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Spherically Symmetric Configurations (Stability — Part II)
Suppose we now want to study the stability of one of the spherically symmetric, equilibrium structures that have been derived elsewhere. The identified set of simplified, time-dependent governing equations will tell us how the configuration will respond to an applied radial (i.e., spherically symmetric) perturbation that pushes the configuration slightly away from its initial equilibrium state.
Assembling the Key Relations
Governing Equations
After combining the Euler equation with the Poisson equation in essentially the manner outlined by the "structural solution strategy" we have called Technique 1, the relevant set of time-dependent governing equations is:
Equation of Continuity
<math>\frac{d\rho}{dt} = - \rho \biggl[\frac{1}{r^2}\frac{d(r^2 v_r)}{dr} \biggr]
= -\rho \biggl[ \frac{dv_r}{dr} + \frac{2v_r}{r} \biggr] </math>
Euler + Poisson Equations
<math>\frac{dv_r}{dt} = - \frac{1}{\rho}\frac{dP}{dr} - \frac{GM_r}{r^2} </math>
Adiabatic Form of the
First Law of Thermodynamics
<math>~\frac{d\epsilon}{dt} + P \frac{d}{dt} \biggl(\frac{1}{\rho}\biggr) = 0</math>
where,
<math>v_r \equiv \frac{dr}{dt}</math> ,
and, as before, the mass enclosed inside radius <math>r</math> is,
<math>M_r \equiv \int_0^r dm_r = \int_0^r 4\pi r^2 \rho dr</math> .
Consistent Lagrangian Formulation
The (Lagrangian) time derivatives in these equations define how a given physical parameter — for example, <math>~\rho</math>, <math>v_r</math>, or <math>~\epsilon</math> — should vary with time in a (Lagrangian) fluid element that is not fixed in space but, rather, is moving along with the flow. However, the radial derivatives describe the spatial variation of various physical parameters as measured at fixed locations in space; that is, as written, the radial derivatives do not track conditions as viewed by a (Lagrangian) fluid element that is moving along with the flow because the position <math>r</math> of each (Lagrangian) fluid element is itself changing with time. A proper Lagrangian representation of the spatial derivatives can be formulated in the case of one-dimensional, spherically symmetric flows by using <math>M_r</math> (or, equivalently, <math>m_r</math>) instead of <math>r</math> as the independent variable. Making the substitution,
<math>\frac{d}{dr} = \frac{dM_r}{dr}\frac{d}{dM_r} = 4\pi \rho r^2 \frac{d}{dM_r}</math>
in the first two equations above gives, respectively,
<math>\frac{d\rho}{dt} = - 4\pi \rho^2 r^2 \frac{dv_r}{dM_r} - \frac{2\rho v_r}{r} </math> ,
and,
<math>\frac{dv_r}{dt} = - 4\pi r^2 \frac{dP}{dM_r} - \frac{GM_r}{r^2} </math>
Supplemental Relations
As has been discussed elsewhere, in any analysis of time-dependent flows, the principal governing equations must be supplemented by adopting an equation of state for the gas and by specifying initial conditions. Here, initial conditions will be given by the structural properties — for example, <math>~\rho</math><math>(M_r)</math> and <math>~P</math><math>(M_r)</math> — of one of our derived, spherically symmetric equilibrium structures — for example, a uniform-density sphere or an n = 1 polytrope. We will adopt an ideal gas equation of state and, specifically, the relation,
<math>~P = (\gamma_\mathrm{g} - 1)\epsilon \rho </math>
As a result, the adiabatic form of the <math>1^\mathrm{st}</math> law of thermodynamics can be written as,
<math> \rho \frac{dP}{dt} - \gamma_\mathrm{g} P \frac{d\rho}{dt} = 0 . </math>
Summary
In summary, the following three one-dimensional ODEs define the physical relationship between the three dependent variables <math>~\rho</math>, <math>~P</math>, and <math>r</math>, each of which should be expressible as a function of the two independent (Lagrangian) variables, <math>~t</math> and <math>M_r</math>:
Equation of Continuity
<math>\frac{d\rho}{dt} = - 4\pi \rho^2 r^2 \frac{d}{dM_r}\biggl(\frac{dr}{dt}\biggr) - \frac{2\rho}{r} \biggl(\frac{dr}{dt}\biggr) </math>
,
Euler + Poisson Equations
<math>\frac{d^2 r}{dt^2} = - 4\pi r^2 \frac{dP}{dM_r} - \frac{GM_r}{r^2} </math>
Adiabatic Form of the
First Law of Thermodynamics
<math>
\rho \frac{dP}{dt} - \gamma_\mathrm{g} P \frac{d\rho}{dt} = 0 .
