# Electromagnetic Lagrangian

## 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:

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

Euler-Lagrange eq:

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

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

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

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

How is this:

$$\frac{q}{c}\sum_j\frac{\partial A_i}{\partial x_j}\dot{x}_j$$

equal to this:

$$\frac{q}{c}(\vec{v}\cdot\vec{\nabla})\cdot\vec{A}$$

??

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

Shouldn't it be:

$$\frac{q}{c}(\vec{\nabla}\cdot\vec{A})\cdot\vec{v}$$?

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

What am I missing?

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fzero
Homework Helper
Gold Member
You're just caught in a confusion over notation. Rearrange

$$\sum_j \frac{\partial A_i}{\partial x_j} \dot{x}_j} = \sum_j v_j \frac{\partial A_i}{\partial x_j} = \left( \sum_j v_j \frac{\partial}{\partial x_j}\right) A_i \rightarrow (\vec{v}\cdot \nabla) \vec{A}.$$

Note that $$(\vec{v}\cdot \nabla)$$ is already a scalar operator so there's no 2nd dot product. The derivative in this operator acts on everything to the right, but not on $$\vec{v}$$.

Now consider

$$(\nabla \cdot \vec{A})\cdot \vec{v}$$

There's a few problems with this formula. First of all, $$\nabla \cdot \vec{A}$$ would be a scalar, so the second dot product is incorrect and confusing. Second, if we dot $$\nabla$$ and $$\vec{A}$$ we get the divergence of $$\vec{A}$$ which involves contracting the vector index on $$A_i$$ which is contrary to the term you derived above.

In

$$\frac{q}{c}\sum_j\frac{\partial A_i}{\partial x_j}\dot{x}_j$$

pay attention to the summation index (it is j). It indicates the scalar product.

Oh! I see now!! The second dot product was probably my error in writing :\

Thanks ^^