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=Uniform-Density Sphere (structure)=
=Spherically Symmetric Configurations (Stability — Part III)=
[[Image:LSU_Structure_still.gif|74px|left]]
[[Image:LSU_Stable.animated.gif|74px|left]]
Here we're going to derive the interior structural properties of a uniform-density sphere using all three [http://www.vistrails.org/index.php/User:Tohline/SphericallySymmetricConfigurations/SolutionStrategies#Solution_Strategies solution strategies].  While deriving essentially the same solution three different ways might seem like a bit of overkill, this approach proves to be instructive because (a) it forces us to examine the structural behavior of a number of different physical parameters, and (b) it illustrates how to work through the different solution strategies for one model whose structure can in fact be derived analytically using any of the techniques. As we shall see when studying other self-gravitating configurations, the three strategies are not always equally fruitful.
Suppose we now want to study the '''stability''' of one of the spherically symmetric, equilibrium structures that have been derived elsewhere.  The [http://www.vistrails.org/index.php/User:Tohline/SSC/Perturbations 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.


==Solution Technique 1==
==The Eigenvalue Problem==


Adopting [http://www.vistrails.org/index.php/User:Tohline/SphericallySymmetricConfigurations/SolutionStrategies#Technique_1 solution technique #1], we need to solve the integro-differential equation,
As has been derived in [http://www.vistrails.org/index.php/User:Tohline/SSC/Perturbations#Eigen_Value_Problem an accompanying discussion], the second-order ODE that defines the Eigenvalue problem is,


<div align="center">
<div align="center">
{{User:Tohline/Math/EQ_SShydrostaticBalance01}}
<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><br />
</div>
</div>
appreciating that,
where, <math>P_0(r_0)</math> and <math>\rho_0(r_0)</math> are the pressure and density distributions in the unperturbed initial equilibrium model and the gravitational acceleration at each radial location in the unperturbed model is,
<div align="center">
<div align="center">
<math> M_r \equiv \int_0^r 4\pi r^2 \rho dr </math> .
<math>  
g_0(r_0) \equiv \frac{GM_r(r_0)}{r_0^2} = - \frac{1}{\rho_0} \frac{dP_0}{dr_0} .
</math><br />
</div>
</div>
For a uniform-density configuration, {{User:Tohline/Math/VAR_Density01}} = <math>\rho_c</math> = constant, so the density can be pulled outside the mass integral and the integral can be completed immediately to give,
<div align="center">
<math> M_r = \frac{4\pi}{3}\rho_c r^3 </math> .
</div>
Hence, the differential equation describing hydrostatic balance becomes,
<div align="center">
<math> \frac{dP}{dr} = - \frac{4\pi G}{3} \rho_c^2 r </math> .
</div>
Integrating this from the center of the configuration &#8212; where <math>r=0</math> and <math>P = P_c</math> &#8212; out to an arbitrary radius <math>r</math> that is still inside the configuration, we obtain,
<div align="center">
<math> \int_{P_c}^P dP = - \frac{4\pi G}{3} \rho_c^2 \int_0^r r dr </math><br />
<math>\Rightarrow ~~~~ P  = P_c - \frac{2\pi G}{3} \rho_c^2 r^2 </math>
</div>
We expect the pressure to drop to zero at the surface of our spherical configuration &#8212; that is, at <math>r=R</math> &#8212; so the central pressure must be,
<div align="center">
<math>P_c = \frac{2\pi G}{3} \rho_c^2 R^2 = \frac{3G}{8\pi}\biggl( \frac{M^2}{R^4} \biggr)</math> ,
</div>
where <math>M</math> is the total mass of the configuration.  Finally, then, we have,
<div align="center">
<math>P(r) = P_c\biggl[1 - \biggl(\frac{r}{R}\biggr)^2 \biggr] </math> .
</div>
==Solution Technique 3==
Adopting [http://www.vistrails.org/index.php/User:Tohline/SphericallySymmetricConfigurations/SolutionStrategies#Technique_3 solution technique #3], we need to solve the ''algebraic'' expression,
<div align="center">
<math>H + \Phi = C_\mathrm{B}</math> .
</div>
in conjunction with the Poisson equation,
<div align="center">
<math>\frac{1}{r^2} \frac{d }{dr} \biggl( r^2 \frac{d \Phi}{dr} \biggr)  = 4\pi G \rho </math> .
</div>
Appreciating that, as shown above, for a uniform density ({{User:Tohline/Math/VAR_Density01}} = <math>\rho_c</math> = constant) configuration,
<div align="center">
<math> M_r = \int_0^r 4\pi r^2 \rho dr = \frac{4\pi}{3}\rho_c r^3 </math> ,
</div>
we can integrate the Poisson equation once to give,
<div align="center">
<math> \frac{d\Phi}{dr} = \frac{4\pi G}{3} \rho_c r </math> ,
</div>
everywhere inside the configuration.  Integrating this expression from any point inside the configuration to the surface, we find that,
<div align="center">
<math> \int_{\Phi(r)}^{\Phi_\mathrm{surf}} d\Phi = \frac{4\pi G}{3} \rho_c \int_r^R r dr </math> <br />
<math>\Rightarrow ~~~~~ \Phi_\mathrm{surf} - \Phi(r) = \frac{2\pi G}{3} \rho_c R^2 \biggl[ 1- \biggl(\frac{r}{R} \biggr)^2 \biggr] </math>
</div>
Turning to the above algebraic condition, we will adopt the convention that {{User:Tohline/Math/VAR_Enthalpy01}} is set to zero at the surface of a barotropic configuration, in which case the constant, <math>C_\mathrm{B}</math>, must be,
<div align="center">
<math>C_\mathrm{B} = (H + \Phi)_\mathrm{surf} = \Phi_\mathrm{surf}</math> .
</div>
Therefore, everywhere inside the configuration {{User:Tohline/Math/VAR_Enthalpy01}} must be given by the expression,
<div align="center">
<math>H(r) = \Phi_\mathrm{surf} - \Phi(r)</math> .
</div>
Matching this with our solution of the Poisson equation, we conclude that, throughout the configuration,
<div align="center">
<math> H(r) = \frac{2\pi G}{3} \rho_c R^2 \biggl[ 1- \biggl(\frac{r}{R} \biggr)^2 \biggr]</math> .
</div>
Comparing this result with the result we obtained using solution technique #1, it is clear that throughout a uniform-density, self-gravitating sphere,
<div align="center">
<math>\frac{P}{H} = \rho</math> .
</div>


