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What is the divergence of a unit vector not in the r direction?

  1. Feb 1, 2013 #1
    Hi guys,

    I've run across a problem. In finding the potential energy between two electrical quadrupoles, I've come across the expression for the energy as follows:

    [itex]U_{Q}=\frac{3Q_{0}}{4r^{4}}\left[(\hat{k}\cdot \nabla)(5(\hat{k}\cdot \hat{r})^3-2(\hat{k}\cdot \hat{r})^2-(\hat{k}\cdot \hat{r}))\right],[/itex]

    where [itex]\hat{k}[/itex] is the orientation of the quadrupoles, and [itex]\hat{r}[/itex] is the direction between the quadrupoles.

    If I let [itex]\hat{r}[/itex] be in the [itex]\hat{z}[/itex]-direction, I get

    [itex]U_{Q}=\frac{3Q_{0}}{4r^{4}}\left[(\hat{k}\cdot \nabla)(5(\cos{\theta})^3-2(\cos{\theta})^2-(\cos{\theta}))\right].[/itex]

    My problem now is, that I don't know what to do about the divergence of the [itex]\hat{k}[/itex]-vector. I would like to do the differentiation in cartesian coordinates, but have them translated into spherical polar coordinates. I know, that the result should probably involve a [itex]\frac{1}{r}[/itex]-factor, but I can't seem to do it right. I've tried to rewrite [itex]\hat{k}[/itex] in polar coordinates and tried using the chain rule on the derivative, but I get 3 as an answer. So I don't know if the initial expression is wrong, or I just dont know how to take the derivative. Can anyone please help?

    Thanks,
     
  2. jcsd
  3. Feb 1, 2013 #2

    chiro

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    Hey SiggyYo.

    Have you tried representing a transformation between spherical and cartesian?

    (Example: for (r,theta) -> (x,y) we have y/x = arctan(theta) and x^2 + y^2 = r^2 which can be used to get (x,y)).
     
  4. Feb 2, 2013 #3
    Thank you chiro for the quick response.

    I am afraid I don't know what you mean. Wouldn't I just obtain the usual
    [itex]x=r\sin{\theta}\cos{\phi}[/itex]
    [itex]y=r\sin{\theta}\sin{\phi}[/itex]
    [itex]z=r\cos{\theta}[/itex]?

    Also, I want [itex]\hat{k}[/itex] to be a unit vector, which gives me [itex]r=1[/itex]. How do I take this into account, when trying to get a result with a factor of [itex]\frac{1}{r}[/itex]? I am really lost on this one :P
     
  5. Feb 2, 2013 #4

    chiro

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    If k is a unit vector, then I don't think you will have any extra terms.

    I'm not really sure what you are doing or trying to say: you have a conversion from polar to R^3 and provided the formula is correct, you should be able to plug these definitions in.

    Also is the r term in your equation related to some vector in polar or is it some other variable?
     
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