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

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, then also plot how $\rho$ varies with $m$ according to the analytic equilibrium structure; the plotted curves should be nearly, but not exactly, the same.)

STEP 3:   Determine how $P_{j-1/2}$ varies with radial shell location, using the above 1^st Law relation; also 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?

STEP 5:   Guess the eigenvector, ${\delta r}_i$, remembering that a reasonably good trial eigenfunction for the core is,

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

$=$

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

where,

$\xi$

$=$

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

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,

$+{\omega }_i^2$

$\approx$

$\frac{1}{{\delta r}_i\cdot r_i^2}[4\pi r_i^4\cdot \frac{dP}{dm}+Gm].$

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

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

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.

See Also