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(→‎Try Mode22 ~(\ell,j) = (2,2): Write out expression for the core's adiabatic exponent)
(→‎Try Mode22 ~(\ell,j) = (2,2): Fill in gamma_c and alpha_c on right of Figure 2)
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<td align="right"><math>~\gamma_c</math></td>
<td align="right"><math>~\gamma_c</math></td>
<td align="center"><math>~</math></td>
<td align="center"><math>~1.0464786</math></td>
</tr>
</tr>
<tr>
<tr>
<td align="right"><math>~\alpha_c</math></td>
<td align="right"><math>~\alpha_c</math></td>
<td align="center"><math>~</math></td>
<td align="center"><math>~-0.822343</math></td>
</tr>
</tr>



Revision as of 22:22, 24 December 2016

Searching for Additional Eigenvectors of Zero-Zero Bipolytropes

This chapter is an extension of two accompanying discussions: The original discovery and detailed derivation; and the more readable, summary.

Whitworth's (1981) Isothermal Free-Energy Surface
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In our accompanying summary, we have demonstrated how analytically specified eigenvectors can be constructed for the mode labeled, <math>~(\ell, j) = (2,1)</math>. This was done by specifying <math>~\gamma_e</math>, then solving a quartic equation for <math>~q</math>. Shortly after completing this summary chapter, we noticed that an alternate approach may be to specify <math>~q</math>, then solve for <math>~\gamma_e</math>; and this path may be simpler because it may only involve solution of a quadratic equation. (Actually, we later have realized that the relevant equation is cubic, rather than quadratic. This is nevertheless simpler than the quartic equation.) If this proves to be the case, then it may also be possible to analytically construct eigenvectors of additional modes. Let's see.

Seek Alternate Solution to Mode21 <math>~(\ell,j) = (2,1)</math>

Setup

According to STEP 4 in our accompanying summary discussion, we need to solve the following "derivative matching" expression:

<math>~\frac{14(1+2q^3)^2}{7(1+2q^3)^2 - 5}</math>

<math>~=</math>

<math>~ \frac{c_0 + (c_0 + 3)A_{21}q^3 + (c_0 + 6)A_{21}B_{21} q^6}{1 + A_{21}q^3 + A_{21}B_{21}q^6} \, , </math>

where, recognizing that, <math>~\alpha_e = c_0(c_0+2) \, ,</math>

<math>~A_{21}</math>

<math>~\equiv</math>

<math>~\biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha_e}\biggr] </math>

 

<math>~=</math>

<math>~\biggl[ \frac{c_0^2 + 5c_0 - (c_0^2 + 17c_0 + 66)}{(c_0^2 + 8c_0 + 15) - (c_0^2+2c_0)}\biggr] </math>

 

<math>~=</math>

<math>~-\biggl( \frac{ 4c_0 + 22}{2c_0 + 5}\biggr) \, ,</math>

<math>~B_{21}</math>

<math>~\equiv</math>

<math>~\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha_e}\biggr] </math>

 

<math>~=</math>

<math>~\biggl[ \frac{(c_0^2 +11c_0 + 24) - (c_0^2 + 17c_0 + 66)}{(c_0^2+14c_0+48) - (c_0^2 + 2c_0)}\biggr] </math>

 

<math>~=</math>

<math>~-\biggl( \frac{c_0 + 7 }{2c_0+8}\biggr) \, . </math>

Here, we assume that <math>~\Chi \equiv q^3</math> is specified and seek the corresponding value of <math>~c_0</math>. Given that the LHS of this matching relation is known once <math>~\Chi</math> has been specified, in order to simplify notation we will also define,

<math>~Q</math>

<math>~\equiv</math>

<math>~\frac{14(1+2\Chi)^2}{7(1+2\Chi)^2 - 5} \, .</math>

Then the matching relation becomes,

<math>~Q</math>

<math>~=</math>

<math>~ \frac{c_0 + (c_0 + 3)A_{21}\Chi + (c_0 + 6)A_{21}B_{21} \Chi^2}{1 + A_{21}\Chi + A_{21}B_{21}\Chi^2} </math>

