Different vlasov approach

[kcdodd]

I am about to give up on my other project. I think it is out of reach for me to try to solve the vlasov totally general. But I do have an idea. If we assume the distribution in momentum space has a certain structure it may not take much information to store. I think our main goal is to not assume the distribution is Maxwellian. But, we might assume other things about it.

So, I was thinking about ellipsoids. It can be described by three numbers, for the semi axes. In spherical coordinates we would have for an ellipsoid surface:

Vx = A*cos(theta)*sin(phi)
Vy = B*sin(theta)*sin(phi)
Vz = C*cos(phi)

Now we need density data, so we could have three more numbers and interpolate them between the axes for the surface of the ellipsoid:

f = (f_a*cos^2(theta) + f_b*sin^2(theta))*sin^2(phi) + f_c*cos^2(phi)

where f_a is the density at the Vx-axis, f_b at Vy-axis, and f_c at Vz-axis.

For the momentum gradient in spherical coordinates (ignoring the radial part for a second) would be:

grad(f) = (1/r*sin(phi)) df/dtheta (theta_hat) + (1/r) df/dphi (phi_hat)

df/dtheta = (-2*f_b*sin(theta)cos(theta) + 2*f_b*sin(theta)cos(theta))*sin^s(phi)

df/dphi = (f_a*cos^2(theta) + f_b*sin^2(theta))*2*sin(phi)cos(phi) - f_c*sin(phi)cos(phi)

Now, for the radial part of the momentum gradient we could have concentric ellipsoids. Simply dividing the difference of density by the radial distance of the two surfaces.

For the spatial part of the gradient we can have "adjacent" ellipsoids. Since we can place these on a regular spatial grid we can have a very nice gradient in cartesian coordinates by just taking the difference of an interpolated density divided by the spatial distances.

We could also make it fancier by having each ellipsoid be oriented in a favorable direction, say Vz is the radial direction in the polywell, or the parallel ocomponent to the B-field.

Since the description of the momentum space is essentially 1-D by doing this the memory requirements might be manageable, allowing much denser meshes. What I haven't figured out yet is how to actually alter the ellipsoids and densities to find a solution. So, yeah it's just a brainstorm right now.

[Art Carlson]

I am about to give up on my other project. I think it is out of reach for me to try to solve the vlasov totally general. But I do have an idea. If we assume the distribution in momentum space has a certain structure it may not take much information to store. I think our main goal is to not assume the distribution is Maxwellian. But, we might assume other things about it.

So, I was thinking about ellipsoids. ...

...

Since the description of the momentum space is essentially 1-D by doing this the memory requirements might be manageable, allowing much denser meshes. What I haven't figured out yet is how to actually alter the ellipsoids and densities to find a solution. So, yeah it's just a brainstorm right now.

The idea is good, but it is also a brazen attempt to tell nature what to do. In other words, you need to be very clear about which phenomena can be captured by your parametric form and which can't. Remember that Todd Rider was criticized (possibly correctly) because he assumed a particular form for the non-Maxwellian distribution function when calculating the recirculating power fraction. His form is plausible and rather general, but the loss rate can possibly be reduced greatly by choosing a different form. Many years before Rider, mirror confinement theory made the same mistake.
I'm thinking of Rick Nebel arguing that the equi-potential surfaces are more nearly spherical than the flux surfaces because of non-Maxwellian effects. That works (if it really works at all) by letting the electrons run along field lines up and down a potential, leaving a hole in the parallel velocity distribution at the bottom of the well. At the same time, the perpendicular velocity distribution would remain (more or less) Maxwellian. You will not be able to describe this distribution by elliptical surfaces in velocity space. A better choice (and nearly as efficient) might be choosing f(v_x,v_y,v_z) = f_x(v_x)*f_y(v_y)*f_z(v_z), where you probably will want to force one axis to be parallel to B and another to be parallel to grad B (or curl B ~ j).
Disclaimer: This is really out of my area of expertise, and I think it's real easy to go astray when attempting kinetic calculations without expertise.

[kcdodd]

Hmm. I'm no expert either, and I don't really understand your distribution. My understanding is the distribution near v=0 should be tiny. So if you are multiplying them together you would get nearly zero density along each axis at all velocity magnitudes, even though independently they may not be zero in their respective directions.

But I was not thinking the surface to be a constant f. Say in the v_x direction would have f_x(v_x), where each v_x would lie on a surface. f_y(v_y) could have a different value even on the same surface. But the densities are interpolated. I mean, there must be velocities between v_x and v_y, and the distribution in that region has to at least be continuous where it meets the axial directions. I see what you are saying though the cut cannot be very sharp with this method. There's probably a way to assign a more complex distribution function to each surface then just linear interpolation (or whatever you call it) which would give sharper transition between directions.

