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

$$\int \mathbf{J} d^3 x = - \int \mathbf{x} ( \boldsymbol{\nabla} \cdot \mathbf{J} ) d^3 x$$

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

$$\boldsymbol{\nabla} (\mathbf{x} \cdot \mathbf{J})$$

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

$$\boldsymbol{\nabla} (\mathbf{x} \cdot \mathbf{J}) = \mathbf{x} \cdot \boldsymbol{\nabla} \mathbf{J} +\mathbf{J} \cdot \boldsymbol{\nabla} \mathbf{x} + \mathbf{x} \times ( \boldsymbol{\nabla} \times \mathbf{J} ) = \mathbf{J} + \sum_a x_a \boldsymbol{\nabla} J_a. $$

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

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  • $\begingroup$ As far as I remember, Jackson has a long list of vector identities close to the back and front cover of his bock. $\endgroup$ – Fabian Feb 6 '15 at 4:34
  • $\begingroup$ I'm not sure your identity is correct. Maybe you did a mistake copying it? $\endgroup$ – Fabian Feb 6 '15 at 4:42
  • $\begingroup$ Peter where's your notes for multiphysics system on peeterjoot.wordpress.com? thanks $\endgroup$ – Olórin Feb 6 '15 at 5:22
  • $\begingroup$ @Math Newb, see: peeterjoot.com/writing . $\endgroup$ – Peeter Joot Feb 6 '15 at 13:31
  • $\begingroup$ @Fabian, there's a short list in my second edition, and I didn't see anything that looked appropriate. $\endgroup$ – Peeter Joot Feb 6 '15 at 13:32
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I found an answer in Griffiths' Introduction to electrodynamics (problem 5.7) which poses a problem of relating the volume integral of $\mathbf{J}$ to the dipole moment, and hints that this can be done by expanding

$$\int \boldsymbol{\nabla} \cdot ( x \mathbf{J} ) d^3 x.$$

That expansion is

$$\int \boldsymbol{\nabla} \cdot ( x \mathbf{J} ) d^3 x= \int (\boldsymbol{\nabla} x \cdot \mathbf{J}) d^3 x+ \int x (\boldsymbol{\nabla} \cdot \mathbf{J}) d^3 x.$$

Doing the same for the other coordinates and summing gives

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

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

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

and then argue that this is zero for localized-enough currents by taking this surface to infinity, where $\mathbf{J}$ is zero. The end result is the relation that Jackson calls "integration by parts".

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