</math>
The Eigenvalue Problem
Here we adopt a notation and presentation very similar to what can be found in §38 of Kippenhahn & Weigart (KW94). In particular, we will use <math>m</math> rather than the more cumbersome <math>M_r</math> to tag each (Lagrangian) mass shell, both initially and at all later times. As is customary in perturbation studies throughout the field of physics, we will assume that the pressure <math>P(m,t)</math>, density <math>\rho(m,t)</math>, and radial position <math>r(m,t)</math> of each mass shell at any time <math>~t</math> can be written in the form,
<math>
P(m,t) = P_0(m) + P_1(m,t) = P_0(m) \biggl[1 + p(m) e^{i\omega t} \biggr] ,
</math>
<math>
\rho(m,t) = \rho_0(m) + \rho_1(m,t) = \rho_0(m) \biggl[1 + d(m) e^{i\omega t} \biggr] ,
</math>
<math>
r(m,t) = r_0(m) + r_1(m,t) = r_0(m) \biggl[1 + x(m) e^{i\omega t} \biggr] ,
</math>
where the subscript 1 denotes the variation of any variable away from its initial value (subscript 0) as drawn from the derived structure of the selected initial equilibrium model. These expressions encompass the hypothesis that, when perturbations away from the initial equilibrium state are sufficiently small — that is, <math>p</math>, <math>d</math>, and <math>x</math> all <math>\ll 1</math> — the perturbation can be treated as a product of functions that are separable in <math>m</math> and <math>~t</math>, and that in general the time-dependent component can be represented by an exponential with an imaginary argument. The task is to solve a linearized version of the coupled set of key relations for the "eigenfunctions" <math>p_i(m)</math>, <math>d_i(m)</math>, and <math>x_i(m)</math> associated with various characteristic "eigenfrequencies" <math>\omega_i</math> of the underlying equilibrium model.
Linearizing the Key Equations
Adiabatic form of the First Law of Thermodynamics
Plugging the perturbed expressions for <math>P(m,t)</math> and <math>\rho(m,t)</math> into the adiabatic form of the First Law of Thermodynamics, we obtain,
<math>
(i\omega) \rho_0(m) \biggl[1 + d(m) e^{i\omega t} \biggr]P_0(m) p(m) e^{i\omega t} - \gamma_\mathrm{g}(i\omega) P_0(m) \biggl[1 + p(m) e^{i\omega t} \biggr] \rho_0(m) d(m) e^{i\omega t} = 0
</math>
<math>\Rightarrow ~~~~~
(i\omega) \rho_0(m)P_0(m) e^{i\omega t} \biggl\{\biggl[1 + d(m) e^{i\omega t} \biggr] p(m) - \gamma_\mathrm{g} \biggl[1 + p(m) e^{i\omega t} \biggr] d(m) \biggr\} = 0
</math>
Since we are seeking solutions that will be satisfied throughout the configuration — that is, for all mass shells <math>m</math> — the expression inside the curly brackets must be zero. Hence,
<math> p(m) - \gamma_\mathrm{g} d(m) + (1 - \gamma_\mathrm{g} ) d(m)p(m) e^{i\omega t} =0 . </math>
