Why do fusors work?

[ab0032]

> So the question seems to revolve around the sources and sinks of ions in steady operation.

Yes, where do the ions come from? Where are they created, where and how are they lost. How does the plasma keep enough energy and not radiate out all its energy?

Of course all the other replies were interesting too, this is more what I want know.

Anything hitting the grid will pollute the plasma. Some ions might be created anywhere half way between the ends of the field, so they have only a share of the full energy.

On the other hand, if things thermalize, we could also see some upscatter, some ions gaining energy, enough to produce fusion.

[ab0032]

Fusion crosssections and beam targets...

As a sceptic I have already doubted the cross section measurements. How where they obtained? I have not found one source, where conditions where specified exactly. How do we know there where no knock on collisions? What densities of ions were used in measurements?

[Dan Tibbets]

It is a fundamental property that items with less potential energy will release energy and visa verse. But to think that this drives fusion is simplistic. If this was the case protons and neutrons in various combinations would very quickly build to Nickel 62, which has the lowest potential energy of any nucleus. The universe- or more specifically, stars would all end up as nickel (or Fe56). But this is not the case. Only those stars massive enough will reach this endpoint. The sun will only reach carbon/ oxygen. Why- because the obtainable temperature is not high enough. Why is that- because of the coulomb barrier. When the products have lower potential energy than the reactants, then excess heat will occur (KE), but this by itself does not drive the speed of a reaction. You can argue that due to statistics and quantum effects a lower potential energy state will eventually happen. But it says nothing about how fast this will occur. Is it 1 second, one year, 100 billion years? This is dependent on the intermediate energy barrier and the availability of applied KE to approach or exceed this barrier.

The same reasoning can be applied to chemical systems. Why does a mixture of Hydrogen and oxygen gas not spontaneously and rapidly burn to water. It is because a threshold has to be exceeded. This is often referred to as Gibbs free energy. There is an intermediate stage that has more potential energy (you have to put in KE) than either the reactants or products. Catalysts work by decreasing this intermediate PE state. And, it is very fortunate that this characteristic of reactions exist, otherwise we would not be here.

In nuclear fusion this intermediate state is due to the tremendously increasing repulsion as protons approach each other. If the KE is great enough (often helped by quantum uncertainty) the protons can approach to ridiculously close distance before the very short range Strong force can overcome this repulsion (and it can only do so to a limited extent. That is why an intermediate nucleus like Nickel has thew lowest PE and Nuclei with more than a few hundred nucleons fall apart. Uranium 235 falls apart into products with less PE with a fairly long half life. Why does it not do so instantly?. Because there is an intermediate state with higher PE that has to be overcome by chance accumulation of additional PE by the nucleus- the extreme high energy tail of a nucleus equivalent of a thermalized gas.
A counter argument might be that heating a gas of uranium does not change the decay half life. Some think you can change this rate by feeding in KE (like X-rays) to change the decay half life of some isomers, but this is a contested viewpoint. and if applicable, it is adding the KE to the nucleus, not just speeding the total nucleus velocity in an external frame of reference. It is like the difference in firing a cannon ball. It speeds up, but the internal energy is not increased. For that you would need to heat the cannon ball separately.

Dan Tibbets

[Chris Bradley]

Dan DT wrote:
> Why does a mixture of Hydrogen and oxygen gas not spontaneously and rapidly burn to water. It is because a threshold has to be exceeded. This is often referred to as Gibbs free energy.

Dan, sorry but I think you need to revisit all of this. Re-hyopthesising interpretations you've made in your auto-didactic activities isn't helpful here, and this is word-salad. My thermodynamics is a bit shakey, but it looks like you are confusing activation energy with Gibbs free energy here. The rest I can barely even follow enough to comment on it - I cannot figure out what your point is.

[Steven Sesselmann]

Dan,

It is indeed fortunate that nuclei heavier and lighter than Ni62 don't spontaneously fall to the lowest potential state, well at least in the observers frame of reference.

You see it is all observer dependent, by definition the observer is always at ground potential (in my opinion around +930.412 MeV, the energy potential of Ni62). All the other elements have potentials either above us or below this, and will under the right circumstances attempt to fall towards ground potential. But thanks to the low permittivity of free space, these elements can't suddenly convert their potential energy to photons, so they hang around and wait for the right opportunity.

Now as you mentioned, it is believed that nuclear decay rates can not be influenced by external factors, but that's not really the case, we already know from relativity, that increasing the potential of matter, causes time dilation, which is in direct relation to it's decay rate.

