SSC/Structure/BiPolytropes/51RenormaizePart2

Radial Oscillations in (n_c,n_e) = (5,1) Bipolytropes

Logically, this chapter extends the discussion — specifically the subsection titled, Try Again — found in the "Ramblings" chapter in which we introduced a total-mass-based renormalization of models along sequences of $(n_c,n_e)=(5,1)$ bipolytropes.

Building Each Model

Basic Equilibrium Structure

Most of the details underpinning the following summary relations can be found here.

New Normalization

$\tilde{\rho }$	$\equiv$	$\rho \left[\right(\frac{K_c}{G}{)}^{3/2}\frac{1}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^{-5};$
$\tilde{P}$	$\equiv$	$P\left[K_c^{-10}{G}^9{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^6\right];$
$\tilde{r}$	$\equiv$	$r\left[\right(\frac{K_c}{G}{)}^{5/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-2}],$
${\tilde{M}}_r$	$\equiv$	$\frac{M_r}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}};$
$\tilde{H}$	$\equiv$	$H\left[K_c^{-5/2}{G}^{3/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}\right];$
$\tilde{t}$	$\equiv$	$t[K_c^{15}{G}^{-13}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-10}{]}^{1/4}.$

Note:  For an n = 5 polytrope (like our bipolytrope's core), the units of the polytropic constant, $K_c$, are $[\frac{{\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}}^{13}}{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\cdot {\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}}^{10}}{]}^{1/5}$.

Quantity	Core $0\le \xi \le {\xi }_i$ $\theta =[1+\frac{1}{3}{\xi }^2{]}^{-1/2}$ $\frac{d\theta }{d\xi }=-\frac{\xi }{3}[1+\frac{1}{3}{\xi }^2{]}^{-3/2}$	Envelope ${\eta }_i\le \eta \le {\eta }_s$ $\varphi =A\left[\frac{\sin (\eta -B)}{\eta }\right]$ $\frac{d\varphi }{d\eta }=-\frac{A}{{\eta }^2}[\sin (\eta -B)-\eta \cos (\eta -B)]$
$\tilde{r}$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^4\right(\frac{3}{2\pi }{)}^{1/2}\xi$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}(\frac{{\mu }_e}{{\mu }_c}{)}^3{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta$
$\tilde{\rho }$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^5\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-10}\right(1+\frac{1}{3}{\xi }^2{)}^{-5/2}$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^5(\frac{{\mu }_e}{{\mu }_c}{)}^{-9}{\theta }_i^5\varphi$
$\tilde{P}$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^6\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}\right(1+\frac{1}{3}{\xi }^2{)}^{-3}$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^6(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}{\theta }_i^6{\varphi }^2$
${\tilde{M}}_r$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}\left(\frac{{\mu }_e}{{\mu }_c}{)}^2\right(\frac{2\cdot 3}{\pi }{)}^{1/2}\left[{\xi }^3\right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}]$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}{\theta }_i^{-1}\left(\frac{2}{\pi }{)}^{1/2}\right(-{\eta }^2\frac{d\varphi }{d\eta })$
Note that, for a given specification of the molecular-weight ratio, ${\mu }_e/{\mu }_c$, and the interface location, ${\xi }_i$, ${\theta }_i$ $=$ $(1+\frac{1}{3}{\xi }_i^2{)}^{-1/2},$ ${\eta }_i$ $=$ $\left(\frac{{\mu }_e}{{\mu }_c}\right)\sqrt{3}{\theta }_i^2{\xi }_i,$ ${\mathrm{\Lambda }}_i$ $=$ $\frac{{\xi }_i}{\sqrt{3}}\left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}\frac{1}{{\theta }_i^2{\xi }_i^2}-1],$ $A$ $=$ ${\eta }_i(1+{\mathrm{\Lambda }}_i^2{)}^{1/2},$ $B$ $=$ ${\eta }_i-\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i),$ ${\eta }_s$ $=$ $\pi +B=\frac{\pi }{2}+{\eta }_i+{\tan }^{-1}({\mathrm{\Lambda }}_i),$ in which case, ${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$ $=$ $(\frac{2}{\pi }{)}^{1/2}\frac{A{\eta }_s}{{\theta }_i},$ ${\tilde{\rho }}_c$ $=$ ${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^5(\frac{{\mu }_e}{{\mu }_c}{)}^{-10},$ $\nu \equiv \frac{M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$ $=$ $\left(\frac{{\mu }_e}{{\mu }_c}{)}^2\sqrt{3}\right[\frac{{\xi }_i^3{\theta }_i^4}{A{\eta }_s}],$ $q\equiv \frac{r_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{R}$ $=$ $\left(\frac{{\mu }_e}{{\mu }_c}\right)\sqrt{3}\left[\frac{{\xi }_i{\theta }_i^2}{{\eta }_s}\right].$

Additional Relations

Core

The analytically prescribed radial pressure gradient in the core can be obtained as follows.

