Editing SSC/Perturbations (section)

===Linearizing the Key Equations===

====Adiabatic form of the First Law of Thermodynamics====

Plugging the perturbed expressions for <math>~P(m,t)</math> and <math>~\rho(m,t)</math> into the adiabatic form of the First Law of Thermodynamics, we obtain,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(i\omega) \rho_0(m) \biggl[1 + d(m) e^{i\omega t} \biggr]P_0(m) p(m) e^{i\omega t} - \gamma_\mathrm{g}(i\omega) P_0(m) \biggl[1 + p(m) e^{i\omega t}  \biggr] \rho_0(m)  d(m) e^{i\omega t}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~~
(i\omega) \rho_0(m)P_0(m) e^{i\omega t} \biggl\{\biggl[1 + d(m) e^{i\omega t} \biggr] p(m)  - \gamma_\mathrm{g} \biggl[1 + p(m) e^{i\omega t}  \biggr]   d(m)  \biggr\}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>
</div>

Because we are seeking solutions that will be satisfied throughout the configuration &#8212; that is, for all mass shells <math>~m</math> &#8212; the expression inside the curly brackets must be zero. Hence,
<div align="center">
<math>
~p(m) - \gamma_\mathrm{g} d(m) + (1 - \gamma_\mathrm{g} ) d(m)p(m) e^{i\omega t} =0  \, .
</math>
</div>
Also, because we are only examining deviations from the initial equilibrium state in which <math>~|d(m)|</math> and <math>~|p(m)|</math> are both <math>\ll 1</math>, then the third term on the left-hand-side of this equation, which contains a product of these two small quantities, must be much smaller than the first two terms.  As is standard in perturbation theory throughout physics, for our stability analysis, we will drop this "quadradic" term and keep only terms that are linear in the small quantities.  This leads to the following algebraic relationship between <math>~d(m)</math> and <math>~p(m)</math>:
<div align="center">
<math>
~p = \gamma_\mathrm{g} d  \, .
</math>
</div>

====Entropy Conservation====
If, instead, we simply demand that the specific entropy remain constant in time, then the expression that relates <math>P</math> to <math>\rho</math> is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\frac{P}{\rho^{\gamma_g}}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\exp\biggl[ \frac{s(\gamma_g - 1)}{\mathfrak{R}/\bar\mu}\biggr]\, .
</math>
  </td>
</tr>
</table>

Plugging the perturbed expressions for <math>P(m,t)</math> and <math>\rho(m,t)</math> into this expression gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\exp\biggl[ \frac{s(\gamma_g - 1)}{\mathfrak{R}/\bar\mu}\biggr]
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{ P_0(m) \biggl[1 + p(m) e^{i\omega t}  \biggr] \biggr\}
\biggl\{\rho_0(m) \biggl[1 + d(m) e^{i\omega t}  \biggr] \biggr\}^{-\gamma_g}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>P_0(m) \rho_0(m)^{-\gamma_g} 
\biggl[1 + p(m) e^{i\omega t}  \biggr]
\biggl[1 + d(m) e^{i\omega t}  \biggr]^{-\gamma_g} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 1</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\biggl[1 + p(m) e^{i\omega t}  \biggr]
\biggl[1 + d(m) e^{i\omega t}  \biggr]^{-\gamma_g} \, .
</math>
  </td>
</tr>
</table>
Now, according to the [[Appendix/Ramblings/PowerSeriesExpressions#Binomial|binomial theorem]],

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>(1 \pm x)^n</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<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>
&nbsp; &nbsp;&nbsp; for <math>~(x^2 < 1)</math>
  </td>
</tr>
</table>

Hence, we see that,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\biggl[ 1 + d(m) e^{i\omega t}\biggr]^{-\gamma_g}</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>
1 ~- ~\gamma_g d(m) e^{i\omega t} 
+ \biggl[\frac{\gamma_g(\gamma_g+1)}{2!}\biggr]\biggl[d(m) e^{i\omega t}\biggr]^2 \, .
</math>
  </td>
</tr>
</table>
Dropping terms of higher order than unity, the relationship between <math>p(m)</math> and <math>d(m)</math> becomes,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>1</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math> 
\biggl[1 + p(m) e^{i\omega t}  \biggr]
\biggl[1 ~- ~\gamma_g d(m) e^{i\omega t} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
1 + p(m) e^{i\omega t} 
~- ~\gamma_g d(m) e^{i\omega t} 
~- ~\gamma_g \cancelto{\mathrm{small}}{d(m)p(m)} e^{2i\omega t}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ p(m)</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>  
\gamma_g d(m)  \, .
</math>
  </td>
</tr>
</table>

