User:Tohline/SSC/Perturbations

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 <math>~n = 1</math> 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)</math>	<math>~=</math>	<math>~P_0(m) + P_1(m,t) = P_0(m) \biggl[1 + p(m) e^{i\omega t} \biggr] \, ,</math>
<math>~\rho(m,t)</math>	<math>~=</math>	<math>~\rho_0(m) + \rho_1(m,t) = \rho_0(m) \biggl[1 + d(m) e^{i\omega t} \biggr] \, ,</math>
<math>~r(m,t)</math>	<math>~=</math>	<math>~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}</math>	<math>~=</math>	<math>~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\}</math>	<math>~=</math>	<math>~0 \, .</math>

Because 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>

Also, because 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} </math>	<math> \rightarrow </math>	<math> (i\omega)\rho_0 d~e^{i\omega t} </math>
<math> \frac{\rho}{r} </math>	<math> \rightarrow </math>	<math> \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} </math>	<math> \rightarrow </math>	<math> (i\omega) r_0 x~e^{i\omega t} </math>
<math> ~\rho^2 r^2 </math>	<math> \rightarrow </math>	<math> \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) </math>	<math> \approx </math>	<math> \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}</math>	<math>~\approx</math>	<math>~- \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 </math>	<math>~\approx</math>	<math>~- 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} </math>	<math>~\approx</math>	<math>~- 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} </math>	<math> \rightarrow </math>	<math> \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} </math>	<math> \rightarrow </math>	<math> 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} </math>	<math> \rightarrow </math>	<math> \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} = - \frac{1}{\rho_0} \frac{dP_0}{dr_0} \, . </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]</math>	<math>~=</math>	<math>~- \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}</math>	<math>~=</math>	<math>~- \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}</math>	<math>~=</math>	<math>~\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}</math>	<math>~=</math>	<math>~- \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 2^nd-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>

Classic Papers that Derive & Use this Relation

Eddington (1926)

Eddington (1926)

To our knowledge, a derivation of this governing 2nd-order ODE was first presented by Eddington (1926) in a book titled, The Internal Constitution of the Stars. (This entire book has been digitally scanned and is now available online.) The derived expression, which appears on p. 188 as equation (127.6) of Eddington's book, is presented in the following, framed image.

Pulsation Equation as Derived and Presented by Eddington (1926)

The similarity between Eddington's expression and the governing 2nd-order ODE that we have derived is immediately apparent. Specifically, simply after inserting Eddington's definition of his composite variable, <math>~\mu</math>, and making the substitutions,

<math>~\xi_1 \rightarrow x \, ,</math>                 <math>\xi_0 \rightarrow r_0 \, ,</math>         and         <math>n^2 \rightarrow \omega^2 \, ,</math>

Eddington's pulsation equation becomes,

<math>~ \frac{d^2 x}{dr_0^2} + \biggl[ \frac{4}{r_0} - \frac{g_0 \rho_0}{P_0} \biggr] \frac{dx}{dr_0} + \frac{\rho_0}{\gamma P_0} \biggl\{\omega^2 + \biggl(4 - 3\gamma\biggr) \frac{g_0 }{r_0} \biggr\}x </math>

<math>~=</math>

<math>~0 \, ,</math>

which exactly matches our derived governing relation.

Ledoux and Pekeris (1941)

Historically, by the 1940s, the expression just derived was a relatively familiar one to astrophysicists. For example, the opening paragraph of a 1941 paper by Ledoux & Pekeris (1941, ApJ, 94, 124), reads:

If we divide their equation (1) through by <math>~Xr = \Gamma_1 P r</math> and recognize that,

<math> \frac{dX}{dr} = \frac{dX}{dm}\frac{dm}{dr} = - \Gamma_1 g_0 \rho \, , </math>

we obtain,

<math> \frac{d^2\xi}{dr^2} + \biggl[ \frac{4}{r} - \frac{g_0 \rho}{P} \biggr] \frac{d\xi}{dr} +\frac{\rho}{\Gamma_1 P} \biggl[ \sigma^2 + (4 - 3\Gamma_1) \frac{g_0}{r} \biggr] \xi = 0 \, . </math>

This is clearly the same <math>2^\mathrm{nd}</math>-order, ordinary differential equation as the one we have derived, but with a more general definition of the adiabatic exponent that allows consideration of a situation where the total pressure is a sum of both gas and radiation pressure.

