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 ${\tilde{M}}_r=(i-1)(\mathrm{\Delta }m{)}_c.$
• Polytropic radial coordinate (see column titled xi from M_r in Table C1):     Given that, $c_m=m_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}({\mu }_e/{\mu }_c{)}^2(6/\pi {)}^{1/2}=0.0619017$, determine the value of $\xi$ associated with each gridline's value of ${\tilde{M}}_r$ from the expression,

  $\xi$

  $=$

  $3^{1/2}\left[3\right(\frac{c_m}{{\tilde{M}}_r}{)}^{2/3}-1{]}^{-1/2}.$

  For example, at the 21^st gridline (associated with the core/envelope interface), this expression gives the expected, ${\xi }_i=2.69697$.

• Given the value of $\xi$ at each gridline, determine the associated values of $\tilde{r},\tilde{\rho },\tilde{P}$ — see the columns in Table C1 titled tilde r, tilde rho, tilde P — using the appropriate analytic expressions for the Core as provided above. For example, at the 21^st gridline (associated with the core/envelope interface), we find, $\tilde{r}=0.003739$, $\tilde{\rho }=2.5555\times 10^5$, and $\tilde{P}=3.0830\times 10^6$.

STEP2:    Building upon the results of STEP1, determine the value of ${\tilde{M}}_r/(4\pi {\tilde{r}}^4)$ at each gridline; see the column of Table C1 titled M/(r pi r^4),

• As stated in the above Takeaway Expression, this will simultaneously provide a precise evaluation of the pressure gradient, $d\tilde{P}/d{\tilde{M}}_r$, at each gridline when the configuration is in equilibrium.
• After a perturbation is introduced into the (initially equilibrium) configuration, both the pressure gradient and the quantity, ${\tilde{M}}_r/(4\pi {\tilde{r}}^4)$, will deviate from their equilibrium values and, quite generally, from each other. Then, as expressed by the above Normalized Euler Equation, the sum of these to perturbed quantities will dictate the strength and direction of the unbalanced acceleration that will be felt by the Lagrangian fluid element at each gridline.

See Also