Making Electricity with the p-B Polywell

[93143]

If I can make a DC particle accelerator that can take a particle from 0 eV to 2 MeV I can make that same apparatus (neglecting power supplies and beam trajectories) with the same voltage gradients decelerate particles from 2 MeV to 0 MeV. Sinks (accelerator grids) turn into sources (same grids now used for deceleration). Only now we neutralize where we used to ionize.

I'm not arguing against that. I'm saying exactly that.

Perhaps it would help if we could see the potentials and fields:

This is what you described. The fusion product emitter (bottom) and the collector (top) are both grounded, and the tubular electrode is at 999999 V. Geometry is axisymmetric, and the external walls are far away (10 on this scale) and insulated.

This is probably what you were thinking of. The collector is now at 999999 V and the tubular electrode is at an even 1 MV. The emitter is still grounded.

This assumes there aren't enough alphas present at any one time to significantly perturb the potential distribution, which is probably a fair assumption. They're just test charges as far as we're concerned.

[charliem]

Electric fields are conservative, like gravitational fields.

That means that the net energy gain or loss of any charge Q going from point A to point B, if there is a potential difference V between them, is QxV.

In a conservative field it doesnt matter neither the trajectory, nor the geometry or even stregth of the field in the regions it has to cross, going from A to B the NET energy gain/loss is always QxV.

http://en.wikipedia.org/wiki/Conservative_field

Many people get confused because they forget to account for the Potential Energy.

I like to trick my students now and then with a perpetual motion machine made in this way and they buy it almost each time. It's fun to see them try to find the explanation (for them as much as for me).

[kcdodd]

MSimon, why are you making the collector at 0v, that makes no sense if the alpha starts off at 0v. An alpha is not going to "drift" from 1MV to 0V, it will accelerate. Not only will the alphas have the same energy they started out with, you would never extract any useful energy. Even a billion amps at 0v is useless. Forget tubes for a second and just write down the law of conservation of energy.

[charliem]

Simon, I did study tubes and I'm afraid that this time it's you who is mistaken.

Any alpha with initial kinetic energy E, that goes from one point al 0V to another at +1MV (the grid you are talking about) will get there with (E-2MeV) of kinetic energy, but going from that point at +1MV (the grid) to a third one at 0V (the collector) will regain 2MeV, so its final kinetic energy is (E-2MeV+2MeV)=E, the same it started with.

[MSimon]

MSimon, why are you making the collector at 0v, that makes no sense if the alpha starts off at 0v. An alpha is not going to "drift" from 1MV to 0V, it will accelerate. Not only will the alphas have the same energy they started out with, you would never extract any useful energy. Even a billion amps at 0v is useless. Forget tubes for a second and just write down the law of conservation of energy.

Brilliant. Particles can only be accelerated by electrostatic forces.

Another Nobel in the offing.

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Let me make this real simple.

If particles can be accelerated by electrostatic forces (at the cost of energy delivered to the electrostatic acceleration grids) they can be decelerated by electrostatic forces (which will deliver energy to the electrostatic grid(s)).

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Physics education in America is in big trouble. Very big trouble. It looks to me like such an education is very big on the math (my weakness - I haven't had to do any serious calculus for 40+ years) and very weak on principles.

I thank my lucky stars I had PSSC Physics in high school where principles were emphasized.

I have run into this before. I once had a boss who didn't understand why you needed to run transformers at about 1/2 their AC rating for capacitor input power supplies. I tried explaining the difference between the RMS heating value of the pulsed current input vs. power out of the DC current delivered. All I got was an argument. Which I did not win. I did the right thing any way and designed in a transformer of adequate capacity for the RMS current drawn. College educated guy. This same guy insisted that I design a one bit computer to run the machine tool being designed. With an very early version of FLASH memory (3 fookin supply voltages). He must have spent at least $100,000 on that development. At a time when you could buy a very nice KIM One micro (6502 based I think) for about $400 (IIRC). Eight bits. I had a number of engineers working with me who thought my execution was brilliant. It was still a stupid idea. But I did get an experience very few engineers have ever had. I designed a one bit computer.

