile throughout the core is,

$[\frac{\rho (\xi )}{{\rho }_0}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

$(1+\frac{1}{3}{\xi }^2{)}^{-5/2},$

where the dimensionless radial coordinate is,

$\xi$

$=$

$\left[\frac{G{\rho }_0^{4/5}}{K_c}{]}^{1/2}\right(\frac{2\pi }{3}{)}^{1/2}r.$

Switching to the normalizations that have been adopted in the broad context of our discussion of configurations in virial equilibrium and inserting the adiabatic index of the core $({\gamma }_c=6/5)$ into all normalization parameters, we have,

$R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}=\left[\right(\frac{G}{K_c})M_{\mathrm{t}\mathrm{o}\mathrm{t}}^{2-\gamma }{]}^{1/(4-3\gamma )}$	$\Rightarrow$	$R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}=(\frac{G^5{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^4}{K_c^5}{)}^{1/2},$
${\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}=\frac{3}{4\pi }[\frac{K_c^3}{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{]}^{1/(4-3\gamma )}$	$\Rightarrow$	${\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}=\frac{3}{4\pi }(\frac{K_c^3}{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{)}^{5/2}.$

Hence, we can rewrite,

$\xi$	$=$	$\left(\frac{r}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)\left(\frac{{\rho }_0}{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{2/5}\right[\frac{G}{K_c}{]}^{1/2}(\frac{2\pi }{3}{)}^{1/2}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^{2/5}$
	$=$	$r^{*}({\rho }_0^{*}{)}^{2/5}[\frac{G}{K_c}{]}^{1/2}\left(\frac{2\pi }{3}{)}^{1/2}\right(\frac{G^5{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^4}{K_c^5}{)}^{1/2}\left(\frac{3}{4\pi }{)}^{2/5}\right(\frac{K_c^3}{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2})$
	$=$	$r^{*}({\rho }_0^{*}{)}^{2/5}[\left(\frac{2\pi }{3}{)}^5\right(\frac{3}{4\pi }{)}^4{]}^{1/10}=r^{*}({\rho }_0^{*}{)}^{2/5}[\frac{\pi }{2^3\cdot 3}{]}^{1/10}.$

Now, following the same approach as was used in our introductory discussion and appreciating that our aim here is to redefine the coordinate, $\xi$, in terms of normalized parameters evaluated in the equilibrium configuration, we will set,

$r^{*}$	$\to$	$x{\chi }_{\mathrm{e}\mathrm{q}};$
${\rho }_0^{*}$	$\to$	$\left[\frac{{\rho }_0}{\overline{\rho }}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\right(\frac{{\overline{\rho }}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}})=[\frac{{\rho }_0}{\overline{\rho }}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\frac{\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}/(q^3{R}_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^3{)}_{\mathrm{e}\mathrm{q}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}/R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}=\frac{\nu }{q^3}[\frac{{\rho }_0}{\overline{\rho }}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{\chi }_{\mathrm{e}\mathrm{q}}^{-3}.$

Then we can set,

$\xi$

$=$

$(3a_{\xi }{)}^{1/2}x,$

in which case,

$[\frac{\rho (x)}{{\rho }_0}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

$(1+a_{\xi }{x}^2{)}^{-5/2},$

where the coefficient,

$(3a_{\xi }{)}^{1/2}$	$\equiv$	${\chi }_{\mathrm{e}\mathrm{q}}\left[\frac{\nu }{q^3}\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{\chi }_{\mathrm{e}\mathrm{q}}^{-3}{]}^{2/5}(\frac{\pi }{2^3\cdot 3}{)}^{1/10}={\chi }_{\mathrm{e}\mathrm{q}}^{-1/5}[\frac{\nu }{q^3}\left(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}^{2/5}\right(\frac{\pi }{2^3\cdot 3}{)}^{1/10}$
$\Rightarrow a_{\xi }$	$\equiv$	$\frac{1}{3}\left\{{\chi }_{\mathrm{e}\mathrm{q}}^{-1/5}\right[\frac{\nu }{q^3}\left(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}^{2/5}\right(\frac{\pi }{2^3\cdot 3}{)}^{1/10}{\}}^2={\chi }_{\mathrm{e}\mathrm{q}}^{-2/5}\left[\frac{\nu }{q^3}\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}^{4/5}(\frac{\pi }{2^3\cdot 3^6}{)}^{1/5}.$

We therefore have,

$M_r{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left[\frac{\nu }{q^3}\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}\underset{0}{\overset{x}{\int }}3(1+a_{\xi }{x}^2{)}^{-5/2}{x}^{2}dx$
	$=$	$M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left[\frac{\nu }{q^3}\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}\left[x^3\right(1+a_{\xi }{x}^2{)}^{-3/2}].$

Note that, when $x\to q$, $M_r{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\to M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}$. Hence, this last expression gives,

$\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}$	$=$	$M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left[\frac{\nu }{q^3}\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}\left[q^3\right(1+a_{\xi }{q}^2{)}^{-3/2}]$
$\Rightarrow \left[\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$(1+a_{\xi }{q}^2{)}^{3/2}.$

