# Issue with the Laplacian in Cylindrical Coordinates

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?

• Hi, you made a typo. In the definition of the pot. the angle is called: t, but in the Laplacian the angle is called theta. Dec 9, 2020 at 20:03
• @DanielHuber I believe t is time, not the angle theta. Dec 9, 2020 at 22:17
• Yup, as @BillWatts said, t refers to time, while θ is the azimuth angle.
– Sid
Dec 10, 2020 at 3:42

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*)

• Oh, thanks a lot! So, if Aθ and Az were non zero as well, would I have to calculate them separately and then add the final results?
– Sid
Dec 10, 2020 at 3:40
• To be honest, I've never seen them added together, and I'm not sure what that would be physically. Assigning the pieces to vector components would be OK if you really need them to form a vector. Dec 10, 2020 at 6:36
• Ah alright. Thank you again
– Sid
Dec 10, 2020 at 17:45
• You are welcome. But I am puzzled at Mathematica's answer to your code. It seems to me that if it cannot calculate the Laplacian for a vector, it should return unevaluated, rather than return the wrong answer. Dec 11, 2020 at 1:20
• @SiddharthChaini As to how Mathematica deals with Laplacian of a vector, you may want to read this: mathematica.stackexchange.com/q/225423/1871 Oct 16, 2022 at 3:35

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

• Hi, actually the azimuth angle is θ, while t refers to time. So the basis will be {r, θ, z} and not {r, t, z}. That still gives an error. However, Bill's answer worked well instead. Thank you nevertheless.
– Sid
Dec 10, 2020 at 3:44