SSC/Structure/UniformDensity - Revision history

174.64.14.12: /* Summary */

2021-09-23T00:06:59Z

Summary

← Older revision		Revision as of 20:06, 22 September 2021
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	: 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,		: 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,
	<div align="center">		<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) = \biggl[ \frac{\pi}{6} G^3 \rho_c^4 M^2 \biggr]^{1/3}</math> ;		<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> ;<br />
			[http://astrowww.phys.uvic.ca/~tatum/celmechs/celm5.pdf J. B. Tatum (2021)] Celestial Mechanics class notes (UVic), §5.13, p. 45, Eq. (5.13.4)
	</div>		</div>

174.64.14.12: /* Solution Technique 1 */

2021-09-23T00:06:08Z

Solution Technique 1

← Older revision		Revision as of 20:06, 22 September 2021
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	<div align="center">		<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> ,~~<br />~~		<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> ,
	~~[http://astrowww.phys.uvic.ca/~tatum/celmechs/celm5.pdf J. B. Tatum (2021)] Celestial Mechanics class notes (UVic), §5.13, p. 45, Eq. (5.13.4)~~
	</div>		</div>

174.64.14.12: /* Solution Technique 1 */

2021-09-23T00:03:44Z

Solution Technique 1

← Older revision		Revision as of 20:03, 22 September 2021
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	<div align="center">		<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> ,		<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> ,<br />
			[http://astrowww.phys.uvic.ca/~tatum/celmechs/celm5.pdf J. B. Tatum (2021)] Celestial Mechanics class notes (UVic), §5.13, p. 45, Eq. (5.13.4)
	</div>		</div>

174.64.14.12: /* Summary */

2021-09-22T23:54:12Z

Summary

← Older revision		Revision as of 19:54, 22 September 2021
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	: 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:		: 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:
	<div align="center">		<div align="center">
	<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> .		<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> .<br />
			[http://astrowww.phys.uvic.ca/~tatum/celmechs/celm5.pdf J. B. Tatum (2021)] Celestial Mechanics class notes (UVic), §5.8.9, p. 36, Eq. (5.8.23)
	</div>		</div>

174.64.8.247: /* Isolated Uniform-Density Sphere */

2021-07-28T17:13:55Z

Isolated Uniform-Density Sphere

← Older revision		Revision as of 13:13, 28 July 2021
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	<br />		<br />
	<br />		<br />

	==Solution Technique 1==		==Solution Technique 1==

174.64.8.247: /* Isolated Uniform-Density Sphere */

2021-07-28T17:03:51Z

Isolated Uniform-Density Sphere

← Older revision		Revision as of 13:03, 28 July 2021
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	=Isolated Uniform-Density Sphere=		=Isolated Uniform-Density Sphere=

	[[~~Image:LSU_Structure_still.gif\|74px~~\|~~left~~]]		{\| class="PGEclass" style="float:left; margin-right: 20px; border-style: solid; border-width: 3px border-color: black"
			\|-
			! style="height: 125px; width: 125px; background-color:#ffff99;" \|
			<font size="-1">[[H_BookTiledMenu#Equilibrium_Structures\|<b>Uniform-Density<br />Sphere</b>]]</font>
			\|}
	Here we derive the interior structural properties of an isolated uniform-density sphere using all three [[SSCpt2/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.		Here we derive the interior structural properties of an isolated uniform-density sphere using all three [[SSCpt2/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.
			<br />
			<br />
			<br />

	==Solution Technique 1==		==Solution Technique 1==

Jet53man at 22:08, 26 July 2021

2021-07-26T22:08:20Z

← Older revision		Revision as of 18:08, 26 July 2021
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	==Solution Technique 1==		==Solution Technique 1==

	Adopting [[~~SphericallySymmetricConfigurations~~/SolutionStrategies#Technique_1\|solution technique #1]], we need to solve the integro-differential equation,		Adopting [[SSCpt2/SolutionStrategies#Technique_1\|solution technique #1]], we need to solve the integro-differential equation,

