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Is the curl of a vector field only defined on $\Bbb R^3$?

I was wondering if the criterion $$\nabla \times \vec{F}=\vec{0} \implies \vec{F} \space\text{is conservative}$$

only applies to three dimensional vector fields or if it also applies to $n$-dimensional vector fields?

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  • $\begingroup$ There is a condition in $n$ dimensions for conservativity of a vector field. However, it involves ${} \gt n$ equations of expressions involving the derivatives of the components. So it is not of the form of a vector equation in $\mathbb R^n$. $\endgroup$
    – GEdgar
    Jul 14, 2015 at 12:25
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    $\begingroup$ Strictly, zero curl is not enough of a restriction to guarantee a conservative field in 3-space. You also need that the domain is simply connected. $\endgroup$
    – Rory Daulton
    Jul 14, 2015 at 12:27
  • $\begingroup$ @RoryDaulton Thanks! I keep forgetting that. $\endgroup$
    – qmd
    Jul 14, 2015 at 12:30

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From wikipedia

"Unlike the gradient and divergence, curl does not generalize as simply to other dimensions; some generalizations are possible, but only in three dimensions is the geometrically defined curl of a vector field again a vector field. This is a similar phenomenon as in the 3 dimensional cross product, and the connection is reflected in the notation ∇ × for the curl."

So you can define $\triangledown \times \vec{F} $ in higher dimensions but it does not have special geometrical properties there.

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  • $\begingroup$ Is this only for higher dimensions or also for lower dimensions, i..e 2d ? thanks. $\endgroup$
    – yourlazyphysicist
    Dec 5 at 19:46

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