Analytics II: Adiabatic energy balance

I spoke with Jerry Jungman.  It seems this idea of holding temperature fixed and attempting to do an energy balance is somewhat flawed.  I am going to make another attempt, this time setting the entropy equal to zero.  Jerry's feedback (paraphrased by me) can  be seen in the previous entry, highlighted in RED.

The mechanical work that is done on the gas by the shell is

$dW=T dS + P dV$

Let us assume that the system evolves adiabatically (which is okay until the very end, when the temperature gets too high).  In that case, as in the previous calculation we have

$dW=P dV$

Now we need a better expression for $P$ than we had before, one that takes into account that the temperature goes up as the system squishes in.  For an adiabatic process

$PV^\gamma=K$  where $K$ is constant, and $\gamma$ is the ratio of specific heats.  For a monatomic ideal gas  (is that what we have ???), $\gamma \approx 5/3$  so

$P=K \left(  \frac{4 \pi r^3}{3}\right)^{-5/3}$

So integrating it to get the work:

$W=\int P dV = K \left( \frac{4 \pi}{3}\right)^{-5/3} \int r^{-5} \frac{dV}{dr} \, dr$

$W=K \left( \frac{4 \pi}{3}\right)^{-5/3}\int 4\pi r^{-3} dr$

$W=  -2 \pi K \left( \frac{4 \pi}{3}\right)^{-5/3} r^{-2} + C$

Dependence of turnaround $r$ on $\delta v$: 

Following the previous line of reasoning (only now since entropy is constant it is much better motivated than before), we will set this work done in slowing the shell to a stop equal to the shell's initial specific kinetic energy (and let us group all the constants into K):

$\frac{(v_i-\delta v)^2}{2}=K \left(  \frac{1}{r_i^2} - \frac{1}{r^2} \right)$

Again we have introduced a $\delta v$ meant to represent some deficit in the kinetic energy associated with the radial velocity, possibly due to some non radial component of the velocity.  We can first solve for K assuming (1) $v_i \approx 3 \times 10^5$ m/s   (2) at $\delta v=0$ $r_i = 30 r$  (3) at turnaround, we expect $r \approx 15 \times 10^{-6}$ m.

$K=\frac{(3 \times 10^5)^2}{2*[(30 \times 15 \times 10^{-6})^{-2} - (15 \times 10^{-6})^{-2}]}$

Next we'd like an equation to plot which shows the turnaround radius as a function of the non-radial component of the velocity $\delta v$.

$r(\delta v) = \left[ \frac{1}{r_i^2} - \frac{(v_i - \delta v)^2}{2K} \right]^{-1/2}$

And the plot of this is here (you can click it to make it a little bigger):

If anything this makes the turnaround radius even less sensitive to small changes in the initial implosion velocity.  (Anna, this is for you) The following output from the code confirms that even a 20 percent change does not change the result by even a micron:



A one percent change in vinit yeilds delta r of: 0.15134424489 microns
A ten percent change in vinit yeilds delta r of: 1.66449516612 microns
A twenty percent change in vinit yeilds delta r of: 3.74414337015 microns

Dependence of $T$ on $\delta v$: 

One thing that might be possible (and nice to know) might be the gas temperature dependence on $\delta v$.  To do this, let's use the following property of adiabatic processes:

$T^\gamma P^{1-\gamma} = K_2$

where $K_2$ is constant.  Using

$P=K \left(  \frac{4 \pi r^3}{3}\right)^{-5/3}$

we get the following scaling of $T$ with $r$.

$T^\gamma=K_2 \left[  K \left( \frac{4 \pi r^3}{3}\right)^{-5/3} \right]^{\gamma-1}$

$T=\left[ K_2 \left[  K \left( \frac{4 \pi r^3}{3}\right)^{-5/3} \right]^{2/3} \right] ^{3/5}
      = K_3 r^{-2}$

We can get a numerical value by using assumption (3) above, and adding the info that at turnaround, the temperature should be about 20 keV.  I am being infernally sloppy with units, I hope it doesn't bite me.  In this case, $K_3= 20/(15 \times 10^{-6})^2$ keV/m^2.  Ergo

$T(\delta v) = K_3 \left[ \frac{1}{r_i^2} - \frac{(v_i - \delta v)^2}{2K} \right] $

(Anna this is for you) The temperature dependence is a bit more sensitive, here are the changes in T at the same three values of $\delta v$:



A one percent change in vinit yeilds delta T of: 0.397557777778 kev
A ten percent change in vinit yeilds delta T of: 3.79577777778 kev
A twenty percent change in vinit yeilds delta T of: 7.192 kev

PLAN:  Next, I think, I'd like to let the gas radiate energy away like a black body, and see what happens, both to the pressure support and to the temperature.