This is identical to the algebraic relationship between <math>p(m)</math> and <math>d(m)</math> that has been derived, [[#Adiabatic_form_of_the_First_Law_of_Thermodynamics|immediately above]].

====Continuity Equation====

Adopting the same approach, we will now "linearize" each term in the continuity equation:
<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
~\frac{d\rho}{dt}  
</math>
  <td align="center">
<math>
\rightarrow
</math>
  </td>
  <td align="left">
<math>
(i\omega)\rho_0 d~e^{i\omega t}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>
\frac{\rho}{r}  
</math>
  </td>
  <td align="center">
<math>
\rightarrow
</math>
  </td>
  <td align="left">
<math>
\frac{\rho_0}{r_0} \biggl[1 + d e^{i\omega t}  \biggr] \biggl[1 + x e^{i\omega t}  \biggr]^{-1} 
\approx  \frac{\rho_0}{r_0}  \biggl[1 + d ~e^{i\omega t}  \biggr]\biggl[1 - x~ e^{i\omega t}  \biggr] \approx 
 \frac{\rho_0}{r_0} \biggl[1 + (d - x) ~e^{i\omega t}  \biggr]
</math>
</tr>
<tr>
  <td align="right">
<math>
\frac{dr}{dt}
</math>
  </td>
  <td align="center">
<math>
\rightarrow
</math>
  </td>
  <td align="left">
<math>
(i\omega) r_0 x~e^{i\omega t}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>
~\rho^2 r^2 
</math>
  </td>
  <td align="center">
<math>
\rightarrow
</math>
  </td>
  <td align="left">
<math>
\rho_0^2 r_0^2 \biggl[1 + d e^{i\omega t}  \biggr]^2 \biggl[1 + x e^{i\omega t}  \biggr]^2 
\approx  \rho_0^2 r_0^2 \biggl[1 + 2d ~e^{i\omega t}  \biggr]\biggl[1 + 2x~ e^{i\omega t}  \biggr] \approx 
\rho_0^2 r_0^2 \biggl[1 + 2(d + x) ~e^{i\omega t}  \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>
\frac{d}{dm}\biggl(\frac{dr}{dt}\biggr)
</math>
  </td>
  <td align="center">
<math>
\approx
</math>
  </td>
  <td align="left">
<math>
\frac{d}{dm}\biggl[(i\omega) r_0 x~e^{i\omega t}\biggr] = (i\omega) e^{i\omega t} \biggl[x\frac{dr_0}{dm} + r_0\frac{dx}{dm} \biggr] = (i\omega) e^{i\omega t} \biggl[\frac{x}{4\pi r_0^2 \rho_0} + r_0\frac{dx}{dm} \biggr]
</math>
  </td>
</tr>
</table>

In the last step of this last expression we have made use of the fact that, in the initial, unperturbed equilibrium model, <math>dr_0/dm = 1/(4\pi r_0^2 \rho_0)</math>.  Combining all of these terms and linearizing the combined expression further, the linearized continuity equation becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(i\omega)\rho_0 d~e^{i\omega t}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~- \frac{2\rho_0}{r_0} \biggl[1 + (d - x) ~e^{i\omega t}  \biggr] (i\omega) r_0 x~e^{i\omega t} - 4\pi \rho_0^2 r_0^2 \biggl[1 + 2(d + x) ~e^{i\omega t}  \biggr](i\omega) e^{i\omega t} \biggl[\frac{x}{4\pi r_0^2 \rho_0} + r_0\frac{dx}{dm} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \rho_0 d </math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~- 3\rho_0 x - 4\pi \rho_0^2 r_0^3 \frac{dx}{dm}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 4\pi \rho_0 r_0^3 \frac{dx}{dm}  </math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~- 3 x - d \, ,</math>
  </td>
</tr>
</table>
<table border="1" cellpadding="8"><tr><td align="center">
NOTE: &nbsp; An alternate approach to deriving this expression can be found in an [[SSC/Structure/BiPolytropes/51RenormaizePart2#STEP4|accompanying <font color="red">ASIDE</font>]].
</td></tr></table>
</div>

