# An apparent paradox (what am I missing?)

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

1. 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?

• Are you sure you're using the vector laplacian? It's not the same as the laplacian of the components. – Javier Nov 21 '20 at 15:02
• @Javier Ah. Griffiths' says that the laplacian of a vector is the laplacian applied to each of the components. That's what I considered. Is that wrong? – PhutureFysicist Nov 21 '20 at 15:07
• Only in cartesian coordinates. – Javier Nov 21 '20 at 17:14

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