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A vector field is called irrotational if its curl is zero. A vector field is called solenoidal if its divergence is zero. If A and B are irrotational, prove that A $ \times $ B is solenoidal.

I'm having a hard time the proof equation that is required, and the steps that would go with it. I am defining V as a vector.

$ \nabla \times V = 0 $ = irrotational
$ \nabla \cdot V = 0 $ = solenoidal

$ \nabla \times A = 0 $
$ \nabla \times B = 0 $

so therefore, ($ \nabla \times A $)+ ($ \nabla \times B) = \nabla \cdot (A \times B) $

would this be a correct setup? I'm having a hard time expanding this to E.S. form.

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    $\begingroup$ Your $(\nabla\times A)+(\nabla \times B)=\nabla\cdot(A\times B)$ does not seem to make much sense -- the left-hand side is a (pseudo)vector whereas the right-hand side is a (pseudo)scalar. $\endgroup$ Commented Mar 12, 2013 at 22:45

2 Answers 2

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This doesn't work; $(\nabla\times A)+(\nabla\times B)$ doesn't correspond to anything, and $\nabla\cdot(A\times B)$ doesn't expand to it (as noted by Henning Makholm in a comment, one of these is effectively a scalar and one effectively a vector). Instead, you want a form of the triple product identity $A\cdot(B\times C) = B\cdot(C\times A) = C\cdot (A\times B)$ - but this is where you need to be at least a little careful, because $\nabla$ isn't 'really' a vector.

As for Einstein summation form, the most important piece to keep in mind is the form for the cross-product: if $A\times B = C$, then $C^i = \epsilon^i\ _{jk}A^jB^k$ where $\epsilon$ is the so-called Levi-Civita symbol which essentially represents the sign of the permutation of its coordinates (i.e., $\epsilon_{ijk}=0$ if any two of $i,j,k$ are pairwise equal, $\epsilon_{012}=\epsilon_{120}=\epsilon_{201}=1$, and $\epsilon_{021}=\epsilon_{210}=\epsilon_{102}=-1$). Writing out the triple-product identity in terms of this notation should make it clear how it works, and then substituting in your hypotheses should show you how to draw your conclusion.

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  • $\begingroup$ so the triple product identity you listed is just the general form of the problem, where A, B, and C are $ \nabla\ $, A, B respectively? i'm not exactly clear on why that triple product is the proof, since it is a combination of the 3 elements in different orders. how did you reach that line? $\endgroup$
    – julesverne
    Commented Mar 12, 2013 at 23:12
  • $\begingroup$ @julesverne I got to the triple product identity because of the form of what you're being asked to prove - namely, that $\nabla\cdot(A\times B)=0$. If you use the TPI to permute the factors, you should be able to see how it easily shows this from the original stipulation you're given - what do you get when you permute with the TPI? $\endgroup$ Commented Mar 12, 2013 at 23:37
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Note that your left side is a vector (which happens to be zero) and the right side is a scalar (which you are hoping to prove is zero) so they can't be equal . The right side is $\varepsilon_{ijk}\frac d{dx_ i}A_jB_k$ while $\nabla \times A = 0$ is $\varepsilon_{ijk}\frac d{dx_i}A_j=0$ but those look a lot the same. If you expand out the product you should be able to find a way to make use of the irrotational nature of $A$ and $B$.

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  • $\begingroup$ ... and by "expand out the product" you mean apply the Leibniz product rule, right? $\endgroup$ Commented Mar 12, 2013 at 22:52
  • $\begingroup$ @HenningMakholm: that is correct. Thanks. $\endgroup$ Commented Mar 12, 2013 at 23:26
  • $\begingroup$ so im attempting to apply the product rule to the right side. But A is listed as the "j-th" component, and B is listed as the "k-th" component. The product rule gives me: (d/dx_i)(B_k) + A_j(dB_k/dx_i), in which case how would I apply the cross product identity? $\endgroup$
    – julesverne
    Commented Mar 12, 2013 at 23:57
  • $\begingroup$ The first term in your result for the product rule should be $\frac {dA_j}{dx_i}B_k$. Now if you collect the terms proportional to $B_k$ you get $(\frac {dA_j}{dx_i}-\frac {dA_i}{dx_j})B_k$ which is the $k$ component of $(\nabla \times A)\cdot B$. You will prove $\nabla \cdot (A \times B)=(\nabla \times A)\cdot B \pm A \cdot (\nabla \times B)$ which you know is zero (and I forget the sign but it doesn't matter here). $\endgroup$ Commented Mar 13, 2013 at 0:08

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