or,
<div align="center">
<math>
r_0 \frac{dx}{dr_0} \approx - 3 x - d ,
</math>
</div>

where, to obtain this last expression, we have switched back from differentiation with respect to <math>~m</math> to differentiation with respect to <math>~r_0</math>.

====Euler + Poisson Equations====
Finally, linearizing each term in the combined "Euler + Poisson" equation gives:

<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
\frac{d^2r}{dt^2}  
</math>
  </td>
  <td align="center">
<math>
\rightarrow
</math>
  </td>
  <td align="left">
<math>
\frac{d}{dt}\biggl[(i\omega) r_0 x~e^{i\omega t}\biggr] = - \omega^2 r_0 x~e^{i\omega t}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>
r^2 \frac{dP}{dm}  
</math>
  </td>
  <td align="center">
<math>
\rightarrow
</math>
  </td>
  <td align="left">
<math>
r_0^2 \biggl[1 + x~ e^{i\omega t}  \biggr]^2 \biggl\{\frac{dP_0}{dm} \biggl[1 + p~ e^{i\omega t}  \biggr]  + P_0~e^{i\omega t} \frac{dp}{dm}   \biggr\} \approx r_0^2 \frac{dP_0}{dm} \biggl[1 + (2x+p)~ e^{i\omega t}  \biggr] + P_0 r_0^2~e^{i\omega t} \frac{dp}{dm}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>
\frac{Gm}{r^2}  
</math>
  </td>
  <td align="center">
<math>
\rightarrow
</math>
  </td>
  <td align="left">
<math>
\frac{Gm}{ r_0^2} \biggl[1 + x~ e^{i\omega t}  \biggr]^{-2} \approx \frac{Gm}{ r_0^2} 
\biggl[1 -2 x~ e^{i\omega t}  \biggr] \, .
</math>
  </td>
</tr>
</table>

Hence, the combined linearized relation is,
<div align="center">
<math>
- \omega^2 r_0 x~e^{i\omega t} \approx -4\pi \biggl\{r_0^2 \frac{dP_0}{dm} \biggl[1 + (2x+p)~ e^{i\omega t}  \biggr] + P_0 r_0^2~e^{i\omega t} \frac{dp}{dm} \biggr\} - \frac{Gm}{ r_0^2} \biggl[1 -2 x~ e^{i\omega t}  \biggr]
</math><br />
<math>
\Rightarrow ~~~~~ e^{i\omega t} \biggl\{(2x + p)4\pi r_0^2 \frac{dP_0}{dm}-2x \frac{Gm}{r_0^2} + 4\pi P_0 r_0^2 \frac{dp}{dm} -\omega^2 r_0 x  \biggr\} \approx - 4\pi r_0^2 \frac{dP_0}{dm} - \frac{Gm}{r_0^2}
</math><br />
<math>
\Rightarrow ~~~~~  4\pi P_0 r_0^2 \frac{dp}{dm}  \approx (4x + p)g_0 + \omega^2 r_0 x \, ,
</math><br />
</div>
<span id="g0">where,</span> in order to obtain this last expression we have made use of the fact that, in the unperturbed equilibrium configuration,
<div align="center" id="g0Defined">
<math> 
g_0(m) \equiv \frac{Gm}{r_0^2} = - 4\pi r_0^2 \frac{dP_0}{dm} = - \frac{1}{\rho_0} \frac{dP_0}{dr_0} \, .
</math><br />
</div>
Switching back from differentiation with respect to <math>~m</math> to differentiation with respect to <math>~r_0</math>, the "Euler + Poisson" combined linearized relation can alternatively be written as,
<div align="center">
<math>
\frac{P_0}{\rho_0} \frac{dp}{dr_0}  \approx (4x + p)g_0 + \omega^2 r_0 x .
</math><br />
</div>