[MSimon]

Simon, I did study tubes and I'm afraid that this time it's you who is mistaken.

Any alpha with initial kinetic energy E, that goes from one point al 0V to another at +1MV (the grid you are talking about) will get there with (E-2MeV) of kinetic energy, but going from that point at +1MV (the grid) to a third one at 0V (the collector) will regain 2MeV, so its final kinetic energy is (E-2MeV+2MeV)=E, the same it started with.

It will gain zero energy if it enters the drift space at some low energy because you can't gain the energy back in drift space. Unless you cut across electrostatic (voltage) field lines.

In the drift space you have a field going from 1 MV at the edge to 0 V in the center (at least near the collector). If the particle is at the 0V gradient when it enters the drift space it is not going to gain any energy. If you are not cutting across the equipotential lines you don't gain energy. It is not possible. Unless the sky is green in your universe.

It may be true that you have studied vacuum tubes. What you do not do is think (neural network) tubes. The first step in understanding a tube is to look at field lines. See them in your mind. The seeing is most critical. Be an electron. Feel the forces. The feeling is also critical since we are best at pattern recognition and the "feeling" is one of the recognition signals (it is also important in training). In fact people with no feelings can't think. (Or they don't do it very well).

[MSimon]

There used to be a tool called resistance paper around when tubes were it.

What you did on the paper was draw lines of copper for the grids and put your various (scaled) voltages to them. Then you could measure the fields with a voltmeter. Very handy. I never used it.

[blaisepascal]

In the drift space you have a field going from 1 MV at the edge to 0 V in the center (at least near the collector). If the particle is at the 0V gradient when it enters the drift space it is not going to gain any energy. If you are not cutting across the equipotential lines you don't gain energy. It is not possible. Unless the sky is green in your universe.

I don't see this.

What I see, in terms of fields, is very strong E-fields between the end of the tube and the emitter, no E-field within the main body of the tube (at least, NOT near the collector) and very strong E-fields between the body of the tube and the collector plate. I envision the E-Field lines terminating perpendicular to the surface of the collector and to the inside surface of the tube, since E-Fields are perpendicular to conductive surfaces.

My schooling tells me that the equipotential surfaces are perpendicular to the E-Field, so I see a very strong gradient (tight spacing of equipotential surfaces) near the collector, no gradient within the main body (drift space) of the tube, and another strong gradient between the tube and the emitter.

When you fire the 2MeV alpha particle from the emitter to the collector, I see it crossing lots of equipotential surfaces as it heads towards the tube to have near-0 kinetic energy as it enters the drift space. I see no electrostatic forces acting on it while in the drift space (and my intuition feels the currents aren't large enough to have significant magnetic effects). But when it gets close to the collector, it is going to be accelerated by the potential gradient, crossing lots of equipotential surfaces, near the collector, and slam into the collector with lots of energy. If we assume the emitter and collector have the same voltage, it'll hit the collector with 2MeV of energy.

I'll freely admit I'm not an expert. I've done lots of book-reading of physics, and passed with flying colors a college-level calculus-based EM course in high school -- 20 years ago. I'm old enough that I find FETs easier to understand that bi-polar transistors because FETs work like tubes, but that doesn't mean I understand tubes.

You seem to claim my analysis is wrong. I welcome an opportunity to sharpen my intuition and knowledge. Where is my mental picture failing me?

[drmike]

Another useful method was a saline bath. You could do a 2D problem really easily by setting up the electrodes in the bath and then measure the potential drop everywhere in between directly with a volt meter.

Computers make it easy keeping your hands dry!

