Difference between revisions of "User:Tohline/Appendix/Ramblings/PowerSeriesExpressions"

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<tr>
   <td align="right">
   <td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - r \biggl(-\frac{dw }{dr}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - r \biggl(\frac{dw }{dr}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
   </td>
   </td>
   <td align="center">
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   <td align="left">
<math>~   
<math>~   
- \biggl[ \frac{\sigma_c^2}{6\gamma}  - \frac{\alpha}{r} \biggl(-\frac{dw }{dr}\biggr)_i \biggr]  x  \, ,
- \biggl[ \frac{\sigma_c^2}{6\gamma}  - \frac{\alpha}{r} \biggl(\frac{dw }{dr}\biggr)\biggr]  x  \, ,
</math>
</math>
   </td>
   </td>
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<tr>
   <td align="right">
   <td align="right">
<math>~-\frac{dw}{dr}</math>
<math>~\frac{dw}{dr}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~-\frac{r}{3} + \frac{r^3}{30} - \frac{r^5}{315} \, .</math>
<math>~\frac{r}{3} - \frac{r^3}{30} + \frac{r^5}{315} \, .</math>
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   </td>
</tr>
</tr>
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<math>~\frac{d^2 x}{dr^2} + \biggl[4 + \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
   </td>
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   <td align="left">
<math>~   
<math>~   
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma}  + 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr)_i \biggr]  x
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma}  - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr]  x \, .
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
- \frac{1}{6} \biggl[\mathfrak{F}  + 2\alpha \biggl(- \frac{r^2}{10} + \frac{r^4}{105}\biggr)_i \biggr]  x \, ,
</math>
</math>
   </td>
   </td>
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</table>
</div>
</div>
where, <math>~\mathfrak{F} \equiv (\sigma_c^2/\gamma + 2\alpha)</math>.  Let's now adopt a power-series expression for the displacement function of the form,
 
Let's now adopt a power-series expression for the displacement function of the form,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
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<tr>
   <td align="right">
   <td align="right">
<math>~
<math>~2b + 6cr + 12dr^2 + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr] </math>
\biggl[2b + 6cr + 12dr^2 \biggr] + \biggl[4 + \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr]  
\biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr]
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   <td align="left">
<math>~   
<math>~   
- \frac{1}{6} \biggl[\mathfrak{F+ 2\alpha \biggl(- \frac{r^2}{10} + \frac{r^4}{105}\biggr)_i \biggr]  \biggl[ 1 + ar + br^2 + cr^3 + dr^4 \biggr\, .
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma- 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr]  \biggl( 1 + ar + br^2 + cr^3 + dr^4 \biggr) \, .
</math>
</math>
   </td>
   </td>
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   <td align="right">
<math>~
<math>~
\biggl[2b + 6cr + 12dr^2 \biggr] 
+ 4\biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr]
+\frac{r^2}{3} \biggl[ \frac{a}{r} + 2b \biggr]
</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~ 
- \frac{\mathfrak{F}}{6} \biggl[ 1 + ar + br^2  \biggr]
+  \biggl[\frac{\alpha r^2}{30}  \biggr] 
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~
2b + 6cr + 12dr^2   
2b + 6cr + 12dr^2   
+ \frac{4a}{r} + 8b + 12 cr + 16dr^2  
+ 4  \biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr]
+\frac{2r}{3}  
- \frac{r^2}{3} \biggl[ \frac{a}{r} + 2b  \biggr]
+\frac{2b r^2}{3}  
</math>
</math>
   </td>
   </td>
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   <td align="left">
   <td align="left">
<math>~   
<math>~   
- \frac{\mathfrak{F}}{6} \biggl[ 1 + ar + br^2 \biggr]
- \frac{\mathfrak{F} }{6} \biggl( 1 + ar + br^2 \biggr)
\biggl[\frac{\alpha r^2}{30}  \biggr
- \frac{\alpha}{3} \biggl(\frac{r^2}{10}  \biggr)
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
where,
<div align="center">
<math>~\mathfrak{F} \equiv \frac{\sigma_c^2}{\gamma} - 2\alpha \, .</math>
</div>
</div>


