Talk:Available energy (particle collision)

Another class of Available energy

I would really like to have some of the people here look at my hypothesis. I am asking here because it involves conversion of “Available energy” into mass.

Every one that says it is impossible to create a D+D fusion reaction in a metallic lattice is correct. If you look at the mobility of H/D/T in Pd, clearly any movement of a hydrogen atom in the lattice will result in columbic repulsion moving hydrogen atoms to the nearest free octahedral location.

Further, I realize that in general evaluating a Hamiltonian results in another function. The function resulting from the evaluation of the Hamiltonian in my hypothesis includes terms not previously included in Hamiltonian formula, and addresses a system far from equilibrium; something not normally addressed using this concept.

In all active FCC transition mettles and specifically Pd, I propose that the octahedral points have a correspondence to the to the empty S5 orbitals.

1. The system achieving the required energy (Hamiltonian) consists of at least 6 FCC lattice elements plus at least one H. The higher loading of D required in PF style cells increases the absolute value of the base line Hamiltonian, increasing the likelihood of obtaining the required energy necessary to create a neutron out of a proton plus an electron.

2. That Hydrogen nuclei appear to be constrained to the octahedral points indicates it is possible to affect the Δq (deviation of location) of the nuclei.

3. If the Δq of the nuclei is affected in the low temperature < 300K state, then under higher phononic activity present at a higher temperature and inclusion of hydrogen nuclei, the random vibrations could conceivably further constrain the hydrogen nuclei.

4. Satyendra Nath Bose and Albert Einstein proposed that by making Δp (deviation of momentum) small enough Δq would become large and a condensate would form. (Bose-Einstein condensate)

5. Robert E. Godes proposes that by making Δq small enough Δp becomes large, providing the required 782KeV, and meeting the constraints of ΔE*Δt > h / 4p. I call this Heisenberg Confinement Energy.

I propose that there is a solution to the function resulting from the evaluation of the Hamiltonian described in my hypothesis. Many solutions result in an absolute value greater than 782KeV. At first glance this may seem preposterous, however it is no more preposterous than 80+GeV for the W bosons, which are well accepted given the confines of ΔE*Δt > h / 4π. The phenomena are largely explained by Rogue waves (superposition of phonons) causing Heisenberg Confinement Energy of the trapped hydrogen nuclei. Please Email me if you have an opinion wiki@profusionenergy.com
--Regsoft (talk) 06:00, 27 April 2008 (UTC)[reply]

Thank you for your email query. I am not a physicist, so on a scientific and technical level, I simply do not know what on Earth you are talking about. However, from the document you have on-line, you seem to have completed Phase 1 of some endeavor, and Phase 2 appears to require that someone give you 0.5 million to 2 million dollars. Based on my extensive background in number comparison, I strongly recommend:

1) the larger of those two figures. Obtaining more money from an unknown source is always better than obtaining less money from an unknown source.
2) a narrower range of dollar amounts. When little Billy asks his mama for 45 to 55 cents for candy money, she'll give it to him, and when little Billy asks his mama for $1.50 to $2.50 for hamburger money, she'll give it to him, but when little Billy asks his mama for something between 50 cents and $2 for a quantum fusion reactor, she thinks he may really be spending it on a girlfriend.

I also recommend a skilled professional editor because it doesn't look good when you spell "metal" as "mettle." If you are interested in further editing and consulting services, please email me again for rates. In this situation I'll either need cash in advance or sufficient information to run a credit and background check. Flying Jazz (talk) 11:03, 27 April 2008 (UTC)[reply]

moving to Available energy (particle physics)

The move was done improperly. If someone gets the redirect deleted, and this page moved properly, I don't mind. I just want the history preserved. However, I'm unsure as to how the new name would be any better (or worse for that matter). Fresheneesz 08:19, 29 June 2006 (UTC)[reply]

Sorry about that. I don't move very often. I'll try to move it properly now. The term "Available energy" in physics has a definition by Gibbs in classical thermodynamics (that is also called exergy). To me, differentiating between the two articles by calling one "(physics)" implies that the other isn't physics. I don't think these two meanings are related because the second law in the exergy sense requires a macroscopic system that isn't needed for a two-particle system. "Available" is just an adjective, and it could be used in lots of different ways, I guess. There's also "available energy states" as described in the Fermi energy article, and that's physics too even though it refers to one particle with no collision, so calling this article "particle physics" doesn't seem right either. I'm going to try moving it to "Available energy (particle collision)." Flying Jazz 12:45, 29 June 2006 (UTC)[reply]

Ah, ok, that makes sense. If you're trying to move it, you may not know that you can't move the page to where you wanted it now - because that page exists (as a redirect). You need to propose deletion, using the {{prod|your reason here}} template, and once the page is deleted, you can move this page to it. I suppose thats why this page is at (particle collision) instead of (particle physics). Fresheneesz 19:20, 29 June 2006 (UTC)[reply]

This doesn't even cover the particle physics usage

This definition implicitly (yuch) assumes a setup where an incoming particle (beam) hits a stationary target, a.k.a. "stationary target" experiments. In the last 30 years, setups where particle beams collide dominate particle physics, and in particular, the energy frontier (though stationary target experiments have a very sizable niche).

So this page is overly specific, and even if it's renamed to "particle physics" it still fails to live up to its title. —Preceding unsigned comment added by 131.225.224.150 (talk) 23:39, 17 April 2008 (UTC)[reply]

Depends on the Collision

I never heard of this term--- the available energy is just the center of mass energy minus the sum of the masses of the most stable particles with the same conserved quantities which can be made by the collision. Is this called available energy?Likebox (talk) 15:06, 16 March 2009 (UTC)[reply]

Possible Reference for the Equation

http://www.physics.lancs.ac.uk/resources/alevel/p3.pdf

The above link contains a reasonably good reference for the main equation in the article. I checked the derivation and it gives the right equation. I think that the E term should be the total enrgy of the moving particle rather than it's kinetic energy. — Preceding unsigned comment added by Davide De Mola (talk • contribs) 11:40, 7 October 2012 (UTC)[reply]