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,

$M^{*}$

$=$

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

See Also