# Thread: Laplacian In Cylindrical Coordinates

1. ## Laplacian In Cylindrical Coordinates

Let $\displaystyle (r, \theta, z)$ be the cylindrical coordinates in $\displaystyle \mathbb{R}^3$. Derive the Laplacian in cylindrical coordinates.

So what I know is that $\displaystyle x=r\cos\theta, y=r\sin\theta, z=z$, so that the last term in the Laplacian doesn't change.

I know where I want to go (I have the expression right in front of me), but I'm not seeing how to derive the other two.

2. Here is a pdf for spherical and polar as well.

http://banach.millersville.edu/~bob/.../Laplacian.pdf

3. Originally Posted by mathematicalbagpiper
Let $\displaystyle (r, \theta, z)$ be the cylindrical coordinates in $\displaystyle \mathbb{R}^3$. Derive the Laplacian in cylindrical coordinates.

So what I know is that $\displaystyle x=r\cos\theta, y=r\sin\theta, z=z$, so that the last term in the Laplacian doesn't change.

I know where I want to go (I have the expression right in front of me), but I'm not seeing how to derive the other two.
Just as a side not with the use of differential forms you can derive the Laplacian and many other operators from vector calculus.

In cylindrical coordinates the basis of 1-forms is

$\displaystyle \hat{r}=dr \quad \hat{\theta}=rd\theta \quad \hat{z}=dz$

and the Laplacian is given by

$\displaystyle *d*d f = \Delta f$

Where * is the Hodge dual

This gives

$\displaystyle dr =f_r dr+f_\theta d\theta +f_zdz=f_r dr+\frac{1}{r}f_\theta (r d\theta) +f_zdz$

taking the Hodge dual gives

$\displaystyle *dr = f_r (rd\theta \wedge dz)-\frac{1}{r}f_\theta( dr \wedge dz) +f_z(dr \wedge rd\theta )$

taking the exterior derivative gives

$\displaystyle d*dr = \frac{\partial }{\partial r}\left( r f_r \right) (dr \wedge d\theta \wedge dz)+\frac{1}{r}f_{\theta \theta }( dr \wedge d\theta \wedge dz) +f_{zz}(dr \wedge rd\theta \wedge dz)=\left[\frac{1}{r} \frac{\partial }{\partial r}\left( r f_r \right) +\frac{1}{r^2}f_{\theta \theta }+f_{zz}\right](dr \wedge rd\theta \wedge dz)$

and taking the Hodge dual again gives

$\displaystyle *d*df=\Delta f=\frac{1}{r} \frac{\partial }{\partial r}\left( r f_r \right) +\frac{1}{r^2}f_{\theta \theta }+f_{zz}$

You can do the same this with spherical coordinates ( or any curvilinear coordinate system) but the basis of 1-forms is

$\displaystyle \hat{r}=dr \quad \hat{\theta}=rd\theta \quad \hat{\phi}=(r\sin(\theta))d\phi$