Difference between revisions of "User:Tohline/SSC/Structure/BiPolytropes/Analytic1 5"

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(→‎Step 6: Envelope Solution: Derive relevant cubic equation)
(→‎Step 6: Envelope Solution: write out root of cubic equation)
Line 391: Line 391:
0 \, .
0 \, .
</math>
</math>
  </td>
</tr>
</table>
</div>
<div align="center">
<table border="1" cellpadding="8" width="80%">
<tr>
  <td align="left">
<font color="red">'''ASIDE:'''</font>  As is well known and documented &#8212; see, for example [http://mathworld.wolfram.com/CubicFormula.html Wolfram MathWorld] or [http://en.wikipedia.org/wiki/Cubic_function Wikipedia's discussion] of the topic &#8212; the roots of any cubic equation can be determined analytically.  In order to evaluate the root(s) of our particular cubic equation, we have drawn from the utilitarian [http://www.math.vanderbilt.edu/~schectex/courses/cubic/ online summary provided by Eric Schechter at Vanderbilt University].  For a cubic equation of the general form,
<div align="center">
<math>~ay^3 + by^2 + cy + d = 0 \, ,</math>
</div>
a real root is given by the expression,
<div align="center">
<math>~
y = p + \{q + [q^2 + (r-p^2)^3]^{1/2}\}^{1/3} + \{q - [q^2 + (r-p^2)^3]^{1/2}\}^{1/3}
\, ,</math>
</div>
where,
<div align="center">
<math>~p \equiv -\frac{b}{3a} \, ,</math> &nbsp;&nbsp;&nbsp;&nbsp;
<math>~q \equiv \biggl[p^3 + \frac{bc-3ad}{6a^2} \biggr] \, ,</math>
&nbsp;&nbsp;&nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;&nbsp;
<math>~r=\frac{c}{3a} \, .</math>
</div>
In our particular case,
<div align="center">
<math>~a =(1+\kappa_i)\, ,</math> &nbsp;&nbsp;&nbsp;&nbsp;
<math>~b =-3\, ,</math> &nbsp;&nbsp;&nbsp;&nbsp;
<math>~c = 3(1+\kappa_i) \, ,</math>
&nbsp;&nbsp;&nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;&nbsp;
<math>~d = - 3 \, .</math>
</div>
Hence,
<div align="center">
<math>~p = \frac{1}{(1+\kappa_i)} \, ,</math> &nbsp;&nbsp;&nbsp;&nbsp;
<math>~r=+1 \, ,</math>
&nbsp;&nbsp;&nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;&nbsp;
<math>~q = p^3 = \frac{1}{(1+\kappa_i)^3} \, ,</math>
</div>
which implies that the real root, <math>~y_\mathrm{root}</math>, is given by the expression,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~(1+\kappa_i)y_\mathrm{root}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 + \biggl\{1 + \biggl[1 + \frac{\kappa_i^3(2+\kappa_i)^3}{(1+\kappa_i)^3}\biggr]^{1/2} \biggr\}^{1/3}
+ \biggl\{1 - \biggl[1 + \frac{\kappa_i^3(2+\kappa_i)^3}{(1+\kappa_i)^3}\biggr]^{1/2} \biggr\}^{1/3}
\, .
</math>
  </td>
</tr>
</table>
</div>
(There is also a pair of imaginary roots, but they are irrelevant in the context of our overarching astrophysical discussion.)
   </td>
   </td>
</tr>
</tr>

Revision as of 18:10, 19 April 2015

BiPolytrope with <math>n_c = 1</math> and <math>n_e=5</math>

Whitworth's (1981) Isothermal Free-Energy Surface
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Here we construct a bipolytrope in which the core has an <math>~n_c=1</math> polytropic index and the envelope has an <math>~n_e=5</math> polytropic index. As in the case of our separately discussed, "mirror image" bipolytropes having <math>~(n_c, n_e) = (5, 1)</math>, this system is particularly interesting because the entire structure can be described by closed-form, analytic expressions. [On 12 April 2015, J. E. Tohline wrote: I became aware of the published discussions of this system by Murphy — and especially the work of Murphy & Fiedler (1985) — (see itemization of additional key references, below) in March of 2015 after searching the internet for previous analyses of radial oscillations in polytropes and, then, reading through Horedt's (2004) §2.8.1 discussion of composite polytropes.]

