SSC/Structure/BiPolytropes/AnalyzeStepFunction

More Careful Examination of Step Function Behavior

The ideas that are captured in this chapter have arisen as an extension of our accompanying "renormalization" of the Analytic51 bipolytrope.

Discontinuous Density Distribution

Expectations

From among the set of governing relations that apply to spherically symmetric configurations, we focus, first, on the combined,

Euler + Poisson Equations

$\frac{d{v}_r}{dt}=-\frac{1}{\rho }\frac{dP}{dr}-\frac{G{M}_r}{r^2}$

At the interface between the core and envelope of an equilibrium bipolytrope, both the core and the envelope must satisfy the relation,

$\frac{1}{\rho }\frac{dP}{dr}$

$=$

$-\frac{G{M}_r}{r^2}.$

Now, the quantity on the right-hand side of this expression must be the same at the interface, when viewed either from the perspective of the core or from the perspective of the envelope. Therefore, at the interface, the equilibrium configuration must obey the relation,

$[\frac{1}{\rho }\frac{dP}{dr}{]}_{\mathrm{e}\mathrm{n}\mathrm{v},i}$

$=$

$[\frac{1}{\rho }\frac{dP}{dr}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e},i}.$

Now, if we set ${\rho }_e=({\mu }_e/{\mu }_c){\rho }_c$ at the interface, then,

$[\frac{dP}{dr}{]}_{\mathrm{e}\mathrm{n}\mathrm{v},i}$

$=$

$\frac{{\mu }_e}{{\mu }_c}[\frac{dP}{dr}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e},i}.$

Check Behavior

In step 4 of our accompanying analysis, we find that from the perspective of the core,

$P^{*}$

$=$

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

and,

$r^{*}$

$=$

$(\frac{3}{2\pi }{)}^{1/2}\xi .$

Hence, at the interface,

$\frac{d{P}^{*}}{d{r}^{*}}{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=\{\frac{d\xi }{d{r}^{*}}\cdot \frac{d{P}^{*}}{d\xi }{\}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$-2\left(\frac{2\pi }{3}{)}^{1/2}\right[(1+\frac{1}{3}{\xi }^2{)}^{-4}\xi {]}_i$
	$=$	$-\left(\frac{4\pi }{3}\right){\theta }_i^8{r}_i^{*}.$

While, in step 8 of that analysis, we find from the perspective of the envelope,

$P^{*}$

$=$

${\theta }_i^6{\varphi }^2,$

and,

$r^{*}$

$=$

$(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta .$

Hence, at the interface,

$\frac{d{P}^{*}}{d{r}^{*}}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}=\{\frac{d\eta }{d{r}^{*}}\cdot \frac{d{P}^{*}}{d\eta }{\}}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$\left\{\right(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^2(2\pi {)}^{1/2}\left\}2{\theta }_i^6\right[\varphi \cdot \frac{d\varphi }{d\eta }{]}_i$
	$=$	$2(2\pi {)}^{1/2}{\theta }_i^8\{\left(\frac{{\mu }_e}{{\mu }_c}\right)\left\}\right[3^{1/2}{\theta }_i^{-3}\left(\frac{d\theta }{d\xi }{)}_i\right]$
	$=$	$-\frac{2}{3}(6\pi {)}^{1/2}{\theta }_i^8(\frac{{\mu }_e}{{\mu }_c}){\xi }_i$
	$=$	$-2\left(\frac{2\pi }{3}{)}^{1/2}{\theta }_i^8\right(\frac{{\mu }_e}{{\mu }_c}\left)\right[\left(\frac{2\pi }{3}{)}^{1/2}{r}_i^{*}\right]$
	$=$	$-\left(\frac{4\pi }{3}\right){\theta }_i^8\left(\frac{{\mu }_e}{{\mu }_c}\right)r_i^{*}.$

Gratifyingly, we find as expected that,

$[\frac{d{P}^{*}}{d{r}^{*}}{]}_{\mathrm{e}\mathrm{n}\mathrm{v},i}$

$=$

$\frac{{\mu }_e}{{\mu }_c}[\frac{d{P}^{*}}{d{r}^{*}}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e},i}.$

Discrete LAWE for Bipolytrope

Summary Set of Nonlinear Governing Relations

In summary, the following three one-dimensional ODEs define the physical relationship between the three dependent variables $\rho$, $P$, and $r$, each of which should be expressible as a function of the two independent (Lagrangian) variables, $t$ and $M_r$:

Equation of Continuity
$\frac{d\rho }{dt}=-4\pi {\rho }^2{r}^2\frac{d}{d{M}_r}\left(\frac{dr}{dt}\right)-\frac{2\rho }{r}\left(\frac{dr}{dt}\right)$
,

Euler + Poisson Equations
$\frac{d^{2}r}{d{t}^2}=-4\pi r^2\frac{dP}{d{M}_r}-\frac{G{M}_r}{r^2}$

Adiabatic Form of the
First Law of Thermodynamics
$\rho \frac{dP}{dt}-{\gamma }_{\mathrm{g}}P\frac{d\rho }{dt}=0.$

March from the Center, Outward

Establish Fidelity of Finite-Difference Model

We know the analytic structure of the equilibrium configuration. Let's choose a Lagrangian grid that is labeled by $(r_0{)}_j$ and the corresponding enclosed mass, $m_j(r_0)$, where the center of the spherical bipolytrope is labeled by $j=0$ while each subsequent grid "line" is labeled $j$. We will identify the mean density of each mass shell by the expression,

${\overline{\rho }}_{j-1/2}$

$=$

$[\frac{\mathrm{\Delta }m}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}}{]}_{j-1/2}=[m_j-m_{j-1}\left]\right[\frac{4\pi }{3}(r_j^3-r_{j-1}^3){]}^{-1}.$

The pressure can be determined from knowledge of the density via knowledge of the (fixed) specific entropy, namely,

$P_{j-1/2}$

$=$

$({\overline{\rho }}_{j-1/2}{)}^{\gamma }\cdot \exp [\frac{\mu (\gamma -1)s}{\mathfrak{\Re }}].$

These two expressions, effectively, originate from the continuity equation and the adiabatic form of the first law of thermodynamics, respectively. They are relations that allow the determination of the mass density and the pressure, given fixed mass shells but varying mass-shell radial locations.

STEP 1:   From the analytically known equilibrium structure of the $(n_c,n_e)=(5,1)$ bipolytrope, create a table that documents how the radial location of each mass shell, $r_j$, varies with the enclosed mass, $m_j$.

STEP 2:   Determine how ${\overline{\rho }}_{j-1/2}$ varies with radial shell location, using the above continuity-equation relation. (Plot $\overline{\rho }$ versus $m$ obtained in this discrete manner — see the red-dotted curve in the following figure — then also plot how $\rho$ varies with $m$ according to the analytic equilibrium structure — see the dark-blue-dotted curve; the plotted curves should be nearly, but not exactly, the same.)

STEP 3A:   Given ${\overline{\rho }}_{j-1/2}$, determine how $P_{j-1/2}$ varies with radial shell location, using the above 1^st Law relation. (Plot the pressure, as determined in this discrete manner, versus $m$ — see the light-green-dotted curve in the following figure — then also plot how $P$ varies with $m$ according to the analytic equilibrium structure — see the purple-dotted curve; the plotted curves should be nearly, but not exactly, the same.)

STEP 3B:   Determine (and plot) how $(dP/dm{)}_j$ varies with $m$.

Now, what can we learn from the "Euler + Poisson Equation"? Well, for the equilibrium state, we should find that,

$\frac{dP}{dm}$

$=$

$-\frac{Gm}{4\pi r^4}.$

STEP 4:   Show how $dP/dm$ varies with $m$, according to this analytic prescription, and compare it against the pressure gradient behavior obtained in STEP 3. Do they match?

