User:Tohline/Appendix/Ramblings/Additional Analytically Specified Eigenvectors for Zero-Zero Bipolytropes

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.

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

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) + 4\Chi[ 13\Chi -14 -Q (\Chi-2) ] + 22\Chi (\Chi - 2)\, , </math> <math>~c</math> <math>~\equiv</math> <math>~ [40 - 26Q]+ 4\Chi[42\Chi -Q (7\Chi-8) -24] + 22\Chi[13\Chi -14 -Q (\Chi-2) ] \, , </math> <math>~d</math> <math>~\equiv</math> <math>~ - 40Q+ 22\Chi[42\Chi -Q (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>

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