Key References

Steps 2 & 3

Based on the discussion presented elsewhere of the structure of an isolated <math>n=1</math> polytrope, the core of this bipolytrope will have the following properties:

<math> \theta(\xi) = \frac{\sin\xi}{\xi} ~~~~\Rightarrow ~~~~ \theta_i = \frac{\sin\xi_i}{\xi_i} ; </math>

<math> \frac{d\theta}{d\xi} = \frac{\cos\xi}{\xi}- \frac{\sin\xi}{\xi^2} ~~~~\Rightarrow ~~~~ \biggl(\frac{d\theta}{d\xi}\biggr)_i = \frac{\cos\xi_i}{\xi_i}- \frac{\sin\xi_i}{\xi_i^2} \, . </math>

The first zero of the function <math>~\theta(\xi)</math> and, hence, the surface of the corresponding isolated <math>~n=1</math> polytrope is located at <math>~\xi_s = \pi</math>. Hence, the interface between the core and the envelope can be positioned anywhere within the range, <math>~0 < \xi_i < \pi</math>.

Step 4: Throughout the core (<math>0 \le \xi \le \xi_i</math>)

Specify: <math>~K_c</math> and <math>~\rho_0 ~\Rightarrow</math>

 

<math>~\rho</math>

<math>~=</math>

<math>\rho_0 \theta^{n_c}</math>

<math>~=</math>

<math>~\rho_0 \biggl( \frac{\sin\xi}{\xi} \biggr)</math>

<math>~P</math>

<math>~=</math>

<math>~K_c \rho_0^{1+1/n_c} \theta^{n_c + 1}</math>

<math>~=</math>

<math>~K_c \rho_0^{2} \biggl( \frac{\sin\xi}{\xi}\biggr)^{2}</math>

<math>~r</math>

<math>~=</math>

<math>~\biggl[ \frac{(n_c + 1)K_c}{4\pi G} \biggr]^{1/2} \rho_0^{(1-n_c)/(2n_c)} \xi</math>

<math>~=</math>

<math>~\biggl[ \frac{K_c}{2\pi G} \biggr]^{1/2} \xi</math>

<math>~M_r</math>

<math>~=</math>

<math>~4\pi \biggl[ \frac{(n_c + 1)K_c}{4\pi G} \biggr]^{3/2} \rho_0^{(3-n_c)/(2n_c)} \biggl(-\xi^2 \frac{d\theta}{d\xi} \biggr)</math>

<math>~=</math>

<math>~4\pi \biggl[ \frac{K_c}{2\pi G} \biggr]^{3/2} \rho_0 \biggl[\sin\xi - \xi \cos\xi \biggr]</math>

Step 5: Interface Conditions

 

Setting <math>~n_c=1</math>, <math>~n_e=5</math>, and <math>~\phi_i = 1 ~~~~\Rightarrow</math>

<math>~\frac{\rho_e}{\rho_0}</math>

<math>~=</math>

<math>~\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{n_c}_i \phi_i^{-n_e}</math>

<math>~=</math>

<math>~\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i </math>

<math>~\biggl( \frac{K_e}{K_c} \biggr) </math>

<math>~=</math>

<math>~\rho_0^{1/n_c - 1/n_e}\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-(1+1/n_e)} \theta^{1 - n_c/n_e}_i</math>

<math>~=</math>

<math>~\biggl[\rho_0^{4}\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-6} \theta^{4}_i\biggr]^{1/5}</math>

