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I have the following integral: $$\int_V ((\vec\nabla.\vec A)(\vec\nabla.\vec B)+\vec B.\vec\nabla(\vec\nabla.\vec A))dV$$

Using Gauss' theorem I can convert this into a surface integral. However, I need to rewrite the integrandum into something of the form $\vec\nabla.\vec V$ in order to apply Gauss.

In short:

$$(\vec\nabla.\vec A)(\vec\nabla.\vec B)+\vec B.\vec\nabla(\vec\nabla.\vec A)=\vec\nabla.\vec V$$

Where I need to find $\vec V$ so that the identity is correct.

I've tried using index notation but since the result is a scalar that doesn't make much sense so I was a bit hesitant to try it. I decided to give it a shot and got the following:
$$(\partial_iA_i)(\partial_jB_j)+B_k\partial_l\partial_lA_k$$
Which leads me nowhere.

I've also tried putting $(\vec\nabla.\vec A)$ in front (applying distributivity) but that's not correct because, while $(\vec\nabla.\vec A)$ does appear in the second term, there is still a $\vec\nabla$ that works in on it.

I honestly don't know what else I can try.

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2 Answers 2

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Write everything in index notation. Your integrand is $$ \frac{\partial A_i}{\partial x_i}\frac{\partial B_j}{\partial x_j} + B_j \hat{e}_j \cdot \left ( \hat{e}_m \frac{\partial}{\partial x_m} \left ( \frac{\partial A_p}{\partial x_p} \right)\right) $$ which reduces, on simplification, to $$ \frac{\partial A_i}{\partial x_i}\frac{\partial B_j}{\partial x_j} + B_j \frac{\partial^2 A_p}{\partial x_j \partial x_p} $$

Now consider $$ \frac{\partial}{\partial x_p} \left( B_p \frac{\partial A_j}{\partial x_j} \right) \\ = \frac{\partial B_p}{\partial x_p}\frac{\partial A_j}{\partial x_j} + B_p \frac{\partial^2 A_j}{\partial x_p \partial x_j} \\ = \frac{\partial A_i}{\partial x_i} \frac{\partial B_j}{\partial x_j} + B_j \frac{\partial^2 A_p}{\partial x_j \partial x_p} $$ where we have interchanged the dummy indices j and p in the last step, and renamed the dummy indices in the first term. This is precisely your integrand above.

So your integrand can be written as $$ \int_V \frac{\partial}{\partial x_p} \left( B_p \frac{\partial A_j}{\partial x_j} \right) dV $$ which is now in a suitable form for application of the divergence theorem.

ETA: If you want everything back in vector form, $$ \frac{\partial}{\partial x_p} \left( B_p \frac{\partial A_j}{\partial x_j} \right) =\nabla \cdot (B (\nabla \cdot A)) $$

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The key is that you have both $B \cdot \nabla$ and $\nabla \cdot B$. This is a tell-tale sign of the product rule being applied. You can use "overdot notation" to write the expression as a total divergence. First, recognize that

$$(B \cdot \nabla) (\nabla \cdot A) + (\nabla \cdot B)(\nabla \cdot A)= (\dot \nabla \cdot B)(\nabla \cdot \dot A) + (\dot \nabla \cdot \dot B)(\nabla \cdot A)$$

The dots denote what is to be differentiated; in $\dot \nabla \cdot B$, $B$ is "held constant", and as such, this is equivalent to $B \cdot\nabla$.

The overdot notation makes it easy to recognize that the product rule has been used to expand, and easy to see how the product rule should be undone:

$$ (\dot \nabla \cdot B)(\nabla \cdot \dot A) + (\dot \nabla \cdot \dot B)(\nabla \cdot A) = \dot \nabla \cdot [\dot B (\nabla \cdot A) + B(\nabla \cdot \dot A)] = \nabla \cdot [B (\nabla \cdot A)]$$

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  • $\begingroup$ I'm not sure what you mean when you say that it's "held constant", but I'd still like to understand your solution since it seems like a simple way to keep track of things $\endgroup$
    – Joshua
    Dec 18, 2014 at 14:36
  • $\begingroup$ I mean that it's not differentiated. For instance, $(\dot \nabla \cdot B)(\nabla \cdot \dot A) = \dot \partial_i B^i (\partial_j \dot A^j)$ in index notation. Because $B^i$ is not being differentiated, you can rearrange this to $B^i \partial_i (\partial_j A^j)$. The overdot notation just gives you the same flexibility you'd have in index notation: the flexibility to differentiate only what should be differentiated. But overdot notation lets you do so with familiar dot and cross products instead of having to resort to index notation. $\endgroup$
    – Muphrid
    Dec 18, 2014 at 16:45

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