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

<math>~=</math>

<math>~[c_0 + (c_0 + 3)A_{21}\Chi + (c_0 + 6)A_{21}B_{21} \Chi^2 ] - Q[1 + A_{21}\Chi + A_{21}B_{21}\Chi^2 ] </math>

 

<math>~=</math>

<math>~\biggl[c_0 - (c_0 + 3)\biggl( \frac{ 4c_0 + 22}{2c_0 + 5}\biggr)\Chi + (c_0 + 6)\biggl( \frac{ 4c_0 + 22}{2c_0 + 5}\biggr)\biggl( \frac{c_0 + 7 }{2c_0+8}\biggr) \Chi^2 \biggr] - Q\biggl[1 - \biggl( \frac{ 4c_0 + 22}{2c_0 + 5}\biggr)\Chi + \biggl( \frac{ 4c_0 + 22}{2c_0 + 5}\biggr)\biggl( \frac{c_0 + 7 }{2c_0+8}\biggr)\Chi^2 \biggr] </math>

 

<math>~=</math>

<math>~\biggl[c_0(2c_0+5)(2c_0+8) - (c_0 + 3)(2c_0+8)( 4c_0 + 22)\Chi + (c_0 + 6)( 4c_0 + 22)(c_0 + 7 ) \Chi^2 \biggr] </math>

 

 

<math>~ - Q\biggl[(2c_0+5)(2c_0+8) - (2c_0+8)( 4c_0 + 22)\Chi + ( 4c_0 + 22)( c_0 + 7 )\Chi^2 \biggr] </math>

 

<math>~=</math>

<math>~\biggl\{c_0(4c_0^2 + 26c_0 + 40) + ( 4c_0 + 22)\Chi [(c_0^2 + 13c_0 + 42 ) \Chi- (2c_0^2 +14c_0 +24)] \biggr\} </math>

 

 

<math>~ - Q\biggl\{ (4c_0^2 + 26c_0 + 40) + ( 4c_0 + 22)\Chi [c_0(\Chi-2) + (7\Chi-8) ] \biggr\} </math>

 

<math>~=</math>

<math>~4c_0^3 +[ 26 - 4Q]c_0^2 + [40 - 26Q]c_0 - 40Q </math>

 

 

<math>~ + ( 4c_0 + 22)\Chi \biggl\{ [\Chi - 2]c_0^2 + [13\Chi -14 -Q (\Chi-2) ]c_0 + [42\Chi -Q (7\Chi-8) -24] \biggr\} </math>

 

<math>~=</math>

<math>~4c_0^3 +[ 26 - 4Q]c_0^2 + [40 - 26Q]c_0 - 40Q </math>

 

 

<math>~ + 4\Chi[\Chi - 2]c_0^3 + 4\Chi[13\Chi -14 -Q (\Chi-2) ]c_0^2 + 4\Chi[42\Chi -Q (7\Chi-8) -24]c_0 </math>

 

 

<math>~ + 22\Chi[\Chi - 2]c_0^2 + 22\Chi[13\Chi -14 -Q (\Chi-2) ]c_0 + 22\Chi[42\Chi -Q (7\Chi-8) -24] \, . </math>

Solve Cubic Equation

Here we draw from a separate discussion of solutions to a cubic equation.