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   </td>
   </td>
   <td align="left">
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<math>~\Rightarrow ~~~ a=0</math>
<math>~\Rightarrow ~~~a = 0 </math>
   </td>
   </td>
</tr>
</tr>
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   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~10b</math>
<math>~2b + 8b</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~-\frac{\mathfrak{F}}{6}</math>
<math>~- \frac{\mathfrak{F}}{6}</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~b = -\frac{\mathfrak{F}}{60}</math>
<math>~\Rightarrow ~~~b = - \frac{\mathfrak{F}}{60}</math>
   </td>
   </td>
</tr>
</tr>
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   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\biggl(18c+\frac{2}{3}\biggr)</math>
<math>~6c + 12c - \frac{a}{3}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~- \frac{a\mathfrak{F}}{6}</math>
<math>~-\frac{a\mathfrak{F}}{6}</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~c = -\frac{1}{27}</math>
<math>~\Rightarrow ~~~c=0</math>
   </td>
   </td>
</tr>
</tr>
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   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\biggl(28d + \frac{2b}{3}\biggr)</math>
<math>~12d + 16d - \frac{2b}{3}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\biggl(\frac{\alpha}{30} - \frac{b\mathfrak{F}}{6}\biggr)</math>
<math>~-\frac{\mathfrak{F}b}{6} - \frac{\alpha}{30}</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~d = + \frac{\mathfrak{F}}{10080}\biggl( \mathfrak{F} + 4 \biggr)</math>
<math>~\Rightarrow ~~~28d = \frac{1}{30}\biggl[ 5b (4- \mathfrak{F} ) - \alpha \biggr] </math>
   </td>
   </td>
</tr>
</tr>

Revision as of 22:44, 27 February 2017

Approximate Power-Series Expressions

Whitworth's (1981) Isothermal Free-Energy Surface
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Broadly Used Mathematical Expressions (shown here without proof)

Binomial

<math>~(1 \pm x)^n</math>

<math>~=</math>

<math>~ 1 ~\pm ~nx + \biggl[\frac{n(n-1)}{2!}\biggr]x^2 ~\pm~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]x^3 + \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]x^4 ~~\pm ~~ \cdots </math>      for <math>~(x^2 < 1)</math>

See also:

Exponential

<math>~e^x</math>

<math>~=</math>

<math>~ 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots </math>


Expressions with Astrophysical Relevance

Polytropic Lane-Emden Function

We seek a power-series expression for the polytropic, Lane-Emden function, <math>~\Theta_\mathrm{H}(\xi)</math> — expanded about the coordinate center, <math>~\xi = 0</math> — that approximately satisfies the Lane-Emden equation,

LSU Key.png

<math>~\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr) = - \Theta_H^n</math>

A general power-series should be of the form,

<math>~\Theta_H</math>

<math>~=</math>

<math>~ \theta_0 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots </math>

First derivative:

<math>~\frac{d\Theta_H}{d\xi}</math>

<math>~=</math>

<math>~ a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5 + 7g\xi^6 + 8h\xi^7 + \cdots </math>

Left-hand-side of Lane-Emden equation:

<math>~\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr)</math>

<math>~=</math>

<math>~ \frac{2a}{\xi} + 2\cdot 3b + 2^2\cdot 3c\xi + 2^2\cdot 5d\xi^2 + 2\cdot 3\cdot 5e\xi^3 + 2\cdot 3\cdot 7f\xi^4 + 2^3\cdot 7g\xi^5 + 2^3\cdot 3^2h\xi^6 + \cdots </math>

Right-hand-side of Lane-Emden equation (adopt the normalization, <math>~\theta_0=1</math>, then use the binomial theorem recursively):

<math>~\Theta_H^n</math>

<math>~=</math>

<math>~ 1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2 ~+~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]F^3 + \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]F^4 ~~+ ~~ \cdots </math>

where,

<math>~F</math>

<math>~\equiv</math>

<math>~ a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots </math>

 

<math>~=</math>

<math>~ a\xi\biggl[1 + \frac{b}{a}\xi + \frac{c}{a}\xi^2 + \frac{d}{a}\xi^3 + \frac{e}{a}\xi^4 + \frac{f}{a}\xi^5 + \frac{g}{a}\xi^6 + \frac{h}{a}\xi^7 + \cdots\biggr] \, . </math>

First approximation:  Assume that <math>~e=f=g=h=0</math>, in which case the LHS contains terms only up through <math>~\xi^2</math>. This means that we must ignore all terms on the RHS that are of higher order than <math>~\xi^2</math>; that is,

<math>~\Theta_H^n</math>

<math>~\approx</math>

<math>~ 1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2 </math>

 

