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Deep-Gas, Deep Hot Biosphere Theory
- Larry Burford
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16 years 11 months ago #18349
by Larry Burford
Replied by Larry Burford on topic Reply from Larry Burford
<b>[Gregg] "If it works, I don't care much about its naturalness."</b>
There must be something obvious about your theory that I have missed. So far I am mostly just waiting for something ... not sure what ... to turn on a light bulb, so to speak.
Here is a summary of what I see in my minds eye, based on my interpretation of what you have said.
I'm pretty sure this is not exactly what you have in mind, so you might want to back up and take another stab at explaining the basics. It also wouldn't hurt to grow your understanding of gravitons and elysons a little.
Unless they really are inconsequential to your theory.
These particles are still theoretical, so comming to grips with what we do know about them won't be a large chore. To a first approximation, you become an expert when you no longer feel a need to publically advertise your lack of understanding. Other experts will then have their opinion (mostly kept private (at least on the serious message boards)) about your expertise.
There must be something obvious about your theory that I have missed. So far I am mostly just waiting for something ... not sure what ... to turn on a light bulb, so to speak.
Here is a summary of what I see in my minds eye, based on my interpretation of what you have said.
Code:
force = non existant
proton(s) = atom
neutron = disguised proton / figment of our imagination
electron = side effect of proton / rapidly evaporates
You have mentioned gravitons and elysons on a few occasions. Mostly you
say that, although you don't really know much about them, they seem
to be involved in some way. Gravitons liquify elysons, then boil them away.
Unless they really are inconsequential to your theory.
These particles are still theoretical, so comming to grips with what we do know about them won't be a large chore. To a first approximation, you become an expert when you no longer feel a need to publically advertise your lack of understanding. Other experts will then have their opinion (mostly kept private (at least on the serious message boards)) about your expertise.
Please Log in or Create an account to join the conversation.
16 years 11 months ago #18299
by Gregg
Replied by Gregg on topic Reply from Gregg Wilson
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Larry Burford</i>
<br /><b>[Gregg] "If it works, I don't care much about its naturalness."</b>
There must be something obvious about your theory that I have missed. So far I am mostly just waiting for something ... not sure what ... to turn on a light bulb, so to speak.
Here is a summary of what I see in my minds eye, based on my interpretation of what you have said.
force = non existant
<hr noshade size="1">
<b>A force has reality when object A, which has mass and velocity, collides with object B.
The atmospheric pressure is a force when the air molecules collide against your body.
There would be no force if these collisions are not happening.
Light waves impacting on a white panel would be a force, though hard to detect.
And that light would be a geometric momentum wave impacting on the white panel.
My definitional purpose is to remove any magic from the phenomena of force. No action at a distance, no attractiveness.</b><hr noshade size="1">
proton(s) = atom
<hr noshade size="1">
<b>A single proton would make up only the simplest atom, hydrogen.
Larger atoms will be made up of a construct of protons,
Most of them being in their "neutron state", where the protons are mated base to base.
These doubled up protons can then be mated side to side.
When radioactive breakdown occurs to this kind of polymer..
Then some of the doubled protons would be separated, base from base.
Once a proton has an open base, it reacts with Elysium and gravitons.
When this interaction occurs, one has the repulsive effect of vaporized elysons.</b><hr noshade size="1">
<hr noshade size="1">
neutron = disguised proton / figment of our imagination
<hr noshade size="1">
<b>If protons are mated base to base, they no longer have a gravity well.
Therefore, there is no place for gravitons to vaporize liquid Elysium.
There is no such thing as a single - stable - neutron.
When released from a nucleus, the neutron releases its hot Elysium.
This makes for high speed and a very short life.
Once the hot, liquid Elysium is expended, it is simply a naked proton.
That is simply a hydrogen atom.</b><hr noshade size="1">
electron = side effect of proton / rapidly evaporates
<hr noshade size="1">
<b>An electron is a phenomena, not necessarily a single particle.
The issue is very similar to the photon. Three scenarios:
1) The vaporizing of Elysium out of the proton well is an electron.
Here, it is an expanding vapor cloud of elysons which <b>push(a force).</b>
2) Liquid Elysium flowing over the outer surfaces of protons.
This is electricity as we describe it in a copper wire.
If it reaches deadends in the proton matrix, it spills into proton wells.
Once that happens, it gets out only through vaporization by gravitons.
3) If a proton well is loaded with liquid Elysium, then....
If there is a vibration, the liquid Elysium could be ejected.
