Lorentz force law from potentials ?

[Widdekind]

is the following derivation correct (ignoring for simplicity's sake a few factors of c):

 

for a single charged particle, instantaneously located at [math]\vec{r}[/math], in the Classical limit, the fields from said particle, at point [math]\vec{r}'[/math]:

 

[math]\Phi(r') \propto \frac{q}{r}[/math]

[math]r = \sqrt{ \left( \vec{r}' - \vec{r} \right) \circ \left( \vec{r}' - \vec{r} \right) }[/math]

 

[math]\vec{A}(r') = \vec{v} \Phi(r')[/math]

 

the potential generate the force fields:

 

 

[math]\vec{E} = -\nabla \Phi - \frac{\partial \vec{A}}{\partial t} = -\nabla \Phi - \vec{a} \Phi - \vec{v} \frac{\partial \Phi}{\partial r} \frac{\partial r}{\partial t}[/math]

[math]\frac{\partial r}{\partial t} = \frac{1}{2 r} \left( 2 \left( x' - x \right) \left( - \frac{\partial x}{\partial t} \right) + \cdots \right) = - \hat{r} \circ \vec{v}[/math]

where [math]\hat{r}[/math] points away from the force-field generating particle, towards the other point. But [math]\hat{r} \frac{\partial \Phi}{\partial r} = \nabla \Phi[/math]. so

[math]\vec{E}=-\nabla\Phi-\vec{a}\Phi+\vec{v}\left(\vec{v}\circ\nabla\Phi\right)[/math]

[math]\vec{B} = \nabla \times \vec{A} = - \vec{v} \times \nabla \Phi[/math]

[math]\vec{v}' \times \vec{B} = \nabla \Phi \left( \vec{v}' \circ \vec{v} \right) - \vec{v} \left( \vec{v}' \circ \nabla \Phi \right)[/math]

the force felt by a test charge [math]q'[/math] at [math]\vec{r}'[/math]:

 

[math]\vec{F}' = q' \left( \vec{E} + \vec{v}' \times \vec{B}\right)[/math]

 

[math]\frac{\vec{F}'}{q'} = -\nabla \Phi \left( 1 - \left( \vec{v}' \circ \vec{v} \right) \right) - \vec{a} \Phi - \vec{v} \left( \left( \vec{v}' - \vec{v} \right) \circ \nabla \Phi \right)[/math]

 

 

 

i think the above is equivalent to

 

http://en.wikipedia.org/wiki/Lorentz_force#Lorentz_force_in_terms_of_potentials