<math>~\frac{\eta_i}{\xi_i}</math>

<math>~=</math>

<math>~\biggl[ \frac{n_c + 1}{n_e+1} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c}\biggr) \theta_i^{(n_c-1)/2} \phi_i^{(1-n_e)/2}</math>

<math>~=</math>

<math>~\biggl( \frac{1}{3} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c}\biggr) </math>

<math>~\biggl( \frac{d\phi}{d\eta} \biggr)_i</math>

<math>~=</math>

<math>~\biggl[ \frac{n_c + 1}{n_e + 1} \biggr]^{1/2} \theta_i^{- (n_c + 1)/2} \phi_i^{(n_e+1)/2} \biggl( \frac{d\theta}{d\xi} \biggr)_i</math>

<math>~=</math>

<math>~\biggl( \frac{1}{3} \biggr)^{1/2} \theta_i^{- 1} \biggl( \frac{d\theta}{d\xi} \biggr)_i</math>

Step 6: Envelope Solution

Following the work of Murphy (1983) and of Murphy & Fiedler (1985a), we will adopt for the envelope's structure the F-Type solution of the <math>~n=5</math> Lane-Emden function discovered by S. Srivastava (1968, ApJ, 136, 680) and described in an accompanying discussion, namely,

<math>~\phi</math>

<math>~=</math>

<math>~\frac{B\sin[\ln(A\eta)^{1/2})]}{\eta^{1/2}\{3-2\sin^2[\ln(A\eta)^{1/2}]\}^{1/2}} \, ,</math>

specifically over the physically viable interval, <math>~e^{2\pi} \ge A\eta \ge \eta_\mathrm{crit} \equiv e^{2\tan^{-1}(1+2^{1/3})} \, .</math> The first derivative of this function is,

<math>~\frac{d\phi}{d\eta}</math>

<math>~=</math>

<math>~ \frac{B[3\cos\Delta-3\sin\Delta + 2\sin^3\Delta] }{2\eta^{3/2}(3-2\sin^2\Delta)^{3/2}} \, , </math>

where,

<math>~\Delta \equiv \ln(A\eta)^{1/2} = \ln A^{1/2} + \ln\eta^{1/2} \, .</math>

From Step 5, above, we know the value of the function, <math>~\phi</math> and its first derivative at the interface; specifically,

<math> \phi_i = 1~~~~\mathrm{and} ~~~~\biggl( \frac{d\phi}{d\eta}\biggr)_i =3^{-1/2} \theta_i^{- 1} \biggl( \frac{d\theta}{d\xi} \biggr)_i~~~~ \mathrm{at}~~~~\eta_i =3^{1/2} \xi_i \biggl( \frac{\mu_e}{\mu_c}\biggr) </math>

From this information we can determine the constants <math>~A</math> and <math>~B</math>.

Let's begin by recognizing that the task of solving for <math>~A</math> can be changed to the task of solving for <math>~\Delta_i</math> — that is, solving for <math>~\Delta</math> at the interface — because, given that <math>~\eta_i</math> is known, the determination of <math>~\Delta_i</math> allows us to immediately deduce that,

<math>~A = \eta_i^{-1} e^{2\Delta_i} \, .</math>

Given the values of <math>~\phi_i</math> and <math>~\eta_i</math>, from the definition of Srivastava's function we have,

<math>~B</math>

<math>~=</math>

<math>~\phi_i \eta_i^{1/2} \biggl[ \frac{(3-2\sin^2\Delta_i)^{1/2}}{\sin\Delta_i} \biggr] \, .</math>

Plugging this expression into the expression for the first derivative of Srivastava's function evaluated at the interface gives,

<math>~\frac{2\eta_i}{\phi_i}\biggl(\frac{d\phi}{d\eta}\biggr)_i</math>

<math>~=</math>

<math>~ \biggl[\frac{3\cos\Delta_i-3\sin\Delta_i + 2\sin^3\Delta_i }{(3-2\sin^2\Delta_i)^{3/2}} \biggr] \biggl[ \frac{(3-2\sin^2\Delta_i)^{1/2}}{\sin\Delta_i} \biggr] </math>

