Issue with the Laplacian in Cylindrical Coordinates

Question

I am using Mathematica 12.1.1 and am unable to get the correct result for a simple laplacian in 3D Cylindrical Coordinates. I want to reproduce the following result on Mathematica:
But, I am only getting the second term from the above result. Here's the code I am using:
APotential = {A0/(k r) Sin[k z - ω t], 0, 0};

Laplacian[APotential, {r, θ, z}, "Cylindrical"]

(* OUTPUT IS: {-((A0 k Sin[k z-t ω])/r),0,0} *)

Is this because of a bug, or am I missing something?

Bill Watts · Accepted Answer

The Laplacian takes a scalar argument, so if you want to take the Laplacian of a vector you need to do each component separately.  This works:
Ar[r_, θ_, z_] = A0/(k r) Sin[k z - ω t]

Laplacian[Ar[r, θ, z], {r, θ, z}, "Cylindrical"]
(*(A0 Sin[k z - t ω])/(k r^3) - (A0 k Sin[k z - t ω])/r*)

cvgmt · Answer

For general vector field {f[r, t, z], g[r, t, z], h[r, t, z]}, the  Laplacian is
Laplacian[{f[r, t, z], g[r, t, z], h[r, t, z]}, {r, t, z}, 
  "Cylindrical"] // Expand

So we write
APotential = {A0/(k r) Sin[k z - ω t], 0, 0};
Laplacian[APotential, {r, t, z}, "Cylindrical"] // Expand

Issue with the Laplacian in Cylindrical Coordinates

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