Replacing the current by the fields in electromagnetism.

by kalish
Tags: current, electromagnetism, fields, replacing
kalish is offline
Jan20-14, 10:22 AM
P: 28
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
$$ \frac{d\vec P}{dt} + q\frac{\partial\vec A}{\partial t} + q \vec V.\vec\nabla \vec A= -q\vec\nabla \phi + q\vec\nabla (\vec A.\vec V) $$

$$ \frac{d\vec P}{dt} =- q\frac{\partial\vec A}{\partial t} -q\vec\nabla \phi + q\vec\nabla \times (\vec \nabla \times\vec A) $$

$$ \vec\nabla (\vec A.\vec V)- \vec V.\vec\nabla \vec A= \vec\nabla \times (\vec \nabla \times\vec A) $$
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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