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!

Showing that Laplace's equation holds

  1. Jan 23, 2013 #1
    1. The problem statement, all variables and given/known data

    (From Plonsey, R. and R. C. Barr, "Bioelectricity: A Quantitative Approach")

    Show
    [tex] \nabla^2 r = 0 [/tex]
    given
    [tex] r = \sqrt{x^2 + y^2 + z^2}. [/tex]

    2. Relevant equations
    [tex] \nabla = \frac{\partial}{\partial x}{\bf i} + \frac{\partial}{\partial y}{\bf j} + \frac{\partial}{\partial z}{\bf k} [/tex] (the "abuse of notation")


    3. The attempt at a solution

    This is my first attempt at a post on this forum, and I'm embarrassed to say I'm really struggling with this question. (Not a homework problem per se, just an in-text side note). It has been a good while since I've studied vector calculus, so if somebody could point out my error, I would appreciate it.

    Beginning with
    [tex]
    \begin{equation}
    r = (x^2 + y^2 + z^2)^{1/2}
    \end{equation}
    [/tex]
    taking the gradient results in
    [tex]
    \begin{equation}
    \begin{aligned}
    \nabla r &= \frac{\partial r}{\partial x} {\bf i} + \frac{\partial r}{\partial y} {\bf j} + \frac{\partial r}{\partial z} {\bf k} \\
    &= \left[ \frac{x}{(x^2 + y^2 + z^2)^{1/2}} \right]{\bf i} + \left[ \frac{y}{(x^2 + y^2 + z^2)^{1/2}} \right]{\bf j} + \left[ \frac{z}{(x^2 + y^2 + z^2)^{1/2}} \right]{\bf k}.
    \end{aligned}
    \end{equation}
    [/tex]
    Now taking the divergence of this gradient [itex] - [/itex] which amounts cumulatively to taking the Laplacian of [itex] r - [/itex] gives
    [tex]
    \begin{equation}
    \begin{aligned}
    \nabla \cdot \nabla r &= \nabla^2 r \\
    &= \frac{\partial}{\partial x}\left[ \frac{x}{(x^2 + y^2 + z^2)^{1/2}} \right] + \frac{\partial}{\partial y}\left[ \frac{y}{(x^2 + y^2 + z^2)^{1/2}} \right] + \frac{\partial}{\partial z}\left[ \frac{z}{(x^2 + y^2 + z^2)^{1/2}} \right] \\
    &= \left[ \frac{1}{(x^2 + y^2 + z^2)^{1/2}} - \frac{x^2}{(x^2 + y^2 + z^2)^{3/2}} \right] + \left[ \frac{1}{(x^2 + y^2 + z^2)^{1/2}} - \frac{y^2}{(x^2 + y^2 + z^2)^{3/2}} \right] + \left[ \frac{1}{(x^2 + y^2 + z^2)^{1/2}} - \frac{z^2}{(x^2 + y^2 + z^2)^{3/2}} \right] \\
    &= \left[ \frac{y^2 + z^2}{(x^2 + y^2 + z^2)^{3/2}} \right] + \left[ \frac{x^2 + z^2}{(x^2 + y^2 + z^2)^{3/2}} \right] + \left[ \frac{x^2 + y^2}{(x^2 + y^2 + z^2)^{3/2}} \right] \\
    &= \frac{2}{(x^2 + y^2 + z^2)^{1/2}}
    \end{aligned}
    \end{equation}
    [/tex]
    which is not zero, although I very much wish it was. (Sorry for the long-winded calculations.)
     
  2. jcsd
  3. Jan 23, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Very good job. You got the right answer. I think the problem is with the original question. The laplacian of r isn't zero. The laplacian of 1/r is zero. Are you sure that's not what it says??
     
  4. Jan 23, 2013 #3
    Ugh, I was starting to suspect I was wasting time because of a typo in the book. I attached the relevant portion of the textbook page as an image, where the above equality is listed along with a couple other vector identities.

    Thanks for the extra input (and sanity check).
     

    Attached Files:

    Last edited: Jan 24, 2013
  5. Jan 23, 2013 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's a typo alright. Pretty bad one too. Jeez. I'm aghast. Shocked, really. That's pretty fundamental, you'd think they would have checked that.
     
    Last edited: Jan 24, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook