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As we know cross product of any two vectors yields a vector perpendicular to plane containing both the vectors so is it same for the vector operator del crossed with a vector ∇ × F (curl of vector field F). if not why?

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  • $\begingroup$ $F$ needs to be a vector field, not a vector. The curl of a vector is zero. $\endgroup$
    – Wintermute
    Commented Dec 27, 2016 at 13:16

3 Answers 3

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If $F = (-y, x, 1)$, then $\nabla \times F = (0, 0, 2)$ is not orthogonal to $F$. The cross product formalism is a mnemonic for remembering the curl formula, not a literal cross product of vectors: The "components" of $\nabla$ act on the components of $F$ by differentiation, not by multiplication.

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    $\begingroup$ thank you so much !! this really helped @andrew D Hwang $\endgroup$
    – Syed Ilyas
    Commented Dec 27, 2016 at 13:20
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No. As an example, consider the vector field $\vec{F} = y \hat{i} - x \hat{j} + \hat{k}$, whose curl is $\vec{\nabla} \times \vec{F} = 2 \hat{k}$. This is obviously not perpendular to $\vec{F}$ itself.

As to why this is not the case, the best answer I can think of is that the value of a function at a particular point and its derivatives at that same point are basically independent of each other, i.e., you can always find a function with a given value and a given set of derivatives at any one point in space. Since the curl of a vector field depends on the field's derivatives, it makes sense that the vector field and its curl could point pretty much any direction relative to each other.

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  • $\begingroup$ Surprisingly similar to the above $\endgroup$ Commented Dec 27, 2016 at 13:21
  • $\begingroup$ Yeah, except I got the sign of the curl wrong. :-) $\endgroup$ Commented Dec 27, 2016 at 13:22
  • $\begingroup$ i got the answer already from andrew D Hwang .but still thank you for drawing my attention to the point " the value of a function at a particular point and its derivatives at that same point are basically independent of each other".i appreciate every bit of knowledge shared @Micheal Seifert $\endgroup$
    – Syed Ilyas
    Commented Dec 27, 2016 at 13:25
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I know this is an old question, but for posterity's sake, something important happens when the curl is orthogonal to the vector field.

If $V$ is a $C^\infty$ vector field and $V\cdot (\nabla \times V) \equiv 0$ on an open set $U$ of $\mathbb{R}^3$, then on $U$, we have that for some nowhere $0$ $C^\infty$ scalar function f, the vector field $fV$ is conservative (see problem 3 of https://secure.math.ucla.edu/gradquals/data/gt-05W.pdf).

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  • $\begingroup$ access forbidden :( $\endgroup$ Commented Jun 21, 2022 at 10:08

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