Using <math>~y</math> in place of <math>~c_0</math>, this "derivative matching" relation can be written in the form of a standard cubic equation. Specifically,

<math>~a y^3 + b y^2 + c y + d</math>

<math>~=</math>

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

where,

<math>~a</math>

<math>~\equiv</math>

<math>~ 4 + 4\Chi(\Chi - 2)\, , </math>

<math>~b</math>

<math>~\equiv</math>

<math>~ ( 26 - 4Q_{21}) + 4\Chi[ 13\Chi -14 - Q_{21} (\Chi-2) ] + 22\Chi (\Chi - 2)\, , </math>

<math>~c</math>

<math>~\equiv</math>

<math>~

[40  - 26 Q_{21}]+ 4\Chi[42\Chi  - Q_{21} (7\Chi-8) -24]  + 22\Chi[13\Chi  -14  - Q_{21}  (\Chi-2) ] \, ,

</math>

<math>~d</math>

<math>~\equiv</math>

<math>~ - 40 Q_{21}+ 22\Chi[42\Chi -Q_{21} (7\Chi-8) -24] \, . </math>

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

where,

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

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


Upon evaluation, we have found that the expression inside of the square root is negative over the region of parameter space that is of most physical interest. Hence, we need to call upon a separate discussion in which the cube root of complex numbers was discussed.


We'll shift to Wolfram's notation; specifically,

<math>~a_0 </math>

<math>~=</math>

<math>~\frac{d}{a}\, ,</math>

<math>~a_1 </math>

<math>~=</math>

<math>~\frac{c}{a}\, ,</math>

<math>~a_2 </math>

<math>~=</math>

<math>~\frac{b}{a}\, ,</math>

<math>~R </math>

<math>~\equiv</math>

<math>~\frac{3^2 a_2 a_1 - 3^3a_0 - 2a_2^3}{2\cdot 3^3}</math>

 

<math>~=</math>

<math>~\frac{bc - 3ad }{6 a^2} - \biggl( \frac{ b}{3 a} \biggr)^3 = z \, ,</math>

<math>~Q </math>

<math>~\equiv</math>

<math>~\frac{3 a_1 - a_2^2}{3^2}</math>

 

<math>~=</math>

<math>~\frac{ c }{3a} - \biggl(\frac{b}{3a}\biggr)^2 = r-p^2 \, .</math>

Then, after defining,

<math>~D \equiv Q^3 + R^2 = z^2 + (r-p^2)^3 \, ;</math>         <math>~S^3 \equiv R+ \sqrt{D} \, ;</math>         and         <math>~T^3 \equiv R- \sqrt{D} \, ;</math>

Wolfram states that the three roots of the cubic equation are (the first one being identical to the "real" root identified above),

<math>~y_1</math>

<math>~=</math>

<math>~p + (S + T) \, ,</math>

<math>~y_2</math>

<math>~=</math>

<math>~p - \frac{1}{2}(S + T) + \frac{1}{2} i \sqrt{3}(S-T) \, ,</math>

<math>~y_2</math>

<math>~=</math>

<math>~p - \frac{1}{2}(S + T) - \frac{1}{2} i \sqrt{3}(S-T) \, .</math>

Now, whenever <math>~D</math> is intrinsically negative, we need to treat both <math>~S^3</math> and <math>~T^3</math> as complex numbers. If we define,

<math>~\mathfrak{r} \equiv (R^2 + |D|)^{1 / 2} \, ,</math>

      and      

<math>~\theta \equiv \tan^{-1}\biggl[ \frac{\sqrt{|D|}}{R} \biggr] \, ,</math>

then we can write,

<math>~S^3 = \mathfrak{r} e^{+i\theta} \, ,</math>

      and      

<math>~T^3 = \mathfrak{r} e^{-i\theta} \, .</math>

As is explained in this online resource, both <math>~S</math> and <math>~T</math> must formally submit to three separate roots tagged by the integer index, <math>~(j=0,1,2)</math>. Working only with the <math>~j=0</math> root for both, we find that the above expressions for the three roots of our cubic equation become,

<math>~y_1</math>

<math>~=</math>

<math>~p + 2 \mathfrak{r}^{1 / 3} \cos\biggl(\frac{\theta}{3}\biggr) \, ,</math>

<math>~y_2</math>

<math>~=</math>

<math>~p - \mathfrak{r}^{1 / 3} \biggl[ \cos\biggl(\frac{\theta}{3}\biggr) + \sqrt{3} \sin\biggl(\frac{\theta}{3}\biggr) \biggr] \, ,</math>

