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Let $\vec v = (a,b,c)$ be a constant vector and let $\vec r = (x,y,z)$ denote the position vector. Consider the vector field $\vec v \times \vec r$. A straightforward calculation (using determinants or a vector identity) shows that $\nabla \times (\vec v \times \vec r) = 2\vec v$. In order to get a better understanding of this result, a sketch of the vector field helps to convince one that indeed its curl should be $\alpha \vec v$ for some positive constant $\alpha$ .

Question: Are there any further insights from physics, geometry, topology, etc. which help to explain why the constant should turn out to be $\alpha$= 2 ?

Siminore
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user2052
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1 Answers1

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We can get some insight if we generalize the expression to other dimensions. Since the cross product is only defined for 3-D, we need to generalize in the language of Geometric Calculus

First, generalize the cross product using

$$ \vec v \times \vec r = (\vec v \wedge \vec r)^*$$

where the $^*$ denotes dualization (geometric-complement). Therefore

$$ \nabla\times(\vec v \times \vec r) = (\nabla\wedge(\vec v \wedge \vec r)^*)^*$$

Now, the RHS is defined for arbitrary dimensions.

Computing it for 2-D, we get $$ (\nabla\wedge(\vec v \wedge \vec r)^*)^* = \vec v$$

In 4-D: $$ (\nabla\wedge(\vec v \wedge \vec r)^*)^* = -3\vec v$$

In general, in n-D (a conjecture at this stage, I haven't written a proof yet): $$ (\nabla\wedge(\vec v \wedge \vec r)^*)^* = (-1)^{\frac{n(n-1)}{2}}(n-1)\vec v$$

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