Hence, finally,

$M_r{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

$\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left(\frac{x^3}{q^3}\right)[\frac{1+a_{\xi }{x}^2}{1+a_{\xi }{q}^2}{]}^{-3/2};$

and, after the equilibrium radius, ${\chi }_{\mathrm{e}\mathrm{q}}$, has been determined from the free-energy analysis, the coefficient, $a_{\xi }$, can be determined via the relation,

${\chi }_{\mathrm{e}\mathrm{q}}^2$

$=$

$\left(\frac{\pi }{2^3\cdot 3^6}\right)\left(\frac{\nu }{q^3}{)}^4\right(1+a_{\xi }{q}^2{)}^6{a}_{\xi }^{-5}.$

MORE USEFUL: Letting, $\ell \equiv \xi /\sqrt{3}$, $a_{\xi }$ $=$ $(\frac{{\ell }_i}{q}{)}^2,$ ${\widetilde{\mathfrak{𝔣}}}_{\mathrm{M}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$ $=$ $(\frac{\overline{\rho }}{{\rho }_c}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=(1+{\ell }_i^2{)}^{-3/2}.$

The Envelope

The envelope has $n_e=1\Rightarrow {\gamma }_e=1+1/n_e=2$. Again, referring to the general relation as established in our accompanying overview, and continuing to use ${\rho }_0$ to represent the central density, we can write,

$(\mathrm{F}\mathrm{o}\mathrm{r}q\le x\le 1)$       $M_r$

$=$

$M_{\mathrm{t}\mathrm{o}\mathrm{t}}\{\nu +(\frac{1-\nu }{1-q^3}\left)\underset{x_i}{\overset{x}{\int }}3\right[\frac{\rho (x)}{\overline{\rho }}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}{x}^{2}dx\}.$

Drawing on the derivation of detailed force-balance models of $(n_c,n_e)=(5,1)$ bipolytropes, the density profile throughout the envelope is,

$[\frac{\rho (\eta )}{{\rho }_0}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$A\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5\left[\frac{\sin (\eta -B)}{\eta }\right],$

where definitions of the constants $A$ and $B$ are given in an accompanying table of parameter values, and the dimensionless radial coordinate is,

$\eta$

$=$

$\left[\frac{G{\rho }_0^{4/5}}{K_c}{]}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^2(2\pi {)}^{1/2}r.$

Using the same radial and mass-density normalizations as defined, above, for the core, we can write,

$\eta$	$=$	$r^{*}({\rho }_0^{*}{)}^{2/5}[\frac{G}{K_c}{]}^{1/2}\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^2(2\pi {)}^{1/2}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^{2/5}$
	$=$	$r^{*}({\rho }_0^{*}{)}^{2/5}(\frac{3}{4\pi }{)}^{2/5}\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^2(2\pi {)}^{1/2}.$

Next, we set,

$r^{*}$	$\to$	$x{\chi }_{\mathrm{e}\mathrm{q}};$
${\rho }_0^{*}$	$\to$	$\left[\frac{{\rho }_0}{\overline{\rho }}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}\right(\frac{{\overline{\rho }}_{\mathrm{e}\mathrm{n}\mathrm{v}}}{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}})=[\frac{{\rho }_0}{\overline{\rho }}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}\frac{(1-\nu )M_{\mathrm{t}\mathrm{o}\mathrm{t}}/[(1-q^3)R_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^3{]}_{\mathrm{e}\mathrm{q}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}/R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}=\frac{1-\nu }{1-q^3}(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}{\chi }_{\mathrm{e}\mathrm{q}}^{-3}.$

Hence, we can write,

$\eta =b_{\eta }x,$

where,

$b_{\eta }$

$\equiv$

${\chi }_{\mathrm{e}\mathrm{q}}^{-1/5}\left[\frac{1-\nu }{1-q^3}\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}{]}^{2/5}\left(\frac{3}{4\pi }{)}^{2/5}\right(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^2(2\pi {)}^{1/2}.$

In which case,

$[\frac{\rho (x)}{{\rho }_0}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$A\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5\left[\frac{\sin (b_{\eta }x-B)}{b_{\eta }x}\right],$

so,

$M_r{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$M_{\mathrm{t}\mathrm{o}\mathrm{t}}\{\nu +[\left(\frac{1-\nu }{1-q^3}\right)\left(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}{]}_{\mathrm{e}\mathrm{q}}\underset{q}{\overset{x}{\int }}3A\right(\frac{{\mu }_e}{{\mu }_c}\left){\theta }_i^5\right[\frac{\sin (b_{\eta }x-B)}{b_{\eta }x}\left]x^{2}dx\right\}$
	$=$	$\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}+M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left[\right(\frac{1-\nu }{1-q^3}\left)\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}{]}_{\mathrm{e}\mathrm{q}}\left(\frac{{\mu }_e}{{\mu }_c}\right)\frac{3A{\theta }_i^5}{b_{\eta }}\underset{q}{\overset{x}{\int }}\sin (b_{\eta }x-B)xdx$
	$=$	$\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}+M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left[\right(\frac{1-\nu }{1-q^3}\left)\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}{]}_{\mathrm{e}\mathrm{q}}\left(\frac{{\mu }_e}{{\mu }_c}\right)\frac{3A{\theta }_i^5}{b_{\eta }}[-\frac{\sin (B-b_{\eta }x)+b_{\eta }x\cos (B-b_{\eta }x)}{b_{\eta }^2}{]}_q^x$
	$=$	$\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}+M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left[\right(\frac{1-\nu }{1-q^3}\left)\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}{]}_{\mathrm{e}\mathrm{q}}\left(\frac{{\mu }_e}{{\mu }_c}\right)\frac{3A{\theta }_i^5}{b_{\eta }^3}[C_1-\sin (B-b_{\eta }x)-x{b}_{\eta }\cos (B-b_{\eta }x)],$