[Art Carlson]

Are we talking about the same thing here. What I mean by f is
f(x,y,z,v_x,v_y,v_z)*dx*dy*dz*dv_x*dv_y*dv_z = the number of particles with
an x-position between x and x+dx,
a y-position between y and y+dy,
a z-position between z and z+dz,
an x-component of the velocity between v_x and v_x+dv_x,
a y-component of the velocity between v_y and v_y+dv_y,
a z-component of the velocity between v_z and v_z+dv_z.
If you integrate f over all three components of the velocity, you get n(x,y,z).

And you?

[kcdodd]

Yes. What I didn't get was the f(v_x,v_y,v_z) = f_x(v_x)*f_y(v_y)*f_z(v_z) part. I thought you were saying I should have three functions which only depend on one variable and multiply them together for my momentum distribution. As opposed to some other approximation to the general multivariable function. I think that would force a certain value along all the axial directions and planes. Breaking it up into single variable functions is sort of what I'm thinking of, but not just multiply them together.

[Art Carlson]

Yes. What I didn't get was the f(v_x,v_y,v_z) = f_x(v_x)*f_y(v_y)*f_z(v_z) part. I thought you were saying I should have three functions which only depend on one variable and multiply them together for my momentum distribution. As opposed to some other approximation to the general multivariable function.

Yes, that is what I meant.

I think that would force a certain value along all the axial directions and planes. Breaking it up into single variable functions is sort of what I'm thinking of, but not just multiply them together.

I don't follow you here. Take something like this:

f(v_x,v_y,v_z) = f_x(v_x)*f_y(v_y)*f_z(v_z)
with
f_x(v_x) = exp(-mv_x^2/2kT)
f_y(v_y) = exp(-mv_y^2/2kT)
f_z(v_z) = exp(-m(v_z-v_0)^2/2kT) + exp(-m(v_z+v_0)^2/2kT)

This is the sum of two Boltzmann distributions with centers displaced along the z axis. In what sense does this function have "a certain value along all the axial directions and planes"?

[kcdodd]

f_z could be nearly zero at v_z = 0, depending on the constants, right? Well, that could punch a hole in the v_x/v_y plane at v_z = 0 when you multiply everything together, meaning there would be no particles with velocity with component v_z=0 regardless of the distribution you set for f_x and f_y. Is that what you intend?

[drmike]

Then change the *'s to +'s.

You are still making an assumption about the over all distribution function. This is fine as long as you know what assumptions match reality. a priori we don't really know what the distribution function is, so any assumption you make will create a system which you can compare to experiment.

I don't see anything wrong with Art's form, if you have a hole in velocity space that's not the same thing as a hole in physical space. To be explicit use a form more like f(x, y, z, vx, vy, vz) = fx(x, y, z, vx)*fy(x, y, z, vy)*fz(x, y, z, vz). An alternative form is f(x, y, z, vx, vy, vz) = fs(x, y, z)*fv(vx, vy, vz). That's what happens with a thermal distribution, so it's probably semi realistic.

Another approach is to attack a small chunk of the problem rather than the whole thing at once. Keep the numerical 6D form, but limit the number of particles and positions to something reasonable, like 10 steps in every direction (1 million phase-space storage locations) and 100 particles. Make half the particles ions and half electrons, start them at random places in phase space and watch for 100,000 steps. do that 100 times and average the distributions (kind of a Monte-Carlo method).

But what ever you do, understand that you are making assumptions. I don't see anything wrong with Art's suggestion, the hole in phase space is just an artifact of the math and fits the description correctly.

[Art Carlson]

f_z could be nearly zero at v_z = 0, depending on the constants, right? Well, that could punch a hole in the v_x/v_y plane at v_z = 0 when you multiply everything together, meaning there would be no particles with velocity with component v_z=0 regardless of the distribution you set for f_x and f_y. Is that what you intend?

Yes. If the electric and magnetic fields are co-parallel and uniform, then the perpendicular velocity is not affected by whatever happens in the parallel direction. If the distribution is Maxwellian at some potential (e.g. the top of the hill), and all particles in the system have at least enough energy to reach this position, then the velocity distribution at some position with lower potential will actually be
f(v_z) = exp(-mv_z^2/2kT) for |v_z| > sqrt(2e(Delta Phi)/m)
f(v_z) = 0 for |v_z| < sqrt(2e(Delta Phi)/m)
that is, a Maxwellian with a hole in it, rather than the sum of two Maxwellian distributions.

I'm also assuming there was a minus sign in the exponent.

Sorry. Fixed that.