2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Henning Makholm Mar 12 '13 at 22:45
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Mar 12 '13 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? $\endgroup$ – Steven Stadnicki Mar 12 '13 at 23:37
1
$\begingroup$

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 $B$.

$\endgroup$
  • $\begingroup$ ... and by "expand out the product" you mean apply the Leibniz product rule, right? $\endgroup$ – Henning Makholm Mar 12 '13 at 22:52
  • $\begingroup$ @HenningMakholm: that is correct. Thanks. $\endgroup$ – Ross Millikan Mar 12 '13 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 Mar 12 '13 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). $\endgroup$ – Ross Millikan Mar 13 '13 at 0:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.