On a semi-log plot … As prescribed in STEP 2: The red-dotted curve shows how the discrete density, ${\overline{\rho }}_{j-1/2}$, varies with enclosed mass, and the dark-blue-dotted curve shows how the analytically determined density varies with enclosed mass. As prescribed in STEP 3A: The light-green-dotted curve shows how the discrete pressure, $P_{j-1/2}$, varies with enclosed mass, and the purple-dotted curve shows how the analytically determined pressure varies with enclosed mass. As prescribed in STEP 3B: The light-blue-dotted curve shows how the discrete pressure gradient, $(dP/dm{)}_j$, varies with enclosed mass, and the orange-dotted curve shows how the analytically determined quantity, $-Gm/(4\pi r^4)$, varies with enclosed mass. Note that, in each case, we have added "1" to the logarithm of the specified quantity in order to shift both curves up in the plot and thereby unclutter the diagram.

Guessing Eigenvector

Step 5

STEP 5:   Guess the eigenvector, ${\delta r}_i$, remembering that a reasonably good trial eigenfunction for the core is one that has a "parabolic dependence on the radius," namely,

$\frac{x}{{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}}$

$=$

$1-\frac{{\xi }^2}{15},$

where,

$\xi$

$=$

$(\frac{2\pi }{3}{)}^{1/2}{r}_0.$

This means,

$x(r_0)$	$=$	${\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}[1-(\frac{2\pi }{45}\left)r_0^2\right],$
$\Rightarrow \frac{dx}{d{r}_0}$	$=$	$-{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}\left(\frac{4\pi }{45}\right)r_0.$

Note as well that,

$\frac{d\ln x}{d\ln r_0}$	$=$	$-{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}\left(\frac{4\pi }{45}\right)\frac{r_0^2}{x}$
	$=$	$\left(\frac{4\pi }{45}\right)\frac{3{\xi }^2}{2\pi }[\frac{{\xi }^2}{15}-1{]}^{-1}.$

Finite-Difference Representation of $\delta r$ Once a "guess" for the fractional displacement vector, $(\delta r{)}_j=[x\cdot r_0{]}_j,$ has been specified, we recognize that the perturbed location of each radial shell is given by the expression, $r_j$ $=$ $(r_0{)}_j+(\delta r{)}_j.$

---

Parabolic Displacement Function w/ ${\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}=-0.001$
Shell	$m$	$r_0$	$\frac{x}{{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}}$	$r_j$	Analytic		FD			$P^{*}$	$-\frac{d{P}^{*}}{dm}$	$-4\pi (r^{*}{)}^2\frac{d{P}^{*}}{dm}=g_0$
					${\rho }_0$	$d_{\mathrm{a}\mathrm{n}\mathrm{a}\mathrm{l}\mathrm{y}\mathrm{t}\mathrm{i}\mathrm{c}}$	$[{\rho }_{j-1/2}{]}_0$	$d_{j-1/2}$	${\rho }_{j-1/2}$
0	0.000000	0.000000	1.000000	0.000000	1.000000	+ 0.003000	---	---	---	1.000000	$\mathrm{\infty }$	0.000000
½	---	---	---	---	---	---	0.995868	0.003004	0.998860	1.000000	$\mathrm{\infty }$	0.000000
1	0.001039	0.062922	0.999447	0.062859	0.993123	0.002997	---	---	---	0.991754	5.275715	0.262476
1½	---	---	---	---	---	---	0.981896	0.002999	0.984840	1.000000	$\mathrm{\infty }$	0.000000
2	0.008211	0.125843	0.997789	0.125718	0.972886	0.002989	---	---	---	0.967552	2.605474	0.518508
2½	---	---	---	---	---	---	0.955465	0.002988	0.958319	1.000000	$\mathrm{\infty }$	0.000000
3	0.027155	0.188765	0.995025	0.188577	0.940420	0.002975	---	---	---	0.928937	1.701968	0.762083
3½	---	---	---	---	---	---	0.917695	0.002971	0.920421	1.000000	$\mathrm{\infty }$	0.000000
4	0.062586	0.251686	0.991155	0.251437	0.897462	0.002956	---	---	---	0.878252	1.241164	0.988001
⋮
95	6.769823	5.977546	- 3.988996	6.001390	0.0002917	- 0.021945	---	---	---	0.000057	0.000422	0.189466
95½	---	---	---	---	---	---	0.0002844	- 0.021852	0.0002782	1.000000	$\mathrm{\infty }$	0.000000
96	6.777942	6.040467	- 4.094580	6.065200	0.0002773	- 0.022473	---	---	---	0.000054	0.000405	0.185762
96½	---	---	---	---	---	---	0.002705	- 0.022366	0.0002644	1.000000	$\mathrm{\infty }$	0.000000
97	6.785828	6.103389	- 4.201270	6.129031	0.0002638	- 0.023006	---	---	---	0.000051	0.000389	0.182163
97½	---	---	---	---	---	---	0.0002574	- 0.022884	0.0002515	1.000000	$\mathrm{\infty }$	0.000000
98	6.793488	6.166310	- 4.309066	6.192881	0.0002511	- 0.023545	---	---	---	0.000048	0.000374	0.178666
98½	---	---	---	---	---	---	0.0002450	- 0.023408	0.0002393	1.000000	$\mathrm{\infty }$	0.000000
99	6.800930	6.229232	- 4.417967	6.256752	0.0002391	- 0.024090	---	---	---	0.000045	0.000359	0.175267

---

Density perturbation

Hence, from the linearized continuity equation,

$d$	$=$	$-r_0\frac{dx}{d{r}_0}-3x$
	$=$	${\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}\left\{\right(\frac{4\pi }{45})r_0^2-3[1-\left(\frac{2\pi }{45}\right)r_0^2\left]\right\}$
	$=$	${\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}\left\{2\right(\frac{2\pi }{45})r_0^2-3+3(\frac{2\pi }{45}\left)r_0^2\right\}$
	$=$	${\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}\left\{\right(\frac{2\pi }{9})r_0^2-3\}.$

The solid black curve in the figure shown here, on the right, displays this analytically specified density perturbation, $d(m)$, for the case ${\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}=-0.001$.

Finite-Difference Representation of $d$ Our finite-difference representation of the mass-density at each radial shell in the equilibrium configuration (subscript "0") is, $[{\overline{\rho }}_{j-1/2}{]}_0$ $=$ $[m_j-m_{j-1}{]}_0[\frac{4\pi }{3}(r_j^3-r_{j-1}^3){]}_0^{-1}.$ After perturbing the radial location of each shell — that is, after setting $r_j=(r_0{)}_j+[x\cdot r_0{]}_j$ — the resulting finite-difference representation of the perturbed mass density of each shell is given by the expression, ${\overline{\rho }}_{j-1/2}$ $=$ $[m_j-m_{j-1}{]}_0[\frac{4\pi }{3}(r_j^3-r_{j-1}^3){]}^{-1}.$ (Note that we retain a subscript "0" on the mass, $m_j$, because it serves as our Lagrangian identifier for each shell.) The fractional density perturbation at each discrete shell is, then, $d_{j-1/2}\equiv [\frac{\delta \rho }{{\rho }_0}{]}_{j-1/2}=\{\frac{{\overline{\rho }}_{j-1/2}}{[{\overline{\rho }}_{j-1/2}{]}_0}-1\}$ $=$ $\frac{(r_j^3-r_{j-1}^3{)}_0}{(r_j^3-r_{j-1}^3)}-1.$ The red dots in the above "density perturbation" figure display how $d_j$ varies with mass shell when $x$ is specified by the parabolic dependence on radius. The dots lie virtually on top of the solid black curve in this figure, indicating that our finite-difference representation of the perturbed mass density matches well the analytically specified perturbed mass density.

