Torque from anisotropic T

These are some calculations I did about two weeks ago, I'm posting them here so I don't have to keep track of the papers themselves anymore.

Question:  Suppose there is some sort of weird anisotropic temperature gradient around the outside of the NIF capsule, is it even possible for a torque to be created?  The NIF capsule will deform, in an asymmetric way presumably, so at first it seems the answer should be yes, however, the force causing the deformation is always normal to the surface, therefore, any deformation will tend to symmetrize itself, so the torque being imparted may indeed be negligible.  Lets consider a 2D model, with a $\theta$ dependent temperature field deforming the capsule.  Let's compute the resulting torque as a function of time.  Here was attempt 1:  This attempt fails because I started describing the ring parametrically, but I should have left the parameter some arbitrary $t$, and not used the angle $theta$ as the parameter.  I am working on updating this attempted calculation.

Here's the faulty one:

Conservation of Angular Momentum

Jim sent me this in an email:

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.  

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.