where, $C_1$ is a constant obtained by evaluating the integral at the interface $(x=x_i=q)$, specifically,

$C_1\equiv \sin (B-b_{\eta }q)+b_{\eta }q\cos (B-b_{\eta }q).$

Now, this expression can be significantly simplified by drawing on earlier results of this section as well as on attributes of the corresponding detailed force-balanced model. First, independent of the specific density profiles that define the structure of a bipolytrope, the ratio of the mean densities of the two structural regions is,

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

$=$

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

Hence the bracketed pre-factor of the second term of the expression for $M_r{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$ may be rewritten as,

$\left[\right(\frac{1-\nu }{1-q^3}\left)\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}{]}_{\mathrm{e}\mathrm{q}}$

$=$

$\left[\frac{\nu }{q^3}\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}.$

But, from the above derivation of the mass profile in the core, we know that,

$\left[\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}$

$=$

$(1+a_{\xi }{q}^2{)}^{3/2}=(1+\frac{1}{3}{\xi }_i^2{)}^{3/2}={\theta }_i^{-3},$

where the final step comes from knowledge of the expression for ${\theta }_i$ drawn from the detailed force-balanced model (see, for example, the associated Parameter Values table). Hence, we can write,

$M_r{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}+M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left(\frac{{\mu }_e}{{\mu }_c}\right)\frac{3\nu A{\theta }_i^2}{(b_{\eta }q{)}^3}[C_1-\sin (B-b_{\eta }x)-x{b}_{\eta }\cos (B-b_{\eta }x)],$

and note that the expression for the coefficient, $b_{\eta }$, becomes simpler as well, specifically,

$b_{\eta }$	$=$	${\chi }_{\mathrm{e}\mathrm{q}}^{-1/5}\left(\frac{3}{4\pi }{)}^{2/5}\right(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^2(2\pi {)}^{1/2}(\frac{\nu }{q^3{\theta }_i^3}{)}^{2/5}$
$\Rightarrow {\chi }_{\mathrm{e}\mathrm{q}}$	$=$	$b_{\eta }^{-5}\left(\frac{3^4\pi }{2^3}{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^5\frac{{\nu }^2{\theta }_i^4}{q^6}.$

Next — and, again, drawing from knowledge of the internal structure of the detailed force-balanced model, in particular, realizing that,

$b_{\eta }q={\eta }_i=3^{1/2}\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^2{\xi }_i,$

— note that the constant, $C_1$, can be rewritten as,

$C_1$	$=$	$b_{\eta }q\cos (b_{\eta }q-B)-\sin (b_{\eta }q-B)$
	$=$	${\eta }_i\cos ({\eta }_i-B)-\sin ({\eta }_i-B)$
	$=$	$\frac{{\eta }_i^2}{A}(\frac{d\varphi }{d\eta }{)}_i=-\frac{1}{A}(\frac{{\mu }_e}{{\mu }_c}{)}^2{3}^{1/2}{\theta }_i^4{\xi }_i^3$
	$=$	$-\frac{1}{A}\left(\frac{{\mu }_e}{{\mu }_c}{)}^2{3}^{1/2}{\theta }_i^4\right[3^{-1/2}(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}{b}_{\eta }q{]}^3$
	$=$	$-\frac{1}{3A{\theta }_i^2}(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}(b_{\eta }q{)}^3,$

which means,

$M_r{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}-\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}\{1-\frac{1}{C_1}[\sin (B-b_{\eta }x)+x{b}_{\eta }\cos (B-b_{\eta }x)\left]\right\}$
	$=$	$\frac{\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{C_1}[\sin (B-b_{\eta }x)+x{b}_{\eta }\cos (B-b_{\eta }x)].$

MORE USEFUL: Letting, $\ell \equiv \xi /\sqrt{3}$, $b_{\eta }={\eta }_s$        and        $b_{\eta }q={\eta }_i=3\left(\frac{{\mu }_e}{{\mu }_c}\right){\ell }_i(1+{\ell }_i^2{)}^{-1},$ ${\widetilde{\mathfrak{𝔣}}}_{\mathrm{M}\mathrm{e}\mathrm{n}\mathrm{v}}\equiv (\frac{\overline{\rho }}{{\rho }_c}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}$ $=$ $\frac{q^3(1-\nu )}{\nu (1-q^3)}\cdot {\widetilde{\mathfrak{𝔣}}}_{\mathrm{M}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

See Also