Editing Appendix/Ramblings/PatrickMotl (section)

===Slope in Semi-log Plots===
We need to change the way we analyze Patrick's hydro-data output.  First, note that Patrick's figure (black background with yellow and red dashed-line segments) is a plot of,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\log_{10}\biggl\{\frac{\rho(t)}{\rho_0}\biggr|_c \biggr\}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; ''vs.''&nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\frac{t}{P_0} \, .</math>
  </td>
</tr>
</table>

But, as has been stated in our [[Appendix/Ramblings/PatrickMotl#Additional_Information_on_Frequencies|earlier comments]], from the standpoint of a linear-perturbation analysis,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl\{ \frac{\rho(t)}{\rho_0} \biggr|_c \biggr\}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 + d_c e^{\sqrt{\omega^2}t} \, .
</math>
  </td>
</tr>
</table>
So we should not expect the plots of <math>~\rho_c(t)</math> in Patrick's (black background) figure to show a linear relationship.  Instead, we should be plotting the quantity,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\log_{10}\biggl\{\frac{\rho(t)}{\rho_0}\biggr|_c - 1\biggr\}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; ''vs.''&nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\frac{t}{P_0} \, .</math>
  </td>
</tr>
</table>

Over the earliest phase of the hydrodynamical simulation while the deviation from the initial model's structure is still quite small &#8212; say, for times, <math>0 \le t/P_0 \le 5</math> &#8212; such a plot should reveal the following linear relationship:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\log_{10}\biggl\{ \frac{\rho(t)}{\rho_0} \biggr|_c - 1\biggr\}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\log_{10}(d_c) + \log_{10} \biggl[ e^{\sqrt{\omega^2}t} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\log_{10}(d_c) + (\log_{10}e) \cdot \ln \biggl[ e^{\sqrt{\omega^2}t} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\log_{10}(d_c) + 0.434294 \cdot \sqrt{\omega^2}t \, .
</math>
  </td>
</tr>
</table>

<table border="1" cellpadding="8" align="center" width="60%">
<tr><td align="left">
Alternatively, we can use natural logarithms throughout, which gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\ln\biggl\{ \frac{\rho(t)}{\rho_0} \biggr|_c - 1\biggr\}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\ln(d_c) + \sqrt{\omega^2}t \, .
</math>
  </td>
</tr>
</table>

</td></tr>
</table>

Converting to <math>~\sigma_c^2</math>, [[#convert|as above]], we therefore have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\log_{10}\biggl\{ \frac{\rho(t)}{\rho_0} \biggr|_c - 1\biggr\}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\log_{10}(d_c) + 0.434294 \cdot \biggl[\frac{\pi}{4}\biggl( \frac{\rho_c}{\bar\rho}\biggr)^{1 / 2}\biggr] \sqrt{\sigma_c^2} \biggr(\frac{t}{T_0}\biggr) \, .
</math>
  </td>
</tr>
</table>

Hence,

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

<tr>
  <td align="left">For <math>~\tilde\xi = 3.25~</math>:&nbsp; &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~\log_{10}\biggl\{ \frac{\rho(t)}{\rho_0} \biggr|_c - 1\biggr\}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\log_{10}(d_c) + 0.434294 \cdot \biggl[0.48847\biggr] \biggr(\frac{t}{T_0}\biggr) \, ; 
</math>&nbsp; and,
  </td>
</tr>

<tr>
  <td align="left">For <math>~\tilde\xi = 3.5~</math>:&nbsp; &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~\log_{10}\biggl\{ \frac{\rho(t)}{\rho_0} \biggr|_c - 1\biggr\}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\log_{10}(d_c) + 0.434294 \cdot \biggl[0.67936\biggr] \biggr(\frac{t}{T_0}\biggr) \, .
</math>
  </td>
</tr>
</table>

<font color="red">'''NOTE:''' &nbsp; Even though we know analytically what the (spherically symmetric) expression is for the equilibrium model's radial density distribution, when this model is introduced into the hydrocode's grid, there inevitably will be some noise.  In particular, the value of the central density at time <math>~t = 0</math> will not be perfectly correct. The parameter, <math>~d_c</math> &#8212; perhaps on the order of 10<sup>-2</sup> or 10<sup>-6</sup> &#8212; represents the amplitude of this ''departure'' from the correct value at the start of a hydrodynamic simulation; we won't know its value until the simulation has been completed and this linear function is fit to the plotted data.</font>  Alternatively, given that we know from Joel's linear stability analysis what the eigenfunction is for the fundamental mode of oscillation, Patrick could introduce this eigenfunction '''with a known, specified amplitude''' into the initial density distribution.  But, first, let's see how the "linear fit" works without introducing a particular initial perturbation.