$\frac{d{\tilde{M}}_r}{d\xi }$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}\left(\frac{{\mu }_e}{{\mu }_c}{)}^2\right(\frac{2\cdot 3}{\pi }{)}^{1/2}\left\{3{\xi }^2\right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}-{\xi }^4(1+\frac{1}{3}{\xi }^2{)}^{-5/2}\}$
	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}\left(\frac{{\mu }_e}{{\mu }_c}{)}^2\right(\frac{2\cdot 3}{\pi }{)}^{1/2}\left\{3{\xi }^2\right(1+\frac{1}{3}{\xi }^2)-{\xi }^4\}(1+\frac{1}{3}{\xi }^2{)}^{-5/2}$
	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}\left(\frac{{\mu }_e}{{\mu }_c}{)}^2\right(\frac{2\cdot 3}{\pi }{)}^{1/2}\left\{3{\xi }^2\right(1+\frac{1}{3}{\xi }^2{)}^{-5/2}\}$
$\Rightarrow \frac{d\xi }{d{\tilde{M}}_r}$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\right(\frac{\pi }{2\cdot 3}{)}^{1/2}\left\{\frac{1}{3{\xi }^2}\right(1+\frac{1}{3}{\xi }^2{)}^{5/2}\}.$

Also,

$\frac{d\tilde{P}}{d\xi }$

$=$

$-{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^6\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}\right\{2\xi (1+\frac{1}{3}{\xi }^2{)}^{-4}\}$

Hence,

$\frac{d\tilde{P}}{d{\tilde{M}}_r}$	$=$	$-{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^6\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}\right\{2\xi (1+\frac{1}{3}{\xi }^2{)}^{-4}\}\cdot {{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\right(\frac{\pi }{2\cdot 3}{)}^{1/2}\left\{\frac{1}{3{\xi }^2}\right(1+\frac{1}{3}{\xi }^2{)}^{5/2}\}$
	$=$	$-{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^7\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-14}\right(\frac{2\pi }{3^3}{)}^{1/2}\frac{1}{\xi }(1+\frac{1}{3}{\xi }^2{)}^{-3/2}.$

For comparison, in hydrostatic balance we expect …

$\frac{dP}{d{M}_r}=\frac{dP}{dr}\cdot \frac{dr}{d{M}_r}$	$=$	$-\frac{G{M}_r\rho }{r^2}\cdot \frac{1}{4\pi r^2\rho }=-\frac{G{M}_r}{4\pi r^4}$
$\Rightarrow \frac{d\tilde{P}}{d{\tilde{M}}_r}=\left[\frac{dP}{d{M}_r}\right]\cdot \left[K_c^{-10}{G}^9{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^6\right]M_{\mathrm{t}\mathrm{o}\mathrm{t}}$	$=$	$-\frac{G{M}_r}{4\pi r^4}\left[K_c^{-10}{G}^9{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^7\right]$
	$=$	$-\frac{{\tilde{M}}_r}{4\pi r^4}\left[K_c^{-10}{G}^{10}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^8\right]$
	$=$	$-\frac{{\tilde{M}}_r}{4\pi {\tilde{r}}^4}$
	$=$	$-\frac{1}{4\pi }{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}\left(\frac{{\mu }_e}{{\mu }_c}{)}^2\right(\frac{2\cdot 3}{\pi }{)}^{1/2}\left[{\xi }^3\right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}]\cdot \{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^4\right(\frac{3}{2\pi }{)}^{1/2}\xi {\}}^{-4}$
	$=$	$-\left\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^7\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-14}\left(\frac{2^2{\pi }^2}{3^2}\right)\left\}\frac{1}{4\pi }\right(\frac{2\cdot 3}{\pi }{)}^{1/2}\left[\frac{1}{\xi }\right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}]$
	$=$	$-\left\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^7\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-14}\left\}\right(\frac{2\cdot \pi }{3^3}{)}^{1/2}\left[\frac{1}{\xi }\right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}].$

This matches our earlier expression, as it should.

