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Divergence of vector field: Del operator/nabla

  1. Mar 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Let ν(x,y,z) = (xi + yj + zk)rk where v, i, j, k are vectors
    The k in rk∈ℝ and r=√(x2+y2+z2).
    Show that ∇.v=λrk except at r=0 and find λ in terms of k.

    2. Relevant equations
    As far as I understand it, ∇.v=∂/∂x i + ∂/∂y j + ∂/∂z k, but this may very well be wrong.

    3. The attempt at a solution
    That would mean that ∇.v = (i+j+k)rk? So λ= i+j+k. And the as to finding λ in terms of k, no idea! I don't know how to manipulate an equation with a ∇.v in it.
     
  2. jcsd
  3. Mar 10, 2015 #2

    BiGyElLoWhAt

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    Hey, trippy name XD
    It seems to me as though there're actually a few issues we need to get through here. One is the definition of the del operator and the other is the definition of the dot product.
    Del is defined (in cartesian coords.) as ##\vec{\nabla} = <\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z},>##
    The dot product is defined (in all vector spaces) as ##\vec{A}\cdot \vec{B} = |A||B|cos(\theta_{A \to B})##
    I wrote del as a vector, which it is, but it's commonly denoted without the vector notation, as it is always a vector (at least to my knowledge[?]).
    so ##\vec{\nabla} <=> \nabla##
     
  4. Mar 10, 2015 #3
    In this case I got ##\nabla \cdot v = r^k + r^k + r^k = 3r^k## so it looks like ##\lambda = 3##. Remember divergence is a "dot product" which is a scalar. The x derivative of the first component plus the y derivative of the 2nd component, etc. What to do from there I'm not sure if it's as simple as ##\frac{\nabla \cdot v}{r^k} = \lambda## or what. The letter k is used in two different ways & it took me a while to sort out what was what. :oldtongue:
     
  5. Mar 10, 2015 #4

    BiGyElLoWhAt

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    Way to hand that out on a silver platter... -.-
    Also, r is a function of x, y, and z, so you must apply the product rule during differentiation. That is not correct for an r dependant on x, y, & z.
     
  6. Mar 10, 2015 #5
    So it's not 3rk? Very annoying. I was just about to say that I see where that came from, so clearly I still don't understand.
    So I am doing the partial differential dotted with (xi + yj + zk)rk?
     
  7. Mar 10, 2015 #6
    So is the partial derivative ∂/∂x rk+xr-k? Because it's ∂(xrk)/∂x, which is x√x2+y2+z2

    Edit: No, I take that back... There should be a k somewhere.
     
  8. Mar 10, 2015 #7

    BiGyElLoWhAt

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    Yes you are, but what you actually have is this:
    ##\vec{v} = (x\sqrt{x^2+y^2+z^2}^k\hat{i}+y\sqrt{x^2+y^2+z^2}^k\hat{j}+z\sqrt{x^2+y^2+z^2}^k\hat{k})##
     
  9. Mar 10, 2015 #8
    Oops I forgot that r depended on x, y & z so it's not as easy as I thought but yes $$\nabla \cdot v = \frac{\partial (xr^k)}{\partial x} + the rest$$
     
  10. Mar 10, 2015 #9

    BiGyElLoWhAt

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    Here, let me give you an example. Fourier had the right answer assuming r^k was independant of x,y,z , but it's not.
     
  11. Mar 10, 2015 #10

    BiGyElLoWhAt

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    What fourier said ^
     
  12. Mar 10, 2015 #11
    Right. I need to sort the partial derivatives first, then dot those with ν.
    ∂/∂x is rk + kx2r-k, I think.
    ∂/∂y is rk + ky2r-k
    ∂/∂z is rk + kz2r-k

    Then the dot product of those with (xi + yj + zk) rk
     
  13. Mar 10, 2015 #12
    I think I did what Fourier said... But then, I didn't get something of the form λrk. Instead:
    xr2k+kx3+yr2k+ky3+zrk+kz3. So my partial derivatives or my dot product is wrong! But only if I actually dot product the derivatives with the vector.
    Wikipedia makes it sound like ∇.ν is just the partial derivatives not dotted with anything, unless I misunderstood it?
     
    Last edited: Mar 10, 2015
  14. Mar 10, 2015 #13

    BiGyElLoWhAt

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    not quite. the first term is right, but the secont term is not so much.
    You need to apply the chain rule.
    ## \frac{\partial}{\partial x} = \frac{\partial}{\partial r}\frac{\partial r}{\partial u}\frac{\partial u}{\partial x}## where ##u=x^2 + y^2 + z^2## for this case only.
     
  15. Mar 10, 2015 #14

    CAF123

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    Hey whatisreality,
    The computation of ##\frac{\partial}{\partial x} \sqrt{(x^2+y^2+z^2)}^k = (x^2+y^2+z^2)^{\frac{k}{2}}## and similarly for the other two derivatives has been done incorrectly. But you are on right track now.

    Alternatively, without converting to cartesian coordinates, just write ##\mathbf v = r^{k+1}\hat r## and compute ##\nabla \cdot \mathbf v## using the product rules for the divergence of a product of a scalar field (here ##r^{k+1}##) and a vector field (here ##\hat r##).
     
  16. Mar 10, 2015 #15
    Should the second term be kx2rk-1 instead?
     
  17. Mar 10, 2015 #16

    BiGyElLoWhAt

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    Yes it should. Good job W.I.R.!
     
  18. Mar 10, 2015 #17
    Got there! You're being so patient, thank you :)
    Now, I multiply ∂/∂x by xr^k, and ∂/∂y by yr^k etc.
     
  19. Mar 10, 2015 #18

    BiGyElLoWhAt

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    Yes you do, but you've already done the x one, so you need the second two terms.
     
  20. Mar 10, 2015 #19
    Shouldn't they be of the same form? As in ∂/∂y = rk+kyrk-1, ∂/∂z = rk + kzrk-1.
     
  21. Mar 10, 2015 #20

    BiGyElLoWhAt

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    They should be in the same form, but you're missing a squared.
     
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