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Griffiths Chapter 10 del cross position vector

  1. May 31, 2013 #1
    I am working through chapter 10 of Griffith’s electrodynamics (for fun and in my spare time). While I don’t have a formal bucket list, getting to an understanding of how Newton’s third law is not as straightforward for electrodynamics has been on my mental bucket list.
    I am an engineer not a physicist. I find myself having to research and review a good bit especially on the mathematics.
    On page 436, Griffith is showing the differentiation and simplification of
    [tex] \nabla V [/tex] (pg. 435)

    Through some product rules and algebra and so forth he is simplifying terms.
    On the bottom of page 436 he has:
    [tex] \nabla \times \mathcal r = \nabla \times \mathtt r- \nabla \times w [/tex]
    (X = cross product)
    W = position function for moving charge
    Script r = r – w
    [tex] \mathcal r = \mathtt r- w [/tex]

    r = position vector
    My question is in the text, the author has the statement:
    [tex] \nabla \times \mathtt r = 0 [/tex]
    No reason, no background just the statement. So I assume I am supposed to know this and it is probably obvious but I don’t see it.
    I wrote out the cross product for del x r in matrix form and turned the crank – nothing canceled or became apparent.
    Can you explain why
    [tex] \nabla \times \mathtt r = 0 [/tex]
    Del cross position vector is zero?
  2. jcsd
  3. May 31, 2013 #2


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    You must have done it wrong. Post your work and we can see what happened.
  4. May 31, 2013 #3
    Its easy to see in Cartesian coordinates. The position vector is
    [itex]\vec{r} = x \hat{e_x} +y \hat{e_y} +z \hat{e_z} [/itex]

    The curl of this vector is

    [itex]\nabla \times \vec{r} = \left( \frac{dz}{dy}-\frac{dy}{dz} \right)\hat {e_x}+\left( \frac{dx}{dz}-\frac{dz}{dx} \right)\hat {e_x}+\left( \frac{dy}{dx}-\frac{dx}{dy} \right)\hat {e_z} = \vec 0[/itex]
  5. May 31, 2013 #4


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    Hi there! Since this is a vector equation, we can evaluate it in any coordinate system and if it holds in one, it will hold in all. Say we choose cartesian coordinates so that ##\mathbf{r} = x\hat{\mathbf{x}} + y\hat{\mathbf{y}} + z\hat{\mathbf{z}}##. Then, ##\nabla\times \mathbf{r} = \begin{vmatrix}
    \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\
    \partial_{x} & \partial_{y} & \partial_{z} \\
    x& y &z
    \end{vmatrix} = (0 - 0)\hat{\mathbf{x}} - (0 - 0)\hat{\mathbf{y}} + (0 - 0)\hat{\mathbf{z}}## because ##\partial_{j}x^{i} = \delta^{i}_{j}##. Hence ##\nabla\times \mathbf{r} = 0## in any coordinate system.

    Equivalently, you can show this in a coordinate-free manner from the start: ##(\nabla\times \mathbf{r})^{i} = \epsilon^{ijk}\partial_{j}x_{k} = \epsilon^{ijk}\delta_{jk} = 0##.
  6. May 31, 2013 #5
    Oh crap - is it this obvious: the derivative of "z" with respect to y (which is no function of z :) )is zero likewise for all of the derivatives?
  7. May 31, 2013 #6


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  8. May 31, 2013 #7
    once again - embarrassing and I couldn't see the forest for the trees

    Thank you all for the help!!
  9. May 31, 2013 #8


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    It's not always easy to see the forest for the trees unless someone else points it out to you. Happens to everyone-what's important is you get the physics. Good luck!
  10. May 31, 2013 #9
    thank you!

    In my "circle" I really don't have any friends or resources I can run question like this by. I do appreciate this forum!

    I do enjoy playing with the physics and it is cool when the light goes on and I get it. When I heard about this Newton's 3rd law interesting deal in electrodynamics, I bought Griffith's text book. I have been slowly going through this material off and on and feel very close to being able to explain it. (something an under grad physics major could do easily but as an older engineer removed from some of the mathematics I am doing this just for fun)

    Thanks again
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