	<div align="center">		<div align="center">
	{{~~User:Tohline/~~Math/EQ_SShydrostaticBalance01}}		{{Math/EQ_SShydrostaticBalance01}}
	</div>		</div>
	appreciating that,		appreciating that,
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	<math> M_r \equiv \int_0^r 4\pi r^2 \rho dr </math> .		<math> M_r \equiv \int_0^r 4\pi r^2 \rho dr </math> .
	</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,		For a uniform-density configuration, {{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">		<div align="center">
	<math> M_r = \frac{4\pi}{3}\rho_c r^3 </math> .		<math> M_r = \frac{4\pi}{3}\rho_c r^3 </math> .
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	==Solution Technique 3==		==Solution Technique 3==

	Adopting [[~~User:Tohline/SphericallySymmetricConfigurations~~/SolutionStrategies#Technique_3\|solution technique #3]], we need to solve the ''algebraic'' expression,		Adopting [[SSCpt2/SolutionStrategies#Technique_3\|solution technique #3]], we need to solve the ''algebraic'' expression,
	<div align="center">		<div align="center">
	<math>~H + \Phi = C_\mathrm{B}</math> .		<math>~H + \Phi = C_\mathrm{B}</math> .
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	<math>\frac{1}{r^2} \frac{d }{dr} \biggl( r^2 \frac{d \Phi}{dr} \biggr) = 4\pi G \rho </math> .		<math>\frac{1}{r^2} \frac{d }{dr} \biggl( r^2 \frac{d \Phi}{dr} \biggr) = 4\pi G \rho </math> .
	</div>		</div>
	Appreciating that, as shown above, for a uniform density ({{~~User:Tohline/~~Math/VAR_Density01}} = <math>\rho_c</math> = constant) configuration,		Appreciating that, as shown above, for a uniform density ({{Math/VAR_Density01}} = <math>\rho_c</math> = constant) configuration,
	<div align="center">		<div align="center">
	<math> M_r = \int_0^r 4\pi r^2 \rho dr = \frac{4\pi}{3}\rho_c r^3 </math> ,		<math> M_r = \int_0^r 4\pi r^2 \rho dr = \frac{4\pi}{3}\rho_c r^3 </math> ,
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	<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>		<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>		</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,		Turning to the above algebraic condition, we will adopt the convention that {{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">		<div align="center">
	<math>~C_\mathrm{B} = (H + \Phi)_\mathrm{surf} = \Phi_\mathrm{surf}</math> .		<math>~C_\mathrm{B} = (H + \Phi)_\mathrm{surf} = \Phi_\mathrm{surf}</math> .
	</div>		</div>
	Therefore, everywhere inside the configuration {{~~User:Tohline/~~Math/VAR_Enthalpy01}} must be given by the expression,		Therefore, everywhere inside the configuration {{Math/VAR_Enthalpy01}} must be given by the expression,
	<div align="center">		<div align="center">
	<math>~H(r) = \Phi_\mathrm{surf} - \Phi(r)</math> .		<math>~H(r) = \Phi_\mathrm{surf} - \Phi(r)</math> .
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	==Solution Technique 2==		==Solution Technique 2==

