Ed,
On Sun, Jun 2, 2013 at 10:45 AM, Edmund Storms <[email protected]>wrote: > > On Jun 2, 2013, at 12:15 AM, Harry Veeder wrote: > > > > On Fri, May 31, 2013 at 9:11 AM, Edmund Storms <[email protected]>wrote: > >> >> On May 30, 2013, at 11:39 PM, Harry Veeder wrote: >> >> On Thu, May 30, 2013 at 11:00 AM, Edmund Storms <[email protected]>wrote: >> >>> Harry, imagine balls held in line by springs. If the end ball is pull >>> away with a force and let go, a resonance wave will pass down the line. >>> Each ball will alternately move away and then toward its neighbor. If >>> outside energy is supplied, this resonance will continue. If not, it will >>> damp out. At this stage, this is a purely mechanical action that is well >>> understood. >>> >>> >> >> >>> In the case of the Hydroton, the outside energy is temperature. The >>> temperature creates random vibration of atoms, which is focused along the >>> length of the molecule. Again, this is normal and well understood behavior. >>> >>> The strange behavior starts once the nuclei can get within a critical >>> distance of each other as a result of the resonance. This distance is less >>> than is possible in any other material because of the high concentration of >>> negative charge that can exist in this structure and environment. The >>> barrier is not eliminated. It is only reduced enough to allow the distance >>> to become small enough so that the two nuclei can "see" and respond. The >>> response is to emit a photon from each nuclei because this process lowers >>> the energy of the system. >>> >>> >> Ed, >> >> With each cycle energy of the system is only lowered if the energy of the >> emitted photon is greater than the work done by the "random vibration of >> atoms" on the system. >> >> >> NO Harry! >> > > Ed, I am trying to help you understand your model. I am not trying to tear > it down. > > > I know and I appreciate the effort. However, I want you to accurately > understand what I'm proposing. Only then can you add a new insight. You are > not accurately describing what I proposing. > > > >> There is no work done by the random vibrations. These are the result of >> normal temperature. The photon is emitted from the nucleus and carries with >> it the excess mass-energy of the nucleus. >> >> > Let us return to your ball and spring model of the hydroton and assume an > ideal spring which doesn't dissipate energy by getting warm during > compressions. If heat energy is the vibration of atoms in the lattice, > then the spring is compressed by atoms from the lattice pushing on the > spring. As the spring is compressed work is done on the spring, however, > the spring will eventually bounce back to its original length so no net > work is done on the spring in the course of one oscillation. The > oscillations will repeat indefinitely with the same amplitude as long as > the temperature remains constant. However, in your model the spring does > not return to its original length. Now for sake argument assume no photon > is emitted. This means some work has been performed on the spring, which > means the spring has effectively turned a little thermal energy into > potential energy and thereby slightly cooled the lattice. Now assume a > photon is emitted. The subsequent temperature of the lattice will depend on > the energy of this emitted photon. If the energy of the photon is less than > the work done (W) then the temperature of the lattice will not return to > the initial the temperature. The cycle can repeat until the protons fuse > but the temperature will gradually decline and the end result can aptly be > described as cold fusion! On the other hand if the energy of the photon is > greater than W then the temperature of the lattice will be greater after > fusion. > > > No analogy is perfect and you are extending my effort to get one idea > understood and applying it to a different idea, which is not correct. The > vibration is like a periodic switch acting on the nucleus. The vibration > itself does not release energy. It has no friction. Energy is totally > conserved during the vibration. However, the vibration causes the nuclei to > emit a proton because the vibration periodically causes them to get within > a critical distance of each other. > > Getting closer _and_ staying closer means work has been done on the system since there is a mutual force of repulsion keeping them apart. The kinetic energy of the lattice is transformed into potential energy of repulsion according to the principle of CoE. Whether the temperature of the environment cools, stays constant or warms depends on whether the energy of the emitted photon is less than / equal to / greater than the work done. Your model at the present time is silent on these possibilities. > All atoms vibrate, but normally in random ways. The Hydroton forces this > vibration into a particular direction. In fact all chemical bonds do this. > For example, in the water molecule, the H-O-H bond vibrates and causes the > molecule to periodically gets slightly longer and shorter, and cause the > angle to change. This process does not cause a nuclear reaction because the > H and O are too far apart. In contrast, the H in the hydroton are close > enough that this vibration periodically causes the nuclei to release > mass-energy. This ability of a bond to do this is very rare. > Nevertheless, I suspect it can happen when the bond with or between H or D > is especially strong. The conditions producing the Hydroton just happen to > be so efficient at producing the rare condition that the effect is easily > detectable, and now has enough attention to be acknowledged when it is > detected. > > Yes, but work is done in the process. > > > > >> The change is analogous to an exothermic chemical reaction which requires >> some activation energy to initiate but the reaction products are in a lower >> energy state. Because of the shape of the coulomb "hill" the hill can only >> be climbed if the energy emitted increases with each cycle. >> >> >> No! The hill height is reduced by an intervening negative charge. As a >> result, the hill height is reduced so that it can be surmounted by the >> vibrations occuring in the Hydroton. Normally, the