Integration by parts, changing vector to moment & divergence

  • #1
305
2

Main Question or Discussion Point

In Jackson's 'classical electrodynamics' he re-expresses a volume integral of a vector in terms of a moment like divergence:

[tex]\begin{align}\int \mathbf{J} d^3 x = - \int \mathbf{x} ( \boldsymbol{\nabla} \cdot \mathbf{J} ) d^3 x\end{align} [/tex]

He calls this change "integration by parts". If this is integration by parts, there must be some form of chain rule (where one of the terms is zero on the boundry), but I can't figure out what that chain rule would be. I initially thought that the expansion of

[tex]\begin{align}\boldsymbol{\nabla} (\mathbf{x} \cdot \mathbf{J})\end{align} [/tex]

might have the structure I was looking for (i.e. something like [itex]\mathbf{x} \boldsymbol{\nabla} \cdot \mathbf{J}+\mathbf{J} \boldsymbol{\nabla} \cdot \mathbf{x}[/itex]), however

[tex]\begin{align}\boldsymbol{\nabla} (\mathbf{x} \cdot \mathbf{J}) =\mathbf{x} \cdot \boldsymbol{\nabla} \mathbf{J}+\mathbf{J} \cdot \boldsymbol{\nabla} \mathbf{x}+ \mathbf{x} \times ( \boldsymbol{\nabla} \times \mathbf{J} )= \mathbf{J} + \sum_a x_a \boldsymbol{\nabla} J_a.\end{align} [/tex]

I tried a few other gradients of various vector products (including [itex]\boldsymbol{\nabla} \times ( \mathbf{x} \times \mathbf{J} )[/itex]), but wasn't able to figure out one that justifies what the author did with this integral.

I am probably missing something obvious (or at least something that is obvious to Jackson) ?
 

Answers and Replies

  • #2
RUber
Homework Helper
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The divergence is the derivative with respect to the outward normal.
Make the standard substitution for integration by parts.
##u = J \quad v = x ##
##du = \nabla \cdot J \quad dv = 1 ##
What you have is ##\int u dv ## clearly this is ## Jx|_{boundary} - \int vdu ##.
You are correct about needing the boundary condition -- this is very common.
 
  • #3
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I'm not certain how to interpret the reply of @RUber, but I resolved this after finding a hint in Griffiths, which poses a problem of relating the volume integral of [itex]\mathbf{J}[/itex] to the dipole moment using by expanding

[tex]\int \boldsymbol{\nabla} \cdot ( x \mathbf{J} ) d^3 x.[/tex]

That expansion is

[tex]\int \boldsymbol{\nabla} \cdot ( x \mathbf{J} ) d^3 x= \int (\boldsymbol{\nabla} x \cdot \mathbf{J}) d^3 x+ \int x (\boldsymbol{\nabla} \cdot \mathbf{J}) d^3 x.[/tex]

Doing the same for the other coordinates and summing gives

[tex]\sum_{i = 1}^3 \mathbf{e}_i \int \boldsymbol{\nabla} \cdot ( x_i \mathbf{J} ) d^3 x=\int \mathbf{J} d^3 x+ \int \mathbf{x} (\boldsymbol{\nabla} \cdot \mathbf{J}) d^3 x.[/tex]

I think the boundary condition argument would be to transform the left hand side using the divergence theorem

[tex]\int \mathbf{J} d^3 x+ \int \mathbf{x} (\boldsymbol{\nabla} \cdot \mathbf{J}) d^3 x=\sum_{i = 1}^3 \mathbf{e}_i \int_{S} ( x_i \mathbf{J} ) \cdot \hat{\mathbf{n}} dS[/tex]

and then argue that this is zero for localized-enough currents by taking this surface to infinity, where [itex]\mathbf{J}[/itex] is zero.
 

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