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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?

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  • $\begingroup$ Are you sure you're using the vector laplacian? It's not the same as the laplacian of the components. $\endgroup$ – Javier Nov 21 '20 at 15:02
  • $\begingroup$ @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? $\endgroup$ – PhutureFysicist Nov 21 '20 at 15:07
  • $\begingroup$ Only in cartesian coordinates. $\endgroup$ – Javier Nov 21 '20 at 17:14
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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.

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  • $\begingroup$ Aha! So, the laplacian of a vector function is the scalar laplacian of the individual components only for the cartesian system. So is the resolution the fact that this is not the case for cylindrical coordinates? Why is that so? $\endgroup$ – PhutureFysicist Nov 21 '20 at 15:22

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