Difference between revisions of "User:Tohline/SSC/UniformDensity"
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=Uniform-Density Sphere (structure)= | =Uniform-Density Sphere (structure)= | ||
==Solution Technique | ==Solution Technique 1== | ||
Adopting [http://www.vistrails.org/index.php/User:Tohline/SphericallySymmetricConfigurations/SolutionStrategies#Technique_1 solution technique #1], we need to solve the integro-differential equation, | Adopting [http://www.vistrails.org/index.php/User:Tohline/SphericallySymmetricConfigurations/SolutionStrategies#Technique_1 solution technique #1], we need to solve the integro-differential equation, | ||
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</div> | </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> | |||
{{LSU_HBook_footer}} | {{LSU_HBook_footer}} |
Revision as of 18:26, 2 February 2010
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Uniform-Density Sphere (structure)
Solution Technique 1
Adopting solution technique #1, we need to solve the integro-differential equation,
appreciating that,
<math> M_r \equiv \int_0^r 4\pi r^2 \rho dr </math> .
For a uniform-density configuration, <math>~\rho</math> = <math>\rho_c</math> = constant, so the density can be pulled outside the mass integral and the integral can be completed immediately to give,
<math> M_r = \frac{4\pi}{3}\rho_c r^3 </math> .
Hence, the differential equation describing hydrostatic balance becomes,
<math> \frac{dP}{dr} = - \frac{4\pi G}{3} \rho_c^2 r </math> .
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,
<math> \int_{P_c}^P dP = - \frac{4\pi G}{3} \rho_c^2 \int_0^r r dr </math>
<math>\Rightarrow ~~~~ P = P_c - \frac{2\pi G}{3} \rho_c^2 r^2 </math>
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,
<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> ,
where <math>M</math> is the total mass of the configuration. Finally, then, we have,
<math>P(r) = P_c\biggl[1 - \biggl(\frac{r}{R}\biggr)^2 \biggr] </math> .
Solution Technique 3
Adopting solution technique #3, we need to solve the algebraic expression,
<math>H + \Phi = C_\mathrm{B}</math> .
in conjunction with the Poisson equation,
<math>\frac{1}{r^2} \frac{d }{dr} \biggl( r^2 \frac{d \Phi}{dr} \biggr) = 4\pi G \rho </math> .
Appreciating that, as shown above, for a uniform density (<math>~\rho</math> = <math>\rho_c</math> = constant) configuration,
<math> M_r = \int_0^r 4\pi r^2 \rho dr = \frac{4\pi}{3}\rho_c r^3 </math> ,
we can integrate the Poisson equation once to give,
<math> \frac{d\Phi}{dr} = \frac{4\pi G}{3} \rho_c r </math> ,
everywhere inside the configuration. Integrating this expression from any point inside the configuration to the surface, we find that,
<math> \int_{\Phi(r)}^{\Phi_\mathrm{surf}} d\Phi = \frac{4\pi G}{3} \rho_c \int_r^R r dr </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>
Turning to the above algebraic condition, we will adopt the convention that <math>~H</math> is set to zero at the surface of a barotropic configuration, in which case the constant, <math>C_\mathrm{B}</math>, must be,
<math>C_\mathrm{B} = (H + \Phi)_\mathrm{surf} = \Phi_\mathrm{surf}</math> .
Therefore, everywhere inside the configuration <math>~H</math> must be given by the expression,
<math>H(r) = \Phi_\mathrm{surf} - \Phi(r)</math> .
Matching this with our solution of the Poisson equation, we conclude that, throughout the configuration,
<math> H(r) = \frac{2\pi G}{3} \rho_c R^2 \biggl[ 1- \biggl(\frac{r}{R} \biggr)^2 \biggr]</math> .
Comparing this result with the result we obtained using solution technique #1, it is clear that throughout a uniform-density, self-gravitating sphere,
<math>\frac{P}{H} = \rho</math> .
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