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Prove the following relation:

$$ \oint f \vec{\bigtriangledown}g \cdot d\vec{l} = -\oint g \vec{\bigtriangledown}f \cdot d\vec{l} $$

where f, g are scalar functions.

I've tried a lot of work, but can't seem to figure this relation out. I initially tried using stokes theorem thinking it would give me some kind of result but I ended up just proving the relation:

$$ \oint f \vec{\bigtriangledown}g \cdot d\vec{l} = \int_S ((\vec{\bigtriangledown}f)\times(\vec{\bigtriangledown}g))$$

Which is correct, apparently, but I still want to prove the top relation.

I tried taking the gradient of g, and distributing f as a scalar, but I can't see any relation that would re-arrange the del operator.

$$\oint f \vec{\bigtriangledown}g \cdot d\vec{l} = \oint \langle f \frac{\partial g}{\partial x},f\frac{\partial g}{\partial y},f\frac{\partial g}{\partial z}\rangle \cdot\langle dxdydz\rangle $$

Everything I've seen related to closed line integrals states that it is equivalent to 0, but I just don't see how to get this result. I've even thought of using the gradient theorem, by replacing with g, and distributing f (this is the general equation):

$$ \int_{a}^{b} (\vec{\bigtriangledown}f)\cdot d\vec{l} = f(\vec{b})-f(\vec{a}) $$

Any help would be appreciated.

SignalProcessed
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1 Answers1

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Notice that you can rewrite the original relation as \begin{equation} \oint (f\cdot \nabla g + \nabla f\cdot g)dl = 0. \end{equation} Also notice that \begin{equation} \nabla(f\cdot g) = f\cdot\nabla g + \nabla f\cdot g. \end{equation} Using the fact that gradients are conservative vector fields, and what you know about integrating conservative vector fields about closed paths, can you make the desired conclusion?

Edit: Here's an alternative to the product rule approach. You said you've shown that \begin{equation} \oint f\cdot\nabla g dl = \int_S((\nabla f)\times (\nabla g)). \end{equation} Presumably this also means that \begin{equation} \oint \nabla f\cdot g dl = \int_S((\nabla g)\times (\nabla f)). \end{equation} But when we reverse order in a cross product we pick up a factor of $-1$, so \begin{equation} \oint f\cdot\nabla g dl = \int_S((\nabla f)\times (\nabla g)) = -\int_S((\nabla g)\times (\nabla f)) = -\oint \nabla f\cdot g dl. \end{equation}

211792
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  • So the g, and f can both be affected by the gradient? Even though it looks as though the operator is ONLY acting on g? – SignalProcessed Sep 01 '16 at 16:47
  • The way I read it, we're integrating $f$ times the gradient of $g$ on the left, and $g$ times the gradient of $f$ on the right, so the gradient operator acts on both $f$ and $g$. Notice that the second equation above is a sort of product rule for gradients. – 211792 Sep 01 '16 at 16:51
  • Okay, I just figured it was like $$ f (\vec{\bigtriangledown} g) $$ and not that it was applied to both, I wasn't sure though. – SignalProcessed Sep 01 '16 at 17:07
  • Gotcha. No, the notation $f\nabla g$ means $f$ (a scalar-valued function) times $\nabla g$ (a vector-valued function). – 211792 Sep 01 '16 at 17:09
  • Should we please state differently for vector operator from scalar function? For instance in $\nabla(f\cdot g) = f\cdot\nabla g + \nabla f\cdot g$? – MathArt Aug 01 '21 at 16:15