[Aero]

Hey guys, I know that this is no fun, but the problem has been worked before so I don't think we'll need to discover any new science here. Of course it is fun to figure out how it really works, keeps us sharp, too. And the power regime of BFRs is totally new, but adapting to that regime is a different problem.

http://ntrs.nasa.gov/search.jsp?R=21685 ... 4294967207

Of course, its always possible that this report is not applicable to our problem or that NASA made a mistake.

[TheRadicalModerate]

Simon:

I understand what you're trying to do with your 0V electrode plus +1MV cylindrical electrode + -1V collector plate, but let's make the gedanken experiment even simpler:

We've got a 0V point source that's (magically) firing 2 MeV alphas at a +1MV infinite-plate electrode. Just before the alphas reach the plate, they are (again, magically) transformed to neutral He4 and whisked away.

So, a list of things we're not worrying about--yet:

1) We aren't worrying about how the alphas get neutralized and collected, and we're assuming that the neutralization current is trivial.

2) We aren't worrying about the fact that the forces experienced by charged particles in a spherical charged shell are 0.

Now, I think we all agree that the plate electrode (which is presumably connected to a voltage source, which is presumably is connected to ground) is effectively an open circuit. And I also think we all agree that our magically appearing and disappearing alphas start at the 0V source with 2 MeV of kinetic energy and end up at the plate with 0 kinetic energy.

Now (and here's where I, with no degree in physics and absolutely no ego invested in this, can be as stupid as I like and you can make fun of me as much as you want), I think you're saying that that kinetic energy has been converted to some incremental potential difference in the plate, which in turn can be bled off as a current connected to a load parallel to the plate and, voila!--power.

But here's what's freakin' me out: When we bleed off that excess voltage as power, we're physically sucking electrons out of ground (0V in our little experiment), through the load, and into plate electrode. However, last time I checked, Kirchoff's Current Law is still on the books. So I ask the question: Where do the electrons go once they've been sucked into the electrode?

Again, all of this stems from the fact that the decelerator is effectively an open circuit. In normal systems, open circuits don't produce power. So, if you could explain why this one is different, I'd be very grateful.

[TheRadicalModerate]

OK, I finally realized why the spherical electrode objection is bogus: You've got an E-field from any charge on the the spherical electrode (which is everywhere zero) and an E-field from the potential difference between the outer shell and the center of the wiffle ball, which looks like the E-field from a point charge at the center. Superpose the two and you get null field + point field = point field.

Right?

[kcdodd]

MSimon, dude, for real. This is not "big bath".

KE + PE = E, and PE = 2*e*V for alpha.

alpha starts at 0V with 2MeV: 2MeV + 2*e*0V = 2MeV.

we are assuming that there are no other collisions with particles before it is collected, so total energy of the alpha is conserved dE/dt = 0.

to find the KE at 1MV we already know that E = 2MeV from initial conditions, KE + 2*e*1MV = 2MeV, solving for KE = 2MeV - 2*e*1MV = 0. At 1MeV it is zero kinetic energy, this is where you should collect the alpha.

If the alpha reached any spot at 0v: KE = 2MeV - 2*e*0V = 2MeV. It will crash at full energy it started out with.

I don't think anyone is arguing that things can't decelerate. It obviousy did to reach 1MV, but why on earth then put a 0V collector. You are just re-accelerating it. Maybe you should stop insulting everyone else for a second because honestly you have put your foot so far down your throat its not funny any more.

[TheRadicalModerate]

I don't think anyone is arguing that things can't decelerate. It obviousy did to reach 1MV, but why on earth then put a 0V collector. You are just re-accelerating it. Maybe you should stop insulting everyone else for a second because honestly you have put your foot so far down your throat its not funny any more.

Why are we arguing about the details of a thought experiment that obviously wasn't intended to be a workable system? It has nothing to do with whether electrostatic induction will actually produce usable power.

[kcdodd]

Conservation of energy is not a "detail". The laws of physics are the same no matter what.


Btw, I have no idea what the thought experiment was originally supposed to show. Maybe I have my foot in my mouth for saying that it wouldn't work in real life.