# Laplace Operator in Polar Coordinates

1. Oct 5, 2008

### dav2008

1. The problem statement, all variables and given/known data
Compute $$\nabla \cdot \nabla f$$ in polar coordinates.

2. Relevant equations

3. The attempt at a solution
It seems like a straightforward dot product yields
$$\nabla \cdot \nabla f = {\partial^2 f \over \partial \rho^2} + {1 \over \rho^2} {\partial^2 f \over \partial \theta^2} + {\partial^2 f \over \partial z^2 }$$
since the 3 basis vectors are mutually orthogonal.

This is obviously not the correct expression, which should be:
$${1 \over \rho} {\partial \over \partial \rho} \left( \rho {\partial f \over \partial \rho} \right) + {1 \over \rho^2} {\partial^2 f \over \partial \theta^2} + {\partial^2 f \over \partial z^2$$

Where is that first term coming from?

It seems like I'm ignoring something simple when calculating the dot product. What am I failing to take into account?

2. Oct 5, 2008

### gabbagabbahey

It comes from the fact that the divergence of a vector $\vec{v}$ in cylindrical coordinates is

$$\vec{\nabla} \cdot \vec{v}=\frac{1}{\rho} \frac{\partial}{\partial \rho} (\rho v_{\rho})+\frac{1}{\rho} \frac{\partial v_{\theta}}{\partial \theta}+ \frac{\partial v_{z}}{\partial z}$$

not just

$$\vec{\nabla} \cdot \vec{v}= \frac{\partial v_{\rho}}{\partial \rho}+\frac{1}{\rho} \frac{\partial v_{\theta}}{\partial \theta}+ \frac{\partial v_{z}}{\partial z}$$

3. Oct 5, 2008

### dav2008

ok, but why is it incorrect to compute the dot product:

$$\left(\frac{\partial }{\partial \rho}\mathbf{e}_\rho+ \frac{1}{\rho}\frac{\partial }{\partial \theta}\mathbf{e}_\theta+ \frac{\partial }{\partial z}\mathbf{e}_z\right) \bullet \left( \frac{\partial }{\partial \rho}\mathbf{e}_\rho+ \frac{1}{\rho}\frac{\partial }{\partial \theta}\mathbf{e}_\theta+ \frac{\partial }{\partial z}\mathbf{e}_z\right)$$

as

$${\partial^2 \over \partial \rho^2} + {1 \over \rho^2} {\partial^2 \over \partial \theta^2} + {\partial^2 \over \partial z^2 }$$

What assumptions in taking the dot product are incorrect?

Edit: I guess I'll just write it in terms of the Cartesian base vectors and work it out that way. I see now that the dot products of the unit vectors don't work out to just be 1 or 0.

4. Oct 5, 2008

### dav2008

Hmm I wrote er and eθ in terms of the cartesian base vectors, did the dot product, and it still comes out the same way...which I guess is expected.

5. Oct 5, 2008

### Ben Niehoff

You need to remember to differentiate the unit vectors also. For example,

$$\frac{\partial}{\partial \rho} \hat e_{\rho} \cdot \frac{\partial}{\partial \rho} \hat e_{\rho} = \frac{\partial^2}{\partial \rho^2} \hat e_{\rho} + \frac{\partial}{\partial \rho} \frac{\partial \hat e_{\rho}}{\partial \rho}$$

It just so happens that

$$\frac{\partial \hat e_{\rho}}{\partial \rho} = 0$$

so this particular term does not contribute. However, the $\theta$ term will give you

$$\frac{\partial \hat e_{\theta}}{\partial \theta}$$

which is nonzero; this should give you your missing term (you will have to collect terms and undo a product rule to get the exact form needed).

6. Oct 5, 2008

### dav2008

Ahh there it is. Thanks.

Edit: Do you mean $$\frac{\partial}{\partial \rho} \hat e_{\rho} \cdot \frac{\partial}{\partial \rho} \hat e_{\rho} = \frac{\partial^2}{\partial \rho^2}} + \hat e_{\rho} \cdot \frac{\partial}{\partial \rho} \frac{\partial \hat e_{\rho}}{\partial \rho}$$

Last edited: Oct 5, 2008
7. Oct 6, 2008

### Ben Niehoff

Right, yes. I forgot to take the dot product. :P