Increasing the potential of matter is easy, you either accelerate it to a higher velocity, elevate it in a field, or move it to a large distance (into the past), all of which increases it's potential with respect to you the observer.

So by accelerating deuterons to large velocities in the hope that this will help to make them fuse, is to further increase the potential of a nucleus that is already suffering from a high potential, in other words pretty stupid!

Which then leaves us in a bit of a pickle, because to make deuterons easier to fuse we need to lower their potential, and trust me that is not so easy, how do you lower the potential of a deuteron already at rest?

This problem gave me quite a few sleepless nights, but there had to be a way, and one night the solution woke me up as if someone slapped me in the face...

For now you are going to have to figure it out for yourself, but soon I hope to demonstrate that man can control fusion, but I fear that fusion might end up controlling man, and that's pretty scary.

Steven

[Chris Bradley]

Steven Sesselmann wrote:
> Now as you mentioned, it is believed that nuclear decay rates can not be influenced by external factors, but that's not really the case, we already know from relativity, that increasing the potential of matter, causes time dilation, which has a direct relation to it's decay rate.
....
> So by accelerating deuterons to large velocities in the hope that this will help you make them fuse, is to further increase the potential of a nucleus that is already suffering from a high potential, in other words pretty stupid!

Sorry, Steven, that does not follow. The time dilation on a 20keV deuteron relative to a 1eV deuteron is negligible and does not account for why it is sooo much more likely to react than the 1 eV deuteron. You'll never get a fusion with such low energy deuterons, yet the time dilation only accouns for a few fractions of percent difference. You will have to come up with another explanation for justifying the 'high potential' argument, wrt DD fusion rates.

You'll also have to account for the fact that at higher energies still, the reactivity drops off. If 'higher potential' is gained by velocity, then how does this figure?

[Carl Willis]

This discussion is losing focus...heading off into La-la Land...

[Steven Sesselmann]

[Chris Bradley Wrote]
Sorry, Steven, that does not follow. The time dilation on a 20keV deuteron relative to a 1eV deuteron is negligible and does not account for why it is sooo much more likely to react than the 1 eV deuteron. You'll never get a fusion with such low energy deuterons, yet the time dilation only accouns for a few fractions of percent difference. You will have to come up with another explanation for justifying the 'high potential' argument, wrt DD fusion rates.

You'll also have to account for the fact that at higher energies still, the reactivity drops off. If 'higher potential' is gained by velocity, then how does this figure?[end quote]

Chris, my ramblings must have been a little confusing, but not quite La La land

I wasn't suggesting that time dilation had anything to do with the fusion rates, what I stated was simply that...

a) Lighter elements are naturally above ground potential
b) Observers are always at ground potential
c) Heavy elements are naturally below ground potential

From this it follows that lighter elements are more inclined to fuse when taken below ground, and heavier elements are more inclined to fission when taken above ground.

That (in my opinion) is why fusors work, and as you know I intend to prove it soon.

Steven

PS: Chris, from my logic above, it also follows that fusion reactivity should fall off with increasing particle energies, ie higher potential.

[ab0032]

Carl Willis:
> This discussion is losing focus...heading off into La-la Land...

Then let me try to bring it back to what I wanted to know.

What if we had an ideal 100% tranparent grid inside, that ions could not hit, what would happen? Would we still see fusion? Or would the plasma simply cool off? Is the grid required to absorb ions, so they can be replaced by new high energy ions?

Cheers,
Alex

[Frank Sanns]

Having no inner grid at all works great. The "Pillar of Fire" with plasma electrodes instead of an inner grid works very well and is patented. Us patent #7550741.

Frank Sanns

[Dan Tibbets]

Steven Sesselmann. Your assertions that nikel 62 is the zero point and fusion and fison exothermic reactions are either pos or negative ( greater or less potential energy) is not accurate. Nickel 62 has the smallest potential energy of any possible nucleus. A proton is not bound so has no potential energy defined . Actually the potential energy often is expresed as the sum of the repulsive energy (electromagnetic repulsion of protons), and the attractive energy of the Strong force. This attractive energy is by convention given a negative value (just like gravity). At Nickel62 the difference between these two forces is at a maximum( the most negative sum is reached). Past Nickel 62 the electromagnetic repulsion is growing faster than the Strong force attraction, as such the potential energy inside the nucleus is increasing (becoming less negative). At some point the net potential energy becomes maximum (~0) and the nucleus quickly falls apart. The position of Nickel 52 at the lowest potential energy state (most stable or most compact) is obvious from the Nuclear Binding Energy per Nucleon graphs. It is even more obvious if you flip the graph upside down (as it would be with a negative value assigned to the attractive Strong force component).