Takeaway Expression $\frac{d\tilde{P}}{d{\tilde{M}}_r}$ $=$ $-\frac{{\tilde{M}}_r}{4\pi {\tilde{r}}^4}$

Envelope

Given that, for the envelope,

${\tilde{M}}_r$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}{\theta }_i^{-1}\left(\frac{2}{\pi }{)}^{1/2}A\right[\sin (\eta -B)-\eta \cos (\eta -B)],$      and,
$\tilde{r}$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}(\frac{{\mu }_e}{{\mu }_c}{)}^3{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta ,$

we deduce that,

$\frac{d\tilde{P}}{d{\tilde{M}}_r}=-\frac{{\tilde{M}}_r}{4\pi {\tilde{r}}^4}$	$=$	$\left(\frac{1}{4\pi }\right){{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}{\theta }_i^{-1}\left(\frac{2}{\pi }{)}^{1/2}A\right[\eta \cos (\eta -B)-\sin (\eta -B)]\cdot [{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}(\frac{{\mu }_e}{{\mu }_c}{)}^3{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta {]}^{-4}$
	$=$	$(\frac{1}{2^4{\pi }^2}\cdot \frac{2}{\pi }\cdot 2^4{\pi }^4{)}^{1/2}{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^7{\theta }_i^{7}A[\eta \cos (\eta -B)-\sin (\eta -B)]\cdot [\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}{\eta }^{-4}\right]$
	$=$	$\left(2\pi {)}^{1/2}{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^7{\theta }_i^7\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}\cdot \frac{A}{{\eta }^4}[\eta \cos (\eta -B)-\sin (\eta -B)]\cdot$

As a cross-check …

$\frac{d\tilde{P}}{d\eta }$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^6\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}{\theta }_i^6\right[2\varphi \cdot \frac{d\varphi }{d\eta }]$
	$=$	$2{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^6(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}{\theta }_i^6\cdot \frac{A^2}{{\eta }^3}\cdot [\eta \cos (\eta -B)-\sin (\eta -B)]\sin (\eta -B),$

and,

$\frac{d{\tilde{M}}_r}{d\eta }$	$=$	$A{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}{\theta }_i^{-1}\left(\frac{2}{\pi }{)}^{1/2}\frac{d}{d\eta }\right[\sin (\eta -B)-\eta \cos (\eta -B)]$
	$=$	$A{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}{\theta }_i^{-1}\left(\frac{2}{\pi }{)}^{1/2}\right\{\eta \sin (\eta -B)\}.$

That is,

$\frac{d\tilde{P}}{d{\tilde{M}}_r}$	$=$	$2{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^6(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}{\theta }_i^6\cdot \frac{A^2}{{\eta }^3}\cdot [\eta \cos (\eta -B)-\sin (\eta -B)]\sin (\eta -B)\{A{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}{\theta }_i^{-1}\left(\frac{2}{\pi }{)}^{1/2}\right[\eta \sin (\eta -B)]{\}}^{-1}$
	$=$	$(2\pi {)}^{1/2}{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^7(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}{\theta }_i^7\cdot \frac{A}{{\eta }^4}\cdot [\eta \cos (\eta -B)-\sin (\eta -B)].$

Correct!

Time-Dependent Euler Equation

We begin with the form of the,

Euler Equation

$\frac{d{v}_r}{dt}=-\frac{1}{\rho }\frac{dP}{dr}-\frac{d\mathrm{\Phi }}{dr}$

that is broadly relevant to studies of radial oscillations in spherically symmetric configurations. Recognizing from, for example, a related discussion that, $v_r=dr/dt$, and that,

$\frac{d\mathrm{\Phi }}{dr}$

$=$

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

we obtain our

Desired Form of the Euler Equation
$\frac{d^{2}r}{d{t}^2}$	$=$	$-\frac{1}{\rho }\frac{dP}{dr}-\frac{G{M}_r}{r^2}.$

Given as well that,

$\frac{d{M}_r}{dr}=4\pi r^2\rho$

we see that,

$\frac{d{M}_r}{4\pi r^2}$	$=$	$\rho dr$
$\Rightarrow \frac{d^{2}r}{d{t}^2}$	$=$	$-4\pi r^2\frac{dP}{d{M}_r}-\frac{G{M}_r}{r^2}$
$\Rightarrow \frac{1}{4\pi r^2}\cdot \frac{d^{2}r}{d{t}^2}$	$=$	$-\frac{dP}{d{M}_r}-\frac{G{M}_r}{4\pi r^4}.$