Now you have electrons being "shot into space" as liquid Elysium balls.</b><hr noshade size="1">
You have mentioned gravitons and elysons on a few occasions. Mostly you
say that, although you don't really know much about them, they seem
to be involved in some way. Gravitons liquify elysons, then boil them away.
I'm pretty sure this is not exactly what you have in mind, so you might want to back up and take another stab at explaining the basics. It also wouldn't hurt to grow your understanding of gravitons and elysons a little.
<hr noshade size="1">
<b>I understand gravitons to be extremely small, extremely fast particles.
They apparently have a very long mean free path, a few thousand light years.
Therefore, at the scale of our solar system, they do not act like a gas.
At our scale, they have no viscosity or friction with each other.
Elysons apparently have a very short mean free path. They collide.
We see waves, so elyson geometry is probably highly relevant.
That may explain the wave behavior.
I have assumed that the Elysium medium is compressible.
So it can be both vapor and liquid; can condense or vaporize.
There would be no condensation of Elysium unless something is in the way.
That would be protons or assemblies of protons.
One cannot have condensation and vaporization at the same position.
That is one reason why I propose that the proton is asymmetric.</b><hr noshade size="1">
Unless they really are inconsequential to your theory.
<hr noshade size="1">
<b>Obviously they are integral to my "theory".</b><hr noshade size="1">
These particles are still theoretical, so comming to grips with what we do know about them won't be a large chore.
<hr noshade size="1">
<b>See above.</b><hr noshade size="1">
To a first approximation, you become an expert when you no longer feel a need to publically advertise your lack of understanding. Other experts will then have their opinion (mostly kept private (at least on the serious message boards)) about your expertise.[/i]<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
<hr noshade size="1">
<b>The above comment is strange. "Experts" never admit their ignorance.
I really don't care about other people's opinion of my status.
Examine the idea, not my pedigree, please.
My peer reviewer is Reality - at plant startups.
My academic education (seven years) has proved to be of limited value.
Most of my useful knowledge has been learned on the job.</b>
Gregg Wilson
<br /><b>[Gregg] "If it works, I don't care much about its naturalness."</b>
There must be something obvious about your theory that I have missed. So far I am mostly just waiting for something ... not sure what ... to turn on a light bulb, so to speak.
Here is a summary of what I see in my minds eye, based on my interpretation of what you have said.
force = non existant
<hr noshade size="1">
<b>A force has reality when object A, which has mass and velocity, collides with object B.
The atmospheric pressure is a force when the air molecules collide against your body.
There would be no force if these collisions are not happening.
Light waves impacting on a white panel would be a force, though hard to detect.
And that light would be a geometric momentum wave impacting on the white panel.
My definitional purpose is to remove any magic from the phenomena of force. No action at a distance, no attractiveness.</b><hr noshade size="1">
proton(s) = atom
<hr noshade size="1">
<b>A single proton would make up only the simplest atom, hydrogen.
Larger atoms will be made up of a construct of protons,
Most of them being in their "neutron state", where the protons are mated base to base.
These doubled up protons can then be mated side to side.
When radioactive breakdown occurs to this kind of polymer..
Then some of the doubled protons would be separated, base from base.
Once a proton has an open base, it reacts with Elysium and gravitons.
When this interaction occurs, one has the repulsive effect of vaporized elysons.</b><hr noshade size="1">
<hr noshade size="1">
neutron = disguised proton / figment of our imagination
<hr noshade size="1">
<b>If protons are mated base to base, they no longer have a gravity well.
Therefore, there is no place for gravitons to vaporize liquid Elysium.
There is no such thing as a single - stable - neutron.
When released from a nucleus, the neutron releases its hot Elysium.
This makes for high speed and a very short life.
Once the hot, liquid Elysium is expended, it is simply a naked proton.
That is simply a hydrogen atom.</b><hr noshade size="1">
electron = side effect of proton / rapidly evaporates
<hr noshade size="1">
<b>An electron is a phenomena, not necessarily a single particle.
The issue is very similar to the photon. Three scenarios:
1) The vaporizing of Elysium out of the proton well is an electron.
Here, it is an expanding vapor cloud of elysons which <b>push(a force).</b>
2) Liquid Elysium flowing over the outer surfaces of protons.
This is electricity as we describe it in a copper wire.
If it reaches deadends in the proton matrix, it spills into proton wells.
Once that happens, it gets out only through vaporization by gravitons.
3) If a proton well is loaded with liquid Elysium, then....
If there is a vibration, the liquid Elysium could be ejected.
