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I'm reading a physics paper and am trying to figure out how a certain expression is derived (If interested, see Appendix of the paper, Eq. (A7), (A8)). The authors skip a lot of derivation steps and at a certain point an expression like the following is encountered. Suppose we have the following:

$$\iint_{S} \left[\nabla^{t}A \times \boldsymbol{B}^{t}\right]^{z} dxdy,$$

where $A$ is a scalar field, $\boldsymbol{B}$ is a vector field, $S$ lies completely in the xy-plane. The superscripts $t$ and $z$ denote the transverse (x, y) and longitudinal (z) components, i.e.:

$$\nabla^{t}M = \left(\frac{\partial M}{\partial x}, \frac{\partial M}{\partial y},0\right)$$

is the transverse part of the gradient, and:

$$\boldsymbol{N}^{t} = \left(N_{x}, N_{y}, 0\right);$$ $$\boldsymbol{N}^{z} = N_{z}.$$

At this point they state that they use "an integration by parts" and they somehow manage to move the nabla so that it is cross-product multiplied with $\boldsymbol{B}$, i.e. the final expression should contain something like this:

$$A(\nabla \times \boldsymbol{B})$$

together with other terms and the correct superscripts, which I have omitted here because I don't know them.

Does anyone know some integration by parts formula that can help me get from the original expression to something looking like this?

Any help and advice is much appreciated. Also, let me know if you would need more details. I've tried to keep things as abstract as I can.

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  • $\begingroup$ By integration by parts they mean Stokes' theorem. Try it for yourself: let $dv=\nabla^tA$, $u=\mathbf{B}$ and just do integration by parts like normal, except the boundary term is a $1$D line integral instead of a $0$D boundary evaluation. It's not that complicated when you practice it a few times. $\endgroup$
    – Ninad Munshi
    Commented May 28 at 9:52
  • $\begingroup$ Also (+1) for an excellent job of translating the physics for a math audience $\endgroup$
    – Ninad Munshi
    Commented May 28 at 9:59
  • $\begingroup$ @NinadMunshi could you please provide more details? I want to be technically correct here with my notation. How do I incorporate taking the z-component in the integration by parts procedure? Also, if u=$\boldsymbol{B}$ then what would du be? PS. Thank you for the +1 and your comment! $\endgroup$
    – RawPasta
    Commented May 28 at 10:01
  • $\begingroup$ In this case you don't need to do anything special for the $z$ component since $S$ is planar. The normal vector was $(0,0,1)$ so you can just use Stokes' theorem as usual. Because you're using Stokes' theorem the derivative would be a curl. Integration by parts is just a way to convert an integral of one part of a product rule to the other part. When I take $\nabla\times(A\mathbf{B})$, what are my two product rule terms? $\endgroup$
    – Ninad Munshi
    Commented May 28 at 10:08
  • $\begingroup$ Please tell me if this is correct: I have $\nabla A \times \boldsymbol{B} = \nabla \times A\boldsymbol{B} - A(\nabla \times \boldsymbol{B})$. Substituting this in the original expression I get: $\iint_{S}[(\nabla^{t}A \times \boldsymbol{B}^{t})]^{z}dS$ = $\iint_{S}(\nabla^{t}A \times \boldsymbol{B}^{t}).d\boldsymbol{S}$ = $\iint_{S}(\nabla^{t} \times (A\boldsymbol{B}^{t})).d\boldsymbol{S}$ - $\iint_{S}A(\nabla^{t} \times \boldsymbol{B}^{t}).d\boldsymbol{S}$ = $\oint_{\partial S} A\boldsymbol{B}^{t}.d\boldsymbol{l}$ - $\iint_{S}A(\nabla^{t} \times \boldsymbol{B}^{t}).d\boldsymbol{S}$. $\endgroup$
    – RawPasta
    Commented May 28 at 10:29

2 Answers 2

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By integration by parts they mean Stokes' theorem. Let $u=\mathbf{B}$ and $dv=\nabla A$ and we get

$$\iint_S (\nabla A \times \mathbf{B})\cdot \mathbf{dS} = \oint_{\partial S}(A\mathbf{B})\cdot \mathbf{dr} - \iint_S A(\nabla\times\mathbf{B})\cdot\mathbf{dS}$$

The transverse and $z$ component notation is redundant since with ($xy$) planar $S$, only the $z$ component of the surface integral term contributes which only comes from the transverse components of the vector fields anyway. Depending on the physics of your situation you may even be able to say that the line integral vanishes.

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  • $\begingroup$ Hey, yes! I literally just commented this as an answer to your comment. I didn't see your answer here until I refreshed the page. Thank you very much Ninad! $\endgroup$
    – RawPasta
    Commented May 28 at 10:31
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The mathematical version is: replace $\nabla A$ by dA and cross products by wedge products.

There is no need to introduce a third dimension in 2d-integrals. Its 19th century physicists workaround of exterior calculus using euclidean formulas from old times of quaternionic vector calculus translated to classical component vector analysis.

$$\int_S \mathbf d A \wedge \sum_k B_k \ \mathbf d x_k =\int_{\partial_S}A \sum_k B_k \ \mathbf d x_k - \int_S A \sum_{k\ l} \partial_{x_l} B_k \ \ \mathbf d x_l \wedge \mathbf d x_k \ $$

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  • $\begingroup$ Hey Roland, thank you for your answer. I've marked Ninad's one as correct since it keeps it consistent with the notation used in the paper. But nevertheless, much appreciated! $\endgroup$
    – RawPasta
    Commented May 28 at 10:35
  • $\begingroup$ @RawPasta: not "correct one"... "accepted one". :-) $\endgroup$
    – André Caldas
    Commented May 28 at 11:25
  • $\begingroup$ My bad, accepted one. Correct! :) $\endgroup$
    – RawPasta
    Commented May 28 at 11:42

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