And, from linearized entropy conservation,

$p$	$=$	${\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}\left\{\right(\frac{2\pi }{9})r_0^2-3\}{\gamma }_c$
	$=$	${\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}\left\{\right(\frac{2\pi }{9})r_0^2-3\}\frac{6}{5}$
$\Rightarrow \frac{dp}{d{r}_0}$	$=$	${\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}\left(\frac{8\pi }{15}\right)r_0.$

Now, from below we find that,

$\frac{P_0}{{\rho }_0}=\frac{P^{*}}{{\rho }^{*}}$	$=$	$(1+\frac{{\xi }^2}{3}{)}^{-3}(1+\frac{{\xi }^2}{3}{)}^{5/2}$
	$=$	$(1+\frac{{\xi }^2}{3}{)}^{-1/2},$
$g_0=\frac{m}{(r^{*}{)}^2}$	$=$	$\left(\frac{2\cdot 3}{\pi }{)}^{1/2}\right[{\xi }^3(1+\frac{1}{3}{\xi }^2{)}^{-3/2}]\left[\right(\frac{3}{2\pi }{)}^{1/2}\xi {]}^{-2}$
	$=$	$\left(\frac{8\pi }{3}{)}^{1/2}\right[\xi (1+\frac{1}{3}{\xi }^2{)}^{-3/2}].$

Hence, the linearized Euler + Poisson Equations expression gives,

${\omega }^2{r}_{0}x$	$=$	$\frac{P_0}{{\rho }_0}\frac{dp}{d{r}_0}-(4x+p)g_0$
	$=$	${\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}(1+\frac{{\xi }^2}{3}{)}^{-1/2}(\frac{8\pi }{15})r_0-\{4{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}[1-(\frac{2\pi }{45}\left)r_0^2\right]+{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}\left[\right(\frac{2\pi }{9})r_0^2-3]\frac{6}{5}\left\}\right(\frac{8\pi }{3}{)}^{1/2}\left[\xi \right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}]$
$\Rightarrow \left(\frac{{\omega }^2}{{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}}\right)(1+\frac{1}{3}{\xi }^2{)}^{3/2}{r}_{0}x$	$=$	$(1+\frac{{\xi }^2}{3})\left(\frac{8\pi }{15}\right)r_0-\{4-(\frac{8\pi }{45})r_0^2+(\frac{12\pi }{45})r_0^2-\frac{18}{5}\}(\frac{8\pi }{3}{)}^{1/2}\xi$
	$=$	$(1+\frac{{\xi }^2}{3})\left(\frac{8\pi }{15}\right)r_0-[\frac{2}{5}+(\frac{4\pi }{45}\left)r_0^2\right](\frac{8\pi }{3}{)}^{1/2}\xi$
	$=$	$(1+\frac{{\xi }^2}{3})\left(\frac{8\pi }{15}\right)(\frac{3}{2\pi }{)}^{1/2}\xi -[\frac{2}{5}+\left(\frac{4\pi }{45}\right)\left(\frac{3}{2\pi }\right){\xi }^2\left]\right(\frac{8\pi }{3}{)}^{1/2}\xi$
	$=$	$(\frac{2^5\pi }{3\cdot 5^2}{)}^{1/2}\xi -\frac{2}{5}(\frac{8\pi }{3}{)}^{1/2}\xi +(\frac{2^5\pi }{3^3\cdot 5^2}{)}^{1/2}{\xi }^3-(\frac{2^5\pi }{3^3\cdot 5^2}{)}^{1/2}{\xi }^3$
	$=$	$0$        Exactly!

Step 6

STEP 6:   Given that when a proper solution has been obtained,

$\frac{d^{2}r}{d{t}^2}$

$\to$

$-{\omega }^2\cdot {\delta r}_i,$

at each radial shell we can determine what the value of ${\omega }^2$ would be as a result of our ${\delta r}_i$ guess by rewriting the

Euler + Poisson Equations
$\frac{d^{2}r}{d{t}^2}=-4\pi r^2\frac{dP}{d{M}_r}-\frac{G{M}_r}{r^2}$

to read,

$+\left[\frac{{\omega }_i^2}{G{\rho }_c}\right]$

$\approx$

$\frac{1}{{\delta r}_i\cdot r_i^2}[4\pi (r_i^{*}{)}^4\cdot \frac{d{P}^{*}}{d{m}^{*}}+m^{*}].$

Euler + Poisson Equations $\left[\frac{K_c^{1/2}}{G^{1/2}{\rho }_c^{2/5}}\right]\frac{d^2{r}^{*}}{d{t}^2}$ $=$ $-4\pi (r^{*}{)}^2[\frac{K_c^{1/2}}{G^{1/2}{\rho }_c^{2/5}}{]}^2\frac{d{P}^{*}}{dm}\left[K_c{\rho }_c^{6/5}\right]\left[\frac{G^{3/2}{\rho }_c^{1/5}}{K_c^{3/2}}\right]-\frac{Gm}{(r^{*}{)}^2}\left[\frac{K_c^{3/2}}{G^{3/2}{\rho }_c^{1/5}}\right][\frac{G^{1/2}{\rho }_c^{2/5}}{K_c^{1/2}}{]}^2$ $\Rightarrow \left[\frac{K_c^{1/2}}{G^{1/2}{\rho }_c^{2/5}}\right]\frac{d^2{r}^{*}}{d{t}^2}$ $=$ $-4\pi (r^{*}{)}^2[K_c^{1/2}{G}^{1/2}{\rho }_c^{3/5}]\frac{d{P}^{*}}{dm}-\frac{m}{(r^{*}{)}^2}[K_c^{1/2}{G}^{1/2}{\rho }_c^{3/5}]$ $\Rightarrow -\left[\frac{1}{G{\rho }_c}\right]\frac{d^2{r}^{*}}{d{t}^2}$ $=$ $4\pi (r^{*}{)}^2\frac{d{P}^{*}}{dm}+\frac{m^{*}}{(r^{*}{)}^2}$ $\frac{d^{2}r}{d{t}^2}$ $\to$ $\frac{d}{dt}[(i\omega )r_{0}x{e}^{i\omega t}]=-{\omega }^2{r}_{0}x{e}^{i\omega t}$ $r^2\frac{dP}{dm}$ $\to$ $r_0^2[1+x{e}^{i\omega t}{]}^2\{\frac{d{P}_0}{dm}[1+p{e}^{i\omega t}]+P_0{e}^{i\omega t}\frac{dp}{dm}\}\approx r_0^2\frac{d{P}_0}{dm}[1+(2x+p)e^{i\omega t}]+P_0{r}_0^2{e}^{i\omega t}\frac{dp}{dm}$ $\frac{Gm}{r^2}$ $\to$ $\frac{Gm}{r_0^2}[1+x{e}^{i\omega t}{]}^{-2}\approx \frac{Gm}{r_0^2}[1-2x{e}^{i\omega t}].$

After introducing a perturbation, we find that …

RHS	$=$	$4\pi [r^{*}{]}^2\frac{d{P}^{*}}{dm}+m^{*}[r^{*}{]}^{-2}$
	$=$	$4\pi r_0^2[1+\frac{\delta r}{r_0}{]}^2\frac{d(P_0+\delta P)}{dm}+\frac{m^{*}}{r_0^2}[1+\frac{\delta r}{r_0}{]}^{-2}$
	$\approx$	$4\pi r_0^2[1+2x]\frac{d(P_0+\delta P)}{dm}+g_0[1-2x]$
	$\approx$	$4\pi r_0^2\left[\frac{d(P_0+\delta P)}{dm}\right]+g_0+4\pi r_0^2\left[2x\right]\frac{d(P_0+{\xcancel{\delta P}}^{\mathrm{s}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{l}})}{dm}-2x{g}_0$
	$\approx$	$4\pi r_0^2\left[\frac{d(P_0+\delta P)}{dm}\right]+g_0+2x[4\pi r_0^2\frac{d{P}_0}{dm}-g_0]$
	$\approx$	$\left\{4\pi r_0^2\right[\frac{d(P_0+\delta P)}{dm}]+g_0\}-4x{g}_0.$

