# Where does Laplace's equation in spherical polars come from

1. Apr 9, 2009

### latentcorpse

Where does Laplace's equation in spherical polars come from

$\frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}=0$

???

i can derive from scratch the expression for the laplacian in spherical polars but this bears no resemblance to the above, even if i decide to ignore the $\phi$ dependence?

2. Apr 9, 2009

### robphy

Re: Laplacian

Those are in 2D-polar coordinates.

3. Apr 9, 2009

### latentcorpse

Re: Laplacian

hmm, im guessing that's meant to be easier but we haven't covered that in class (despite doing the 3D case) - could you possibly run over the derivation or post a link to a website (whenever i google it i only get the 3d case) please?

4. Apr 9, 2009

5. Apr 9, 2009

### Ben Niehoff

Re: Laplacian

When deriving it from scratch, remember that in spherical coordinates, the unit vectors are not constants anymore!

$$\vec \nabla = \hat r \frac{\partial}{\partial r} + \hat \theta \frac1{r} \frac{\partial}{\partial \theta} + \hat \phi \frac1{r \sin \theta} \frac{\partial}{\partial \phi}$$

what you want to find is

$$\vec \nabla \cdot \vec \nabla$$

but you must take into account the fact that the differential operators also act on the unit vectors.

6. May 10, 2009

### ber70

Re: Laplacian

From my weblog