hill is too high for >> such small vibrations to have any effect. The hill is reduced in height as >> a result of the Hydroton forming. As a result, it is the unique condition >> required to make CF work. All the theories use something similar, but >> without a clear description. >> >> > The barrier is reduced by the electron but I think the net effect only > reduces the force of repulsion by 1/2. > > > Perhaps. I have not estimated the amount yet because the concept is the > focus now. > > However, this is not a problem since you have theoretically enlarged > the total energy of a p-e-p association (or molecule as you call > it) to include all the excess mass-energy as well as the electrostatic > energy of the association. Therefore the p-e-p association can shrink in > size by entering a lower energy through the conversion of mass into a > photon. > > > Yes, that is correct. However, until the H get within the critical > distance, the system does not "know" that it has excess mass-energy. Until > this distance can be reached, the system is like any other chemical that > only knows about chemical energy, not potential nuclear energy. For > example, the ordinary D2 molecule has potential nuclear energy, but it > cannot release this because the conditions do not permit its release, or > you can say the molecule is ignorant of this potential. > > I would say as soon as the bond is made the system knows and the process begins. You don't need to wait for a critical distance to be reached. > > > > >> This is like a ball rolling between two hills. It rolls down the side of >> one hill, through the valley and up the other side. In the process, it >> picks up a little energy from the surroundings (temperature in this case) >> to reach the top, where it throws a switch and turns on a light for a brief >> time. Immediately, it starts to roll back down and returns to the first >> hill where it again reaches the top and turns on a light for a brief time. >> This back and forth continues until the battery powering the light is >> exhausted and the hills disappear. The light has no relationship to the >> motion of the ball. The ball only throws the switch. >> >> > This will not warm the lattice as I explained above. > > > The photon warms the lattice as it is absorbed by the well known process > of photon absorption by materials. Because the photons have very little > energy, they are completely absorbed by the material within the apparatus, > which they heat. Too few escape to make detection easy, although they have > been detected when efforts are made. In addition, the frequency of the > vibration determines how long the switch is closed, i.e. how long the light > is on. This determines how energetic the photon will be. This vibration > frequency depends on the isotopic composition of the Hydroton. > Consequence, the energy of the photon emitted from the nucleus will be > variable, with a few having enough energy to get out of the apparatus and > be detected. > > Whether or not it will warm the lattice will depend on the ratio of energy of the emitted photon to the work done with each resonant cycle. If the ratio < 1 the lattice cools. If the ratio = 1 the temperature is constant. If the ratio > 1 the lattice warms. > That does not mean I think your model is wrong. It means your > understanding of your model is incomplete. Whether a > creator is sculpting, inventing ecats or models, a creator does not > immediately grasp his creation. > > > You might consider, Harry, that it is you who do not understand the model. > > I am just applying the concept of potential energy to your model. > > > > >> >> >>> The Hydroton allows the Coulomb barrier to be reduced enough for the >>> nuclei to respond and emit excess energy. Because the resonance immediately >>> increases the distance, the ability or need to lose energy is lost before >>> all the extra energy can be emitted. If the distance did not increased, hot >>> fusion would result. The distance is again reduced, and another small burst >>> of energy is emitted. This process continues until ALL energy is emitted >>> and the intervening electron is sucked into the final product. >>> >>> >> In your model, the coulomb barrier appears to be like a hill in a uniform >> gravitational field. >> >> >> Yes, see above >> >> It is possible to climb such a barrier in steps by emitting the same >> amount of energy with each cycle, but this barrier does not correspond with >> the actual barrier that exists between protons. Climbing a genuine >> coulomb barrier requires more energy with each cycle, so that requires more >> energy be emitted with each cycle. The extra energy emitted heats the >> lattice even more and produces more powerful vibrations of the lattice >> which can push the protons even closer together. >> >> >> No, the Coulomb barrier is slowly reduced in height as mass-energy is >> lost, thereby allowing the nuclei to get closer each time the cycle >> repeats. Finally, the Coulomb barrier disappears and the two nuclei fuse, >> but very little excess mass-energy is present when this happens. >> Consequently, when the electron is absorbed, the resulting neutrino has >> very little energy to carry away. >> >> > The coulomb barrier only gets smaller if charge is being consumed in > process. However, this is not necessary, if as I said above the total > energy potential includes both the excess mass-energy and the electrostatic > energy, progressively lowering the energy state will overcome the coulomb > barrier if the energy of the photons progressively increases. > Anyway, before we consider the realistic coulomb barriers, we > should keep talking about the simplified barrier. > > > The Coulomb barrier is a complicated structure. It is related to charge > but, according to QM, it can also be affected by other factors; thus the > idea of tunneling has been introduced. I do not believe we know enough at > this stage to discuss the details of the barrier. I'm simply proposing that > the nuclei gradually get closer together as photons are lost so that they > are in contact when all excess mass-energy is gone. This requirement is > based on simple logic. > > Ed Storms > > Harry