This "RHS" expression must be paired with …

LHS

$=$

$\left[\frac{{\omega }_i^2}{G{\rho }_c}\right]\delta r_i=\left[\frac{{\omega }_i^2}{G{\rho }_c}\right]x{r}_0.$

The term inside the curly braces on the RHS is easy to evaluate inside our finite-difference scheme. For our example parabolic displacement function, we expect this term to give,

$\left\{4\pi r_0^2\right[\frac{d(P_0+\delta P)}{dm}]+g_0\}$	$\approx$	$\frac{P_0}{{\rho }_0}\cdot \frac{dp}{d{r}_0}-p{g}_0=4x{g}_0.$
	$=$	${\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}(1+\frac{{\xi }^2}{3}{)}^{-1/2}(\frac{8\pi }{15})r_0-\{{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}\left[\right(\frac{2\pi }{9})r_0^2-3]\frac{6}{5}\left\}\right(\frac{8\pi }{3}{)}^{1/2}\left[\xi \right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}]$
	$=$	$-{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}\left[\xi \right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}\left]r_0^2\right(\frac{2^7{\pi }^3}{3^3\cdot 5^2}{)}^{1/2}$
	$=$	${\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}\left(\frac{2^5\cdot 3^3\pi }{5^2}{)}^{1/2}\xi \right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}+{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}\left(\frac{2^3\pi }{3\cdot 5}\right)(1+\frac{{\xi }^2}{3}{)}^{-1/2}{r}_0-{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}{r}_0^2(\frac{2^7{\pi }^3}{3^3\cdot 5^2}{)}^{1/2}\xi (1+\frac{1}{3}{\xi }^2{)}^{-3/2}$
	$=$	${\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}(1+\frac{1}{3}{\xi }^2{)}^{-3/2}\{(\frac{2^5\cdot 3^3\pi }{5^2}{)}^{1/2}\xi +(\frac{2^5\pi }{3\cdot 5^2}{)}^{1/2}(1+\frac{{\xi }^2}{3})\xi -\left(\frac{2^5\pi }{3\cdot 5^2}{)}^{1/2}{\xi }^3\right\}$
	$=$	${\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}(1+\frac{1}{3}{\xi }^2{)}^{-3/2}\{\left[\right(\frac{2^5\cdot 3^3\pi }{5^2}{)}^{1/2}+\left(\frac{2^5\pi }{3\cdot 5^2}{)}^{1/2}\right]\xi +\left[\right(\frac{2^5\pi }{3^3\cdot 5^2}{)}^{1/2}-\left(\frac{2^5\cdot 3^2\pi }{3^3\cdot 5^2}{)}^{1/2}\right]{\xi }^3\}$
	$=$	${\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}(1+\frac{1}{3}{\xi }^2{)}^{-3/2}\{(\frac{2^7\pi }{3}{)}^{1/2}\xi -2(\frac{2^5\pi }{3^3\cdot 5^2}{)}^{1/2}{\xi }^3\}.$

For comparison, note that this expression is identical to the expression for $4x{g}_0$, namely,

$4x{g}_0$	$=$	$4{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}[1-\frac{{\xi }^2}{15}]\left(\frac{8\pi }{3}{)}^{1/2}\right[\xi (1+\frac{1}{3}{\xi }^2{)}^{-3/2}]$
	$=$	${\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}(1+\frac{1}{3}{\xi }^2{)}^{-3/2}\{(\frac{2^7\pi }{3}{)}^{1/2}\xi -(\frac{2^7\pi }{3^3\cdot 5^2}{)}^{1/2}{\xi }^3\}.$

Step 7

STEP 7:   Alternatively, from our summary set of linearized equations, we expect …

${\omega }^2{r}_{0}x$	$\approx$	$\frac{P_0}{{\rho }_0}\frac{dp}{d{r}_0}-(4x+p)g_0$
	$=$	$\frac{P_0}{{\rho }_0}\cdot \frac{d(\delta P/P_0)}{d{r}_0}-(\frac{4\delta r}{r_0}+\frac{\delta P}{P_0})\frac{Gm}{r_0^2}$
	$=$	$\frac{P_0}{{\rho }_0}[\frac{1}{P_0}\cdot \frac{d(\delta P)}{d{r}_0}-\frac{\delta P}{P_0^2}\frac{d{P}_0}{d{r}_0}]+\left[\right(1-\frac{4\delta r}{r_0})-(1+\frac{\delta P}{P_0}\left)\right]\frac{Gm}{r_0^2}$
	$=$	$\frac{1}{{\rho }_0}[\frac{d(\delta P)}{d{r}_0}-\frac{\delta P}{P_0}\frac{d{P}_0}{d{r}_0}]+(1-\frac{4\delta r}{r_0})\frac{Gm}{r_0^2}+(1+\frac{\delta P}{P_0})\frac{1}{{\rho }_0}\cdot \frac{d{P}_0}{d{r}_0}$
	$=$	$\frac{1}{{\rho }_0}\left[\frac{d(P_0+\delta P)}{d{r}_0}\right]+[1+\frac{\delta r}{r_0}{]}^{-4}\frac{Gm}{r_0^2}$
	$=$	$\frac{1}{{\rho }_0}\left[\frac{dP}{d{r}_0}\right]+\frac{r_0^4}{r^4}\frac{Gm}{r_0^2}$
	$=$	$\frac{r_0^2}{r^4}[4\pi r^4\cdot \frac{dP}{dm}+Gm]$
$\Rightarrow {\omega }^2$	$\approx$	$\frac{1}{\delta r\cdot r^2}[4\pi r^4\cdot \frac{dP}{dm}+Gm]\frac{r_0^2}{r^2}$

Step 8

STEP 8:   Finally, we should guess a new eigenvector (then guess again, and again, and again …) until ${\omega }^2$ settles down to have the same value at all radial locations.

Try Again, With Detailed Example

Model Amodel2

Under the "AModel2FD" tab of an Excel spreadsheet titled, "qAndNuMaxAug21", we have constructed a discrete model of the core of a bipolytrope that has the following properties:

Model A
${\mu }_e/{\mu }_c$	0.31
$\nu \equiv \frac{M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$	0.337217
$q\equiv \frac{r_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{R_{\mathrm{t}\mathrm{o}\mathrm{t}}}$	0.0755023
Core Properties (at interface)
${\xi }_i$	9.0149598
${\theta }_i$	0.1886798
$\frac{d\theta }{d\xi }{|}_i$	-0.0201845
$M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=\left(\frac{2\cdot 3}{\pi }{)}^{1/2}\right[{\xi }^3(1+\frac{1}{3}{\xi }^2{)}^{-3/2}{]}_i$	6.8009303
$r_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=(\frac{3}{2\pi }{)}^{1/2}{\xi }_i$	6.2292317

We have divided this model into 100 equally spaced radial zones (center and surface included), that is, each with the following radial-shell thickness:   $\delta \xi \equiv {\xi }_i/99=0.0910602$, and $\delta r=r_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}/99=0.0629215$. Analytic expressions providing the physical properties of our $(n_c,n_e)=(5,1)$ bipolytropes at each radial shell have been drawn from an accompanying chapter — for example …

$r^{*}$	$=$	$(\frac{3}{2\pi }{)}^{1/2}\xi ,$
${\rho }^{*}$	$=$	$(1+\frac{{\xi }^2}{3}{)}^{-5/2},$
$P^{*}$	$=$	$(1+\frac{{\xi }^2}{3}{)}^{-3},$
$m$	$=$	$\left(\frac{2\cdot 3}{\pi }{)}^{1/2}\right[{\xi }^3(1+\frac{1}{3}{\xi }^2{)}^{-3/2}].$

We note as well that, in equilibrium,

$g_0\equiv \frac{m}{(r^{*}{)}^2}$

$=$

$-4\pi (r^{*}{)}^2\cdot \frac{d{P}^{*}}{dm}=-\frac{1}{{\rho }^{*}}\cdot \frac{d{P}^{*}}{d{r}^{*}}.$

