An apparent paradox (what am I missing?)

Question

Griffiths' Introduction to Electrodynamics, 4th Ed. features the following question:

What current density ($\vec{J}$) would produce the vector potential $\vec{A} = k \hat{\phi}$, where k is a constant, in cylindrical coordinates?

Now, there are two equivalent approaches to the problem:

1. Using the definition of the magnetic field, followed by Ampere's law

$$\vec{B} = \nabla \times \vec{A},$$ so $$\nabla\times(\nabla\times\vec{A}) = \mu_{0}\vec{J}.$$

2. The second method (which is basically the first) is to write $$\nabla\times(\nabla\times\vec{A}) = \nabla(\nabla\cdot\vec{A})-\nabla^{2}\vec{A} = \mu_{0}\vec{J}.$$

Here's where my problem arises. My calculations suggest that the second method gives zero, while the first one (that applies the cross product twice) gives $\frac{k}{\mu_{0}s^{2}}\hat{\phi}$.

A cursory glance over $\vec{A}$ shows that its divergence and laplacian (taken in cylindrical coordinates) should be zero. However, the first curl gives us $\frac{k}{s}\hat{z}$, while the second curl gives us $\frac{k}{\mu_{0}s^{2}}\hat{z}$.

Where is my (perhaps obvious) error?

Answer 1

Where is my (perhaps obvious) error?

The second method doesn't give zero. Look here for expression:

$$\nabla\times (\nabla\times \mathbf{A})=\nabla(\nabla\cdot\mathbf{A})-\nabla^2\mathbf{A}$$

The second term has a nonzero component given by $$\nabla^2\mathbf{A}=-\frac{A_\phi}{\rho^2}\hat\phi=-\frac{k}{\rho^2}\hat\phi$$

The first term is zero. This is the same as the one from the first method.