It's not hard to show that the function:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]g = \frac{1}{2} (c \times r)[/tex]

is a "vector potential" function for the constant vector "c". That is, that:

[tex]\nabla \times g = c[/tex]

The calculation is straightforward to carry out in Cartesian coordinates, and I won't reproduce it here.

However, my question concerns the following. It is also a standard result in vector algebra that we have:

[tex]a \times (b \times c) = (a \cdot c) b - (a \cdot b) c[/tex]

My question is, when taking the "curl" of the vector potential above, why can't I just treat the "del" operator as though it were a vector, and write something like the following:

[tex]\nabla \times (c \times r) = (\nabla \cdot r) c - (\nabla \cdot c) r [/tex]

[tex]\nabla \times (c \times r) = 3c [/tex]

where the last equality follows because (a) the divergence of a constant vector is 0; and (b) the divergence of the radial vector (i.e., the vector [x,y,z]) is 3.

However, when factoring back in the factor of (1/2) to obtain the final "answer", the "answer" obtained using this method is (3/2)c, rather than the "correct" answer of c.

Why is this?

Is it because you can't treat the "del" operator as though it's a "vector" in this case?

Or am I doing something else wrong?

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# Vector Potentials

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