Hence, in equilibrium we have,

$\frac{d{P}^{*}}{dm}$	$=$	$\frac{1}{4\pi (r^{*}{)}^2{\rho }^{*}}\left(\frac{\xi }{r^{*}}\right)\frac{d{P}^{*}}{d\xi }$
	$=$	$\frac{\xi }{4\pi (r^{*}{)}^3{\rho }^{*}}\cdot [-3(1+\frac{{\xi }^2}{3}{)}^{-4}\frac{2\xi }{3}]$
	$=$	$\frac{1}{4\pi }\left[\right(\frac{3}{2\pi }{)}^{1/2}\xi {]}^{-3}\left[\right(1+\frac{{\xi }^2}{3}{)}^{-5/2}{]}^{-1}[-2{\xi }^2(1+\frac{{\xi }^2}{3}{)}^{-4}]$
	$=$	$-\frac{1}{2\pi }\left[\right(\frac{3}{2\pi }{)}^{-3/2}{\xi }^{-3}\left]\right[(1+\frac{{\xi }^2}{3}{)}^{5/2}]\left[{\xi }^2\right(1+\frac{{\xi }^2}{3}{)}^{-4}]$
	$=$	$-(\frac{2^3{\pi }^3}{3^3}\cdot \frac{1}{2^2{\pi }^2}{)}^{1/2}(1+\frac{{\xi }^2}{3}{)}^{-3/2}{\xi }^{-1}$
	$=$	$-\left(\frac{2\pi }{3^3}{)}^{1/2}\right(1+\frac{{\xi }^2}{3}{)}^{-3/2}{\xi }^{-1};$

and,

$4\pi (r^{*}{)}^2\cdot \frac{d{P}^{*}}{dm}$	$=$	$-2^2\pi \left[\right(\frac{3}{2\pi }{)}^{1/2}\xi {]}^2\left(\frac{2\pi }{3^3}{)}^{1/2}\right(1+\frac{{\xi }^2}{3}{)}^{-3/2}{\xi }^{-1}$
	$=$	$-2\left(\frac{2\pi }{3}{)}^{1/2}\right(1+\frac{{\xi }^2}{3}{)}^{-3/2}\xi ;$

and,

$4\pi (r^{*}{)}^4\cdot \frac{d{P}^{*}}{dm}$	$=$	$-\left[\right(\frac{3}{2\pi }{)}^{1/2}\xi {]}^{2}2\left(\frac{2\pi }{3}{)}^{1/2}\right(1+\frac{{\xi }^2}{3}{)}^{-3/2}\xi$
	$=$	$-\left(\frac{2\cdot 3}{\pi }{)}^{1/2}\right(1+\frac{{\xi }^2}{3}{)}^{-3/2}{\xi }^3.$

The following table provides a sample of variable values in the central region (shells 0 - 4) and near the interface (shells 95 - 99) of the equilibrium configuration.

Analytically Determined Physical Properties at Various Radial Shells
Shell	$\xi$	$m=-4\pi (r^{*}{)}^4\frac{d{P}^{*}}{dm}$	$r^{*}$	${\rho }^{*}$	$P^{*}$	$-\frac{d{P}^{*}}{dm}$	$-4\pi (r^{*}{)}^2\frac{d{P}^{*}}{dm}=g_0$
0	0.000000	0.000000	0.000000	1.000000	1.000000	$\mathrm{\infty }$	0.000000
1	0.091060	0.001039	0.062922	0.993123	0.991754	5.275715	0.262476
2	0.182120	0.008211	0.125843	0.972886	0.967552	2.605474	0.518508
3	0.273181	0.027155	0.188765	0.940420	0.928937	1.701968	0.762083
4	0.364241	0.062586	0.251686	0.897462	0.878252	1.241164	0.988001
⋮
95	8.650719	6.769823	5.977546	0.000292	0.000057	0.000422	0.189466
96	8.741779	6.777942	6.040467	0.000277	0.000054	0.000405	0.185762
97	8.832839	6.785828	6.103389	0.000264	0.000051	0.000389	0.182163
98	8.923900	6.793488	6.166310	0.000251	0.000048	0.000374	0.178666
99	9.014960	6.800930	6.229232	0.000239	0.000045	0.000359	0.175267

This last expression is precisely the same (in magnitude, but opposite in sign) as the expression that we have presented for $m$. If we therefore return to STEP 6, we appreciate that the right-hand side of the "eigenvector" expression,

$+\left[\frac{{\omega }^2}{G{\rho }_c}\right]\left(\frac{\delta r^{*}}{r^{*}}\right)(r^{*})$

$\approx$

$[4\pi (r^{*}{)}^2\cdot \frac{d{P}^{*}}{dm}+\frac{m}{(r^{*}{)}^2}],$

goes to zero at every radial shell if the configuration is in equilibrium.

ASIDE:  Next, we will focus on building a discrete, finite-difference representation of each equilibrium model, as well as of this "eigenvector" expression. In doing so, we will immediately find that the two terms on the right-hand-side do not exactly sum to zero, even for an equilibrium configuration. We should nevertheless try to construct a finite-difference representation for which the two terms cancel to a relatively high degree of precision. We have considered rewriting this "eigenvector" expression in the form, $+\left[\frac{{\omega }^2}{G{\rho }_c}\right]\left(\frac{\delta r^{*}}{r^{*}}\right)(r^{*}{)}^3$ $\approx$ $[4\pi (r^{*}{)}^4\cdot \frac{d{P}^{*}}{dm}+m],$ because this form isolates the Lagrangian mass, $m$, which is time-invariant. But as the numbers in the third column of the above table illustrate, the terms on the right-hand-side vary by several orders of magnitude as we move from shell 1 to shell 100. If it is written instead in the form that we have initially suggested, namely, $+\left[\frac{{\omega }^2}{G{\rho }_c}\right]\left(\frac{\delta r^{*}}{r^{*}}\right)(r^{*})$ $\approx$ $[4\pi (r^{*}{)}^2\cdot \frac{d{P}^{*}}{dm}+\frac{m}{(r^{*}{)}^2}],$ then, as is illustrated by the numbers in the last column of the table and by the following figure, the terms on the right-hand-side vary by no more than one order of magnitude as we move from the center to the surface of the configuration. This choice should facilitate cancellation to a higher degree of precision in our finite-difference-based model.

Even Simpler Core

STEP 0:   We construct a finite-difference representation of the initial (unperturbed) equilibrium configuration by dividing the model into 99 radial zones that are equally spaced in $0\le \xi \le {\xi }_i$. The initial radial coordinate of each zone and the corresponding initial enclosed mass are given, respectively, by the expressions …

$[r_j{]}_0$

$=$

$(\frac{3}{2\pi }{)}^{1/2}{\xi }_j,$

and,

$m_j$

$=$

$\left(\frac{2\cdot 3}{\pi }{)}^{1/2}\right[{\xi }^3(1+\frac{1}{3}{\xi }^2{)}^{-3/2}].$

The mass, $m_j$, will serve as our Lagrangian coordinate; that is, we will perturb the model by modifying the radial location of each shell while fixing the enclosed mass.

