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Stellar Splitting and pairing NEW Black holes foun
16 years 1 month ago #15460
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi Jim, you need to explain what you want to do with the electron. It's the fundermental particle that people believe they know the most about. There's going to be no great rush to give it up.
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16 years 1 month ago #15461
by Jim
Replied by Jim on topic Reply from
Sloat, Lets turn this around if thats ok with you. How do you have a zero energy zone (or vacuum or what ever you call the space between the atoms)when the space is full of electrons of great mass and energy orbiting around in the space? The electron was originally just a focus of energy but somehow it morphed into a great mass that does magic tricks. The charge is real for sure so maybe you have a bigger problem fitting it in your model than I have. I figure the electron is a modeling function that makes the model work which is a good thing as long as its not confused with real events.
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16 years 1 month ago #15462
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi Jim, sorry if I havent explained myself very well, its not that easy a concept to get across.
Ive said that an electron; any particle, including the neutron; hides half of its mass inside a negative refractive index core at the Shwartzchild radius. For a particle this is tiny, much smaller than h. So theres much more energy per unit volume in this core. The electromagnetic energy, which is a half of the total energy of the particle, has an angular momentum of h and a angular velocity of c. The core on the other hand is spinning with an angular velocity of the speed of gravity. Most of the cores energy s down to the fact that its spinning at such an incredible rate.
This has to mean that we have two sorts of space, electromagnetic and gravitational. If we could release all of the gravitational energy of a proton say, we would have over a hundred tonnes of electromagnetic mass energy equivalent. However that core in gravitational space only has the mass of the particle, a proton mass in this case. Thats because the core is in a different space.
So lets say that we could make a movie of one second showing how electrons move in a strip of copper. This movie needs to have 1E 45 frames to catch the collision rate.
We then need a second movie to show how the cores in gravitational space behave. A test frame would show them to be too far apart, it needs to be scaled by a huge amount. We would have to screen the electromagnetic movie, using light to project it, then wed need to screen the core movie with its projector thousands of light years away. This would then need faster than light photons to project it. The images would arrive at our screen at the same time.
Change the movie to a holographic system. Our copper wire then would have half of the information required to build it, coming in in a highly compressed form. This seemingly very weak force of gravity is actually choreographing the impacts of electrons.
Ive said that an electron; any particle, including the neutron; hides half of its mass inside a negative refractive index core at the Shwartzchild radius. For a particle this is tiny, much smaller than h. So theres much more energy per unit volume in this core. The electromagnetic energy, which is a half of the total energy of the particle, has an angular momentum of h and a angular velocity of c. The core on the other hand is spinning with an angular velocity of the speed of gravity. Most of the cores energy s down to the fact that its spinning at such an incredible rate.
This has to mean that we have two sorts of space, electromagnetic and gravitational. If we could release all of the gravitational energy of a proton say, we would have over a hundred tonnes of electromagnetic mass energy equivalent. However that core in gravitational space only has the mass of the particle, a proton mass in this case. Thats because the core is in a different space.
So lets say that we could make a movie of one second showing how electrons move in a strip of copper. This movie needs to have 1E 45 frames to catch the collision rate.
We then need a second movie to show how the cores in gravitational space behave. A test frame would show them to be too far apart, it needs to be scaled by a huge amount. We would have to screen the electromagnetic movie, using light to project it, then wed need to screen the core movie with its projector thousands of light years away. This would then need faster than light photons to project it. The images would arrive at our screen at the same time.
Change the movie to a holographic system. Our copper wire then would have half of the information required to build it, coming in in a highly compressed form. This seemingly very weak force of gravity is actually choreographing the impacts of electrons.
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16 years 1 month ago #15464
by Jim
Replied by Jim on topic Reply from
Sloat, This is very confusing-you say the angular momentum is equal to "h" and angular velocity is "c". Assuming r=1(or if not how would "r" fit in?) would mass then be equal to h/c? So h=mc? Then the mass would be very odd wouldn't it? Would you please correct this to include "r" so angular momentum equals mvr? I assume you mean 10E45 not 1E45 or is this wrong?
