linguistic:paraphrasing, structural:scholar_edition, information:extraction
Lorentz force in terms of potentials

The E and B fields can be replaced by the magnetic vector potential A and (scalar) electrostatic potential ϕ by E = − ∇ ϕ − ∂ A ∂ t B = ∇ × A {\displaystyle {\begin{aligned}\mathbf {E} &=-\nabla \phi -{\frac {\partial \mathbf {A} }{\partial t}}\\\mathbf {B} &=\nabla \times \mathbf {A} \end{aligned}}} where ∇ is the gradient, ∇⋅ is the divergence, and ∇× is the curl.

The force becomes F = q. {\displaystyle \mathbf {F} =q\left.}

Using an identity for the triple product this can be rewritten as, F = q, {\displaystyle \mathbf {F} =q\left,}

(Notice that the coordinates and the velocity components should be treated as independent variables, so the del operator acts only on A {\displaystyle \mathbf {A} }, not on v {\displaystyle \mathbf {v} }; thus, there is no need of using Feynman's subscript notation in the equation above). Using the chain rule, the total derivative of A {\displaystyle \mathbf {A} } is: d A d t = ∂ A ∂ t + (v ⋅ ∇) A {\displaystyle {\frac {\mathrm {d} \mathbf {A} }{\mathrm {d} t}}={\frac {\partial \mathbf {A} }{\partial t}}+(\mathbf {v} \cdot \nabla)\mathbf {A} } so that the above expression becomes: F = q. {\displaystyle \mathbf {F} =q\left.}

With v = ẋ, we can put the equation into the convenient Euler–Lagrange form

F = q {\displaystyle \mathbf {F} =q\left}

where ∇ x = x ^ ∂ ∂ x + y ^ ∂ ∂ y + z ^ ∂ ∂ z {\displaystyle \nabla _{\mathbf {x} }={\hat {x}}{\dfrac {\partial }{\partial x}}+{\hat {y}}{\dfrac {\partial }{\partial y}}+{\hat {z}}{\dfrac {\partial }{\partial z}}} and ∇ x ˙ = x ^ ∂ ∂ x ˙ + y ^ ∂ ∂ y ˙ + z ^ ∂ ∂ z ˙. {\displaystyle \nabla _{\dot {\mathbf {x} }}={\hat {x}}{\dfrac {\partial }{\partial {\dot {x}}}}+{\hat {y}}{\dfrac {\partial }{\partial {\dot {y}}}}+{\hat {z}}{\dfrac {\partial }{\partial {\dot {z}}}}.}