When light nuclei fuse you are harvesting the excess strong force energy. When you fission heavy elements you are harvesting the excess electromagnetic energy (compared to Nickel 62). The missing mass is important, but even more important is the balance of the two forces that make up this missing mass. primarily the repulsive Electromagnetic and attractive Strong forces.

I don't know how this would affect your relativity arguements . If we are at the baseline, increased speed of particles to relativistic speeds would not change their decay or reaction ratesin their frame of reference. From our frame of reference the reaction rates would seem to slow down - a good example is the survival of the Muon as it reaches the Earths surface. Without relativistic effects it would decay much too fast to survive it's several hundred mile trip through the atmosphere, from where it was born to the surface of the Earth. This would seem to be the reverse of what you are proposing. Traveling slower than us would result in faster reaction as percieved from our frame of reference, but how to do this ?

Dan Tibbets

[Chris Bradley]

Dan DT wrote:
> the potential energy often is expresed as the sum of the repulsive energy (electromagnetic repulsion of protons), and the attractive energy of the Strong force. This attractive energy is by convention given a negative value (just like gravity). At Nickel62 the difference between these two forces is at a maximum( the most negative sum is reached). Past Nickel 62 the electromagnetic repulsion is growing faster than the Strong force attraction

If the strong force does not balance the electromagnetic force perfectly, in any extant nucleus, then what other forces are at work to hold the nucleus together?

This just doesn't add up (whichever sign you attribute to each!!)

[Steven Sesselmann]

With all due respect to 300 years of science, if it can't answer Alex's simple question it's no b...dy good.
.

I would like to ask Alex a question, what is hypothetically the most negative charge a grid can have?

Steven

[Steven Sesselmann]

Dan Tibbets Wrote...
"Traveling slower than us would result in faster reaction as percieved from our frame of reference, but how to do this ?"

Dan..., you are knocking at the door, now you just have to find the key.

Steven

[Dan Tibbets]

Chris Bradly, i t seems few understand the Binding energy per nucleon relatio0nship (I cannot guarentee I'm not included in that group).

When the electromagnetic repulsion balances against the strong attraction, then the nucleus is balancing on the cusp of falling apart. This condition is approached as heavier nuclei are reached. It becomes prevalent beyond lead and just past plutonium there are no isotopes that can hold themselves together for long (with the strong force). Of course it varies with neutron count due to weak force and Paul exclusion processes, nucleon paring, etc, but the gross balance competition is between the strong force and the electromagnetic force (with a few variations like helium 4). The most stable, most dense nucleus, and the nucleus with the least potential energy is Nickel62. This is not where the two forces balance out. This is where the strong force has the greatest dominance. A nucleus that is ~ 4 nucleons wide is reaching the limiting range where the strong force will increase the attractive binding energy ( negative potential energy) by a greater amount than the acumulating repulsive electromagnetic repulsionby only a relative small amount compared to light nuclei where ALL of the nucleons are close enough together for the very short range strong force to reach across them all. With increasing nuclear diameter, the neighboring nucleons still attract each other, but not so much with nucleons on the other side of the nucleus. It is different for the electromagnetic repulsion because of the greater range of the force. I might add that using a negative value for the attractive force is a convention, but it is not necessary, so long as you keep track of the opposing effects. The attractive energy builds up , but saturates. The repulsive force continuously builds up and it saturates also, but less rapidly. A graph of the two opposing forces have different slopes and at Nickel 62 they cross. Beyond that point the repulsive force increases faster than the attractive force, but the attractive force had a head start because it was increasing faster (becoming more negative by convention) with the lighter nuclei. It takes it a while for the repulsive force to catch up (heaviest elements).
The Nickel 62 is the turn around point, the lowest potential energy possible, but that does not mean it is the point where a balance in established. It is the point where the trends reverse.
Another analogy. A balance scale. One one side is a small pan (representing the volume where the strong force can collect) and a much bigger pan on the opposite side (which represents the much larger volume where the repulsive electromagnetic force can collect).
The strong force lead pellets are added to the small pan at the same rate as lithium pellets are added to the large pan. Initially the scale will swing to the left side( strong force pan side) But the pan is small, additional lead pellets will fall over the side. Now, as additional pellets are added, the lithium pellets are still accumulating and the scale starts swinging to the right. At some point a balance will be reached where the arms are level. This would be ~ lead or beyond. The Nickel 62 is where the lead pan is filled up. This is a simplified analogy. It doesn't take into account that the strong force always continues to increase in a logarithmic fasion ( as does the electromagnetic force, but since this increase becomes progressively less with each step, it approaches a level line (like additional lead pellets falling out of the pan).

