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.  

Analytics 1: Isothermal energy balance (not a good idea)

I'd asked Jim if there were anything possible to be done with analytics, and at first he said no, but then he stopped by my office today and suggested a very simple calculation.  Nobody laugh, I am brand new at this stuff, so everybody keep in mind that we are just talking about a cartoon here.

Suppose you ignore all inward and outward going shocks, then if you have an inward imploding shell (in 1D) coasting inward at a nearly constant initial velocity, it will encounter a repulsive potential.  It will decelerate, stall, and bounce.  As a simple place to start, lets say the pressure is proportional to the inverse volume with constant of proportionality $\alpha$ (here we tacitly assume that the T is constant which can I just say doesn't seem like a great idea, however...):

$PV=NK_bT$ so $P \propto \frac{1}{V}$

$P=\frac{\alpha}{V} = \alpha \frac{3 \alpha}{4 \pi r^3}$

The force experienced by the shell is given by

$dF = -P dA$

JERRY SAYS: this next step is confused.  Force and area are both vectors, but I've somehow lost track of what they mean, there should not be an integral over r to get the force... He says if we want to do a force balance, the least confusing thing is to balance force per unit area (the pressure) with mass per unit area times acceleration.  We have to keep track of the fact that the mass per unit area of the shell is changing, because the area is changing.   

$F=-\int P(r) \, \frac{dA}{dr} \, dr$

$A=4\pi r^2$ so $\frac{dA}{dr}=8 \pi r$

thus
$F=-\int \frac{3 \alpha}{4 \pi r^3} \,  8\pi r \, dr= -\int \frac{6 \alpha}{r^2} \, dr
= \frac{6 \alpha}{r} + C$

We can write the force in terms of a potential, and we can recognize that the potential will have certain symmetries, reducing this to a 1D problem.  So a gradient goes to $\partial/\partial r$.

$F =\nabla \phi = \frac{\partial \phi}{\partial r}$.

$\phi = \int \left(  \frac{6 \alpha}{r} + C \right) dr = 6\alpha \ln r + C_1r + C_2$

This potential is the work per unit mass done on the shell, as far as I can tell...  What puzzles me is the following, we can do the computation another way because we also have an expression for the work:

$dW= P dV$

so

$W=\int P \, \frac{dV}{dr} \, dr  $

using the same sort of trick

JERRY SAYS:  So far this is okay, but I have tacitly set $T dS = 0$ in the equation above for $dW$.  This means we cannot in fact simultaneously hold the temperature fixed.  I will attempt to fix this in the next notebook entry.  

$V=\frac{4 \pi r^3}{3}$  so  $\frac{dV}{dr}=4 \pi r^2$

thus

$W=\int \frac{3 \alpha}{4 \pi r^3} \, 4 \pi r^2 dr$

$W= \int \frac{3\alpha}{r} \, dr = 3\alpha \ln r + C$

So the first term has the same r dependence, but is off by a factor of two, and there is no term that is simply proportional to $r$.  Maybe there is a way to insist that $C_1=0$ ??  Also, since the potential was supposed to be the work per unit mass, there seems to be a unit problem between the two as well...

But let's press on anyhow.   Let's ignore the linear term, and just use the log term, since it was the same in both calculations.  Let's equate this work to the specific energy of the shell (since I don't know the mass of the shell for now), and see what happens functionally to the turn around radius if we allow some component of the initial shell velocity to be non-radial.  We will ignore any influence of the pressure on this non-radial component, just reduce the initial velocity by some amount $\delta v$.  Let's also assume that the shell starts at some initial radius $r_i$ (turning the dW integral into a definite integral from r_i to r).

$\frac{(v-\delta v)^2}{2} = 3\alpha \ln \left( \frac{r}{r_i}\right)$

ack, there must be a sign error some place, since we should definitely have $r<<r_i$

$\frac{r}{r_i} = e^{\frac{-(v-\delta v)^2}{6 \alpha}}$

so

$r( \delta v) = r_i \, e^{\frac{-(v-\delta v)^2}{6 \alpha}}$

alright, so lets make a plot of this... we know if $\delta v = 0$ that $r/r_i \approx 1/30$, and also that the initial velocity should be roughly $v\approx 3 \times 10^5 $ m/s (and so I don't lose this number though it's not relevant now, the inner turn around radius is supposed to be about $r \approx 15 \times 10^{-6}$ m).

$e^{\frac{-(3\times10^5)^2}{6 \alpha}}=1/30$

so (and I have no idea what is up with the units at this point...)

$\alpha = \frac{9 \times 10^{10}}{6 \ln 30} \approx 4.4 \times 10^{9}$

So, here is a plot then, of the turnaround $r$ versus $\delta v/v$

NOTE: This plot is not correct because I have erroneously held both  the temperature and the entropy fixed.

Note: you can click on the plot to see it in a larger size.  Upshot: a percent error seems roughly tolerable, but a few tens of percent change in the imploding velocity of the shell yields a factor of 2 difference in the turnaround radius.

Next on the agenda, clean up the loose ends, and start thinking about angular momentum.