Now you have electrons being "shot into space" as liquid Elysium balls.</b><hr noshade size="1">
You have mentioned gravitons and elysons on a few occasions. Mostly you
say that, although you don't really know much about them, they seem
to be involved in some way. Gravitons liquify elysons, then boil them away.
I'm pretty sure this is not exactly what you have in mind, so you might want to back up and take another stab at explaining the basics. It also wouldn't hurt to grow your understanding of gravitons and elysons a little.
<hr noshade size="1">
<b>I understand gravitons to be extremely small, extremely fast particles.
They apparently have a very long mean free path, a few thousand light years.
Therefore, at the scale of our solar system, they do not act like a gas.
At our scale, they have no viscosity or friction with each other.
Elysons apparently have a very short mean free path. They collide.
We see waves, so elyson geometry is probably highly relevant.
That may explain the wave behavior.
I have assumed that the Elysium medium is compressible.
So it can be both vapor and liquid; can condense or vaporize.
There would be no condensation of Elysium unless something is in the way.
That would be protons or assemblies of protons.
One cannot have condensation and vaporization at the same position.
That is one reason why I propose that the proton is asymmetric.</b><hr noshade size="1">
Unless they really are inconsequential to your theory.
<hr noshade size="1">
<b>Obviously they are integral to my "theory".</b><hr noshade size="1">
These particles are still theoretical, so comming to grips with what we do know about them won't be a large chore.
<hr noshade size="1">
<b>See above.</b><hr noshade size="1">
To a first approximation, you become an expert when you no longer feel a need to publically advertise your lack of understanding. Other experts will then have their opinion (mostly kept private (at least on the serious message boards)) about your expertise.[/i]<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
<hr noshade size="1">
<b>The above comment is strange. "Experts" never admit their ignorance.
I really don't care about other people's opinion of my status.
Examine the idea, not my pedigree, please.
My peer reviewer is Reality - at plant startups.
My academic education (seven years) has proved to be of limited value.
Most of my useful knowledge has been learned on the job.</b>
Gregg Wilson
Please Log in or Create an account to join the conversation.
16 years 11 months ago #18260
by Gregg
Replied by Gregg on topic Reply from Gregg Wilson
If the general question is about whether or not a Grand Unified Theory has been proposed here, that requires an analysis of what is meant by GUT. Standard definition:
"A theory of elementary forces that unites the weak, strong, electromagnetic, and gravitational interactions into one field theory and views the known interactions as low-energy manifestations of a single unified interaction."
This unification is supposed to happen at "1000 billion billion volts" which is undoubtedly very hot. I am quite sure this would result in a single, beautiful equation which explains everything. All the forces become one. What is the utility of all this nonsense?
Actually Meta Research has almost completely come up with a unification of the forces. But this is conceptual; it is not a physical happening:
By first stating that energy actually consists of particles which have mass and velocity. Then that interactions between particles are actual collisions, not "attractive" forces from afar. Dr. Van Flandern has already stated that the energy exhibited by a star ultimately comes from the gravitational flux. One might characterize a graviton as no mass and all velocity. (I have exactly the opposite problem.)
If we introduce geometry as a relevant property, then all the forces or "energies" derive their capacity from the gravitational flux:
1)Strong nuclear energy, or force, is simply Elysium trapped inside protons, which has been made extremely hot by obtaining its momentum from equally trapped gravitons.
2)Weak nuclear energy, or force, is simply a much smaller amount of the strong nuclear energy, or force.
3) Electrical energy can either be liquid Elysium flowing along proton surfaces - pushed by the gravitons - or ejected liquid Elysium traveling through the vapor Elysium (and gravitational flux.)
4) Electromagnetic waves would simply be vaporized Elysium. Being waves is a property of Elysium - as yet not defined.
5) Magnetism is simply a gravitational flux that has been colinearized - goes in one direction. This is possible only if the random gravitational flux has been altered by proton assembly geometries and then emitted.
Contrary to Stephen Hawking's declaration that all is understood and "we can close the book now", we have barely opened the book.
Gregg Wilson
"A theory of elementary forces that unites the weak, strong, electromagnetic, and gravitational interactions into one field theory and views the known interactions as low-energy manifestations of a single unified interaction."
This unification is supposed to happen at "1000 billion billion volts" which is undoubtedly very hot. I am quite sure this would result in a single, beautiful equation which explains everything. All the forces become one. What is the utility of all this nonsense?
