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Electromagnetic Lagrangian

  • Thread starter dingo_d
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  • #1
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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?
 

Answers and Replies

  • #2
fzero
Science Advisor
Homework Helper
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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.
 
  • #3
1,444
4
In

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

pay attention to the summation index (it is j). It indicates the scalar product.
 
  • #4
211
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Oh! I see now!! The second dot product was probably my error in writing :\

Thanks ^^
 

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