STEP 1:   Guess the eigenvector, ${\delta r}_i$, remembering that a reasonably good trial eigenfunction for the core is one that has a "parabolic dependence on the radius," namely,

$\frac{x}{{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}}$

$=$

$1-\frac{{\xi }^2}{15},$

where,

$\xi$

$=$

$(\frac{2\pi }{3}{)}^{1/2}{r}_0.$

This means,

$x(r_0)$	$=$	${\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}[1-(\frac{2\pi }{45}\left)r_0^2\right],$
$\Rightarrow \frac{dx}{d{r}_0}$	$=$	$-{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}\left(\frac{4\pi }{45}\right)r_0.$

Once a "guess" for the fractional displacement vector, $(\delta r{)}_j=[x\cdot r_0{]}_j,$ has been specified, we recognize that the perturbed location of each radial shell is given by the expression,

$r_j$

$=$

$(r_0{)}_j[1+x_j].$

STEP 2:   Our finite-difference representation of the mass-density at each radial shell in the equilibrium configuration (subscript "0") is,

$[{\overline{\rho }}_{j-1/2}{]}_0$

$=$

$[m_j-m_{j-1}{]}_0[\frac{4\pi }{3}(r_j^3-r_{j-1}^3){]}_0^{-1}.$

After perturbing the radial location of each shell — that is, after setting $r_j=(r_0{)}_j+[x\cdot r_0{]}_j$ — the resulting finite-difference representation of the perturbed mass density of each shell is given by the expression,

${\overline{\rho }}_{j-1/2}$

$=$

$[m_j-m_{j-1}{]}_0[\frac{4\pi }{3}(r_j^3-r_{j-1}^3){]}^{-1}.$

(Note that we retain a subscript "0" on the mass, $m_j$, because it serves as our Lagrangian identifier for each shell.)

STEP 3:   The pressure can be determined in the equilibrium configuration (subscript "0") and after the perturbation from knowledge of the density and the chosen adiabatic index, ${\gamma }_c=6/5$, via knowledge of the (fixed) specific entropy, namely,

$[P_{j-1/2}{]}_0$

$=$

$[{\overline{\rho }}_{j-1/2}{]}_0^{{\gamma }_c}\cdot \exp [\frac{\mu (\gamma -1)s}{\mathfrak{\Re }}],$

and,

$P_{j-1/2}$

$=$

$[{\overline{\rho }}_{j-1/2}{]}^{{\gamma }_c}\cdot \exp [\frac{\mu (\gamma -1)s}{\mathfrak{\Re }}].$

The pressure perturbation can therefore be obtained from the simple difference,

$[\delta P{]}_{j-1/2}$

$=$

$P_{j-1/2}-[P_{j-1/2}{]}_0.$

Displacement Function

STEP 4:   If, in the context of our above discussion of the perturbed "Euler + Poisson Equations", we set LHS = RHS, we obtain,

$\left[\frac{{\omega }_i^2}{G{\rho }_c}\right]x{r}_0$	$\approx$	$\left\{4\pi r_0^2\right[\frac{d(P_0+\delta P)}{dm}]+g_0\}-4x{g}_0$
	$=$	$\left\{4\pi r_0^2\right[\frac{d(\delta P)}{dm}\left]\right\}-4x{g}_0$
$\Rightarrow x$	$\approx$	$\left\{4\pi r_0^2\right[\frac{d(\delta P)}{dm}\left]\right\}\left\{\right[\frac{{\omega }_i^2}{G{\rho }_c}]r_0+4g_0{\}}^{-1}.$

So for a specified value of the square of the oscillation frequency, $[{\omega }^2/(G{\rho }_c)]$ — same value for all shells — the strategy should be to (a) guess $x_j$; (b) evaluate the right-hand-side of this last expression; (c) if the RHS does not equal the "guessed" eigenvector, $x_j$, then you need to guess a new eigenvector; (d) repeat!

In the figure shown here on the right, the black curve displays the variation with mass, $m$, of the (parabolic-shaped) displacement function, $x/{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}$, that served as our initial "guess;" while the red dots show how the right-hand side of this last expression varies with $m$ for the case, ${\omega }_i^2=0$. They lie almost exactly on top of one another, as hoped/expected.

Transition

In transitioning from the core to the envelope, all of the above (green) STEPS will remain the same except for the 1^st Law's treatment of the pressure. Following along the lines of our Ramblings idea exchange with Patrick Motl, the mass-density is generically related to the pressure via the expression,

$\frac{s}{\mathrm{\Re }/\overline{\mu }}$	$=$	$\frac{1}{({\gamma }_g-1)}\ln \left[\frac{P}{({\gamma }_g-1){\rho }^{{\gamma }_g}}\right]$
$\Rightarrow \frac{s\overline{\mu }({\gamma }_g-1)}{\mathrm{\Re }}$	$=$	$\ln \left[\frac{P}{({\gamma }_g-1){\rho }^{{\gamma }_g}}\right]$
$\Rightarrow P$	$=$	$({\gamma }_g-1){\rho }^{{\gamma }_g}\exp \left[\frac{s\overline{\mu }({\gamma }_g-1)}{\mathrm{\Re }}\right].$

Given that, in polytropic configurations for which we make the association, ${\gamma }_g=(n+1)/n$, the pressure is related to the density via the expression, $P=K{\rho }^{{\gamma }_g}$, we appreciate that,

$K$

$=$

$({\gamma }_g-1)\exp \left[\frac{s\overline{\mu }({\gamma }_g-1)}{\mathrm{\Re }}\right].$

Core

For the core, we set ${\gamma }_g=6/5$. Hence,

$P$

$=$

$K_c{\rho }^{6/5}.$

Normalizing the pressure and the density as we have in a closely related discussion of the structure of bipolytropes, we have throughout the core,

$P^{*}=\frac{P}{K_c{\rho }_0^{6/5}}$

$=$

$({\rho }^{*}{)}^{6/5}.$

Now, in this normalized expression we see that the polytropic constant for the core is, $K_c^{*}=1$. This means that the specific entropy of all the fluid in the core is given by the expression,

$1$	$=$	$\frac{1}{5}\exp \left[\frac{s_c^{*}{\overline{\mu }}_c}{5\mathrm{\Re }}\right]$
$\Rightarrow \frac{s_c^{*}{\overline{\mu }}_c}{\mathrm{\Re }}$	$=$	$5\ln (5).$

Envelope

For the envelope, we set ${\gamma }_g=2$. Hence,

$P$

$=$

$K_e{\rho }^2.$

Adopting the same pressure and density normalizations as used in the core, we have throughout the envelope,

$P^{*}=\frac{P}{K_c{\rho }_0^{6/5}}$

$=$

$\left(\frac{K_e}{K_c}\right){\rho }_0^{4/5}({\rho }^{*}{)}^2.$

Now, at the interface of the equilibrium model, we know that,

$\frac{K_e}{K_c}$

$=$

${\rho }_0^{-4/5}(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}={\rho }_0^{-4/5}(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}[1+\frac{{\xi }_i^2}{3}{]}^2.$

Hence, throughout the envelope,

$P^{*}$

$=$

$\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\right[1+\frac{{\xi }_i^2}{3}{]}^2({\rho }^{*}{)}^2,$

in which case we find,

$K_e^{*}$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\right[1+\frac{{\xi }_i^2}{3}{]}^2$
$\Rightarrow \frac{s_e^{*}{\overline{\mu }}_e}{\mathrm{\Re }}$	$=$	$\ln \left\{\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}[1+\frac{{\xi }_i^2}{3}{]}^2\}.$

STEP 3 Clarification

It is critically important to appreciate that the manner in which the pressure is determined at discrete locations in our finite-difference model must be different in the envelope and the core. Keeping in mind that, in general,

$P$

$=$

$({\gamma }_g-1){\rho }^{{\gamma }_g}\exp \left[\frac{s\overline{\mu }({\gamma }_g-1)}{\mathrm{\Re }}\right],$

for $(n_c,n_e)=(5,1)$ bipolytropes, we have …

Core $({\gamma }_g=6/5)$:       $P_{j-1/2}$	$=$	$\frac{1}{5}[{\overline{\rho }}_{j-1/2}{]}^{6/5}\cdot \exp [\frac{{\mu }_c{s}_c^{*}}{5\mathfrak{\Re }}]=[{\overline{\rho }}_{j-1/2}{]}^{6/5}.$
Envelope $({\gamma }_g=2)$:       $P_{j-1/2}$	$=$	$[{\overline{\rho }}_{j-1/2}{]}^2\cdot \exp [\frac{{\mu }_e{s}_e^{*}}{\mathfrak{\Re }}]=(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}[1+\frac{{\xi }_i^2}{3}{]}^2[{\overline{\rho }}_{j-1/2}{]}^2.$

Interface

Drawing from our chapter in which the bipolytrope is constructed, we note the following determination of the density and pressure at the interface, as viewed from the perspective of the core and, separately, from the perspective of the envelope.

