I've got a question I've seen on slashdot and want to make sure I'm seeing it right. I'm hoping someone here can help me out if nobody there steers me straight. Here's the objections raised:
"Thermalisation alone can NEVER take you from a maxwellian to a non-maxwellian energy distribution. It doesn't matter how you shape your potential well. It follows directly from statistical mechanics."
and
"Its the 2nd law of thermodynamics. Anything not in equilibrium (in this case Maxwellian velocity distribution) then is relaxes to this state. In this case it relaxes really fast. Electric fields change nothing. Magnetic fields change nothing."
Now, I'm not nearly up on plasma physics, but isn't the initial condition with all ions at the same potential height a maxwellian energy distribution? Thus, up/down scatter caused by collisions in the core are actually shifting the ion energy distribution away from maxwellian? In other words, thermalization in the outer edge IS restoring a maxwellian energy distribution. Am I missing something fundamental in considering the mono-energetic initial state as being a maxwellian distribution?
Maxwellian distribution question
Re: Maxwellian distribution question
Well the terminology used is not precise. What we mean by non-maxwellian is really narrow distribution of energy. Maxwellian is broad.bcglorf wrote:I've got a question I've seen on slashdot and want to make sure I'm seeing it right. I'm hoping someone here can help me out if nobody there steers me straight. Here's the objections raised:
"Thermalisation alone can NEVER take you from a maxwellian to a non-maxwellian energy distribution. It doesn't matter how you shape your potential well. It follows directly from statistical mechanics."
and
"Its the 2nd law of thermodynamics. Anything not in equilibrium (in this case Maxwellian velocity distribution) then is relaxes to this state. In this case it relaxes really fast. Electric fields change nothing. Magnetic fields change nothing."
Now, I'm not nearly up on plasma physics, but isn't the initial condition with all ions at the same potential height a maxwellian energy distribution? Thus, up/down scatter caused by collisions in the core are actually shifting the ion energy distribution away from maxwellian? In other words, thermalization in the outer edge IS restoring a maxwellian energy distribution. Am I missing something fundamental in considering the mono-energetic initial state as being a maxwellian distribution?
A low energy (in the eV range) is non-maxwellian with reference to the ultimate accelerated particle energies (100s of KeV).
The idea is to keep the 100s of KeV particles distributed in a narrow range (+/- 100 eV would be very good) .
So if you can get the particles to thermalize at low energy it narrows the spread of particle energy.
The whole thread
The original post was at
http://hardware.slashdot.org/article.pl ... 27/1813225
The place where the quote originated was
http://hardware.slashdot.org/comments.p ... d=21155331
If you haven't already read the whole thread, it's worth looking at, starting here:
http://hardware.slashdot.org/comments.p ... d=21143503
One question I had about the polywell design is whether the ions will generate synchrotron radiation losses as they go through the magnetic fields.
http://hardware.slashdot.org/article.pl ... 27/1813225
The place where the quote originated was
http://hardware.slashdot.org/comments.p ... d=21155331
If you haven't already read the whole thread, it's worth looking at, starting here:
http://hardware.slashdot.org/comments.p ... d=21143503
One question I had about the polywell design is whether the ions will generate synchrotron radiation losses as they go through the magnetic fields.
Re: Maxwellian distribution question
The Maxwellian distribution is essentially a Gaussian normal distribution in each of the three velocity components, centered on zero (no average velocity in any direction, but of course the average scalar speed is not zero because the odds of all three velocity components being zero at the same time are vanishingly small). The standard deviation of this distribution is proportional to the square root of temperature, meaning that kinetic energy is linear with temperature. The Maxwellian arises naturally from random collisions between particles.
There is a distribution called the "drifting Maxwellian", which is just a Maxwellian with a finite net velocity superimposed. A good example is a cylinder of gas on a moving train.
The edge region in the Polywell is (roughly) a low-temperature Maxwellian (small standard deviation). As the particles drop into the intermediate acceleration region, the distribution can be described as a drifting Maxwellian with the same standard deviation as the edge region but a large "drift" velocity. It is met at every point by a somewhat more spread out distribution returning from the core at the same average speed in the opposite direction, so the distribution of the accelerating particles spreads out a bit due to high-speed collisions.
This intermediate region is (obviously) characterized by a predominance of head-on collisions. Near the core, where the two "drift" velocities are high, this makes a good amount of fusion.
The core is where these drifting Maxwellians converge from all directions and result in a weird high-density distribution with zero average velocity but a very narrow spread (basically that low-temperature Maxwellian from the edge) about a very large average speed. This wants to relax to a high-temperature Maxwellian.
The key to Bussard's explanation is that the particles aren't in the core long enough to fully Maxwellianize on one pass, but since at the top of the well they have to slow down to a stop (barring angular momentum considerations) and turn around, they ARE in the edge region long enough to cancel out the core effect with collisional relaxation back down to the low-temperature distribution.
This is because if the density is correct, higher-energy particles that would normally leave the reactor plow through the high-density cold edge region and slow down, and lower-energy particles that wouldn't normally reach the edge tend to get rear-ended, which evens out the radial velocities except in the case of a perfect dead-on hit. Obviously this won't result in an infinitely thin edge with a perfect Maxwellian distribution, but Bussard is claiming that it's close enough, at least with the fusion process removing old ions at a sufficient rate.
My gas kinetics isn't strong enough to figure out on my own whether he's right or not.
MSimon said basically the same thing in a lot fewer words...
There is a distribution called the "drifting Maxwellian", which is just a Maxwellian with a finite net velocity superimposed. A good example is a cylinder of gas on a moving train.
The edge region in the Polywell is (roughly) a low-temperature Maxwellian (small standard deviation). As the particles drop into the intermediate acceleration region, the distribution can be described as a drifting Maxwellian with the same standard deviation as the edge region but a large "drift" velocity. It is met at every point by a somewhat more spread out distribution returning from the core at the same average speed in the opposite direction, so the distribution of the accelerating particles spreads out a bit due to high-speed collisions.
This intermediate region is (obviously) characterized by a predominance of head-on collisions. Near the core, where the two "drift" velocities are high, this makes a good amount of fusion.
The core is where these drifting Maxwellians converge from all directions and result in a weird high-density distribution with zero average velocity but a very narrow spread (basically that low-temperature Maxwellian from the edge) about a very large average speed. This wants to relax to a high-temperature Maxwellian.
The key to Bussard's explanation is that the particles aren't in the core long enough to fully Maxwellianize on one pass, but since at the top of the well they have to slow down to a stop (barring angular momentum considerations) and turn around, they ARE in the edge region long enough to cancel out the core effect with collisional relaxation back down to the low-temperature distribution.
This is because if the density is correct, higher-energy particles that would normally leave the reactor plow through the high-density cold edge region and slow down, and lower-energy particles that wouldn't normally reach the edge tend to get rear-ended, which evens out the radial velocities except in the case of a perfect dead-on hit. Obviously this won't result in an infinitely thin edge with a perfect Maxwellian distribution, but Bussard is claiming that it's close enough, at least with the fusion process removing old ions at a sufficient rate.
My gas kinetics isn't strong enough to figure out on my own whether he's right or not.
MSimon said basically the same thing in a lot fewer words...