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A closer look at Jeans Instability plots

First is a plot of the total density at a peak (i.e., a peak is a region that is growing in density over simulation). The analytical function for the density is:

\rho = \rho_0 + \rho ' = \rho_0 + 0.001 \rho_0 Cos(kx) e^{\Gamma t}

where \rho_0 is the background density, \rho ' is the density perturbation, and \Gamma is the growth rate, calculated to be ~628.8 (in computational units).

At the peak, the Cos(x) function is just 1, and so in scaled units, the function becomes:

\rho = \rho_0 / rscale + \rho ' / rscale = 1 + 0.001 e^{628.8 t}

Since this is not an exact exponential, it does not appear as a straight line on a log-linear plot (see for reference). Here is the data output from astrobear compared to the analytical function, on a linear-linear plot:

Now, since the perturbation \rho ' is just an exponential (given by \rho - 1 above), it looks like this on a semi-log plot:

The slope of this line, on a semi-log plot, gives the power 'e' is raised to. That is, since \rho ' \propto e^{628.8t}, log \rho ' \propto 628.8 ~log(e) ~t. Here the log is base 10, NOT e, so log(e) DNE 1. This is an equation of a line with slope = log(e)*628.8, or ~273. Thus, if one plots the perturbation on a semi-log plot, and calculates the slope — it should give back this number - m=273.

When calculating the average slope using log(y2)-log(y1)/x2-x1, I get a number very close to 273. This is good. But perhaps more descriptive is a plot of the instantaneous slope of this line - to see any deviations of astrobear's output from the analytical curve. To make this plot, I calculated the derivative of the log of the analytical function:

\frac{d log\rho '}{dt} = 628.8 log_{10} e = const. = 273

(as expected), and plotted this against the instantaneous derivative of the data (calculated using log(i+1)-log(i) / log(j+1) - log(j) in my spreadsheet). This is what I get:

As we see, initially the data from astrobear lines up very well with the analytical result, but over a couple of e-folding times, the rate of change increases compared to the analytical curve. By 5 e-folding times, the difference is ~10% (easily seen in the first plot). If we consider the error from the non-linear term to grow roughly as (0.001)^2*(e^{628.8*t})^2, and take t = 5/628.8 (5 e-folding times), then the error is ~2%. This seems to match the data. This means that by a few e-folding times, the non-linear term becomes significant, and so we can no longer assume a linear perturbation to govern evolution.

Posted: 11 years ago (Updated: 11 years ago)
Author: Erica Kaminski
Categories: (none)

Attachments (3)

totalJeans.png (16.4 KB) - added by Erica Kaminski 11 years ago.
slopeJeans.png (16.7 KB) - added by Erica Kaminski 11 years ago.
densityJeans.png (15.1 KB) - added by Erica Kaminski 11 years ago.

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Comments

1. Jonathan -- 11 years ago

I would rescale x-axis to be in e-folding times… Also (and this is not necessary - but more for curiosity) if you include the first non-linear term and take \frac{d log(g)}{dt} where g = \frac{\rho_1}{\rho_0} you get something that goes like

g = \epsilon e^{\alpha t} + \left(\epsilon e^{\alpha t} \right)^2 = \epsilon e^{\alpha t} \left (1+\epsilon e^{\alpha t} \right )

\log g = \alpha t + \epsilon e^{\alpha t}

\frac{d \log g}{d t} = \alpha+\epsilon \alpha e^{\alpha t} = \alpha \left ( 1 + \epsilon e^{\alpha t} \right )

Just curious if you plot this function for the instantaneous growth rate - how well it compares with AstroBEAR. A rough estimate puts it fairly close…

2. Jonathan -- 11 years ago

Should have said that \log g = \log \epsilon e^{\alpha t} + \log \left (1 + \epsilon e^{\alpha t} \right ) \approx \log \epsilon + \alpha t + \epsilon e^{\alpha t} where I used the taylor expansion for \log(1+x) = x - \frac{1}{2}x^2 + ...

and the \log \epsilon gets dropped when you take the derivative

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