## Making Electricity with the p-B Polywell

Discuss how polywell fusion works; share theoretical questions and answers.

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93143
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MSimon wrote:Nope. By the time the alphas hit the -1V collector inside the 1 MV cylindrical grid they have -2eV of energy.

They lost 2 MeV getting to the -1V collector. According to the potential gradient.

Any way I see why you are confused.
Two things: a) -1V relative to what? b) -1 V adds 2 eV of kinetic energy, not -2 eV (negative kinetic energy!?), because it's downhill. This last is a minor point; (a) is the big one...

Sure, going to the 1 MV electrode costs them 2 MeV. But going from there to a -1 V electrode (assuming they're both relative to the emitter at 0 V), they gain it back.

More likely they would never see the full 1 MV potential because there would be a minimum-potential path around it, so they wouldn't slow to a stop at any point.

MSimon
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93143 wrote:
MSimon wrote:Nope. By the time the alphas hit the -1V collector inside the 1 MV cylindrical grid they have -2eV of energy.

They lost 2 MeV getting to the -1V collector. According to the potential gradient.

Any way I see why you are confused.
Two things: a) -1V relative to what? b) -1 V adds 2 eV of kinetic energy, not -2 eV (negative kinetic energy!?), because it's downhill. This last is a minor point; (a) is the big one...

Sure, going to the 1 MV electrode costs them 2 MeV. But going from there to a -1 V electrode (assuming they're both relative to the emitter at 0 V), they gain it back.

More likely they would never see the full 1 MV potential because there would be a minimum-potential path around it, so they wouldn't slow to a stop at any point.
-1V relative to zero. I can see why you are confused.

And yeah the sign is wrong.

Actually most likely they will head in another direction because of the low quality of the neighborhood. They won't want to get involved. At least that is true for the smart ones. People think alphas have no sense of taste, but it is not true. Some do.

So let me see. The alphas gain back 2 MeV in the drift region. Nobel quality work. If it is true.

Don't they teach kids ANYTHING about vacuum tubes any more?
Engineering is the art of making what you want from what you can get at a profit.

MSimon
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93143,

Did you actually study physics? Did you pay for it?

I'd demand a refund. You have been cheated.

This is so depressing.

It is a fookin' particle accelerator in reverse. Grids that supply energy in an accelerator are collectors of energy in a decelerator.

Look. This is useless. You are not paying attention. I'm not going to give any more. Send me a private message when you get it. And I don't want any slop about why you are correct. Contact me when you agree with me.

You want to know how bad it is? You are not even wrong. 2 MeV of energy gain in the drift space - isn't there a strictly enforced law about that? Gauss or something? I don't think you have the political clout to get that one changed.

I hope this has been helpful for the lurkers. Because, it does not seem to be helping you.
Engineering is the art of making what you want from what you can get at a profit.

93143
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My M.Sc. was in electrohydrodynamics. Currently doing transcritical spray simulation for a Ph.D.

So yes, I know some physics.

For the benefit of the lurkers:

One of the things I learned in physics was that E = -grad V. That is, if there is a difference in potential between two objects, there MUST BE an electric field between them.

I assume that when MSimon talks about "drift space" he means constant velocity for charged particles; that is, zero field.

If there's a 1 MV electrode next to a -1 V electrode, it's NOT drift space.

MSimon: we're not done here. If you can see why I'm confused, go ahead and tell me. All I can see so far is you selectively ignoring potential differences.
Last edited by 93143 on Wed Jul 09, 2008 12:37 am, edited 1 time in total.

kcdodd
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I've been lurking but I still have no idea what you guys are arguing about. But if an alpha has 2MeV anywhere at 0V, it will still have 2MeV any time it is at 0V. And it will have 2.000002MeV any time it is at -1V, and it will have 0eV any time it is at 1MV. And the only place it can end up at -1V with 0eV is in imagination land.
Carter

93143
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kcdodd wrote:But if an alpha has 2MeV anywhere at 0V, it will still have 2MeV any time it is at 0V. And it will have 2.000002MeV any time it is at -1V, and it will have 0eV any time it is at 1MV. And the only place it can end up at -1V with 0eV is in imagination land.
THANK you.

Jccarlton
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You have to understand that particle accelerator devices are not exactly operating like the electrical devices we are used to . When you measure a particle's energy in electron volts you are measuring the total energy including the kinetic energy as well as any charge the particle may have. So when you see a particle with an energy of 2mev that does not mean that there are two electric fields 2mv apart. It's sometimes done that way put particles can acquire energy in other ways. Most that energy will in fact represent kinetic energy. Now in order to extract the energy we need to slow the particle down with some sort of induced electric field which creates a current across it. The alphas will still have their charge but they play no part other than giving up their kinetic energy until the hit a plate at -1v and drop to neutral. If I recall when I was playing with FEL none of the various components acted like anything other than static voltages. I don't think the beam dump had anything other than being connected to ground.

MSimon
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OK. I can see you still don't get it.

Alpha with 2 MeV of energy. Let us make the decelerator +999,999 V. Collector plate inside the cylindrical grid at 0 V.

Alpha starts at the 0V (some distance away from the cylindrical grid and its internal collector) point with 2 MeV of energy aimed so that the alpha enters the drift space and the fields are perfectly symmetrical, the aiming is perfect - right down the center of the drift tube, the collector plate is a point that takes zero space, the alpha travels the 0V gradient, yadda, yadda.

When the alpha enters the drift space it has 2 eV of energy. The alpha hits the 0V plate and is collected (neutralized) while in the drift space.

Where did the alpha energy go?

=============

But I don't care. Tell me how an electrostatic (DC) particle accelerator works. Then tell me how the same apparatus would work if it was a sink for high energy particles rather than a source.

BTW you wuz robbed.

=============

Seriously. Back when I was a kid any one who knew vacuum tubes well could see this intuitively. Tom Ligon mentioned this: if you know tubes what is going on seems obvious. If tubes are a mystery it requires more effort.

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What I don't get is that you are obviously smart enough to see how an electrostatic accelerator works, but can't see how a decelerator would work. It boggles the mind.

Tubes are DC accelerators. What we are proposing is just the reverse.

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.
Engineering is the art of making what you want from what you can get at a profit.

MSimon
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I am encouraged by one thing. Bussard planned to use decelerator grids to extract energy.
Engineering is the art of making what you want from what you can get at a profit.

93143
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MSimon wrote: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
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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
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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.
Carter

charliem
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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
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kcdodd wrote: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.

=====

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)).

=====

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.
Engineering is the art of making what you want from what you can get at a profit.

MSimon
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charliem wrote: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).
Engineering is the art of making what you want from what you can get at a profit.