Schwarzschild (1941)

In the same volume of The Astrophysical Journal, Schwarzschild (1941) published work on "Overtone Pulsations for the Standard [Stellar] Model." The following excerpt has been drawn from the first page of this article.

3A. S. Eddington (1930), The Internal Constitution of the Stars, pp. 188 and 192.

The similarity between Schwarzschild's "pulsation equation" and the governing 2nd-order ODE that we have derived is immediately apparent; for example, the eigenfrequency, <math>~\omega</math>, is the same in both,

<math>~\xi_1 \leftrightarrow x</math>         and         <math>\xi_0 \leftrightarrow r_0 \, .</math>

But the two equations are not exactly the same. To show this, we begin by comparing the last term on the lefthand-side in both expressions and presume that Schwarzschild's <math>~u</math> is related to the state variables in our equation as,

<math>u = \frac{\gamma_g P_0}{\rho_0} \, .</math>

Restricting the discussion to only polytropic equations of state — that is, <math>~P_0 = K\rho_0^{(n+1)/n}</math> — we can write,

<math>~u</math>

<math>~=</math>

<math>~\gamma_g K\rho_0^{1/n} \, ,</math>

which means that,

<math>~\frac{du}{d\xi_0}</math>

<math>~=</math>

<math>~\biggl( \frac{\gamma_g K}{n} \biggr) \rho_0^{(1-n)/n} \frac{d\rho_0}{d\xi_0} = \biggl( \frac{\gamma_g}{n+1} \biggr) \frac{1}{\rho_0} \frac{dP_0}{d\xi_0} = - \biggl( \frac{\gamma_g}{n+1} \biggr) g_0 \, ,</math>

where the last step results from recalling that, by our definition above, the unperturbed gravitational acceleration is,

<math>g_0 = - \frac{1}{\rho_0} \frac{dP_0}{dr_0} \, .</math>

Hence, when considering polytropic configurations, the following substitutions are appropriate between the two equations:

<math>\frac{du}{d\xi_0} \leftrightarrow - \biggl( \frac{\gamma_g}{n+1} \biggr) g_0 </math>         and         <math>\frac{1}{u} \frac{du}{d\xi_0} \leftrightarrow - \frac{1}{(n+1)} \frac{g_0 \rho_0}{P_0} \, .</math>

Making these substitutions into Schwarzschild's pulsation equation gives,

<math> \frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \frac{4}{(n+1)}\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 - \biggl( \frac{4a\gamma_g}{n+1} \biggr)\frac{g_0}{r_0} \biggr] x = 0 \, . </math>

Appreciating that, in Schwarzschild's expression, <math>~a \equiv - (4-3\gamma_g)/\gamma_g</math>, we see that our expression matches his if and only if <math>~n=3</math>. This is, indeed, precisely the "standard model" that Schwarzschild was considering.

Dimensionless Expression

Let's write this governing, 2nd-order 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.

