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12-29-2005, 01:59 PM
In light of natural asymmetry, does spherical symmetry really exist for any given vector field? This symmetry seems exclusively for all scalar fields since they don’t have to define direction. However, if the metric property is removable from both groups of vector and scalar field then they all become equivalent. Examples of vector field are EM field, gravity field, strong-weak nuclear field, electroweak field and for scalar field there are Higgs field, true-false vacuum field, thermal field, entropy field, density field, volume field, mass field, charge field, and time field. To say that they are equivalent without the spacetime metric property would be a gross understatement.
Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
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04-06-2007, 08:01 PM
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Originally Posted by AntonioLao  In light of natural asymmetry, does spherical symmetry really exist for any given vector field? This symmetry seems exclusively for all scalar fields since they don’t have to define direction. However, if the metric property is removable from both groups of vector and scalar field then they all become equivalent. Examples of vector field are EM field, gravity field, strong-weak nuclear field, electroweak field and for scalar field there are Higgs field, true-false vacuum field, thermal field, entropy field, density field, volume field, mass field, charge field, and time field. To say that they are equivalent without the spacetime metric property would be a gross understatement. | Yes it exists,in fact there is no other?All energy emerges from one source and returns
to said source in perfect spherical symmetry.
regards michael.
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04-07-2007, 12:52 PM
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Originally Posted by mkirkpatrick Yes it exists,in fact there is no other | It exists for scalar fields: mass, energy, density, temperature, volume. But not for vector fields: acceleration, velocity, momentum, force, time, position, spin, helicity, and all other vectors.
Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
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04-07-2007, 01:53 PM
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Originally Posted by AntonioLao It exists for scalar fields: mass, energy, density, temperature, volume. But not for vector fields: acceleration, velocity, momentum, force, time, position, spin, helicity, and all other vectors. | If not on a returning cycle,what happens to the released energy,is it transported to another phase expression?
regards michael.
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04-10-2007, 03:44 PM
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what happens to the released energy,is it transported to another phase expression?
| If these energies are backward then they could not possibly become useful for us living in a forward time universe. Another name for backward energy is negative energy and positive energy for forward time energy.
Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
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04-10-2007, 08:00 PM
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Originally Posted by AntonioLao If these energies are backward then they could not possibly become useful for us living in a forward time universe. Another name for backward energy is negative energy and positive energy for forward time energy. | Thats another way of expressing Ying and Yang! one spiralling foward,the other backward.
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04-12-2007, 03:34 PM
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Originally Posted by mkirkpatrick Thats another way of expressing Ying and Yang! one spiralling foward,the other backward. | I happened to look up some textbooks on quantum mechanics about spherical symmetry and what I found is that this symmetry is what is needed in order to describe the first 3 quantum numbers: azimuthal, radial, and magnetic.
Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
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04-12-2007, 06:50 PM
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Originally Posted by AntonioLao I happened to look up some textbooks on quantum mechanics about spherical symmetry and what I found is that this symmetry is what is needed in order to describe the first 3 quantum numbers: azimuthal, radial, and magnetic. | The order you placed these numbers spell A.R.M.which at the end of each is a hand?
Whose hand therefore is reflected within the symmetry of motion?
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04-13-2007, 01:22 PM
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Originally Posted by mkirkpatrick Whose hand therefore is reflected within the symmetry of motion | Unfortunately we live in a left-handed universe. All the right-handed particles cannot be found even by creative searching methods together with bulk antimatter.
Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
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04-13-2007, 07:09 PM
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Originally Posted by AntonioLao Unfortunately we live in a left-handed universe. All the right-handed particles cannot be found even by creative searching methods together with bulk antimatter. | That is because we are limited as yet in consciousness that trancends the relative universe,and fail to see the overall "big-picture" where all these "apparent" anomalies
are reconciled.
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