Actually Meta Research has almost completely come up with a unification of the forces. But this is conceptual; it is not a physical happening:
By first stating that energy actually consists of particles which have mass and velocity. Then that interactions between particles are actual collisions, not "attractive" forces from afar. Dr. Van Flandern has already stated that the energy exhibited by a star ultimately comes from the gravitational flux. One might characterize a graviton as no mass and all velocity. (I have exactly the opposite problem.)
If we introduce geometry as a relevant property, then all the forces or "energies" derive their capacity from the gravitational flux:
1)Strong nuclear energy, or force, is simply Elysium trapped inside protons, which has been made extremely hot by obtaining its momentum from equally trapped gravitons.
2)Weak nuclear energy, or force, is simply a much smaller amount of the strong nuclear energy, or force.
3) Electrical energy can either be liquid Elysium flowing along proton surfaces - pushed by the gravitons - or ejected liquid Elysium traveling through the vapor Elysium (and gravitational flux.)
4) Electromagnetic waves would simply be vaporized Elysium. Being waves is a property of Elysium - as yet not defined.
5) Magnetism is simply a gravitational flux that has been colinearized - goes in one direction. This is possible only if the random gravitational flux has been altered by proton assembly geometries and then emitted.
Contrary to Stephen Hawking's declaration that all is understood and "we can close the book now", we have barely opened the book.
Gregg Wilson
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- neilderosa
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16 years 11 months ago #20505
by neilderosa
Replied by neilderosa on topic Reply from Neil DeRosa
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">The questions I pose for this post are about the weak nuclear force. Gregg, do you see this as being essentially the same thing as the chemical bond that holds molecules together? And related to that, do you see any significant difference between elements and molecules, (e.g., between C and CH4), aside from the number and configuration of the protons, and of course how they play out in the physical qualities of the substance at our scale. [ND]
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Chemical bonds are weaker than this "weak" nuclear force. Most of the radioactive decays involve - from my point of view - the breaking off of a helium nucleus from the parent nucleus. This gives the weakest energy release in radioactive decay: alpha. And this would be the release of any liquid Elysium encased between proton sides.
There is an obvious difference between the atomic structure and molecular structure in terms of the strengths of the bonds. Otherwise, alchemists would have succeeded in turning lead into gold. [Gregg]
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
I want to consider this answer again for a moment. In my paper on the topic of this thread, I reported the following from T. Gold’s research:
“It used to be thought that hydrocarbons would dissociate, or break down to their elemental components at between 300°C and 600°C. This can be demonstrated by a simple [and naïve] stovetop experiment. Moreover it is easily demonstrated that methane gas quickly oxidizes to CO2 and H2O once it comes into contact with the abundant free oxygen of the earth’s atmosphere. We placed the word “naïve” in brackets because in the deep earth, pressure increases with depth… Russian Geoscientist E. B. Chekaliuk’s thermodynamic calculations…determined that methane, for example, would be stable at 300 km, provided the temperature (the geotherm) of the earth at that depth did not exceed 2,000°C.”
So, as you have put it elsewhere, the stability of methane is a “contest between temperature and pressure.” As I understand your model, whether or not an element is stable depends on its “geometry.” After the original combination of matter (protons) in sunspots as you speculated, the heavier matter can break off and form planets (presumably by means of TVF’s fission model mechanism). Afterward within this mega-element, if protons have their bases not exposed to the outside environment, the coulomb repulsive force consisting of vaporizing Elysium, cannot prevent “attack” from free protons or presumably other combinations of protons (i.e., other elements) and smaller elements break off in a random manner, thus forming the various elements of the periodic table, plus their isomers, etc. This means by fission and radioactivity.
Now, it seems that molecule formation in your model is not much different, that is from my naïve viewpoint. Here (you seem to be saying) elements or single protons are pushed into position by heat or pressure, into the same “available spaces” where the coulomb repulsive force does not prevent it.
Can you explain the difference better? I’m either missing something or am a slow learner. [Neil]
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Chemical bonds are weaker than this "weak" nuclear force. Most of the radioactive decays involve - from my point of view - the breaking off of a helium nucleus from the parent nucleus. This gives the weakest energy release in radioactive decay: alpha. And this would be the release of any liquid Elysium encased between proton sides.