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16 years 1 month ago #15465
by Jim
Replied by Jim on topic Reply from
Sloat, Maybe a better question would be: How long does it take the electron to complete an orbit around the nucleus? That would establish the radius of the orbit and then you need to explain how all the electrons orbit at the same radius. And then why all the electrons have different ion energy. It gets very messy don't you think?
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16 years 1 month ago #15466
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi Jim, we can talk about a unit radius but we would here be talking about an electron with a radius of one metre, which of course it doesnt have.
Taking h as the angular velocity. (Edited, oops that should be momentum rather than velocity) h = mvr for an electron 9.1093897E-31 * 2.99792458E 08 * r divide h by mv to get r. That gives us a radius of 2.42631060001E-12 metres which is the Compton wavelength. Hows about a proton? 1.32140999301E-15 metres. We are talking about conserved angular momentum and conserved angular velocity, the energy density changes. Change the energy density of the vacuum and it can crush an electron down to the radius of a proton, or perhaps an anti proton for now.
(On that point about notation. If you see the number 10^3 in an equation, you write it as 1E 3 in sci notation. It just makes life easier when using a calculator. That 1E 45 could be written as
10E 44 but that can cause confusion)
(Also, we would have hf / c^2 = m Which doesnt give us a very odd mass)
Now for where we do have a problem. A spinning electron contains a lot of energy and if its spinning round the nucleus then we have to add energy to it. All electrons do not have the same radius by the way. We can even have elliptical orbits which dont break quantum mechanical rules. Leave that for now though.
Whats supposed to happen is that the electron, as a point (ignore for the moment that this implies a zero radius, and therefor infinite mass) hurtles round the nucleus like an elastic thread round a golf ball. This is at the electrons frequency, so is is shifting some. Does this happen? I very much doubt it.
Atomic hydrogen might have the electron rotating about the nucleus but it sounds plain stupid to see it that way for neutral hydrogen. Two circular race tracks next to each other. Where they meet theres a little cafe. Every time you walk in two drivers, electrons, are sitting in there drinking tea. They claim that most of the time they are out there driving round their respective tracks. So you try to catch them out, vary your times. Still there, point an accusing finger, you two never go out onto the tracks. They smile condescendingly, we go round instantaneously, to allow us to spend all of our time in here. Sounds like a good job to me. Can I join their trade union.
Taking h as the angular velocity. (Edited, oops that should be momentum rather than velocity) h = mvr for an electron 9.1093897E-31 * 2.99792458E 08 * r divide h by mv to get r. That gives us a radius of 2.42631060001E-12 metres which is the Compton wavelength. Hows about a proton? 1.32140999301E-15 metres. We are talking about conserved angular momentum and conserved angular velocity, the energy density changes. Change the energy density of the vacuum and it can crush an electron down to the radius of a proton, or perhaps an anti proton for now.
(On that point about notation. If you see the number 10^3 in an equation, you write it as 1E 3 in sci notation. It just makes life easier when using a calculator. That 1E 45 could be written as
10E 44 but that can cause confusion)
(Also, we would have hf / c^2 = m Which doesnt give us a very odd mass)
Now for where we do have a problem. A spinning electron contains a lot of energy and if its spinning round the nucleus then we have to add energy to it. All electrons do not have the same radius by the way. We can even have elliptical orbits which dont break quantum mechanical rules. Leave that for now though.
Whats supposed to happen is that the electron, as a point (ignore for the moment that this implies a zero radius, and therefor infinite mass) hurtles round the nucleus like an elastic thread round a golf ball. This is at the electrons frequency, so is is shifting some. Does this happen? I very much doubt it.
Atomic hydrogen might have the electron rotating about the nucleus but it sounds plain stupid to see it that way for neutral hydrogen. Two circular race tracks next to each other. Where they meet theres a little cafe. Every time you walk in two drivers, electrons, are sitting in there drinking tea. They claim that most of the time they are out there driving round their respective tracks. So you try to catch them out, vary your times. Still there, point an accusing finger, you two never go out onto the tracks. They smile condescendingly, we go round instantaneously, to allow us to spend all of our time in here. Sounds like a good job to me. Can I join their trade union.
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