<math>~\approx</math>

<math>~ 1 ~+ ~n(a\xi+b\xi^2) + \biggl[\frac{n(n-1)}{2!}\biggr]a^2\xi^2 </math>

 

<math>~\approx</math>

<math>~ 1 ~+~na\xi + ~\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]\xi^2\, . </math>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>. Remembering to include a negative sign on the RHS, we find:

Term LHS RHS Implication

<math>~\xi^{-1}:</math>

<math>~2a</math>

<math>~0</math>

<math>~\Rightarrow ~~~a=0</math>

<math>~\xi^{0}:</math>

<math>~2\cdot 3 b</math>

<math>~-1</math>

<math>~\Rightarrow ~~~b=- \frac{1}{6}</math>

<math>~\xi^{1}:</math>

<math>~2^2\cdot 3 c</math>

<math>~-na</math>

<math>~\Rightarrow ~~~c=0</math>

<math>~\xi^{2}:</math>

<math>~2^2\cdot 5 d</math>

<math>~-\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]</math>

<math>~\Rightarrow ~~~d=+\frac{n}{120}</math>

By including higher and higher order terms in the series expansion for <math>~\Theta_H</math>, and proceeding along the same line of deductive reasoning, one finds:

  • Expressions for the four coefficients, <math>~a, b, c, d</math>, remain unchanged.
  • The coefficient is zero for all other terms that contain odd powers of <math>~\xi</math>; specifically, for example, <math>~e = g = 0</math>.
  • The coefficients of <math>~\xi^6</math> and <math>~\xi^8</math> are, respectively,

<math>~f</math>

<math>~=</math>

<math>~- \frac{n}{378}\biggl(\frac{n}{5}-\frac{1}{8} \biggr) \, ;</math>

<math>~h</math>

<math>~=</math>

<math>~\frac{n(122n^2 -183n + 70)}{3265920} \, .</math>


In summary, the desired, approximate power-series expression for the polytropic Lane-Emden function is:

<math>~\theta</math>

<math>~=</math>

<math>~ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \biggl[ \frac{n(122n^2 -183n + 70)}{3265920} \biggr] \xi^8 + \cdots </math>

Isothermal Lane-Emden Function

Here we seek a power-series expression for the isothermal, Lane-Emden function — expanded about the coordinate center — that approximately satisfies the isothermal Lane-Emden equation; making the variable substitution (sorry for the unnecessary complication!), <math>~\psi(\xi) \leftrightarrow w(r)</math>, the governing ODE is,

<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr} </math>

<math>~=</math>

<math>~e^{-w} \, . </math>

A general power-series should be of the form,

<math>~w</math>

<math>~=</math>

<math>~ w_0 + ar + br^2 + cr^3 + dr^4 + er^5 + fr^6 + gr^7 + hr^8 +\cdots </math>

Derivatives:

<math>~\frac{dw}{dr}</math>

<math>~=</math>

<math>~ a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 +\cdots \, ; </math>

<math>~\frac{d^2w}{dr^2}</math>

<math>~=</math>

<math>~ 2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6 +\cdots \, . </math>

Put together, then, the left-hand-side of the isothermal Lane-Emden equation becomes:

<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr} </math>

<math>~=</math>

<math>~ 2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6 + \frac{2}{r}\biggl[ a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 \biggr] + \cdots </math>

 

<math>~=</math>

<math>~\frac{2a}{r} + r^0(6b) + r^1(2^2\cdot 3c) + r^2(2^2\cdot 3d + 2^3d) + r^3(2^2\cdot 5e + 2\cdot 5e) + r^4(2\cdot 3\cdot 5 f + 2^2\cdot 3f) + r^5(2\cdot 3\cdot 7 g+ 2\cdot 7g) + r^6(2^3 \cdot 7 h + 2^4 h) + \cdots </math>

Drawing on the above power-series expression for an exponential function, and adopting the convention that <math>~w_0 = 0</math>, the right-hand-side becomes,

<math>~e^{-w}</math>

<math>~=</math>

<math>~ e^{0}\cdot e^{-ar} \cdot e^{-br^2} \cdot e^{-cr^3} \cdot e^{-dr^4} \cdot e^{-er^5} \cdot e^{-fr^6} \cdot e^{-gr^7} \cdot e^{-hr^8} \cdots </math>

 