There is an obvious difference between the atomic structure and molecular structure in terms of the strengths of the bonds. Otherwise, alchemists would have succeeded in turning lead into gold. [Gregg]
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
I want to consider this answer again for a moment. In my paper on the topic of this thread, I reported the following from T. Gold’s research:
“It used to be thought that hydrocarbons would dissociate, or break down to their elemental components at between 300°C and 600°C. This can be demonstrated by a simple [and naïve] stovetop experiment. Moreover it is easily demonstrated that methane gas quickly oxidizes to CO2 and H2O once it comes into contact with the abundant free oxygen of the earth’s atmosphere. We placed the word “naïve” in brackets because in the deep earth, pressure increases with depth… Russian Geoscientist E. B. Chekaliuk’s thermodynamic calculations…determined that methane, for example, would be stable at 300 km, provided the temperature (the geotherm) of the earth at that depth did not exceed 2,000°C.”
So, as you have put it elsewhere, the stability of methane is a “contest between temperature and pressure.” As I understand your model, whether or not an element is stable depends on its “geometry.” After the original combination of matter (protons) in sunspots as you speculated, the heavier matter can break off and form planets (presumably by means of TVF’s fission model mechanism). Afterward within this mega-element, if protons have their bases not exposed to the outside environment, the coulomb repulsive force consisting of vaporizing Elysium, cannot prevent “attack” from free protons or presumably other combinations of protons (i.e., other elements) and smaller elements break off in a random manner, thus forming the various elements of the periodic table, plus their isomers, etc. This means by fission and radioactivity.
Now, it seems that molecule formation in your model is not much different, that is from my naïve viewpoint. Here (you seem to be saying) elements or single protons are pushed into position by heat or pressure, into the same “available spaces” where the coulomb repulsive force does not prevent it.
Can you explain the difference better? I’m either missing something or am a slow learner. [Neil]
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16 years 11 months ago #20506
by Gregg
Replied by Gregg on topic Reply from Gregg Wilson
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by neilderosa</i>
Now, it seems that molecule formation in your model is not much different, that is from my naïve viewpoint. Here (you seem to be saying) elements or single protons are pushed into position by heat or pressure, into the same “available spaces” where the coulomb repulsive force does not prevent it.
Can you explain the difference better? I’m either missing something or am a slow learner. [Neil]
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Actually, your question is legitimate. In my speculation, the polymerization of deuterium would occur only within a "sea" of liquid Elysium. This action would not involve collisions but a relative slow compaction under great pressure. Because there would not be open base protons, there would be no source of re-vaporizing Elysium.
I don't think that a portion of this polynuclear, deuterium mass would break off and then spiral out of the Sun. I think, instead, the remaining Sun matter would collapse inward, leaving this nuclear mass in a stable orbit, exposed to "open" space. At this point, it would no longer have the "protection" of liquid Elysium.
Visualize a bullet fired by a rifle with a 3,000 feet per second muzzle velocity. As long as the bullet is in air, it more or less maintains that velocity. If you have seen bullets hit into water, they penetration only about 4 to 5 feet before coming to a halt.
So, if the nuclear mass is suddenly exposed to a vapor Elysium, high velocity proton masses can impact it and initiate radioactive decay. Keep in mind that the various radioactive decays typically have a velocity, which is a discernable fraction of the speed of light.
Once overall macroscopic radioactive decay has produced "normal" nuclei, they are more or less encased within a vapor Elysium. Velocities of these nuclei are far less than the velocity of alpha particles, beta "particles", fissioning neutrons. So these nuclei have regained open proton base "repulsion" and they are exposed to one another at much lower relative velocities. The result is that nuclear binding and nuclear fission are no longer the common actions.
Instead, we get "chemistry". The reactants are not coming together by attraction. They are being pushed together by the vapor Elysium, which in turn, is pushed by the gravitational flux. The hydrogen atoms would be pushed into the carbon atom where the repulsiveness of open protons are not located. They would come to rest against the "hardness" of the carbon nucleus. The resistance of the nucleus, plus compressed Elysium, would balance out against the outer push of vapor Elysium and gravitons. Once these four hydrogen atoms have "combined" with the carbon nucleus - <b>with their open, repulsive bases facing outward</b> - we would have a close approximation to a Noble gas nucleus.
The same for oxygen nuclei coming together. An approximation of a Noble gas nucleus.
But not perfect enough. If methane is mixed with oxygen - and high temperature causes some collisions - the nuclei will be re-arranged for a more "perfect" fit. The results will be CO2 and H2O. Some Elysium - probably liquid like - is forced to take a hike and find a new home. We have a release of light, heat, thermal radiation. In fact, those effects are pretty much a dead giveaway for vaporized liquid Elysium, which was previously "hiding" within the methane and oxygen molecules.