Next, if as above, we multiply through by $\left[K_c^{-10}{G}^9{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^7\right]$, we obtain the relevant,

Normalized Euler Equation
$\frac{1}{4\pi {\tilde{r}}^2}\cdot \frac{d^2\tilde{r}}{d{\tilde{t}}^2}$	$=$	$-\frac{d\tilde{P}}{d{\tilde{M}}_r}-\frac{{\tilde{M}}_r}{4\pi {\tilde{r}}^4},$

where, as a reminder, the dimensionless time is,

$\tilde{t}$

$\equiv$

$t[K_c^{15}{G}^{-13}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-10}{]}^{1/4}.$

CAUTION!   Regarding Our Chosen Lagrangian Fluid Marker If we were to use $\tilde{r}$ as our primary Lagrangian fluid marker, we would be in a position to analytically specify the function, ${\tilde{M}}_r(\tilde{r})$. Here, however, we will call upon ${\tilde{M}}_r$ rather than $\tilde{r}$ to serve as the primary Lagrangian fluid marker because mass facilitates our efforts to highlight a variety of important physical properties of bipolytropic configurations. We will therefore need to specify the function, $\tilde{r}({\tilde{M}}_r)$ instead of ${\tilde{M}}_r(\tilde{r})$. For the core, this choice does not introduce any particularly difficult computational challenges because we can invert the ${\tilde{M}}_r(\tilde{r})$ relationship analytically to obtain … ${\xi }^2$ $=$ $3\left[3\right(\frac{c_m}{{\tilde{M}}_r}{)}^{2/3}-1{]}^{-1},$ where, $c_m$ $\equiv$ $m_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}\left(\frac{{\mu }_e}{{\mu }_c}{)}^2\right(\frac{2\cdot 3}{\pi }{)}^{1/2}.$ This is not the case for the envelope, however; we will not be able to analytically specify $\tilde{r}({\tilde{M}}_r)$. This is unfortunate, as a numerical (rather than analytic) specification will necessarily introduce additional errors into our solution of the displacement function — which already is a small and error-prone quantity. We will nevertheless proceed along this line.

Example Models Along BiPolytrope Sequence 0.3100

For the case of $(n_c,n_e)=(5,1)$ and ${\mu }_e/{\mu }_c=0.3100$, we consider here the examination of models with three relatively significant values of the core/envelope interface:

• Model D $({\xi }_i,\overline{\rho }/{\rho }_c,q,\nu )\approx (2.06061,1.1931E+02,0.16296,0.13754)$: Approximate location along the sequence of the model with the maximum fractional core radius.
• Model C $({\xi }_i,\overline{\rho }/{\rho }_c,q,\nu )\approx (2.69697,3.0676E+02,0.15819,0.19161)$: Approximate location along the sequence of the onset of fundamental-mode instability.
• Model A $({\xi }_i,\overline{\rho }/{\rho }_c,q,\nu )\approx (9.0149598,1.1664E+06,0.075502255,0.337217006)$: Exact location along the sequence of the model with the maximum fractional core mass.

Model C Finite-Difference Representation

Here we examine a discrete representation of a model along the ${\mu }_e/{\mu }_c=0.31$ sequence whose core/envelope interface is located at ${\xi }_i=2.69697$; whose core mass-fraction is $\nu =0.19161$; and for which, $m_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}=2.145465292$.

Table C1

STEP1:    Divide the core into $(N_c+1)$ grid lines — that is, into $N_c$ radial zones — associating the first "grid line" with the center of the core and the last grid line with the radial location of the core/envelope interface; in Table C1, we have set $N_c=20$. Choosing $0\le {\tilde{M}}_r\le \nu$ as the principal Lagrangian coordinate, and using the available analytic expressions, assign values to the following physical quantities at each grid line:

• Mass (see column titled tilde M_r in Table C1):     Set $(\mathrm{\Delta }m{)}_c=\nu /(N_c)$; then, for $i=1\mathrm{t}\mathrm{h}\mathrm{r}\mathrm{u}(N_c+1)$, set ${\tilde{M}}_r=(i-1)(\mathrm{\Delta }m{)}_c.$

See Also