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## Homework Statement

When writing down the Lagrangian and the writing down Euler-Lagrange equation I'm having some difficulties with reasoning something.

## Homework Equations

Lagrangian is:

[tex]\mathcal{L}=\frac{1}{2}mv^2-q\phi+\frac{q}{c}\vec{v}\cdot\vec{A}.[/tex]

Euler-Lagrange eq:

[tex]\frac{\partial \mathcal{L}}{\partial x_i}=-q\frac{\partial \phi}{\partial x_i}+\frac{q}{c}\frac{\partial}{\partial x_i}(\vec{v}\cdot\vec{A})[/tex]

[tex]\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial \dot{x}_i}\right)=m\ddot{x}_i+\frac{q}{c}\frac{\partial}{\partial t}A_i+\frac{q}{c}\sum_j\frac{\partial A_i}{\partial x_j}\dot{x}_j[/tex]

Now my teaching assistant wrote that back in form of vectors, rather then component wise, and there was my puzzlement (or huh? moment):

[tex]\frac{d}{dt}(m\vec{v})+\frac{q}{c}\frac{\partial}{\partial t}\vec{A}+\frac{q}{c}(\vec{v}\cdot\vec{\nabla})\cdot\vec{A}+q\vec{\nabla}\phi-\frac{q}{c}\vec{\nabla}(\vec{v}\cdot\vec{A})=0[/tex]

How is this:

[tex]\frac{q}{c}\sum_j\frac{\partial A_i}{\partial x_j}\dot{x}_j[/tex]

equal to this:

[tex]\frac{q}{c}(\vec{v}\cdot\vec{\nabla})\cdot\vec{A}[/tex]

??

Doesn't the derivative in the sum acts on A? And then the whole thing is multiplied with v?

Shouldn't it be:

[tex]\frac{q}{c}(\vec{\nabla}\cdot\vec{A})\cdot\vec{v}[/tex]?

Because it is not the same if nabla acts on A and v acts on nabla... Or is it? :\

What am I missing?