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angular momentum of bosons
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angular momentum of bosons - 03-13-2007, 12:37 PM

For Bose liquids although their linear momentum could have the value of zero, their angular momentum could not unless the spins of half of the fluid particles are equal and opposite of the other half. Furthermore, at the quantum level angular momentum takes on only two values +nħ or –nħ where n is integer for bosons and half integer for fermions.

Deuteron is a composite of one proton and one neutron. Both are fermions and each with spin angular momentum of +ħ/2 giving a total angular momentum of +1ħ, which is now a boson’s spin. Likewise, for Cooper pair of two electrons, their individual spin is +ħ/2. Two electrons have a total spin of +1ħ, a boson.


Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
  
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Smile Re: angular momentum of bosons - 03-13-2007, 01:18 PM

Quote:
Originally Posted by AntonioLao View Post
For Bose liquids although their linear momentum could have the value of zero, their angular momentum could not unless the spins of half of the fluid particles are equal and opposite of the other half. Furthermore, at the quantum level angular momentum takes on only two values +nħ or –nħ where n is integer for bosons and half integer for fermions.

Deuteron is a composite of one proton and one neutron. Both are fermions and each with spin angular momentum of +ħ/2 giving a total angular momentum of +1ħ, which is now a boson’s spin. Likewise, for Cooper pair of two electrons, their individual spin is +ħ/2. Two electrons have a total spin of +1ħ, a boson.
Containment of such a heady mixture may prove difficult.



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Re: angular momentum of bosons - 03-13-2007, 02:37 PM

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Containment of such a heady mixture may prove difficult.
The stars have been doing it for billions of years. Exactly how remains a mystery although the nuclear reactions are known completely as described by Hans Bethe http://en.wikipedia.org/wiki/Hans_Bethe.


Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
  
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Smile Re: angular momentum of bosons - 03-13-2007, 05:27 PM

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The stars have been doing it for billions of years. Exactly how remains a mystery although the nuclear reactions are known completely as described by Hans Bethe http://en.wikipedia.org/wiki/Hans_Bethe.

The trick is to produce a natural phenomena within a un-natural setting,then convert this back a natural phenomena?


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Re: angular momentum of bosons - 03-14-2007, 12:12 PM

Quote:
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The trick is to produce a natural phenomena within a un-natural setting
What is natural for us might not be natural for an electron, a planet, or a star. Astronomers or astrophysicists are constantly searching for the birthplaces of stars. Naturally, they believe these could not be found in blackness of "empty space" yet to believe in the reality of true vacuum would suggest that stars are something that came from nothing.


Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
  
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Smile Re: angular momentum of bosons - 03-14-2007, 03:28 PM

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What is natural for us might not be natural for an electron, a planet, or a star. Astronomers or astrophysicists are constantly searching for the birthplaces of stars. Naturally, they believe these could not be found in blackness of "empty space" yet to believe in the reality of true vacuum would suggest that stars are something that came from nothing.
What comes from nothing is zero with a huge whole in it?Strange attractors are lonely
particles looking for company!As they team up the party begins,and the temperature
starts to rise.



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Re: angular momentum of bosons - 03-14-2007, 03:37 PM

Quote:
Originally Posted by mkirkpatrick
What comes from nothing is zero with a huge whole in it?
This has a ring to it. And it does. The symbol for zero is really a ring. Take out the circumscribe perimetric circumference and what remains is the nothing that it really is.


Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
  
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Smile Re: angular momentum of bosons - 03-14-2007, 08:50 PM

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This has a ring to it. And it does. The symbol for zero is really a ring. Take out the circumscribe perimetric circumference and what remains is the nothing that it really is.
From within that very circle a vortex is formed,and a transfer occurs.



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Re: angular momentum of bosons - 03-15-2007, 03:37 PM

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Originally Posted by mkirkpatrick
From within that very circle a vortex is formed,and a transfer occurs
The single vortex is unstable. However, two vortices linked together will last forever since they feed on each other essence, a constant give and take vice versa.


Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
  
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Smile Re: angular momentum of bosons - 03-15-2007, 03:46 PM

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The single vortex is unstable. However, two vortices linked together will last forever since they feed on each other essence, a constant give and take vice versa.
Where there is one vortex,another is not far off.

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