Replacing the current by the fields in electromagnetism.by kalish Tags: current, electromagnetism, fields, replacing 

#1
Jan2014, 10:22 AM

P: 28

hello,
In electromagnetism from the Lagrangian formulation, one can defines a canonical momentum and applying Euler's equation to that momentum gives the lorentz law. To get the Lorentz law, one uses a vector identity, only available because it is said the speed V does not 'feel' the differential operator, for example, that equation $$ \frac{d\vec P}{dt} + q\frac{\partial\vec A}{\partial t} + q \vec V.\vec\nabla \vec A= q\vec\nabla \phi + q\vec\nabla (\vec A.\vec V) $$ becomes $$ \frac{d\vec P}{dt} = q\frac{\partial\vec A}{\partial t} q\vec\nabla \phi + q\vec\nabla \times (\vec \nabla \times\vec A) $$ because $$ \vec\nabla (\vec A.\vec V) \vec V.\vec\nabla \vec A= \vec\nabla \times (\vec \nabla \times\vec A) $$ in that case. The problem is, if I replace $$ q\vec V = \vec \nabla \times \vec B \frac{\partial \vec E}{\partial t}$$ from the Maxwell's equation (sorry for the permittivity and permeability let's say = 1) I can perfectly apply a differential operator on that object. So why we should not apply a differential operator on $$ q\vec V$$ ? 


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