Einj said:
Ok, I get that. I basically write down the metric tensor and so I get my denominators. Now I am a bit confused on how to go from the gradient to the laplacian. In particular, since the basis vectors are no more constant in space I am confused on how to derivate them when we compute \nabla\cdot\nabla.
Here's an example with cylindrical coordinates:
We may split up this in three terms to be summed:
\vec{i}_{r}\frac{\partial}{\partial{r}}\cdot{(\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta} \frac{\partial}{r\partial{theta}}+\vec{i}_{z}\frac{\partial}{\partial{z}})}
\vec{i}_{z}\frac{\partial}{\partial{z}}\cdot{(\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta} \frac{\partial}{r\partial{\theta}}+\vec{i}_{z}\frac{\partial}{\partial{z}})}
\vec{i}_{\theta} \frac{\partial}{r\partial{\theta}}\cdot{(\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{theta} \frac{\partial}{r\partial{theta}}+\vec{i}_{z} \frac{\partial}{\partial{z}})}
Now, the two first lines don't present any trouble, because the basis vectors only depend on the angle, not on either r or z! Furthermore, orthogonality ensures that most of the the vector dot products are zero, and the two expressions yield the contribution to the Laplacian:
\frac{\partial^{2}}{\partial{r^{2}}}+\frac{\partial^{2}}{\partial{z^{2}}}
The last line in the three major terms are much nastier, because the polar basis vectors DO depend on on the angle!
In particular, we see we get the contribution (PRIOR to dotting!) from \frac{\partial\vec{i}_{r}}{r\partial\theta}=\frac{1}{r}\vec{i}_{\theta}
Collecting the non-zero terms when dotted with the angular basis vector yields us the final two contributions to the Laplacian:
\frac{\partial^{2}}{r^2\partial{\theta^{2}}}+\frac{1}{r}\frac{\partial}{\partial{r}}
The same process is valid for any computation of the laplacian