Complementary Approach

Looking back at the set of three, coupled, linearized equations — but using <math>~\Delta</math> to denote the density fluctuation, rather than <math>~d</math>, in order to avoid confusion with the differentiation operators,

Linearized Equation of Continuity <math> r_0 \frac{dx}{dr_0} = - 3 x - \Delta , </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} \Delta \, , </math>

let's combine them into a 2^nd-order ODE that governs the eigenfunction of the density perturbation, <math>~\Delta</math>, rather than of the radial displacement, <math>~x</math>. We begin by using the second equation to obtain an expression for <math>~x</math>:

<math>~ x</math>

<math>~=</math>

<math>~(4g_0 + \omega^2 r_0)^{-1}\biggl[ \frac{P_0}{\rho_0} \frac{dp}{dr_0} - pg_0 \biggr] \, .</math>

From this, we determine that,

<math>~ (4g_0 + \omega^2 r_0)\frac{dx}{dr_0}</math>	<math>~=</math>	<math>~\frac{d}{dr_0} \biggl[ \frac{P_0}{\rho_0} \frac{dp}{dr_0} - pg_0 \biggr] + (4g_0 + \omega^2 r_0)\biggl[ \frac{P_0}{\rho_0} \frac{dp}{dr_0} - pg_0 \biggr]\frac{d}{dr_0}(4g_0 + \omega^2 r_0)^{-1} </math>
	<math>~=</math>	<math>~\frac{P_0}{\rho_0} \frac{d^2p}{dr_0^2} + \frac{dp}{dr_0} \cdot \frac{d}{dr_0} \biggl( \frac{P_0}{\rho_0} \biggr) -p \frac{dg_0}{dr_0} - g_0\frac{dp}{dr_0} - (4g_0 + \omega^2 r_0)^{-1}\biggl[ \frac{P_0}{\rho_0} \frac{dp}{dr_0} - pg_0 \biggr] \biggl[ \omega^2 + 4\frac{dg_0}{dr_0}\biggr] </math>
	<math>~=</math>	<math>~\frac{P_0}{\rho_0} \frac{d^2p}{dr_0^2} + \frac{dp}{dr_0} \biggl\{ -g_0 + \frac{d}{dr_0} \biggl( \frac{P_0}{\rho_0} \biggr) - (4g_0 + \omega^2 r_0)^{-1}\biggl[ \frac{P_0}{\rho_0} \biggr] \biggl[ \omega^2 + 4\frac{dg_0}{dr_0}\biggr] \biggr\} + p \biggl\{ (4g_0 + \omega^2 r_0)^{-1}\biggl[ g_0 \biggr] \biggl[ \omega^2 + 4\frac{dg_0}{dr_0}\biggr] -\frac{dg_0}{dr_0} \biggr\} </math>

Next, we substitute this expression for <math>~x</math> into the first equation — gradually replacing <math>~p</math> with <math>~\gamma_g \Delta</math>, as well — and carry out the differentiation:

<math>~ -\Delta - 3\biggl\{(4g_0 + \omega^2 r_0)^{-1}\biggl[ \frac{P_0}{\rho_0} \frac{dp}{dr_0} - \gamma_g \Delta g_0 \biggr] \biggr\} </math>	<math>~=</math>	<math>~r_0 \frac{d}{dr_0} \biggl\{(4g_0 + \omega^2 r_0)^{-1}\biggl[ \frac{P_0}{\rho_0} \frac{dp}{dr_0} - pg_0 \biggr] \biggr\} </math>
<math>~ \Rightarrow~~- \Delta \biggl[ \frac{g_0}{r_0} (4 - 3\gamma_g)+ \omega^2 \biggr] - \frac{3P_0}{r_0 \rho_0} \frac{dp}{dr_0} </math>	<math>~=</math>	<math>~ \frac{d}{dr_0} \biggl[ \frac{P_0}{\rho_0} \frac{dp}{dr_0} - pg_0 \biggr] - (4g_0 + \omega^2 r_0)^{-1} \biggl[ \frac{P_0}{\rho_0} \frac{dp}{dr_0} - pg_0 \biggr]\frac{d}{dr_0} \biggl[(4g_0 + \omega^2 r_0) \biggr] </math>

See Also

• Part I of Spherically Symmetric Configurations: Simplified Governing Equations

© 2014 - 2021 by Joel E. Tohline |   H_Book Home   |   YouTube   | Appendices: | Equations | Variables | References | Ramblings | Images | myphys.lsu | ADS | Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation