SR/Piston

Piston Model

KW94 Piston Model

Here we draw principally from the discussion of a simple piston model as presented in §2.7 and §6.6 of [KW94].

An ideal gas of mass $m^{*}$ is held in a vertical container with a movable piston resting on top of — and confining — the gas; the mass of the piston is $M^{*}$. A vertically directed gravitational acceleration, $g$, acts on the piston, in which case the weight of the piston is given by the expression,

$G^{*}=g{M}^{*}.$

"In the case of hydrostatic equilibrium, the gas pressure $P$ adjusts in such a way that the weight per unit area is balanced by the pressure:"

$P=\frac{G^{*}}{A}.$

"If the forces do not compensate each other, the piston is accelerated in the vertical direction according to the equation of motion,"

$M^{*}\frac{d^{2}h}{d{t}^2}=-G^{*}+PA.$

Bipolytropes

If we consider only the structure and oscillations of the core, we should set the "external" pressure, $P_e$, equal to the pressure, $P_i$, at the core-envelope interface as viewed from the perspective of the envelope.

Extra Relations

Keep in mind that, in hydrostatic balance,

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

Otherwise,

Euler Equation
(Momentum Conservation)

$\frac{d\overrightarrow{v}}{dt}=-\frac{1}{\rho }\mathrm{\nabla }P-\mathrm{\nabla }\mathrm{\Phi }$

In equilibrium, the pressure at the core-envelope interface is,

$P_i=K_c{\rho }_c^{6/5}(1+\frac{{\xi }_i^2}{3}{)}^{-3}$.

$P$

$=$

$K_n{\rho }^{1+1/n}$

Use Surface Area

According to the "piston model," it should be true that,

$P_e=\frac{G^{*}}{A}$

$=$

$\frac{g_i{M}^{*}}{4\pi r_i^2},$

where (the magnitude of) the acceleration at the interface is,

$g_i$

$=$

$\frac{G{M}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{r_i^2}.$

This means that,

$M^{*}$

$=$

$\frac{4\pi r_i^2{P}_e}{g_i}=\frac{4\pi r_i^4{P}_e}{G{M}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}.$

Insert Interface Details

Now, according to our analysis of the bipolytrope having $(n_c,n_e)=(5,1),$ we have,

$P_i$	$=$	$K_c{\rho }_c^{6/5}(1+\frac{{\xi }_i^2}{3}{)}^{-3};$
$r_i$	$=$	$\frac{K_c^{1/2}}{G^{1/2}{\rho }_c^{2/5}}(\frac{3}{2\pi }{)}^{1/2}{\xi }_i;$
$M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$\frac{K_c^{3/2}}{G^{3/2}{\rho }_c^{1/5}}(\frac{6}{\pi }{)}^{1/2}(\xi \theta {)}^3.$

Hence, we find that,

$M^{*}$	$=$	$\frac{4\pi }{G}\left[r_i^4\right]\left[P_i\right][M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}^{-1}$
	$=$	$\frac{4\pi }{G}\left[\frac{K_c^{1/2}}{G^{1/2}{\rho }_c^{2/5}}\right(\frac{3}{2\pi }{)}^{1/2}{\xi }_i{]}^4\left[K_c{\rho }_c^{6/5}\right(1+\frac{{\xi }_i^2}{3}{)}^{-3}\left]\right[\frac{K_c^{3/2}}{G^{3/2}{\rho }_c^{1/5}}(\frac{6}{\pi }{)}^{1/2}(\xi \theta {)}^3{]}^{-1}$
	$=$	$\frac{4\pi }{G}\left[\frac{K_c^2}{G^2{\rho }_c^{8/5}}\right(\frac{3}{2\pi }{)}^2{\xi }_i^4\left]\right[K_c{\rho }_c^{6/5}(1+\frac{{\xi }_i^2}{3}{)}^{-3}]\left[\frac{G^{3/2}{\rho }_c^{1/5}}{K_c^{3/2}}\right(\frac{\pi }{6}{)}^{1/2}(\xi \theta {)}^{-3}]$
	$=$	$\frac{4\pi }{G}\left[\frac{K_c^2}{G^2{\rho }_c^{8/5}}\right]\left[K_c{\rho }_c^{6/5}\right]\left[\frac{G^{3/2}{\rho }_c^{1/5}}{K_c^{3/2}}\right]\left(\frac{3}{2\pi }{)}^2{\xi }_i^4\right(1+\frac{{\xi }_i^2}{3}{)}^{-3}(\frac{\pi }{6}{)}^{1/2}(\xi \theta {)}^{-3}$
	$=$	$4\pi \left[\frac{K_c^{3/2}}{G^{3/2}{\rho }_c^{1/5}}\right]\left(\frac{3^3}{2^5{\pi }^3}{)}^{1/2}\right(1+\frac{{\xi }_i^2}{3}{)}^{-3}\xi {\theta }^{-3}.$

Hydrostatic Balance

Drawing principally from an accompanying discussion, we understand that hydrostatic balance throughout a self-gravitating sphere is given by the key relation,

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

where,

$M_r=\underset{0}{\overset{r}{\int }}4\pi r^2\rho dr$ .

Now, according to our analysis of the bipolytrope having $(n_c,n_e)=(5,1),$ we have throughout the core, $\theta =(1+{\xi }^2/3{)}^{-1/2}$ and,

$r$	$=$	$\frac{K_c^{1/2}}{G^{1/2}{\rho }_c^{2/5}}(\frac{3}{2\pi }{)}^{1/2}\xi ;$
$\rho$	$=$	${\rho }_c{\theta }^5={\rho }_c[1+\frac{{\xi }^2}{3}{]}^{-5/2};$
$P$	$=$	$K_c{\rho }_c^{6/5}(1+\frac{{\xi }^2}{3}{)}^{-3};$
$M_r$	$=$	$\frac{K_c^{3/2}}{G^{3/2}{\rho }_c^{1/5}}(\frac{6}{\pi }{)}^{1/2}(\xi \theta {)}^3=\frac{K_c^{3/2}}{G^{3/2}{\rho }_c^{1/5}}\left(\frac{6}{\pi }{)}^{1/2}{\xi }^3\right[1+\frac{{\xi }^2}{3}{]}^{-3/2}.$

See Also