Prove the given formula.
So far I have $f\textbf{F}=(f\textbf{F}_1, f\textbf{F}_2, f\textbf{F}_3)$, but I'm not sure where to go from there. Could anyone give me some pointers?
Thank you.
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3Just use the definition ... And the product rule. – Ted Shifrin May 27 '14 at 02:40
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1What exactly are you stuck on, given @rschweib's answer? What is unclear? – Alexander Gruber Jun 03 '14 at 01:56
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After taking the cross product, what do you factor out and add to get the $+(\nabla f) x \textbf{F}$? I feel like I'm missing something obvious... – user7000 Jun 03 '14 at 04:46
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@user7000 What do you mean "after taking the cross product"? You mean "after computing the curl of $fF$? – rschwieb Jun 03 '14 at 20:17
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@rschwieb Yes that's what I meant. – user7000 Jun 03 '14 at 23:53
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@user7000 ok, well that suggests again that you are simply not adhering to valid rules of differentiation :) If you wrote out your work I could tell you exactly where the mistake is. – rschwieb Jun 04 '14 at 00:02
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Now compute the curl of that, being sure to apply the product rule properly.
It may also help if you compute the other side of the identity also, and then reorganize stuff to make things match.
Here is a foothold for you to start with. By definition, $\mathrm{curl}(fF_1,fF_2,fF_3):=\left(\left(\frac{\partial fF_3}{\partial y} - \frac{\partial fF_2}{\partial z}\right),\left(\frac{\partial fF_1}{\partial z} - \frac{\partial fF_3}{\partial x}\right) , \left(\frac{\partial fF_2}{\partial x} - \frac{\partial fF_1}{\partial y}\right) \right)$.
Each of the partial derivatives here requires you to take the derivative of a product of functions.
To add a little more detail, let's do the first term. The second line is obtained by the first by simply applying the product rule.
$$\frac{\partial fF_3}{\partial y} - \frac{\partial fF_2}{\partial z}=\\ \frac{\partial f}{\partial y}F_3+\frac{\partial F_3}{\partial y}f - \left(\frac{\partial f}{\partial z}F_2+\frac{\partial F_2}{\partial z}f\right)=\\ \frac{\partial F_3}{\partial y}f-\frac{\partial F_2}{\partial z}f+ \frac{\partial f}{\partial y}F_3-\frac{\partial f}{\partial z}F_2=\\ f\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right) + \left(\frac{\partial f}{\partial y}F_3-\frac{\partial f}{\partial z}F_2\right) $$
Notice that the left hand term is the $x$ component of $f\cdot\mathrm{curl}(F)$ and the right half is the $x$ component of $(\nabla f)\times F$.
The other two pieces proceed exactly the same way.
rschwieb
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1Thank you for your answer. I don't understand how the product rule applies to the curl. Why is the answer not just $f$ curl $\textbf{F}$? – user7000 May 27 '14 at 03:42
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@user7000 It sounds like you might be making the mistake of forgetting that $f$ is a function of $x,y$ and $z$, and is usually not constant. It won't factor out of the curl so easily. – rschwieb May 27 '14 at 10:33