please try this.
If I invoke conservation of angular momentum with a 1 percent theta term I get
$p =m(v_r + 0.01v_t)$
$r_i = 0.08 cm$
$L = (0.08)*(0.01)*3x10^7 cm^2/s$
$ = 2.4 x10^4 cm^2/s$
if $L$ is conserved I can solve for $v_t$ at a final radius of $r_f=0.003 cm$
$v_t = L/r_f $= 0.8 x 10^7 cm/s
this is about 25% of the initial velocity at 30 microns
What does this mean?
The following extremely simple calculation tells us about what happens if we have a small angular momentum component that needs to be conserved in the collapse.
The angular momentum for a rotating body is given by
$L=I\omega$
where
$\omega=\frac{r \times v}{|r|^2} = \frac{v}{r}$ since $v$ is perpendicular to $r$.
The rotational kinetic energy of such a body is
$T=I\omega^2$.
The moment of inertia of a spherical shell at radius R is given by (see e.g. this page for a derivation):
$I=\frac{2}{3}MR^2$
The value of $I$ will decrease significantly as the ablator shell is compressed, and thus $\omega$ will compensate to conserve $L$. The increase in $\omega$ results in a larger value of $T$ at the smaller radii.
Assume for the moment that the ablator shell is extremely thin, and that only the ablator shell (and not the solid fuel or the gaseous fuel) have any angular momentum (we can relax this assumption later and estimate the impact). The conservation of angular momentum will affect the turn-around radius.
In the previous post, we computed the work done by the gas in stopping the shell, and equated that to the specific kinetic energy associated with the shell's initial radial velocity. The conservation of angular momentum generates rotational kinetic energy in the shell during the collapse, and this must be now be subtracted from the amount of work to be done by the gas. (Note: we don't need the shell mass in $I$ for the specific rotational kinetic energy.)
$\frac{(v-\delta v)^2}{2}-(I_2\omega_2^2 - I_1\omega_1^2) = K\left(\frac{1}{r_i^2} - \frac{1}{r^2}\right)$
Using angular momentum conservation we can express $I \omega^2$ in terms of the variables $r_i$ and $\delta v$. First we note that
$I_1 \omega_1 = I_2 \omega_2$
$\frac{2}{3}r_i^2 \frac{\delta v }{r_i} = \frac{2}{3}r^2 \frac{v_2 }{r} $
where $v_2$ is the final rotational velocity of the shell. Thus,
$r_i \delta v = r v_2 $
The rotational kinetic energies are therefore given by
$I_1 \omega_1^2 = \frac{2}{3} r_i^2 \left( \frac{\delta v}{r_i}\right)^2 = \frac{2}{3}\delta v^2$
$I_2 \omega_2^2 = \frac{2}{3}r^2 \left(\frac{v_2}{r}\right)^2 = \frac{2}{3} r^2 \left(\frac{\delta v \, r_i}{r^2}\right)^2 = \frac{2}{3} \frac{\delta v^2 r_i^2}{r^2}$
Plugging these into the energy balance expression, and solving for $r$ in terms of $r_i$ and $\delta v$ we get:
$\frac{(v-\delta v)^2}{2} + \frac{2}{3} \delta v^2 = K \left[ \frac{1}{r_i^2}-\frac{1}{r^2}\right]
+ \frac{2}{3} \frac{\delta v^2 r_i^2}{r^2}$
$r=\left[ \frac{\frac{1}{2K}\left(v-\delta v\right)^2 + \frac{2}{3K}\delta v^2 - r_i^{-2}}{\frac{2}{3K} \delta v^2 \, r_i^2 -1}\right]^{-1/2}$
The effect of this will be to increase the turnaround radius that can be achieved, due to the momentum barrier. Using the same values of $K$ and $r_i$ from the previous post, here is a plot of r versus $\delta v$.
The blue curve is the same as in the previous post, and the green curve is the new curve with angular momentum conservation of the shell only. And a few numbers for comparison, printed out by the code:
A one percent change in vinit yeilds delta r of: 1.03263181609 microns
A ten percent change in vinit yeilds delta r of: 44.5662817245 microns
A twenty percent change in vinit yeilds delta r of: 111.001318027 microns
This translates into a bigger effect on the temperature:
And more numbers from the code:
A one percent change in vinit yeilds delta T of: 2.4933570134 kev
A ten percent change in vinit yeilds delta T of: 18.7317305499 kev
A twenty percent change in vinit yeilds delta T of: 19.7165592179 kev
The next thing, I think, is to find out whether the gas ice layer is rigidly attached to the shell, because if it is, it may also be gaining angular momentum, and trying to conserve L will further increase the turnaround $r$ and decrease the inner temperature.