<math>~=</math>

<math>~ \biggl[ 1 -ar + \frac{a^2r^2}{2!} - \frac{a^3r^3}{3!} + \frac{a^4r^4}{4!} - \frac{a^5r^5}{5!} + \frac{a^6r^6}{6!} + \cdots \biggr] </math>

 

 

<math>~ \times \biggl[ 1 -br^2 + \frac{b^2r^4}{2!} - \frac{b^3r^6}{3!} + \cdots \biggr] \times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2!} + \cdots \biggr] \times \biggl[1 - dr^4\biggr] \times \biggl[1 - er^5\biggr]\times \biggl[1 - fr^6\biggr] </math>

 

<math>~\approx</math>

<math>~ \biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr] \times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2} -br^2 + bcr^5 + \frac{b^2r^4}{2} - \frac{b^3r^6}{6} \biggr] \times \biggl[1 - dr^4 - er^5 - fr^6\biggr] </math>

 

<math>~\approx</math>

<math>~\biggl\{ \biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr] - dr^4 \biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr] - er^5 \biggl[ 1 -ar \biggr] - fr^6 \biggr\} </math>

 

 

<math>~ \times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr] </math>

 

<math>~\approx</math>

<math>~\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} - dr^4 + adr^5 - \frac{a^2d r^6}{2} - er^5 + aer^6 - fr^6 \biggr] </math>

 

 

<math>~ \times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr] </math>

 

<math>~\approx</math>

<math>~\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr) + r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr) \biggr] \times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr] </math>

 

<math>~\approx</math>

<math>~ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr) + r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr) </math>

 

 

<math>~-br^2\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) \biggr] -cr^3 \biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} \biggr] + \frac{b^2r^4}{2}\biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr] + bcr^5\biggl[1 -ar \biggr] + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) </math>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~r</math>. Beginning with the highest order terms, we initially find,

Term LHS RHS Implication

<math>~r^{-1}:</math>

<math>~2a</math>

<math>~0</math>

<math>~\Rightarrow ~~~a=0</math>

<math>~r^{0}:</math>

<math>~6b</math>

<math>~1</math>

<math>~\Rightarrow ~~~b = + \frac{1}{6}</math>

<math>~r^{1}:</math>

<math>~2^2\cdot 3c</math>

<math>~-a</math>

<math>~\Rightarrow ~~~c = -\frac{a}{2^2\cdot 3} =0</math>

<math>~r^{2}:</math>

<math>~(2^2\cdot 3d + 2^3d)</math>

<math>~\frac{a^2}{2} - b</math>

<math>~\Rightarrow ~~~d = \frac{1}{20}\biggl( \frac{a^2}{2} - b \biggr) = - \frac{1}{120}</math>

With this initial set of coefficient values in hand, we can rewrite (and significantly simplify) our approximate expression for the RHS, namely,

<math>~e^{-w}</math>

<math>~\approx</math>

<math>~ 1 -d r^4 -e r^5 -f r^6 -br^2 ( 1 -d r^4 ) + \frac{b^2r^4}{2} - \frac{b^3r^6}{6} </math>

 

<math>~=</math>

<math>~ 1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^5 +r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, . </math>

Continuing, then, with equating terms with like powers on both sides of the equation, we find,

Term LHS RHS Implication

<math>~r^{3}:</math>

<math>~30e</math>

<math>~0</math>

<math>~\Rightarrow ~~~e=0</math>

<math>~r^{4}:</math>

<math>~(2\cdot 3\cdot 5 f + 2^2\cdot 3f)</math>

<math>~\biggl(\frac{b^2}{2} -d \biggr) </math>

<math>~\Rightarrow ~~~f = \frac{1}{2\cdot 3\cdot 7}\biggl(\frac{1}{2^3\cdot 3^2}+\frac{1}{2^3\cdot 3 \cdot 5}\biggr) = \frac{1}{2\cdot 3^3\cdot 5 \cdot 7}</math>

<math>~r^{5}:</math>

<math>~(2\cdot 3\cdot 7 g+ 2\cdot 7g)</math>

<math>~-e</math>

<math>~\Rightarrow ~~~g = 0</math>

<math>~r^{6}:</math>

<math>~(2^3 \cdot 7 h + 2^4 h)</math>

<math>~\biggl( bd - \frac{b^3}{6} -f \biggr)</math>

<math>~\Rightarrow ~~~ h = -\frac{1}{2^3\cdot 3^2}\biggl( \frac{1}{2^4\cdot 3^2 \cdot 5} + \frac{1}{2^4\cdot 3^4} + \frac{1}{2\cdot 3^3\cdot 5\cdot 7}\biggr) = -\frac{61}{2^{6} \cdot 3^6\cdot 5\cdot 7} </math>


Result:

<math>~w(r) </math>

<math>~=</math>

<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>


See also:

  • Equation (377) from §22 in Chapter IV of C67).