When a pure substance exhibits both a vapor phase and a liquid phase, this is a special circumstance existing only over a very limited range of pressure and temperature. Over the vast majority of the pressure and temperature span, the substance will simply be a fluid. It will be compressible - responding to a "contest" between pressure and temperature.
I am not surprised that methane would remain stable at a very high pressure and up to 2,000 degrees Celsius. Thermal breakdown of hydrocarbons is first done in an Atmospheric Column. The bottoms from this column then goes to a Vacuum Column, where the pressure is only about 1 - 2 psia.
Gregg Wilson
Now, it seems that molecule formation in your model is not much different, that is from my naïve viewpoint. Here (you seem to be saying) elements or single protons are pushed into position by heat or pressure, into the same “available spaces” where the coulomb repulsive force does not prevent it.
Can you explain the difference better? I’m either missing something or am a slow learner. [Neil]
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Actually, your question is legitimate. In my speculation, the polymerization of deuterium would occur only within a "sea" of liquid Elysium. This action would not involve collisions but a relative slow compaction under great pressure. Because there would not be open base protons, there would be no source of re-vaporizing Elysium.
I don't think that a portion of this polynuclear, deuterium mass would break off and then spiral out of the Sun. I think, instead, the remaining Sun matter would collapse inward, leaving this nuclear mass in a stable orbit, exposed to "open" space. At this point, it would no longer have the "protection" of liquid Elysium.
Visualize a bullet fired by a rifle with a 3,000 feet per second muzzle velocity. As long as the bullet is in air, it more or less maintains that velocity. If you have seen bullets hit into water, they penetration only about 4 to 5 feet before coming to a halt.
So, if the nuclear mass is suddenly exposed to a vapor Elysium, high velocity proton masses can impact it and initiate radioactive decay. Keep in mind that the various radioactive decays typically have a velocity, which is a discernable fraction of the speed of light.
Once overall macroscopic radioactive decay has produced "normal" nuclei, they are more or less encased within a vapor Elysium. Velocities of these nuclei are far less than the velocity of alpha particles, beta "particles", fissioning neutrons. So these nuclei have regained open proton base "repulsion" and they are exposed to one another at much lower relative velocities. The result is that nuclear binding and nuclear fission are no longer the common actions.
Instead, we get "chemistry". The reactants are not coming together by attraction. They are being pushed together by the vapor Elysium, which in turn, is pushed by the gravitational flux. The hydrogen atoms would be pushed into the carbon atom where the repulsiveness of open protons are not located. They would come to rest against the "hardness" of the carbon nucleus. The resistance of the nucleus, plus compressed Elysium, would balance out against the outer push of vapor Elysium and gravitons. Once these four hydrogen atoms have "combined" with the carbon nucleus - <b>with their open, repulsive bases facing outward</b> - we would have a close approximation to a Noble gas nucleus.
The same for oxygen nuclei coming together. An approximation of a Noble gas nucleus.
But not perfect enough. If methane is mixed with oxygen - and high temperature causes some collisions - the nuclei will be re-arranged for a more "perfect" fit. The results will be CO2 and H2O. Some Elysium - probably liquid like - is forced to take a hike and find a new home. We have a release of light, heat, thermal radiation. In fact, those effects are pretty much a dead giveaway for vaporized liquid Elysium, which was previously "hiding" within the methane and oxygen molecules.
When a pure substance exhibits both a vapor phase and a liquid phase, this is a special circumstance existing only over a very limited range of pressure and temperature. Over the vast majority of the pressure and temperature span, the substance will simply be a fluid. It will be compressible - responding to a "contest" between pressure and temperature.
I am not surprised that methane would remain stable at a very high pressure and up to 2,000 degrees Celsius. Thermal breakdown of hydrocarbons is first done in an Atmospheric Column. The bottoms from this column then goes to a Vacuum Column, where the pressure is only about 1 - 2 psia.
Gregg Wilson
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16 years 11 months ago #18309
by Stoat
Replied by Stoat on topic Reply from Robert Turner
I was making some 3d animation's of the toroidal electron; actually a loxodrome model, for this guy called Dave Thomson. Who's an advocate of the idea. So I went to see what the Russian Ph. M. Kanarev had to say, as he's written extensively on the subject. Now, he's looked at the energy levels of electrons and can't find any orbital component. His model of the hydrogen atom is, to a scale, where the toroidal electron is a metre across, the proton is one millimetre across. The distance between them is one hundred metres. In effect the hydrogen atom is like a child's spinning top. If we add the Le Sage shadow we might be getting somewhere with the creation of a model of any atom.
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