- #1
Meselwulf
- 126
- 0
I'm definitely messing up somewhere :P
The Larmor Equation is
[tex]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)[/tex]
I derived an equation stated as
[tex]\nabla \times \vec{F}_{ij} \cdot \hat{n} = \nabla \times \frac{\partial V(r_{ij})}{\partial r_{ij}}[/tex]
How is put in a simpler way from the original derivation, is imagine we have what we began with
[tex]\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n}[/tex]
Take the dot product of the unit vector on both sides gives
[tex]\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/tex]
Now taking the curl of F is
[tex]\nabla \times \vec{F}_{ij} \cdot \hat{n} = \nabla \times \frac{\partial V(r_{ij})}{\partial r_{ij}}[/tex]
In light of this equation however
[tex]\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/tex]
I could rewrite the Larmor energy as
[tex]\Delta H = \frac{2\mu}{\hbar Mc^2 e}(\vec{F}_{ij} \cdot \hat{n}) L \cdot S[/tex]
But I am sure many agree that is not very interesting.
(and this is really starting to make my brain cells burst), taking the curl of a force gives a F/length, however, the unit vector in this equation
[tex]\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/tex]
cancels these out and what I have again is the force again... yes? ... very circular... since this would be true, then we know what the force is anyway given earlier:
[tex]\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n} [/tex]
Now, just a moment ago I found the Larmor energy written as
[tex]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{1}{r}\frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)[/tex]
Notice the 1/r term which is not in my original case. If that where true, and we plug in my force example again
[tex]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{1}{r}\frac{\partial V(r_{ij})}{\partial r_{ij}}\hat{n} ( L \cdot S)[/tex]
(If I am doing this right) the unit vector would cancel out with the radius term and what would be left with is
[tex]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)[/tex]
Now if that wasn't confusing I don't know what is.
I've either had it right from the beginning, or I have made a tiny mistake which is making a huge impact on my understanding of my own equation. Any help would be gladly appreciated.
The Larmor Equation is
[tex]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)[/tex]
I derived an equation stated as
[tex]\nabla \times \vec{F}_{ij} \cdot \hat{n} = \nabla \times \frac{\partial V(r_{ij})}{\partial r_{ij}}[/tex]
How is put in a simpler way from the original derivation, is imagine we have what we began with
[tex]\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n}[/tex]
Take the dot product of the unit vector on both sides gives
[tex]\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/tex]
Now taking the curl of F is
[tex]\nabla \times \vec{F}_{ij} \cdot \hat{n} = \nabla \times \frac{\partial V(r_{ij})}{\partial r_{ij}}[/tex]
In light of this equation however
[tex]\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/tex]
I could rewrite the Larmor energy as
[tex]\Delta H = \frac{2\mu}{\hbar Mc^2 e}(\vec{F}_{ij} \cdot \hat{n}) L \cdot S[/tex]
But I am sure many agree that is not very interesting.
(and this is really starting to make my brain cells burst), taking the curl of a force gives a F/length, however, the unit vector in this equation
[tex]\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/tex]
cancels these out and what I have again is the force again... yes? ... very circular... since this would be true, then we know what the force is anyway given earlier:
[tex]\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n} [/tex]
Now, just a moment ago I found the Larmor energy written as
[tex]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{1}{r}\frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)[/tex]
Notice the 1/r term which is not in my original case. If that where true, and we plug in my force example again
[tex]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{1}{r}\frac{\partial V(r_{ij})}{\partial r_{ij}}\hat{n} ( L \cdot S)[/tex]
(If I am doing this right) the unit vector would cancel out with the radius term and what would be left with is
[tex]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)[/tex]
Now if that wasn't confusing I don't know what is.
I've either had it right from the beginning, or I have made a tiny mistake which is making a huge impact on my understanding of my own equation. Any help would be gladly appreciated.