==Solution Technique 2==
==Uniform-Density Configuration==
 
From our derived [http://www.vistrails.org/index.php?title=User:Tohline/SSC/UniformDensity Structure of a uniform-density sphere], the required functions are,
Adopting [http://www.vistrails.org/index.php/User:Tohline/SphericallySymmetricConfigurations/SolutionStrategies#Technique_2 solution technique #2], we need to solve the following single, <math>2^\mathrm{nd}</math>-order ODE:
<div align="center">
<math>\frac{1}{r^2} \frac{d }{dr} \biggl( r^2 \frac{d H}{dr} \biggr)  = - 4\pi G \rho </math> .
</div>
Appreciating again that, for a uniform density ({{User:Tohline/Math/VAR_Density01}} = <math>\rho_c</math> = constant) configuration,
<div align="center">
<math> M_r = \int_0^r 4\pi r^2 \rho dr = \frac{4\pi}{3}\rho_c r^3 </math> ,
</div>
we can integrate the <math>2^\mathrm{nd}</math>-order ODE once to give,
<div align="center">
<math> \frac{dH}{dr} = -\frac{4\pi G}{3} \rho_c r </math> ,
</div>
everywhere inside the configuration.  Integrating this expression from any point inside the configuration to the surface &#8212; where, again, we adopt the convention that {{User:Tohline/Math/VAR_Enthalpy01}} = 0 &#8212; we find that,
<div align="center">
<math> \int_{H(r)}^{0} dH = - \frac{4\pi G}{3} \rho_c \int_r^R r dr </math> <br />
<math>\Rightarrow ~~~~~ H(r) = \frac{2\pi G}{3} \rho_c R^2 \biggl[ 1- \biggl(\frac{r}{R} \biggr)^2 \biggr] </math> .
</div>


<font color="darkblue">
<font color="darkblue">

Revision as of 23:49, 14 February 2010

Whitworth's (1981) Isothermal Free-Energy Surface
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Spherically Symmetric Configurations (Stability — Part III)

LSU Stable.animated.gif

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.