Properties at the (Unperturbed) Interface
Core $\frac{s_c^{*}{\overline{\mu }}_c}{\mathrm{\Re }}$ $=$ $5\ln (5)$	Envelope $\frac{s_e^{*}{\overline{\mu }}_e}{\mathrm{\Re }}$ $=$ $\ln \left\{\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}\}$
$({\rho }^{*}{)}_i$ $=$ ${\theta }_i^5$ $(P^{*}{)}_i$ $=$ $\frac{1}{5}({\rho }^{*}{)}_i^{6/5}\exp [\frac{s_c{\overline{\mu }}_c}{5\mathrm{\Re }}]$   $=$ $({\rho }^{*}{)}_i^{6/5}$   $=$ ${\theta }_i^6$	$({\rho }^{*}{)}_i$ $=$ $\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5$ $(P^{*}{)}_i$ $=$ $({\rho }^{*}{)}_i^2\exp [\frac{s_e{\overline{\mu }}_e}{\mathrm{\Re }}]$   $=$ $({\rho }^{*}{)}_i^2(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}$   $=$ ${\theta }_i^6$

---

In what follows, we will continue to focus on the interface but, for simplicity, will drop the "i" subscript; instead, we will use a "0" subscript to indicate a property at the unperturbed interface.

Properties at the Perturbed Interface
Core $\frac{s_c^{*}{\overline{\mu }}_c}{\mathrm{\Re }}$ $=$ $5\ln (5)$	Envelope $\frac{s_e^{*}{\overline{\mu }}_e}{\mathrm{\Re }}$ $=$ $\ln \left\{\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}\}$
${\rho }_c^{*}=({\rho }_c^{*}{)}_0+(\delta \rho {)}_c$ $=$ ${\theta }_i^5+(\delta \rho {)}_c$ $P_c^{*}=(P_c^{*}{)}_0+(\delta P{)}_c$ $=$ $[{\rho }_c^{*}{]}^{6/5}$   $=$ $[{\theta }_i^5+(\delta \rho {)}_c{]}^{6/5}$   $\approx$ ${\theta }_i^6[1+\frac{6(\delta \rho {)}_c}{5{\theta }_i^5}]$ $\Rightarrow (\delta P{)}_c$ $\approx$ $\frac{6{\theta }_i(\delta \rho {)}_c}{5}$ $\Rightarrow p_c\equiv \frac{(\delta P{)}_c}{(P_c^{*}{)}_0}$ $\approx$ $\frac{6(\delta \rho {)}_c}{5{\theta }_i^5}$	${\rho }_e^{*}=({\rho }_e^{*}{)}_0+(\delta \rho {)}_e$ $=$ $\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5+(\delta \rho {)}_e$ $P_e^{*}=(P_e^{*}{)}_0+(\delta P{)}_e$ $=$ $[{\rho }_e^{*}{]}^2\exp [\frac{s_e{\overline{\mu }}_e}{\mathrm{\Re }}]$   $=$ $\left[\right(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^5+(\delta \rho {)}_e{]}^2(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}$   $\approx$ ${\theta }_i^6[1+2(\delta \rho {)}_e\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-5}\right]$ $\Rightarrow (\delta P{)}_e$ $\approx$ $2(\delta \rho {)}_e(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i$ $\Rightarrow p_e\equiv \frac{(\delta P{)}_e}{(P_e^{*}{)}_0}$ $\approx$ $\frac{2(\delta \rho {)}_e}{{\theta }_i^5}(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}$

Now, in order for $(\delta P{)}_c=(\delta P{)}_e$ at the interface, we must have,

$2(\delta \rho {)}_e(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i$	$=$	$\frac{6{\theta }_i(\delta \rho {)}_c}{5}$
$\Rightarrow (\delta \rho {)}_e$	$=$	$\frac{3}{5}\left(\frac{{\mu }_e}{{\mu }_c}\right)(\delta \rho {)}_c$
$\Rightarrow (\frac{{\mu }_e}{{\mu }_c}{)}^{-1}\frac{(\delta \rho {)}_e}{{\theta }_i^5}$	$=$	$\left(\frac{3}{5}\right)\frac{(\delta \rho {)}_c}{{\theta }_i^5}$
$\Rightarrow [\frac{(\delta \rho )}{{\rho }_0^{*}}{]}_e$	$=$	$\left(\frac{3}{5}\right)[\frac{(\delta \rho )}{{\rho }_0^{*}}{]}_c$
$\Rightarrow d_e$	$=$	$\left(\frac{{\gamma }_c}{{\gamma }_e}\right)d_c.$

We also know that, according to the linearized equation of continuity,

$[r_0\frac{dx}{d{r}_0}+3x+d{]}_e$

$=$

$[r_0\frac{dx}{d{r}_0}+3x+d{]}_c.$

Given that, at the interface, $x_e=x_c$, this means that,

$[r_0\frac{dx}{d{r}_0}{]}_e$	$=$	$[r_0\frac{dx}{d{r}_0}{]}_c-[\left(\frac{{\gamma }_c}{{\gamma }_e}\right)-1]d_c$
	$=$	$[r_0\frac{dx}{d{r}_0}{]}_c+[\left(\frac{{\gamma }_c}{{\gamma }_e}\right)-1\left]\right\{[r_0\frac{dx}{d{r}_0}{]}_c+3x\}$
	$=$	$3x\left[\right(\frac{{\gamma }_c}{{\gamma }_e})-1]+\left(\frac{{\gamma }_c}{{\gamma }_e}\right)[r_0\frac{dx}{d{r}_0}{]}_c;$

or, dividing through by $x$,

$[\frac{d\ln x}{d\ln r_0}{]}_e$

$=$

$3\left[\right(\frac{{\gamma }_c}{{\gamma }_e})-1]+\left(\frac{{\gamma }_c}{{\gamma }_e}\right)[\frac{d\ln x}{d\ln r_0}{]}_c.$

This exactly matches the interface constraint presented in a related discussion in a separate chapter.

Now, as has already been stated in Step 5, above, for our analytic guess for the core's displacement function,

$[\frac{d\ln x}{d\ln r_0}{]}_c$	$=$	$\left(\frac{2}{15}\right){\xi }^2[\frac{{\xi }^2}{15}-1{]}^{-1}=2.4526969;$
$\Rightarrow [\frac{d\ln x}{d\ln r_0}{]}_e$	$=$	$0.2716182.$

where the numerical evaluation has been presented specifically for the core/envelope interface of the critical "Model A" configuration in which, ${\xi }_i=9.0149598$. Also, given that,

$\frac{x}{r_0}$

$=$

$\frac{{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}}}{\xi }\left(\frac{2\pi }{3}{)}^{1/2}\right[1-\frac{{\xi }^2}{15}]=(-{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}})\cdot 0.7092314,$

we find that,

$[\frac{dx}{d{r}_0}{]}_c$	$=$	$(-{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}})\cdot 1.7395297;$
$[\frac{dx}{d{r}_0}{]}_e$	$=$	$(-{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}})\cdot 0.1926402.$

Mapping Between Lagrangian Coordinates (Mass and Radius)

Core

$r^{*}$	$=$	$c_r\xi ,$
$M^{*}$	$=$	$c_M{\xi }^3(1+\frac{{\xi }^2}{3}{)}^{-3/2},$

where,

$c_r$	$\equiv$	$(\frac{3}{2\pi }{)}^{1/2},$
$c_M$	$\equiv$	$(\frac{2\cdot 3}{\pi }{)}^{1/2}.$

Establish the derivative of mass with respect to radius:

