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I More elegant way to solve divergence of arbitrary dotproduct

  1. Jul 13, 2018 #1
    This is more of a general question, but I've encountered this kind of exercises a lot in my current preperations for my exam:

    There are two cases but the excercise is pretty much the same:

    $$(1) \space \operatorname{div}\vec{A}(\vec{r}) \qquad , where \thinspace \vec{A}(\vec{r})=(\vec{a} \cdot \vec{r}) \vec{r} \qquad \vec{a} = const.$$
    $$(2) \space \operatorname{div}\vec{A}(\vec{r}) \qquad , where \thinspace \vec{A}(\vec{r})=(\vec{a} \times \vec{r})\times \vec{r} \qquad \vec{a} = const.$$

    My attempt to these was to always write down the expression that would come out. I will just elaborate this for (1) but I think one gets the gist of it:

    $$\nabla \cdot \vec{A}(\vec{r})= \frac{\partial}{\partial x}(a_xr_x^2+a_yr_yr_x+a_zr_zr_x)+\frac{\partial}{\partial y}(...)+\frac{\partial}{\partial z}(...)$$

    As one can see this is very tedious and exhausting to do, especially during the exam, also the results are very vagues, since it depends on how the derivative of r looks like. So I am wondering, if there is a more elegant way to solve these kind of exercises.

    My assumption is, that there are some neat little tricks concering the problem of having a dotproduct/crossproduct of the same vector and a constant vector. Sadly I couldn't really find anything in my lecture notes, nor on the internet.

    Thanks in advance :)

    P.S.: I hope this is the right place to post this question since I wasn't sure if this is rather a vector analysis or a calculus question
    Last edited: Jul 13, 2018
  2. jcsd
  3. Jul 13, 2018 #2
    Revise your equations first:
    In (1). a.r is a scalar and you have a dot product between this scalar and another r.
    In(2), the left-hand side is a scalar and right hand-side is a vector.
  4. Jul 13, 2018 #3
    I am sorry. I corrected it.
  5. Jul 13, 2018 #4
    Be happy now! Your problem is simple. At first I though you have partial differential equations.

    Use the divergence formula in Cylindrical coordinate system( or Spherical depending on the definition of r). Also there are vector calculus identity for divergence of product of a vector vector and a scalar field (1) or the cross product of two vector fields (2). Find them in the following page:
  6. Jul 13, 2018 #5


    Staff: Mentor

    @maxknrd, you said you edited your post, but the first equation still doesn't make sense. As @Hassan2 already said, you can't take the dot product of a scalar and a vector.
  7. Jul 13, 2018 #6
    The second dot was supposed to be a normal multiplication point. I assumed that this was a normal notation for multiplication of a scalar and a vector
  8. Jul 13, 2018 #7
    This page is brilliant. The problem I have though, is that I don't know anything about r. I just know, that a is constant. The identities are a great help, but with that I would just endup writing long equations like the one I started in the original post and couldn't come to a clear solution imo
  9. Jul 13, 2018 #8
    You would still need to follow some procedure to get the final expression but remember that your vector field depends on r only and in (1) it has one component only. So you only need to consider the derivation with respect to r. This makes the derivation much easier.
    Last edited: Jul 13, 2018
  10. Jul 13, 2018 #9
    Okey so I will then probably have to write it down. Thank you.
    I just really hoped there are some relations of taking a dotproduct/crossproduct with the vector itself
  11. Jul 13, 2018 #10


    Staff: Mentor

    No, it isn't. The usual notation for multiplying a vector by a scalar is juxtaposition (placing the scalar next to the vector, with no intervening operator), like this:
    ##(\vec a \cdot \vec r) \vec r##

    For a simpler example, ##c\vec v## is the scalar multiple of ##\vec v## by the scalar c.
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