1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Del operator in different coordinate

  1. Dec 24, 2008 #1

    KFC

    User Avatar

    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

    [tex]
    \nabla = \frac{\partial }{\partial x}\hat{x} + \frac{\partial }{\partial y}\hat{y} + \frac{\partial }{\partial z}\hat{z}
    [/tex]

    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

    [tex]
    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)
    [/tex]

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

    [tex]
    \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)
    [/tex]

    Similarly, I apply the same transformation to Del operator

    [tex]
    \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)
    [/tex]

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

    [tex]
    \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}
    [/tex]

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

    [tex]\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}.[/tex]

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

    [tex]\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)[/tex]

    But I remember the Laplace operator in spherical coordinate is

    [tex] \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}.[/tex]

    I don't know what's wrong here! :(
     
    Last edited: Dec 24, 2008
  2. jcsd
  3. Dec 25, 2008 #2
    Doesn't it basically amount to solving simultaneous equations to get it in that form, for 1)
     
  4. Dec 25, 2008 #3
    Here's how I basically started:

    [tex] x = r sin(\theta) cos(\phi) [/tex]
    [tex] y = r sin(\theta) sin(\phi) [/tex]
    [tex] z = r cos(\theta) [/tex]

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

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

    Similarly for y and z we get

    [tex] \frac{\partial}{\partial y} = sin(\theta) sin(\phi) \frac{\partial}{\partial r} + ... [/tex]

    [tex] \frac{\partial}{\partial z} = cos(\theta) \frac{\partial}{\partial r} + ... [/tex]

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

    [tex] 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) [/tex]

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

    [tex] \frac{\partial}{\partial r} r [/tex]
     
  5. Dec 25, 2008 #4
    For 2), do your parentheses indicate "argument of" or multiplication? I'm fairly certain in theirs they mean argument of
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Del operator in different coordinate
  1. Del operator (Replies: 7)

Loading...