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Given the vector field $\vec A ( \vec r ) = \begin {pmatrix} 3x \\ -z \\ 2y \end {pmatrix}$, I have to prove that the vector field's curl in cartesian coordinates is the same as in spherical coordinates.

In cartesian coordinates: $\nabla \times \vec A(\vec r) = \begin {pmatrix} 3 \\ 0 \\ 0 \end {pmatrix}$

In spherical coordinates: $ \begin {pmatrix} \partial /\partial_r \\ \frac1r \partial / \partial_\theta \\ \frac{1}{r \sin \theta} \partial/\partial_\phi \end{pmatrix} \times \begin {pmatrix} 3r \sin\theta \cos \phi \\ -r \cos \theta \\ 2r \sin \theta \sin \phi\end{pmatrix} = \begin {pmatrix} 2 \cos \theta \sin \phi \\ -3 \sin \phi - 2 \sin \phi \sin \theta \\ -\cos \theta - 3 \cos\theta\cos\phi\end{pmatrix} $

But if I use Wikipedia's formula for curl, I receive $\begin {pmatrix} 4\sin\phi\cos\theta \\ -3\sin\phi -4\sin\theta\sin\phi \\ -2\cos\theta -3 \cos\theta\cos\phi\end {pmatrix}$

Now I got 2 questions about that:

First, where is my mistake?

Second, if I transform it back with $\phi = \arctan (\frac yx) $, I get $\phi = 0$ and thus $x=0$, which is wrong for both. Where am I wrong?

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First, writing the vectors in component form (as column vectors) has made it less obvious that you need to use the product rule here.

Let $\hat x, \hat y, \hat z$ be the usual Cartesian basis. The curl takes the form

$$\nabla \times A = (\nabla A^x) \times \hat x + A_x \nabla \times \hat x + \ldots$$

But since $\hat x$ is constant, $\nabla \times \hat x$ is zero. Hence, the formula for curl in Cartesian can be written

$$\nabla \times A = (\nabla A^x) \times \hat x + (\nabla A^y) \times \hat y + (\nabla A^z) \times \hat z$$

Once you do the cross products, you get $(\partial_y A^z - \partial_z A^y) \hat x$ and so on, as you usually would expect.

In spherical, however, the basis vectors depend on position. $\nabla \times \hat \theta$ isn't zero, for instance. That's where the terms in Wikipeida's form come from.

Second, when you converted $3x \hat x - z \hat y + 2 y \hat z$, you converted $x, y, z$ to spherical coordinates, but you didn't convert the basis. It's quite clear that you wrote

$$ 3x \hat x - z \hat y + 2 y \hat z = 3 r \sin \theta \cos \phi \hat x - r \cos \theta \hat y + 2r \sin \theta \sin \phi \hat z$$

You need to write this in terms of the spherical basis vectors in order to apply the formula for curl in spherical properly.


In summary, remember that curl in a general coordinate system is not as simple as it looks in Cartesian. You can always derive the correct formula for a given coordinate basis by using the product rule. If $\alpha$ is a scalar field and $F$ a vector field, then

$$\nabla \times (\alpha F)= (\nabla \alpha) \times F + \alpha \nabla \times F$$

make $\alpha$ the components and $F$ the basis vectors to derive the correct curl formula for your coordinate system.

Second, make sure you write all vectors and vector operators in the same basis.

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  • $\begingroup$ Ok, I have to admit I don't get the first part of your answer, but the transformation of the basis vectors has helped me. However, now I am stuck in another exercise :/ $\endgroup$ – Christian Jan 11 '15 at 14:18

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