$d{r}^{*}$	$=$	$c_{r}d\xi$
$d{M}^{*}$	$=$	$c_M\left\{3{\xi }^2\right(1+\frac{{\xi }^2}{3}{)}^{-3/2}-\left(\frac{3}{2}\right){\xi }^3(1+\frac{{\xi }^2}{3}{)}^{-5/2}\cdot \frac{2\xi }{3}\}d\xi$
	$=$	$c_M(1+\frac{{\xi }^2}{3}{)}^{-5/2}\{3{\xi }^2(1+\frac{{\xi }^2}{3})-{\xi }^4\}d\xi$
	$=$	$c_M(1+\frac{{\xi }^2}{3}{)}^{-5/2}\{3{\xi }^2\}d\xi$
	$=$	$\frac{3c_M}{c_r}(1+\frac{{\xi }^2}{3}{)}^{-5/2}.{\xi }^{2}d{r}^{*}$

Hence,

$\frac{d{r}^{*}}{d{M}^{*}}$	$=$	$\frac{c_r}{3c_M}(1+\frac{{\xi }^2}{3}{)}^{5/2}{\xi }^{-2}$
	$=$	$\frac{1}{6}(1+\frac{{\xi }^2}{3}{)}^{5/2}{\xi }^{-2}=8.5761836$
$\Rightarrow \frac{dx}{d{M}^{*}}=\frac{d{r}^{*}}{d{M}^{*}}\cdot \frac{dx}{d{r}_0}$	$=$	$(-{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}})\cdot 1.7395297\times 8.5761836=(-{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}})\cdot 14.9185261.$

ASIDE:    Following along the lines of a separate but similar discussion, let's invert the $M^{*}(\xi )$ function to obtain a desired $r^{*}(M^{*})$ function. $M^{*}$ $=$ $c_M{\xi }^3(1+\frac{{\xi }^2}{3}{)}^{-3/2}$ $\Rightarrow (1+\frac{{\xi }^2}{3})$ $=$ $(\frac{c_M}{M^{*}}{)}^{2/3}{\xi }^2$ $\Rightarrow 3+{\xi }^2$ $=$ $3(\frac{c_M}{M^{*}}{)}^{2/3}{\xi }^2$ $\Rightarrow \left[3\right(\frac{c_M}{M^{*}}{)}^{2/3}-1]{\xi }^2$ $=$ $3$ $\Rightarrow \xi$ $=$ $\left[\right(\frac{c_M}{M^{*}}{)}^{2/3}-\frac{1}{3}{]}^{-1/2}$ $\Rightarrow r^{*}$ $=$ $c_r\left[\right(\frac{c_M}{M^{*}}{)}^{2/3}-\frac{1}{3}{]}^{-1/2}.$ Now, differentiate this expression … $\frac{d{r}^{*}}{d{M}^{*}}$ $=$ $\frac{c_r}{2}\left[\right(\frac{M^{*}}{c_M}{)}^{-2/3}-\frac{1}{3}{]}^{-3/2}\left(\frac{2c_M^{2/3}}{3}\right)(M^{*}{)}^{-5/3}$   $=$ $\frac{c_r}{3c_M}\left[\right(\frac{M^{*}}{c_M}{)}^{-2/3}-\frac{1}{3}{]}^{-3/2}(\frac{M^{*}}{c_M}{)}^{-5/3}.$ Or, given that, $(\frac{M^{*}}{c_M}{)}^{-1/3}$ $=$ $\frac{1}{\xi }(1+\frac{{\xi }^2}{3}{)}^{1/2},$ we can also write, $\frac{d{r}^{*}}{d{M}^{*}}$ $=$ $\frac{c_r}{3c_M}\left[\frac{1}{{\xi }^2}\right(1+\frac{{\xi }^2}{3})-\frac{1}{3}{]}^{-3/2}\frac{1}{{\xi }^5}(1+\frac{{\xi }^2}{3}{)}^{5/2}$   $=$ $\frac{c_r}{3c_M}\frac{1}{{\xi }^2}(1+\frac{{\xi }^2}{3}{)}^{5/2}$   $=$ $\left(\frac{1}{6}\right)\frac{1}{{\xi }^2}(1+\frac{{\xi }^2}{3}{)}^{5/2}.$ This exactly matches the expression for this derivative that we derived immediately above.

Envelope

$r^{*}$	$=$	$e_r\eta ,$
$M^{*}$	$=$	$e_M(-{\eta }^2\frac{d\varphi }{d\eta })$
	$=$	$e_M\cdot A[\sin (\eta -B)-\eta \cos (\eta -B)],$

where,

$e_r$	$\equiv$	$(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}(2\pi {)}^{-1/2},$
$e_M$	$\equiv$	$\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}\right(\frac{2}{\pi }{)}^{1/2},$
$\Rightarrow \frac{e_r}{e_M}$	$=$	$(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}(2\pi {)}^{-1/2}\left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}(\frac{2}{\pi }{)}^{1/2}{]}^{-1}$
	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}\right(\frac{1}{2\pi }{)}^{1/2}\left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^2{\theta }_i\left(\frac{\pi }{2}{)}^{1/2}\right]$
	$=$	$\frac{1}{2}\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^{-1},$

and from the parameter values associated with the equilibrium configuration,

${\mathrm{\Lambda }}_i$	$\equiv$	$\frac{1}{{\eta }_i}+(\frac{d\varphi }{d\eta }{)}_i=(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}\frac{1}{3^{1/2}{\theta }_i^2{\xi }_i}-\frac{{\xi }_i}{3^{1/2}}=\frac{1}{3^{1/2}{\xi }_i}\left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}(1+\frac{{\xi }_i^2}{3})-{\xi }_i^2],$
$A$	$\equiv$	${\eta }_i(1+{\mathrm{\Lambda }}_i^2{)}^{1/2}=(\frac{{\mu }_e}{{\mu }_c}\left)3^{1/2}{\xi }_i\right(1+\frac{{\xi }_i^2}{3}{)}^{-1}(1+{\mathrm{\Lambda }}_i^2{)}^{1/2},$
$B$	$\equiv$	${\eta }_i-\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i)=\left(\frac{{\mu }_e}{{\mu }_c}\right)3^{1/2}{\xi }_i(1+\frac{{\xi }_i^2}{3}{)}^{-1}-\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i).$

Note, as well, that $A$ $\equiv$ $\frac{{\eta }_i}{\sin ({\eta }_i-B)}.$

Establish the derivative of mass with respect to radius:

$d{r}^{*}$	$=$	$e_{r}d\eta$
$d{M}^{*}$	$=$	$e_M\cdot A[\eta \sin (\eta -B)]d\eta$
	$=$	$\frac{e_M\cdot A}{e_r}[\eta \sin (\eta -B)]d{r}^{*}$
$\Rightarrow d{M}^{*}$	$=$	$\frac{e_M\cdot A}{e_r}[\eta \sin (\eta -B)]d{r}^{*}$
$\Rightarrow \frac{d{r}^{*}}{d{M}^{*}}$	$=$	$\frac{e_r}{e_M\cdot A}[\eta \sin (\eta -B){]}^{-1}.$

Hence, precisely at the interface,

$[\frac{d{r}^{*}}{d{M}^{*}}{]}_i$	$=$	$\frac{e_r}{e_M\cdot {\eta }_i^2}$
	$=$	$\frac{1}{6}\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^{-1}\left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}{\xi }_i^{-2}]$
	$=$	$\frac{1}{6}(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-5}{\xi }_i^{-2}.$

Given that,

$[\frac{dx}{d{r}_0}{]}_e$

$=$

$(-{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}})\cdot 0.1926402,$

we find,

$[\frac{dx}{d{M}^{*}}{]}_e=[\frac{d{r}^{*}}{d{M}^{*}}{]}_e\cdot [\frac{dx}{d{r}_0}{]}_e$	$=$	$(-{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}})\cdot 0.1926402\cdot \left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}\right[8.5761836]$
	$=$	$(-{\alpha }_{\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{e}})\cdot 5.2394120.$

See Also