Editing Appendix/Ramblings/PatrickMotl (section)

==December 31, 2020 (from Patrick)==

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I am attaching plots of density for xi = 3.5, 3.25.

The dashed lines are for exp(sigma_c t) where t is measured in free fall times and I took sigma from your animated gif. 0.19907 for xi of 3.25 and 0.25550 for  xi of 3.5.

I don’t think it is the growth of just one mode.
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  <td align="center">[[File:Motl n5 ln rho all.png|500px|Patrick's initial log10 plot]]</td>
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Joel's interpretation:  Measuring the slopes of the dashed lines directly from the figure, I obtain the following slopes.  First, the starting and ending points of the red dashed-line segment ''appear to be'', respectively, <math>~(\rho_c, t/P_0) = (1.00, 0.00)</math> and <math>~(\rho_c, t/P_0) = (5.00, 6.25)</math>.  Hence &hellip;

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  <td align="left">For <math>~\tilde\xi = 3.5~</math>: &nbsp; &nbsp;&nbsp; &nbsp;</td>
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<math>~\frac{\Delta \ln\rho_c}{\Delta t/P_0}</math>
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<math>~=</math>
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<math>~
\frac{\ln(5.0) - \ln(1.0)}{6.25 - 0} = \frac{1.6094}{6.25} = 0.2575 \, .
</math>
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This is consistent with <math>~\sigma_c = \sqrt{0.065282} = 0.25555</math>, which is the value that I said comes from the linear-stability analysis.   Second, the starting and ending points of the yellow dashed-line segment ''appear to be'', respectively, <math>~(\rho_c, t/P_0) = (1.00, 0.00)</math> and <math>~(\rho_c, t/P_0) = (5.00, 8.05)</math>.  Hence &hellip;

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  <td align="left">For <math>~\tilde\xi = 3.25~</math>: &nbsp; &nbsp;&nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\Delta \ln\rho_c}{\Delta t/P_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\ln(5.0) - \ln(1.0)}{8.05 - 0} = \frac{1.6094}{8.05} = 0.1999 \, .
</math>
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This is consistent with <math>~\sigma_c = \sqrt{0.039629} = 0.1991</math>, which is the value that I said comes from the linear-stability analysis.  Okay.  I understand Patrick's plot.