Displacement Function for Polytropic LAWE

Displacement Function for Isothermal LAWE

The LAWE for isothermal spheres may be written as,

<math>~\frac{d^2 x}{dr^2} + \biggl[4 - r \biggl(\frac{dw }{dr}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>

<math>~=</math>

<math>~ - \biggl[ \frac{\sigma_c^2}{6\gamma} - \frac{\alpha}{r} \biggl(\frac{dw }{dr}\biggr)\biggr] x \, , </math>

where, <math>~w(r)</math> is the isothermal Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate power-series expression for the isothermal Lane-Emden function is,

<math>~w(r) </math>

<math>~=</math>

<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>

Hence,

<math>~\frac{dw}{dr}</math>

<math>~\approx</math>

<math>~\frac{r}{3} - \frac{r^3}{30} + \frac{r^5}{315} \, .</math>

Therefore, near the center of the configuration, the LAWE may be written as,

<math>~\frac{d^2 x}{dr^2} + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>

<math>~\approx</math>

<math>~ - \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr] x \, . </math>

Let's now adopt a power-series expression for the displacement function of the form,

<math>~x</math>

<math>~=</math>

<math>~ 1 + ar + br^2 + cr^3 + dr^4 + \cdots </math>

<math>~\Rightarrow ~~~ \frac{1}{r}\frac{dx}{dr}</math>

<math>~=</math>

<math>~ \frac{a}{r} + 2b + 3 cr + 4dr^2 + \cdots </math>

and,

<math>~\frac{d^2x}{dr^2}</math>

<math>~=</math>

<math>~ 2b + 6cr + 12dr^2 + \cdots </math>

Substituting these expressions into the LAWE gives,

<math>~2b + 6cr + 12dr^2 + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr] </math>

<math>~\approx</math>

<math>~ - \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr] \biggl( 1 + ar + br^2 + cr^3 + dr^4 \biggr) \, . </math>

Keeping terms only up through <math>~r^2</math> leads to the following simplification:

<math>~ 2b + 6cr + 12dr^2 + 4 \biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr] - \frac{r^2}{3} \biggl[ \frac{a}{r} + 2b \biggr] </math>

<math>~\approx</math>

<math>~ - \frac{\mathfrak{F} }{6} \biggl( 1 + ar + br^2 \biggr) - \frac{\alpha}{3} \biggl(\frac{r^2}{10} \biggr) </math>

where,

<math>~\mathfrak{F} \equiv \frac{\sigma_c^2}{\gamma} - 2\alpha \, .</math>

Finally, balancing terms of like powers on both sides of the equation leads us to conclude the following:

Term LHS RHS Implication

<math>~r^{-1}:</math>

<math>~4a</math>

<math>~0</math>

<math>~\Rightarrow ~~~a = 0 </math>

<math>~r^{0}:</math>

<math>~2b + 8b</math>

<math>~- \frac{\mathfrak{F}}{6}</math>

<math>~\Rightarrow ~~~b = - \frac{\mathfrak{F}}{60}</math>

<math>~r^{1}:</math>

<math>~6c + 12c - \frac{a}{3}</math>

<math>~-\frac{a\mathfrak{F}}{6}</math>

<math>~\Rightarrow ~~~c=0</math>

<math>~r^{2}:</math>

<math>~12d + 16d - \frac{2b}{3}</math>

<math>~-\frac{\mathfrak{F}b}{6} - \frac{\alpha}{30}</math>

<math>~\Rightarrow ~~~28d = \frac{1}{30}\biggl[ 5b (4- \mathfrak{F} ) - \alpha \biggr] </math>


In summary, the desired, approximate power-series expression for the isothermal displacement function is:

<math>~x(r)</math>

<math>~=</math>

<math>~ 1 - \frac{\mathfrak{F}}{60} r^2 + \frac{\mathfrak{F}}{10080}\biggl( \mathfrak{F} + 4 \biggr)r^4 + \cdots </math>


Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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