	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:		Adopting [[SSCpt2/SolutionStrategies#Technique_2\|solution technique #2]], we need to solve the following single, <math>2^\mathrm{nd}</math>-order ODE:
	<div align="center">		<div align="center">
	<math>\frac{1}{r^2} \frac{d }{dr} \biggl( r^2 \frac{d H}{dr} \biggr) = - 4\pi G \rho </math> .		<math>\frac{1}{r^2} \frac{d }{dr} \biggl( r^2 \frac{d H}{dr} \biggr) = - 4\pi G \rho </math> .
	</div>		</div>
	Appreciating again that, for a uniform density ({{~~User:Tohline/~~Math/VAR_Density01}} = <math>\rho_c</math> = constant) configuration,		Appreciating again that, for a uniform density ({{Math/VAR_Density01}} = <math>\rho_c</math> = constant) configuration,
	<div align="center">		<div align="center">
	<math> M_r = \int_0^r 4\pi r^2 \rho dr = \frac{4\pi}{3}\rho_c r^3 </math> ,		<math> M_r = \int_0^r 4\pi r^2 \rho dr = \frac{4\pi}{3}\rho_c r^3 </math> ,
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	<math> \frac{dH}{dr} = -\frac{4\pi G}{3} \rho_c r </math> ,		<math> \frac{dH}{dr} = -\frac{4\pi G}{3} \rho_c r </math> ,
	</div>		</div>
	everywhere inside the configuration. Integrating this expression from any point inside the configuration to the surface — where, again, we adopt the convention that {{~~User:Tohline/~~Math/VAR_Enthalpy01}} = 0 — we find that,		everywhere inside the configuration. Integrating this expression from any point inside the configuration to the surface — where, again, we adopt the convention that {{Math/VAR_Enthalpy01}} = 0 — we find that,
	<div align="center">		<div align="center">
	<math> \int_{H(r)}^{0} dH = - \frac{4\pi G}{3} \rho_c \int_r^R r dr </math> <br />		<math> \int_{H(r)}^{0} dH = - \frac{4\pi G}{3} \rho_c \int_r^R r dr </math> <br />
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	For the ''isolated'' uniform-density sphere, discussed above, the surface of the configuration was identified as the radial location where the pressure drops to zero. Here we embed the sphere in a hot, tenuous medium that exerts a confining "external" pressure, <math>~P_e</math>, and ask how the configuration's equilibrium radius, <math>~R_e</math>, changes in response to this applied external pressure, for a given (fixed) total mass and central pressure, <math>~P_c</math>.		For the ''isolated'' uniform-density sphere, discussed above, the surface of the configuration was identified as the radial location where the pressure drops to zero. Here we embed the sphere in a hot, tenuous medium that exerts a confining "external" pressure, <math>~P_e</math>, and ask how the configuration's equilibrium radius, <math>~R_e</math>, changes in response to this applied external pressure, for a given (fixed) total mass and central pressure, <math>~P_c</math>.

	Following [[~~User:Tohline/~~SSC/Structure/UniformDensity#Solution_Technique_1\|solution technique #1]], the derivation remains the same up through the integration of the hydrostatic balance equation to obtain the relation,		Following [[SSC/Structure/UniformDensity#Solution_Technique_1\|solution technique #1]], the derivation remains the same up through the integration of the hydrostatic balance equation to obtain the relation,
	<div align="center">		<div align="center">
	<math>P = P_c - \frac{2\pi G}{3} \rho_c^2 r^2 \, .</math>		<math>P = P_c - \frac{2\pi G}{3} \rho_c^2 r^2 \, .</math>

Jet53man: Created page with " =Isolated Uniform-Density Sphere= Image:LSU_Structure_still.gif|7..."

2021-07-26T22:04:33Z

Created page with "  =Isolated Uniform-Density Sphere= Image:LSU_Structure_still.gif|7..."

New page

=Isolated Uniform-Density Sphere=

[[Image:LSU_Structure_still.gif|74px|left]]
Here we derive the interior structural properties of an isolated uniform-density sphere using all three [[SSCpt2/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.

==Solution Technique 1==

Adopting [[SphericallySymmetricConfigurations/SolutionStrategies#Technique_1|solution technique #1]], we need to solve the integro-differential equation,

<div align="center">
{{User:Tohline/Math/EQ_SShydrostaticBalance01}}
</div>
appreciating that,
<div align="center">
<math> M_r \equiv \int_0^r 4\pi r^2 \rho dr </math> .
</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 — where <math>~r=0</math> and <math>~P = P_c</math> — out to an arbitrary radius <math>~r</math> that is still inside the configuration, we obtain,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~ \int_{P_c}^P dP </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- \frac{4\pi G}{3} \rho_c^2 \int_0^r r dr </math>
</td>
</tr>

<tr>
<td align="right">
<math>~ \Rightarrow ~~~ P </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~P_c - \frac{2\pi G}{3} \rho_c^2 r^2 \, .</math>
</td>
</tr>
</table>

We expect the pressure to drop to zero at the surface of our spherical configuration — that is, at <math>~r=R</math> — 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 [[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==

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 — where, again, we adopt the convention that {{User:Tohline/Math/VAR_Enthalpy01}} = 0 — 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">
==Summary==
</font>

From the above derivations, we can describe the properties of a uniform-density, self-gravitating sphere as follows:
* <font color="red">Mass</font>:
: Given the density, <math>\rho_c</math>, and the radius, <math>R</math>, of the configuration, the total mass is,
<div align="center">
<math>M = \frac{4\pi}{3} \rho_c R^3 </math> ;
</div>

