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 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. 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

According to our accompanying summary discussion, we need to solve the following "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 will assume that <math>~\Chi \equiv q^3</math> is specified, and we 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, to simply notation, let's 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>

Using <math>~z</math> in place of <math>~c_0</math>, this can be written in the form of a standard cubic equation. Specifically,

<math>~a z^3 + b z^2 + c z + 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>

Related Discussions

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