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!

Divergence of a Gradient

  1. Dec 29, 2014 #1
    1. The problem statement, all variables and given/known data
    image024.gif


    2. Relevant equations

    3. The attempt at a solution
    (a)
    Divergence of a gradient is a Laplacian.

    (b) I suppose to do it in Cartesian coordinates.

    Let [itex]\nabla=\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}[/itex]

    and [itex]\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}, \mid r\mid=\sqrt{x^{2}+y^{2}+z^{2}}[/itex],

    [itex]\hat{r}=\hat{i}+\hat{j}+\hat{k}[/itex]

    First calculate [itex]\nabla\left(\frac{1}{\overrightarrow{r}}\right)=\nabla\left(\frac{1}{\mid r\mid\hat{r}}\right)=\nabla\left(\frac{1}{\mid r\mid}\hat{r}\right)=\left[\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right]\left[\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}\right][/itex]

    [itex] = \hat{i}\frac{\partial}{\partial x}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{-1}{2}}+\hat{j}\frac{\partial}{\partial y}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{-1}{2}}+\hat{k}\frac{\partial}{\partial z}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{-1}{2}} [/itex]

    Am I doing it the right way?

    PS. cross posted at
    https://www.physicsforums.com/threads/laplacian-of-a-vector.789514/#post-4959007
     
  2. jcsd
  3. Dec 29, 2014 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    No. ##\hat{r}## is a vector, so ##\nabla(\hat{r}/r)## does not make sense: you need to take the gradient of a scalar, not of a vector. Anyway, why do you define ##\hat{r}=\hat{i}+\hat{j}+\hat{k}##? I cannot see its relation to anything in the problem. If by ##\hat{r}## you mean the unit vector in the direction of ##\vec{r}##, then that is most certainly not equal to what you wrote.

    Also, never, never write something like ##\frac{1}{|r|\hat{r}}## because that is meaningless: it is a fraction of the form 1/vector, and those things do not exist in any usual form.
     
  4. Dec 30, 2014 #3

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Dear Isaac,

    If you yourself write ##\vec r = \hat {\bf r} | {\bf r}|## you really should go back to ##\vec {r}=x\hat{i}+y\hat{j}+z\hat{k}, \mid r\mid=\sqrt{x^{2}+y^{2}+z^{2}}## and correct the hideous ##\hat{r}=\hat{\imath}+\hat{\jmath}+\hat{k}## to ##\hat{r}={x\over \sqrt{x^{2}+y^{2}+z^{2}}} \hat{\imath} + {y\over \sqrt{x^{2}+y^{2}+z^{2}}}\hat{\jmath} + {z\over \sqrt{x^{2}+y^{2}+z^{2}}}\hat{k}##

    [edit] sorry, pressed wrong button.

    But, as Ray (And TSny) indicate, you don't need ##\hat r##

    In your defence: The original problem formulation is confusing because it uses a boldface r in the 1/r. Many of us are conditioned to see that as a vector.
     
  5. Dec 31, 2014 #4
    Dear BvU, I think Author referred it as a vector. In the book (Mathematical Methods for Physicists by Tai L. Chow) scalars are mentioned as non bold, such as in same exercise page other problems are

    image004.gif
    image020.gif
     
  6. Dec 31, 2014 #5

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I count it as a misprint. What could possibly be the interpretation of ##1/\;\vec {\bf r}## ?
     
  7. Dec 31, 2014 #6
    Ok, thx for the input.

    1. The problem statement, all variables and given/known data

    Show that [itex]\nabla^{2}\left(\frac{1}{r}\right)=0[/itex]

    2. Relevant equations

    Let [itex]\nabla=\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}[/itex]

    and [itex]\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}, \mid r\mid=\sqrt{x^{2}+y^{2}+z^{2}}[/itex]

    3. The attempt at a solution

    [itex]\nabla\left(\frac{1}{r}\right)=\nabla\left(\frac{1}{\mid r\mid}\right)=\left[\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right]\left[\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}\right][/itex]

    [itex]\nabla\left(\frac{1}{r}\right)= \hat{i}\frac{\partial}{\partial x}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}+\hat{j}\frac{\partial}{\partial y}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}+\hat{k}\frac{\partial}{\partial z}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}[/itex]

    [itex]\nabla\left(\frac{1}{r}\right)=-\hat{i}\frac{2x}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-\hat{j}\frac{2y}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-\hat{k}\frac{2z}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}[/itex]

    [itex]\nabla.\nabla\left(\frac{1}{r}\right)=\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right).\left(-\hat{i}x\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-\hat{j}y\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-\hat{k}z\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}\right)[/itex]

    [itex]\nabla^{2}\left(\frac{1}{r}\right)=\frac{\partial}{\partial x}\left(-x\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}\right)+\frac{\partial}{\partial y}\left(-y\left(x^{2}+y^{2}+z^{2}\right)\right)+\frac{\partial}{\partial z}-\left(z\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}\right)[/itex]

    [itex]\nabla^{2}\left(\frac{1}{r}\right)=-x\left(2x\right)\left(-\frac{3}{2}\right)\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-y\left(2y\right)\left(-\frac{3}{2}\right)\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}[/itex]
    [itex]- \left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-z\left(2z\right)\left(-\frac{3}{2}\right)\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}[/itex]

    [itex]\nabla^{2}\left(\frac{1}{r}\right)=3x^{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}+3y^{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}[/itex]
    [itex]+3z^{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}[/itex]

    [itex]\nabla^{2}\left(\frac{1}{r}\right)=3\left(x^{2}+y^{2}+z^{2}\right)\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-3\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}[/itex]

    [itex]\nabla^{2}\left(\frac{1}{r}\right)=3\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}+1}-3\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}[/itex]

    [itex]\nabla^{2}\left(\frac{1}{r}\right)=3\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-3\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}[/itex]


    [itex]\nabla^{2}\left(\frac{1}{r}\right)=0[/itex]

    Hence Showed
     
    Last edited: Dec 31, 2014
  8. Jan 1, 2015 #7

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

  9. Jan 2, 2015 #8
    Yes, but we've to it in both - Cartesian and Spherical coordinates.
     
  10. Jan 3, 2015 #9
    ^ solve it
     
  11. Jan 3, 2015 #10
    image024.gif

    Solving part (c). As suggested above it's also a misprint. It should be
    [itex] \overrightarrow{r}.\left(\nabla.\overrightarrow{r}\right)\neq\left(r\nabla\right)r [/itex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Divergence of a Gradient
  1. Question on Gradient (Replies: 11)

  2. Curl of a gradient (Replies: 3)

  3. Gradient problem (Replies: 5)

Loading...