<math>~y_3</math>

<math>~=</math>

<math>~p - \mathfrak{r}^{1 / 3} \biggl[ \cos\biggl(\frac{\theta}{3}\biggr) - \sqrt{3} \sin\biggl(\frac{\theta}{3}\biggr) \biggr] \, .</math>

We have deduced empirically that <math>~y_3</math> is the root that is physically relevant in our case. That is to say, for a given <math>~0 < q < 1</math>,

<math>~c_0 = p - \mathfrak{r}^{1 / 3} \biggl[ \cos\biggl(\frac{\theta}{3}\biggr) - \sqrt{3} \sin\biggl(\frac{\theta}{3}\biggr) \biggr] \, .</math>

The light-blue diamonds in the right-hand panel of Figure 1 trace this <math>~c_0(q)</math> function, which is a solution to the above cubic equation. In the left-hand panel, we re-display a plot that has been discussed in an accompanying chapter. It contains a plot (blue markers) of the same function, but this time it is <math>~q(c_0)</math>, as determined from the root of a quartic equation. In order to illustrate more clearly that the two curves are the same, we have plotted the quartic solution on top of the cubic solution in the right-hand panel. The <math>~q(\alpha_e)</math> curve in the right-hand panel extends to smaller values of <math>~q</math> than does the corresponding curve in the left-hand panel. The curve on the left has been purposely truncated because, in the context of its original presentation, lower values of q would be associated with unrealistic values of the adiabatic exponent of the core.

Figure 1:  Comparing Roots to Quartic and Cubic Equations

quartic solution quartic solution


Try Mode22 <math>~(\ell,j) = (2,2)</math>

In this case we need to replace the core eigenfunction segment that was specified in STEP 3 in our separate discussion of mode <math>~(\ell,j) = (2,1)</math> with the following:

<math>~x_{j=2} |_\mathrm{core}</math>

<math>~=</math>

<math>~ \frac{35 - 126 (1+2q^3)^2 \xi^2 + 99 (1+2q^3)^4 \xi^4 }{35 - 126 (1+2q^3)^2 + 99 (1+2q^3)^4 } \, .</math>

So, following the procedure outlined in STEP 4 of our separate discussion, the requirement that the first derivatives match at the interface becomes,

<math>~\frac{ - 252 (1+2q^3)^2 + 396 (1+2q^3)^4 }{35 - 126 (1+2q^3)^2 + 99 (1+2q^3)^4 }</math>

<math>~=</math>

<math>~ \frac{c_0 + (c_0 + 3)A_{21}q^3 + (c_0 + 6)A_{21}B_{21} q^6}{1 + A_{21}q^3 + A_{21}B_{21}q^6} \, . </math>

So, as described above, in order to determine <math>~c_0</math> for any specified value of <math>~q</math> — or, equivalently, <math>~\Chi \equiv q^3</math> — we simply need to replace <math>~Q_{21}</math> with <math>~Q_{22}</math> in the definition of the coefficient of each term in the cubic expression, where,

<math>~Q_{22}</math>

<math>~\equiv</math>

<math>~ \frac{ - 252 (1+2\Chi)^2 + 396 (1+2\Chi)^4 }{35 - 126 (1+2\Chi)^2 + 99 (1+2\Chi)^4 } \, . </math>

Figure 2: Results for mode <math>~(\ell,j) = (2,2)</math>

Results for (ell,j) = (2,2)

<math>~q</math> <math>~0.768375</math>
<math>~c_0</math> <math>~-0.16913291</math>
<math>~\gamma_e</math> <math>~1.208583401</math>
<math>~\alpha_e</math> <math>~-0.30965989</math>
<math>~n_e</math> <math>~4.7942454</math>
<math>~\nu</math> <math>~0.6357658</math>
<math>~\rho_e/\rho_c</math> <math>~0.4756979</math>
<math>~\gamma_c</math> <math>~1.0464786</math>
<math>~\alpha_c</math> <math>~-0.822343</math>


