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Surprising Gradient not 'Surprising Enough'

  1. May 28, 2008 #1
    [SOLVED] Surprising Gradient not 'Surprising Enough'

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

    Q] Sketch the vector function and

    [tex]
    v = \frac{\hat{r}}{r^2}
    [/tex]

    and compute it's divergence. The answer may surprise you... can you explain it?

    ['r' is the position vector in the Euclidean space]

    2. Relevant equations

    [tex]
    \nabla \cdot v = \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} + \frac{\partial v_z}{\partial z}
    [/tex]

    3. The attempt at a solution

    [tex]
    \mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k}
    [/tex]

    and,

    [tex]
    \mathbf{\hat{r}} = \frac{x\hat{i} + y\hat{j} + z\hat{k}}{\sqrt{x^2 + y^2 + z^2}}
    [/tex]

    Hence,

    [tex]
    \frac{\mathbf{\hat{r}}}{r^2} = \frac{x\hat{i} + y\hat{j} + z\hat{k}}{(x^2 + y^2 + z^2)^{\frac{3}{2}}}
    [/tex]

    [tex]
    \nabla \cdot v = \frac{(x^2 + y^2 + z^2)^{\frac{1}{2}}((x^2 + y^2 + z^2)^3 - 3x^2)}{(x^2 + y^2 + z^2)^3} + \frac{(x^2 + y^2 + z^2)^{\frac{1}{2}}((x^2 + y^2 + z^2)^3 - 3y^2)}{(x^2 + y^2 + z^2)^3} + \frac{(x^2 + y^2 + z^2)^{\frac{1}{2}}((x^2 + y^2 + z^2)^3 - 3z^2)}{(x^2 + y^2 + z^2)^3}
    [/tex]

    on using Simplify in Mathematica:

    [tex]
    \nabla \cdot v = \frac{3(-1 + x^4 + y^4 + 2y^2 z^2 + z^4 + 2x^2 y^2 + 2x^2 z^2))}{(x^2 + y^2 + z^2)^{\frac{3}{2}}}
    [/tex]

    But there was nothing 'surprising' about this result. What am I doing it wrong?

    Also, this problem is from D. Griffith's 'Introduction to Electrodynamics'. And I am at the 2nd chapter which deals with Calculus used in Electrodynamics. The thing is that I am heavily confused between a Gradient, Divergence and Curl. Hence, to overcome that, I think the only way is practicing a lot of questions. Could anyone please show me any source where I could get lots of questions on Gradient, Divergence and Curl to practice [it'd be nice if it were a free resource].

    thanks
     
  2. jcsd
  3. May 28, 2008 #2

    Dick

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    The divergence should be zero. Would that be surprising? There is something going wrong. Since you now know it's wrong, can you correct it?
     
  4. May 29, 2008 #3
    well yes.. a zero divergence would be surprising.. but i have no idea as to what I have done wrong...
     
  5. May 29, 2008 #4

    Dick

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    I was hoping I wouldn't have to check the gory details. It's
    d/dx(x/r^3)+d/dy(y/r^3)+d/dz(z/r^3). Is that what you did? There's a good physical reason for it being zero. I'm glad you are surprised.
     
  6. May 29, 2008 #5
    Yes.. that is what I did... And is it zero because the direction of the vector is always radially outward?
     
  7. May 29, 2008 #6

    nicksauce

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    Why should you be surprised? Try calculating the flux of that vector field through the unit sphere first using the divergence theorem (saying that the divergence is 0) and then calculate the flux directly as a surface integral.

    Edit: But first maybe let's check your original calculation...
     
  8. May 29, 2008 #7

    Dick

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    No, it's because it's the electric field outside of a point charge and divergence measures the density of source for the charge. So outside of the point charge, there is no charge density, hence divergence is zero. Do I really have to check the details, can't you do it? You have mathematica!
     
    Last edited: May 29, 2008
  9. May 29, 2008 #8
    well yes.. got it.. I had calculated the partial derivative wrongly.

    Thanks for the quick reply...
     
  10. May 29, 2008 #9

    Dick

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    d/dx(x/r^3)=(1*r^3-x*3*r^2*dr/dx)/r^6. dr/dx=x/r. Analogously for y and z. Check that.
     
  11. May 29, 2008 #10

    Dick

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    I knew you could do it.
     
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