Dan Tibbets

[Dan Tibbets]

Concerning transparent grids, if they were transparent/ vertual, the ions could pass back and forth indefinitely in a connectionless plasma, or a plasma that maintained a mono energetic temperature. The would traverse the potential well indefinitely. The problem is that a collisionless plasma is useless. You cannot have fusion without collisions. And scattering Coulomb collisions will always occur more frequently than fusion collisions. You can come close with D-T fusions at ~ 100 KeV (~ 1:10 ratio) and with D-D fusions at several MeV. The problem with the D-D though is that at these energies losses are increasing faster than the fusion rate. Bremsstrulung in particular is a problem as it increases faster than the fusion crossection curve once you are over several hundred KeV. Also other unwanted nuclear reactions become more prevalent.. Even if you can recover the Bremsstrulung losses through a heat engine (steam plant) you can only reach ~ 30% efficiency, and that is not enough to overcome the fusion/ Bremsstrulung rates at these temperatures. If you had a 99% heat to electricity conversion efficiency, thing would change a lot. Concerning conversion, if you have to accelerate your ions to 2 MeV, and you get ~ 3 MeV per fusion, , you would get ~ 5 MeV of heat, converted at 30% efficiency would result in ~ 1.5 MeV of output. This is less than the energy you put into the fuel ions. You are losing ground. Because of this there are compromises needed. The best monoenergetic temperature would probably be somewhere around 100 to 200 KeV. Bussard preferred ~ 80 KeV, but I don't know if he was planing on only burning D-D or if this incorporated recovering the tritium and helium 3 and also burning those secondary fuels where the three fusion crossection curves are different.

Back to thermalization. Upscattering will result in the ions reaching the wall and losing their energy. I don't know the relative rates but it would not be far behind grid collisions, perhaps 1-2 orders of magnitude at best, depending on the temperature - again there would be a compromise between the fusion and scattering crossections. If the gridded fusor can reuse the ions for ~ 10-20 passes, then with a transparent/ virtual/ magnetically shielded cathode (or anode with an Elmor Tuck Watson fusor), them perhaps 100-2000 passes might be achievable. Bussard , etel found that ~ 10,000 or more ion passes was the minimum required ( or > ~ 100,000 electron passes). ~10-100 times more would be needed to achieve good fuel burn up. You need something more. . Without going into details, this is embodied in the Polywell design. There is also a group of MIT graduates types working on something similar.

Dan Tibbets

[Dan Tibbets]

Steve Sesselmann, It is generally accepted that we line in fairly flat space. If we lived near the event horizon of a black hole the relative fusion rates would be faster away from the black hole from out perspective, but I'm not sure how we would interface with this. As far as speed, we are traveling at ~ known velocities relative to distant galaxy clusters, and our movement with the expansion is also known and at least for nearby (several billion light years) the speeds are not reletivistic. . Perhaps if we located a reactor on the opposite side of the observable universe, it would have a much faster percieved fusion rate. But, the transport of the electricity to us might ba a problem.

As for the rest mass/ energy of a proton (or neutron?), if this is actually a result of a velocity in a different frame of reference, I do not see the link to our reality/ reference frame. To a photon the rest mass / velocity in some other dimension may be zero or very close to it. but to increase the fusion rate of these particles from our perspective we would have to speed our selves up, not slow the particles down- that would be decreasing their energy to almost nothing ( in our reality/ frame of reference) and thus eliminating any energy yield for us.
Mmm... I might be knocking on the door, but I don't know what is behind it and I don't have a key.

Dan Tibbets

[Carl Willis]

I feel this thread has been given a fair shake, but the chain of responses is mostly characterized by word salad. We've got wide-ranging and off-topic speculation. A dearth of credible references, or new theoretical developments closely accountable to observed reality. And more than a lion's share of barking-mad nonsense.

It's never been my understanding that off-the-cuff scifi-convention bathroom banter amounts to "Theory." We don't have a general banter forum. (That idea has been proposed, and shot down, several times.) If we did have such, that's where 90% of this thread would belong. But as it is, this is a "Theory" forum.

To get to the point at last: thread's closed.

-Carl