Finally, as explained in our summary discussion, in order for the two (dimensional) eigenfrequencies to match, the adiabatic exponent for the core must be,

<math>~ \gamma_c </math>

<math>~=</math>

<math>~\frac{1}{[ 6 + 2j(2j+5)] } \biggl\{ 8 + \gamma_e \biggl[2\alpha_e + (c_0 + 3\ell)(c_0 + 3\ell +5) \biggr]\frac{\rho_e}{\rho_c} \biggr\} \, . </math>

Try Mode31 <math>~(\ell,j) = (3,1)</math>

In this case we need to replace the envelope eigenfunction segment that was specified in STEP 2 in our separate discussion of mode <math>~(\ell,j) = (2,1)</math> with the following: For the case of <math>~\ell=2</math>, this means that, throughout the envelope, the eigenfunction is,

<math>~x_{\ell=3} |_\mathrm{env}</math>

<math>~=</math>

<math>~ \xi^{c_0}\biggl[ \frac{ 1 + q^3 A_{31} \xi^{3} + q^6 A_{31}B_{31}\xi^{6} + q^9 A_{31}B_{31}C_{31}\xi^{9} }{ 1 + q^3 A_{31} + q^6 A_{31}B_{31} + q^9 A_{31}B_{31}C_{31} }\biggr] \, , </math>

where, the values of the newly introduced coefficients,

<math>~A_{31}</math>

<math>~\equiv</math>

<math>~\biggl[ \frac{c_0(c_0+5) - (c_0 + 9)(c_0 + 14)}{(c_0 + 3)(c_0+5) - \alpha_e}\biggr] </math>

 

<math>~=</math>

<math>~\biggl[ \frac{c_0^2 + 5c_0 - (c_0^2 + 23c_0 + 126)}{(c_0^2 + 8c_0 + 15) - (c_0^2 + 2c_0)}\biggr] </math>

 

<math>~=</math>

<math>~\frac{- 6(c_0 + 7)}{2c_0 + 5 } \, ,</math>

<math>~B_{31}</math>

<math>~\equiv</math>

<math>~\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 9)(c_0 + 14)}{(c_0 + 6)(c_0+8) - \alpha_e}\biggr] </math>

 

<math>~=</math>

<math>~\biggl[ \frac{(c_0^2 + 11c_0 + 24) - (c_0^2 + 23c_0 + 126)}{(c_0^2 + 14c_0 +48) - (c_0^2 + 2c_0) }\biggr] </math>

 

<math>~=</math>

<math>~\frac{ - (6c_0 + 51)}{ 6(c_0 +4) } \, ,</math>

<math>~C_{31}</math>

<math>~\equiv</math>

<math>~\biggl[ \frac{(c_0+6)(c_0+11) - (c_0 + 9)(c_0 + 14)}{(c_0 + 9)(c_0+14) - \alpha_e}\biggr] </math>

 

<math>~=</math>

<math>~\biggl[ \frac{(c_0^2 +17c_0 + 66) - (c_0^2 + 23 c_0 + 126)}{(c_0^2 + 23 c_0 + 126) - (c_0^2 + 2c_0) }\biggr] </math>

 

<math>~=</math>

<math>~\frac{- 2( c_0 + 10)}{7( c_0 + 6) } \, .</math>

Then the matching relation becomes,

<math>~Q</math>

<math>~=</math>

<math>~ \frac{c_0 + (c_0 + 3)A_{31}\Chi + (c_0 + 6)A_{31}B_{31} \Chi^2 + (c_0 + 9)A_{31}B_{31} C_{31}\Chi^3 }{1 + A_{31}\Chi + A_{31}B_{31}\Chi^2 + A_{31}B_{31}C_{31}\Chi^3 } </math>

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

<math>~=</math>

<math>~ [c_0 + (c_0 + 3)A_{31}\Chi + (c_0 + 6)A_{31}B_{31} \Chi^2 + (c_0 + 9)A_{31}B_{31} C_{31}\Chi^3 ] - Q[1 + A_{31}\Chi + A_{31}B_{31}\Chi^2 + A_{31}B_{31}C_{31}\Chi^3 ] </math>

 

<math>~=</math>

<math>~ \biggl[c_0 - (c_0 + 3)\cdot \frac{6(c_0 + 7)}{2c_0 + 5 } \cdot \Chi + (c_0 + 6) \cdot \frac{ 6(c_0 + 7)}{2c_0 + 5 } \cdot \frac{ (6c_0 + 51)}{ 6(c_0 +4) } \cdot \Chi^2 - (c_0 + 9)\cdot \frac{ 6(c_0 + 7)}{2c_0 + 5 } \cdot \frac{ (6c_0 + 51)}{ 6(c_0 +4) } \cdot \frac{ 2( c_0 + 10)}{7( c_0 + 6) } \cdot\Chi^3 \biggr] </math>

 

 

<math>~ - Q \biggl[1 - \frac{ 6(c_0 + 7)}{2c_0 + 5 } \cdot \Chi + \frac{ 6(c_0 + 7)}{2c_0 + 5 } \cdot \frac{ (6c_0 + 51)}{ 6(c_0 +4) } \cdot \Chi^2 - \frac{ 6(c_0 + 7)}{2c_0 + 5 } \cdot \frac{ (6c_0 + 51)}{ 6(c_0 +4) } \cdot \frac{ 2( c_0 + 10)}{7( c_0 + 6) } \cdot \Chi^3 \biggr] </math>

 

<math>~=</math>

<math>~ \biggl[7c_0(2c_0 + 5)(c_0 +4)( c_0 + 6) - (c_0 + 3)(c_0 +4)7( c_0 + 6)\cdot 6(c_0 + 7) \cdot \Chi + (c_0 + 6) 7( c_0 + 6)\cdot (c_0 + 7) \cdot (6c_0 + 51) \cdot \Chi^2 - (c_0 + 9)\cdot (c_0 + 7) \cdot (6c_0 + 51)\cdot 2( c_0 + 10) \cdot\Chi^3 \biggr] </math>

 

 

<math>~ - Q \biggl[ 7(2c_0 + 5)(c_0 +4)( c_0 + 6) - 42(c_0 + 7)(c_0 +4)( c_0 + 6) \cdot \Chi + 7( c_0 + 6)(c_0 + 7) \cdot (6c_0 + 51) \cdot \Chi^2 - 2(c_0 + 7) \cdot (6c_0 + 51) ( c_0 + 10) \cdot \Chi^3 \biggr] </math>

 

<math>~=</math>

<math>~ \biggl[7c_0(2c_0 + 5)(c_0^2 + 10c_0 + 24) - 42 (c_0 + 3)(c_0^2 + 10c_0 + 24)(c_0 + 7) \cdot \Chi + 7(c_0 + 6) ( c_0 + 6)\cdot (6c_0^2+93c_0 + 357)\cdot \Chi^2 - 2(c_0 + 9)\cdot (6c_0^2+93c_0 + 357)( c_0 + 10) \cdot\Chi^3 \biggr] </math>

<math>~ </math>

 

<math>~ - Q \biggl[ 7(2c_0 + 5)(c_0^2 + 10c_0 + 24) - 42(c_0 + 7)(c_0^2 + 10c_0 + 24) \cdot \Chi + 7( c_0 + 6)(6c_0^2+93c_0 + 357) \cdot \Chi^2 - 2(6c_0^2+93c_0 + 357)( c_0 + 10) \cdot \Chi^3 \biggr] </math>

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Whitworth's (1981) Isothermal Free-Energy Surface

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