Since we are only examining deviations from the initial equilibrium state in which <math>d(m)</math> and <math>p(m)</math> are both <math>\ll 1</math>, then the third term on the left-hand-side of this equation, which contains a product of these two small quantities, must be much smaller than the first two terms. As is standard in perturbation theory throughout physics, for our stability analysis, we will drop this "quadradic" term and keep only terms that are linear in the small quantities. This leads to the following algebraic relationship between <math>d(m)</math> and <math>p(m)</math>:
<math> p = \gamma_\mathrm{g} d . </math>
Continuity Equation
Adopting the same approach, we will now "linearize" each term in the continuity equation:
<math>
\frac{d\rho}{dt} \rightarrow (i\omega)\rho_0 d~e^{i\omega t}
</math>
<math>
\frac{\rho}{r} \rightarrow \frac{\rho_0}{r_0} \biggl[1 + d e^{i\omega t} \biggr] \biggl[1 + x e^{i\omega t} \biggr]^{-1}
\approx \frac{\rho_0}{r_0} \biggl[1 + d ~e^{i\omega t} \biggr]\biggl[1 - x~ e^{i\omega t} \biggr] \approx
\frac{\rho_0}{r_0} \biggl[1 + (d - x) ~e^{i\omega t} \biggr]
</math>
<math>
\frac{dr}{dt} \rightarrow (i\omega) r_0 x~e^{i\omega t}
</math>
<math>
\rho^2 r^2 \rightarrow \rho_0^2 r_0^2 \biggl[1 + d e^{i\omega t} \biggr]^2 \biggl[1 + x e^{i\omega t} \biggr]^2
\approx \rho_0^2 r_0^2 \biggl[1 + 2d ~e^{i\omega t} \biggr]\biggl[1 + 2x~ e^{i\omega t} \biggr] \approx
\rho_0^2 r_0^2 \biggl[1 + 2(d + x) ~e^{i\omega t} \biggr]
</math>
<math>
\frac{d}{dm}\biggl(\frac{dr}{dt}\biggr) \approx \frac{d}{dm}\biggl[(i\omega) r_0 x~e^{i\omega t}\biggr] = (i\omega) e^{i\omega t} \biggl[x\frac{dr_0}{dm} + r_0\frac{dx}{dm} \biggr] = (i\omega) e^{i\omega t} \biggl[\frac{x}{4\pi r_0^2 \rho_0} + r_0\frac{dx}{dm} \biggr]
</math>
In the last step of this last expression we have made use of the fact that, in the initial, unperturbed equilibrium model, <math>dr_0/dm = 1/(4\pi r_0^2 \rho_0)</math>. Combining all of these terms and linearizing the combined expression further, the linearized continuity equation becomes,
<math>
(i\omega)\rho_0 d~e^{i\omega t} \approx - \frac{2\rho_0}{r_0} \biggl[1 + (d - x) ~e^{i\omega t} \biggr] (i\omega) r_0 x~e^{i\omega t} - 4\pi \rho_0^2 r_0^2 \biggl[1 + 2(d + x) ~e^{i\omega t} \biggr](i\omega) e^{i\omega t} \biggl[\frac{x}{4\pi r_0^2 \rho_0} + r_0\frac{dx}{dm} \biggr]
</math>
<math>
\Rightarrow ~~~ \rho_0 d \approx - 3\rho_0 x - 4\pi \rho_0^2 r_0^3 \frac{dx}{dm}
</math>
<math>
\Rightarrow ~~~ 4\pi \rho_0 r_0^3 \frac{dx}{dm} \approx - 3 x - d
</math>
or,
<math>
r_0 \frac{dx}{dr_0} \approx - 3 x - d ,
</math>
where, to obtain this last expression, we have switched back from differentiation with respect to <math>m</math> to differentiation with respect to <math>r_0</math>.