: and, expressed as a function of <math>M</math>, the mass that lies interior to radius <math>r</math> is,
<div align="center">
<math>\frac{M_r}{M} = \biggl(\frac{r}{R} \biggr)^3</math> .
</div>
* <font color="red">Pressure</font>:
: 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,
<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) = \biggl[ \frac{\pi}{6} G^3 \rho_c^4 M^2 \biggr]^{1/3}</math> ;
</div>

: and, expressed in terms of the central pressure <math>P_c</math>, the variation with radius of the pressure is,
<div align="center">
<math>P(r) = P_c \biggl[ 1 -\biggl(\frac{r}{R} \biggr)^2 \biggr]</math> .
</div>

* <span id="UniformSphereEnthalpy"><font color="red">Enthalpy</font>:</span>
: Throughout the configuration, the enthalpy is given by the relation,
<div align="center">
<math>H(r) = \frac{P(r)}{ \rho_c} = \frac{GM}{2R} \biggl[ 1 -\biggl(\frac{r}{R} \biggr)^2 \biggr]</math> .
</div>

* <span id="UniformSpherePotential"><font color="red">Gravitational potential</font>: </span>
: Throughout the configuration — that is, for all <math>r \leq R</math> — the gravitational potential is given by the relation,
<div align="center">
<math>\Phi_\mathrm{surf} - \Phi(r) = H(r) = \frac{G M}{2R} \biggl[ 1- \biggl(\frac{r}{R} \biggr)^2 \biggr] </math> .
</div>
: 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,
<div align="center">
<math>\Phi(r) = - \frac{GM}{r} </math> .
</div>
: 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:
<div align="center">
<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> .
</div>

* <font color="red">Mass-Radius relationship</font>:
: We see that, for a given value of <math>\rho_c</math>, the relationship between the configuration's total mass and radius is,
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<math>M \propto R^3 ~~~~~\mathrm{or}~~~~~R \propto M^{1/3} </math> .
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* <font color="red">Central- to Mean-Density Ratio</font>:
: Because this is a uniform-density structure, the ratio of its central density to its mean density is unity, that is,
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<math>\frac{\rho_c}{\bar{\rho}} = 1 </math> .
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=Uniform-Density Sphere Embedded in an External Medium=
For the ''isolated'' uniform-density sphere, discussed above, the surface of the configuration was identified as the radial location where the pressure drops to zero. Here we embed the sphere in a hot, tenuous medium that exerts a confining "external" pressure, <math>~P_e</math>, and ask how the configuration's equilibrium radius, <math>~R_e</math>, changes in response to this applied external pressure, for a given (fixed) total mass and central pressure, <math>~P_c</math>.

Following [[User:Tohline/SSC/Structure/UniformDensity#Solution_Technique_1|solution technique #1]], the derivation remains the same up through the integration of the hydrostatic balance equation to obtain the relation,
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<math>P = P_c - \frac{2\pi G}{3} \rho_c^2 r^2 \, .</math>
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Now we set <math>~P = P_e</math> at the surface of our spherical configuration — that is, at <math>~r=R_e</math> — so we can write,
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<math>P_c - P_e = \frac{2\pi G}{3} \rho_c^2 R_e^2 = \frac{3G}{8\pi}\biggl( \frac{M^2}{R_e^4} \biggr)</math>

<math>\Rightarrow ~~~~~ P_c \biggl( 1 - \frac{P_e}{P_c} \biggr) = \frac{3G}{8\pi}\biggl( \frac{M^2}{R_e^4} \biggr) \, ,</math>
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where <math>M</math> is the total mass of the configuration. Solving for the equilibrium radius, we have,
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<math> R_e = \biggl[ \biggl( \frac{3}{2^3\pi} \biggr) \frac{G M^2}{P_c} \biggl( 1 - \frac{P_e}{P_c} \biggr)^{-1} \biggr]^{1/4} \, .</math>
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As it should, when the ratio <math>~P_e/P_c \rightarrow 0</math>, this relation reduces to the one obtained, above, for the isolated uniform-density sphere, namely,
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<math> R_e^4 = \biggl( \frac{3}{8\pi} \biggr) \frac{G M^2}{P_c} \, .</math>
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