The Eigenvalue Problem

As has been derived in an accompanying discussion, the second-order ODE that defines the Eigenvalue problem is,

<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>

where, <math>P_0(r_0)</math> and <math>\rho_0(r_0)</math> are the pressure and density distributions in the unperturbed initial equilibrium model and the gravitational acceleration at each radial location in the unperturbed model is,

<math> g_0(r_0) \equiv \frac{GM_r(r_0)}{r_0^2} = - \frac{1}{\rho_0} \frac{dP_0}{dr_0} . </math>

Uniform-Density Configuration

From our derived Structure of a uniform-density sphere, the required functions are,

Summary

From the above derivations, we can describe the properties of a uniform-density, self-gravitating sphere as follows:

  • Mass:
Given the density, <math>\rho_c</math>, and the radius, <math>R</math>, of the configuration, the total mass is,

<math>M = \frac{4\pi}{3} \rho_c R^3 </math> ;

and, expressed as a function of <math>M</math>, the mass that lies interior to radius <math>r</math> is,

<math>\frac{M_r}{M} = \biggl(\frac{r}{R} \biggr)^3</math> .

  • Pressure:
Given values for the pair of model parameters <math>( \rho_c , R )</math>, or <math>( M , R )</math>, or <math>( \rho_c , M )</math>, the central pressure of the configuration is,

<math>P_c = \frac{2\pi G}{3} \rho_c^2 R^2 = \frac{3G}{8\pi}\biggl( \frac{M^2}{R^4} \biggr) = \biggl[ \frac{\pi}{6} G^3 \rho_c^4 M^2 \biggr]^{1/3}</math> ;

and, expressed in terms of the central pressure <math>P_c</math>, the variation with radius of the pressure is,

<math>P(r) = P_c \biggl[ 1 -\biggl(\frac{r}{R} \biggr)^2 \biggr]</math> .

  • Enthalpy:
Throughout the configuration, the enthalpy is given by the relation,

<math>H(r) = \frac{P(r)}{ \rho_c} = \frac{GM}{2R} \biggl[ 1 -\biggl(\frac{r}{R} \biggr)^2 \biggr]</math> .

  • Gravitational potential:
Throughout the configuration — that is, for all <math>r \leq R</math> — the gravitational potential is given by the relation,

<math>\Phi_\mathrm{surf} - \Phi(r) = H(r) = \frac{G M}{2R} \biggl[ 1- \biggl(\frac{r}{R} \biggr)^2 \biggr] </math> .

Outside of this spherical configuration— that is, for all <math>r \geq R</math> — the potential should behave like a point mass potential, that is,

<math>\Phi(r) = - \frac{GM}{r} </math> .

Matching these two expressions at the surface of the configuration, that is, setting <math>\Phi_\mathrm{surf} = - GM/R</math>, we have what is generally considered the properly normalized prescription for the gravitational potential inside a uniform-density, spherically symmetric configuration:

<math>\Phi(r) = - \frac{G M}{R} \biggl\{ 1 + \frac{1}{2}\biggl[ 1- \biggl(\frac{r}{R} \biggr)^2 \biggr] \biggr\} = - \frac{3G M}{2R} \biggl[ 1 - \frac{1}{3} \biggl(\frac{r}{R} \biggr)^2 \biggr] </math> .

  • Mass-Radius relationship:
We see that, for a given value of <math>\rho_c</math>, the relationship between the configuration's total mass and radius is,

<math>M \propto R^3 ~~~~~\mathrm{or}~~~~~R \propto M^{1/3} </math> .

  • Central- to Mean-Density Ratio:
Because this is a uniform-density structure, the ratio of its central density to its mean density is unity, that is,

<math>\frac{\rho_c}{\bar{\rho}} = 1 </math> .


Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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