Euler + Poisson Equations
Finally, linearizing each term in the combined "Euler + Poisson" equation gives:
<math>
\frac{d^2r}{dt^2} \rightarrow \frac{d}{dt}\biggl[(i\omega) r_0 x~e^{i\omega t}\biggr] = - \omega^2 r_0 x~e^{i\omega t}
</math>
<math>
r^2 \frac{dP}{dm} \rightarrow r_0^2 \biggl[1 + x~ e^{i\omega t} \biggr]^2 \biggl\{\frac{dP_0}{dm} \biggl[1 + p~ e^{i\omega t} \biggr] + P_0~e^{i\omega t} \frac{dp}{dm} \biggr\} \approx r_0^2 \frac{dP_0}{dm} \biggl[1 + (2x+p)~ e^{i\omega t} \biggr] + P_0 r_0^2~e^{i\omega t} \frac{dp}{dm}
</math>
<math>
\frac{Gm}{r^2} \rightarrow \frac{Gm}{ r_0^2} \biggl[1 + x~ e^{i\omega t} \biggr]^{-2} \approx \frac{Gm}{ r_0^2}
\biggl[1 -2 x~ e^{i\omega t} \biggr]
</math>
Hence, the combined linearized relation is,
<math>
- \omega^2 r_0 x~e^{i\omega t} \approx -4\pi \biggl\{r_0^2 \frac{dP_0}{dm} \biggl[1 + (2x+p)~ e^{i\omega t} \biggr] + P_0 r_0^2~e^{i\omega t} \frac{dp}{dm} \biggr\} - \frac{Gm}{ r_0^2} \biggl[1 -2 x~ e^{i\omega t} \biggr]
</math>
<math>
\Rightarrow ~~~~~ e^{i\omega t} \biggl\{(2x + p)4\pi r_0^2 \frac{dP_0}{dm}-2x \frac{Gm}{r_0^2} + 4\pi P_0 r_0^2 \frac{dp}{dm} -\omega^2 r_0 x \biggr\} \approx - 4\pi r_0^2 \frac{dP_0}{dm} - \frac{Gm}{r_0^2}
</math>
<math>
\Rightarrow ~~~~~ 4\pi P_0 r_0^2 \frac{dp}{dm} \approx (4x + p)g_0 + \omega^2 r_0 x
</math>
where, in order to obtain this last expression we have made use of the fact that, in the unperturbed equilibrium configuration,
<math>
g_0(m) \equiv \frac{Gm}{r_0^2} = - 4\pi r_0^2 \frac{dP_0}{dm} .
</math>
Switching back from differentiation with respect to <math>m</math> to differentiation with respect to <math>r_0</math>, the "Euler + Poisson" combined linearized relation can alternatively be written as,
<math>
\frac{P_0}{\rho_0} \frac{dp}{dr_0} \approx (4x + p)g_0 + \omega^2 r_0 x .
</math>
Summary
In summary, the following three linearized equations describe the physical relationship between the three dimensionless perturbation amplitudes <math>p(r_0)</math>, <math>d(r_0)</math> and <math>x(r_0)</math>, for various characteristic eigenfrequencies, <math>\omega</math>:
Linearized
Equation of Continuity
<math>
r_0 \frac{dx}{dr_0} = - 3 x - d ,
</math>
Linearized
Euler + Poisson Equations
<math>
\frac{P_0}{\rho_0} \frac{dp}{dr_0} = (4x + p)g_0 + \omega^2 r_0 x ,
</math>
Linearized
Adiabatic Form of the
First Law of Thermodynamics
<math>
p = \gamma_\mathrm{g} d .
</math>
It is customary to combine these three relations to obtain a single, second-order ODE in terms of the fractional displacement, <math>x</math> as follows. Using the third expression to replace <math>d</math> by <math>p</math> in the first expression, then differentiating the first expression with respect to <math>r_0</math> generates,
<math>
\frac{d}{dr_0} \biggl[ r_0 \frac{dx}{dr_0}\biggr] = - \frac{d}{dr_0}\biggl[ 3 x + \frac{p}{\gamma_\mathrm{g}} \biggr]
</math>
<math>
\Rightarrow ~~~~~ r_0 \frac{d^2x}{dr_0^2} + 4 \frac{dx}{dr_0} = - \frac{1}{\gamma_\mathrm{g}} \frac{dp}{dr_0}
</math>
Similarly, replacing <math>p</math> by <math>d</math> in the second expression, then using the first expression to eliminate <math>d</math> gives,
<math>
\frac{P_0}{\rho_0} \frac{dp}{dr_0} = \biggl[4x + \gamma_\mathrm{g}\biggl( -3x -r_0\frac{dx}{dr_0} \biggr) \biggr] g_0 + \omega^2 r_0 x ,
</math>
<math>
\Rightarrow ~~~~~ \frac{1}{\gamma_\mathrm{g}} \frac{dp}{dr_0} = - \frac{dx}{dr_0} \biggl(\frac{r_0 g_0 \rho_0}{P_0}\biggr) + \biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr] \biggl(\frac{r_0 \rho_0}{\gamma_\mathrm{g} P_0} \biggr) x ,
</math>
Finally, then, combining these two expressions gives the desired second-order ODE,
<math>
\frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0} + \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr] x = 0 .
</math>
Dimensionless Expression
Let's write this governing ODE and the key physical variables as dimensionless expressions. First, multiply through by <math>R^2</math> and define the dimensionless radius as,
<math>
\chi_0 \equiv \frac{r_0}{R}
</math>
to obtain,
<math>
\frac{d^2x}{d\chi_0^2} + \biggl[\frac{4}{\chi_0} - \biggl(\frac{R g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{d\chi_0} + \biggl(\frac{R^2 \rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{R \chi_0} \biggr] x = 0 .
</math>
Now normalize <math>P_0</math> to <math>P_c</math> and <math>\rho_0</math> to <math>\rho_c</math> to obtain,
<math>
\frac{d^2x}{d\chi_0^2} + \biggl[\frac{4}{\chi_0} - \biggl(\frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{P_c}{P_0}\biggr) \biggl(\frac{R g_0 \rho_c}{P_c}\biggr) \biggr] \frac{dx}{d\chi_0} + \biggl(\frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{P_c}{P_0}\biggr) \biggl(\frac{R^2 \rho_c}{\gamma_\mathrm{g} P_c} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{R \chi_0} \biggr] x = 0 .
</math>
The characteristic time for dynamical oscillations in spherically symmetric configurations (SSC) appears to be,
<math>
\tau_\mathrm{SSC} \equiv \biggl[ \frac{R^2 \rho_c}{P_c} \biggr]^{1/2} ,
</math>
and the characteristic gravitational acceleration appears to be,
<math>
g_\mathrm{SSC} \equiv \frac{P_c}{R \rho_c} .
</math>
So we can write,
<math>
\frac{d^2x}{d\chi_0^2} + \biggl[\frac{4}{\chi_0} - \biggl(\frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{P_0}{P_c}\biggr)^{-1} \biggl(\frac{g_0}{g_\mathrm{SSC}}\biggr) \biggr] \frac{dx}{d\chi_0} + \biggl(\frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{P_0}{P_c}\biggr)^{-1} \biggl(\frac{1}{\gamma_\mathrm{g}} \biggr)\biggl[\tau_\mathrm{SSC}^2 \omega^2 + (4 - 3\gamma_\mathrm{g})\biggl(\frac{g_0}{g_\mathrm{SSC}}\biggr) \frac{1}{\chi_0} \biggr] x = 0 .
</math>
This is the governing relation that we will use to analyze the stability against radial pulsations of spherically symmetric, self-gravitating configurations.
See Also
- Part I of Spherically Symmetric Configurations: Simplified Governing Equations
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