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How we can use the property that $$A×(B×C) = B(A.C)- C(A.B)$$ to prove the relation: $$a×(∇×a) = ∇ (a^2/2) -(a.∇)a.$$ When I use it, the result directly appear to be $$∇(|a|^2 )-(a.∇)a$$ instead of the correct one. Any help please?

joriki
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israa sharbeh
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    $\nabla$ is not a regular vector. That identity is only valid for vectors. – zzchan Aug 01 '15 at 18:51
  • Don't forget that derivative of a product requires a product rule. – GEdgar Aug 01 '15 at 18:54
  • Yes ok. But any other way to prove the above formula ? – israa sharbeh Aug 01 '15 at 19:00
  • From the title, and from the identity you wish to apply, it seems that where it says $\nabla a$ you meant $\nabla\times a$? – joriki Aug 01 '15 at 19:09
  • No, ∇a is the gradient of the vector a – israa sharbeh Aug 01 '15 at 19:16
  • Actually i need to prove the last formula apart from the way used – israa sharbeh Aug 01 '15 at 19:17
  • I see now that this was an error introduced through an edit by another user. I'm not sure why you supported that error when you'd originally posted the relation correctly, but never mind. You posted the curl, and the question only makes sense with the curl, so let's assume you meant the curl. – joriki Aug 01 '15 at 19:35
  • The user who introduced the mistake had tried to correct it, but whereas the erroneous edit was approved, the correction was rejected. I corrected it now; if you agree, we can now remove all these comments to reduce the clutter. – joriki Aug 01 '15 at 19:53
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    Have a look at my answers to this question and this question, both of which show a simple but powerful method that helps to prove vector identities, such as the one you want, in just a few lines. – wltrup Aug 01 '15 at 20:04

2 Answers2

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The problem arises because vector identites are usually derived for "normal" vectors, where the order of the vectors either doesn't matter or, in the case of the cross product, merely affects the sign. If one of the "vectors" is a differential operator, these vector identities still hold in a certain sense and can be very useful, but need to be applied with care, since the differential operator must always operate on what originally came after it and never on what originally came before it – this must be taken into account when the vector identity being applied changes the order of the vectors.

Naively applying the double cross product identity yields

$$ a\times(\nabla\times a)=\nabla(a\cdot a)-a(a\cdot \nabla)\;. $$

Both terms are wrong – the first because $\nabla$ is now acting on both instances of $a$ whereas it should only be acting on one, as on the left-hand side, and the second because $\nabla$ is at the very right and is no longer acting on anything. The second problem is readily corrected by switching the order of multiplication. The first problem isn't quite as easy to correct. The actions of $\nabla$ on both instances of $a$ yield the same result, so if we only wanted one of them, we can correct for having both by dividing by two. Applying both corrections yields

$$ a\times(\nabla\times a)=\frac12\nabla(a\cdot a)-(a\cdot \nabla)a\;. $$

If you intially find these things confusing, it often helps to write them out in coordinates – that makes them seem less mysterious, and over time you'll develop a feel for how the vector identities, applied with care, summarize what you would automatically have done correctly in coordinates.

joriki
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    That is an exceedingly clear explanation. Kudos! – wltrup Aug 01 '15 at 20:11
  • Easily take one of that ∇(a.a) ? this idea is not obvious. Please demonstrate further. – israa sharbeh Aug 01 '15 at 21:13
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    @israasharbeh: For a scalar $y$, we have $\frac{\mathrm d}{\mathrm dx} (y\cdot y) = \frac{\mathrm dy}{\mathrm dx}y+y\frac{\mathrm dy}{\mathrm dx}=2y\frac{\mathrm dy}{\mathrm dx}$, so $y\frac{\mathrm dy}{\mathrm dx}=\frac12\frac{\mathrm d}{\mathrm dx} (y\cdot y)$. The factor $2$ arises because the differential operator acts on both instances of $y$. We're doing the same thing here, the only difference being that we can't explicitly write the equivalent of $y\frac{\mathrm dy}{\mathrm dx}$ in the vector case because our vector notation won't allow it. – joriki Aug 01 '15 at 21:18
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    If it's still not clear, I suggest that before asking further questions you write it out in components, make sure you understand it in components (where the problem with the vector notation doesn't arise), and then try to see how the vector version corresponds to what you understand in components. – joriki Aug 01 '15 at 21:20
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    It's all about how our vector notation and our notation for differential operators interact; keeping clear in your mind what aspects of the ordering relate to the requirements of the vector notation, which aspects of the ordering relate to the requirements of the differential operator notation, and writing things such that both sets of requirements are fulfilled. – joriki Aug 01 '15 at 21:23
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    Thx now it is clear – israa sharbeh Aug 02 '15 at 16:37
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Using tensor notation we have \tag $$\begin{align} \vec a\times \nabla \times \vec a&=(\hat x_ia_i)\times (\hat x_j\partial_j \times \hat x_ka_k)\\\\ &=(\hat x_j\delta_{ik}-\hat x_k\delta_{ij})a_i\partial_j(a_k)\tag 1 \\\\ &=\hat x_j(a_i)\partial_j(a_i)-\hat x_ka_i\partial_i(a_k) \tag2 \\\\ &=\hat x_j\frac12\partial_j(a_i^2)-a_i\partial_i(\hat x_ka_k) \\\\ &=\frac12\nabla |\vec a|^2-(\vec a\cdot \nabla) \vec a \end{align}$$

In arriving at $(1)$, we used the vector triple product rule. Note that $\delta_{ij}$ is the Kronecker Delta where $\delta_{ij}=1$ for $i=j$ and $0$ otherwise.

In going from $(1)$ to $(2)$, we used the sifting property of the Kronecker Delta.

Mark Viola
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