# Del operator in different coordinate

1. Dec 24, 2008

### KFC

1. The problem statement, all variables and given/known data
I would like to transform the Del operator form rectangular coordinate system to spherical coordinate system. The find the Laplace operator in spherical coordinate.

2. The attempt at a solution
1) In rectangular coordinate system, Del operator is given by

$$\nabla = \frac{\partial }{\partial x}\hat{x} + \frac{\partial }{\partial y}\hat{y} + \frac{\partial }{\partial z}\hat{z}$$

I know if you want to transform a vector in one coordinate system to a corresponding vector in other coordinate system, you got to know the transformation (matrix). For spherical coordinate system, the transformation matrix is given

$$M = \left( \begin{matrix} \sin\theta\cos\varphi & \sin\theta\sin\varphi & \cos\theta \\ \cos\theta\cos\varphi & \cos\theta\sin\varphi & -\sin\theta \\ -\sin\varphi & \cos\varphi & 0 \end{matrix} \right)$$

So that any vector $$\vec{r} = u_x \hat{x} + u_y \hat{y} + u_z \hat{z}$$ will be transformed as

$$\left( \begin{matrix} u_r \\ u_\theta \\ u_\varphi \end{matrix} \right) = M \left( \begin{matrix} u_x \\ u_y \\ u_x \end{matrix} \right)$$

Similarly, I apply the same transformation to Del operator

$$\left( \begin{matrix} \frac{\partial}{\partial r} \\ \\ \frac{\partial}{\partial \theta} \\ \\ \frac{\partial}{\partial \varphi} \end{matrix} \right) = M \left( \begin{matrix} \frac{\partial}{\partial x} \\ \\ \frac{\partial}{\partial y} \\ \\ \frac{\partial}{\partial z} \end{matrix} \right)$$

But if you multiply all terms out, for example, the first component of the result vector

$$\frac{\partial}{\partial r} = \sin\theta\cos\varphi \frac{\partial}{\partial x} + \sin\theta\sin\varphi \frac{\partial}{\partial y} + \cos\theta \frac{\partial}{\partial z}$$

I don't know what to do next. How can I get the following result?

$$\nabla = \boldsymbol{\hat r}\frac{\partial}{\partial r} + \boldsymbol{\hat \theta}\frac{1}{r}\frac{\partial}{\partial \theta} + \boldsymbol{\hat \varphi}\frac{1}{r \sin\theta}\frac{\partial}{\partial \varphi}.$$

2) Suppose the Del operator in spherical coordinate is in above form. To find Laplace operator, just dot product the Del operator with itself

$$\nabla\cdot\nabla = \nabla^2 = \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial \theta}\left(\frac{1}{r}\frac{\partial}{\partial \theta}\right) + \frac{1}{r \sin\theta}\frac{\partial}{\partial \varphi}\left(\frac{1}{r \sin\theta}\frac{\partial}{\partial \varphi}\right)$$

But I remember the Laplace operator in spherical coordinate is

$$\nabla^2 = {1 \over r^2} {\partial \over \partial r} \left( r^2 {\partial \over \partial r} \right) + {1 \over r^2 \sin \theta} {\partial \over \partial \theta} \left( \sin \theta {\partial \over \partial \theta} \right) + {1 \over r^2 \sin^2 \theta} {\partial^2 \over \partial \phi^2}.$$

I don't know what's wrong here! :(

Last edited: Dec 24, 2008
2. Dec 25, 2008

### NoMoreExams

Doesn't it basically amount to solving simultaneous equations to get it in that form, for 1)

3. Dec 25, 2008

### NoMoreExams

Here's how I basically started:

$$x = r sin(\theta) cos(\phi)$$
$$y = r sin(\theta) sin(\phi)$$
$$z = r cos(\theta)$$

So let's say we are trying to get the first term, let's differentiate each piece

$$\frac{\partial}{\partial x} = sin(\theta) cos(\phi) \frac{\partial}{\partial r} + ...$$ some terms with partials w.r.t. theta and phi

Similarly for y and z we get

$$\frac{\partial}{\partial y} = sin(\theta) sin(\phi) \frac{\partial}{\partial r} + ...$$

$$\frac{\partial}{\partial z} = cos(\theta) \frac{\partial}{\partial r} + ...$$

Now look at your definition for gradient in Cartesian and just multiply and group so for our $$\frac{\partial}{\partial r}$$ term we get

$$sin(\theta) cos(\phi) \frac{\partial}{\partial r} \cdot r sin(\theta) cos(\phi) + sin(\theta) sin(\phi) \frac{\partial}{\partial r} \cdot r sin(\theta) sin(\phi) + cos(\theta) \frac{\partial}{\partial r} \cdot r cos(\theta)$$

Now if you group and use trig identities you are left with:

$$\frac{\partial}{\partial r} r$$

4. Dec 25, 2008

### NoMoreExams

For 2), do your parentheses indicate "argument of" or multiplication? I'm fairly certain in theirs they mean argument of