How Do Euler-Lagrange Equations Apply in Electromagnetic Theory?

[dingo_d]

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?

 

[fzero]

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.

 

[arkajad]

In

<br /> \frac{q}{c}\sum_j\frac{\partial A_i}{\partial x_j}\dot{x}_j<br />

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

 

[dingo_d]

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

Thanks ^^