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.