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Electromagnetic Lagrangian

  1. Sep 28, 2010 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    Lagrangian is:


    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:



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

    Shouldn't it be:


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

    What am I missing?
  2. jcsd
  3. Sep 28, 2010 #2


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    You're just caught in a confusion over notation. Rearrange

    [tex]\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}.[/tex]

    Note that [tex](\vec{v}\cdot \nabla) [/tex] 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 [tex]\vec{v}[/tex].

    Now consider

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

    There's a few problems with this formula. First of all, [tex]\nabla \cdot \vec{A}[/tex] would be a scalar, so the second dot product is incorrect and confusing. Second, if we dot [tex]\nabla[/tex] and [tex]\vec{A}[/tex] we get the divergence of [tex]\vec{A}[/tex] which involves contracting the vector index on [tex]A_i[/tex] which is contrary to the term you derived above.
  4. Sep 28, 2010 #3

    \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.
  5. Sep 28, 2010 #4
    Oh! I see now!! The second dot product was probably my error in writing :\

    Thanks ^^
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