 

<math>~=</math>

<math>~ \frac{3\cot\Delta_i-3+ 2\sin^2\Delta_i }{(3-2\sin^2\Delta_i)} \, . </math>

Now, defining the known constant,

<math>~\kappa_i \equiv \frac{2\eta_i}{\phi_i}\biggl(\frac{d\phi}{d\eta}\biggr)_i \, ,</math>

and, as in our separate discussion of the properties of Srivastava's function, adopting the shorthand notation,

<math>~y_i \equiv \tan\Delta_i \, ,</math>

this condition becomes,

<math>~\kappa_i</math>

<math>~=</math>

<math>~ \frac{3y_i^{-1} -3+ 2y_i^2(1+y_i^2)^{-1} }{3-2y_i^2(1+y_i^2)^{-1}} </math>

 

<math>~=</math>

<math>~ \frac{3(1+y_i^2) -3y_i(1+y_i^2)+ 2y_i^3}{3y_i(1+y_i^2)-2y_i^3} </math>

 

<math>~=</math>

<math>~ \frac{3 - 3y_i +3y_i^2 - y_i^3}{3y_i+y_i^3} </math>

<math>~\Rightarrow~~~~\kappa_i(3y_i+y_i^3)</math>

<math>~=</math>

<math>~ 3 - 3y_i +3y_i^2 - y_i^3 </math>

<math>~\Rightarrow~~~~ y_i^3(1+\kappa_i) -3 y_i^2 + 3(1+\kappa_i)y_i -3</math>

<math>~=</math>

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

ASIDE: As is well known and documented — see, for example Wolfram MathWorld or Wikipedia's discussion of the topic — the roots of any cubic equation can be determined analytically. In order to evaluate the root(s) of our particular cubic equation, we have drawn from the utilitarian online summary provided by Eric Schechter at Vanderbilt University. For a cubic equation of the general form,

<math>~ay^3 + by^2 + cy + d = 0 \, ,</math>

a real root is given by the expression,

<math>~ y = p + \{q + [q^2 + (r-p^2)^3]^{1/2}\}^{1/3} + \{q - [q^2 + (r-p^2)^3]^{1/2}\}^{1/3} \, ,</math>

where,

<math>~p \equiv -\frac{b}{3a} \, ,</math>      <math>~q \equiv \biggl[p^3 + \frac{bc-3ad}{6a^2} \biggr] \, ,</math>      and      <math>~r=\frac{c}{3a} \, .</math>

In our particular case,

<math>~a =(1+\kappa_i)\, ,</math>      <math>~b =-3\, ,</math>      <math>~c = 3(1+\kappa_i) \, ,</math>      and      <math>~d = - 3 \, .</math>

Hence,

<math>~p = \frac{1}{(1+\kappa_i)} \, ,</math>      <math>~r=+1 \, ,</math>      and      <math>~q = p^3 = \frac{1}{(1+\kappa_i)^3} \, ,</math>

which implies that the real root, <math>~y_\mathrm{root}</math>, is given by the expression,

<math>~(1+\kappa_i)y_\mathrm{root}</math>

<math>~=</math>

<math>~ 1 + \biggl\{1 + \biggl[1 + \frac{\kappa_i^3(2+\kappa_i)^3}{(1+\kappa_i)^3}\biggr]^{1/2} \biggr\}^{1/3} + \biggl\{1 - \biggl[1 + \frac{\kappa_i^3(2+\kappa_i)^3}{(1+\kappa_i)^3}\biggr]^{1/2} \biggr\}^{1/3} \, . </math>

(There is also a pair of imaginary roots, but they are irrelevant in the context of our overarching astrophysical discussion.)

Related Discussions


Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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