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Divergence and solenoidal vector fields

  1. Nov 28, 2004 #1
    I want to find which values of n make the vector field

    [tex]\underline{F} = {|\underline{r}|}^n\underline{r}[/tex] solenoidal.

    So I have to evaluate the divergence of this vector field I think, then show for which values of n it is zero?

    Im starting by substituting:

    [tex]\underline{r} = \sqrt{x^2 + y^2 + z^2}[/tex]

    getting..

    [tex] \underline{F} = {(x^2 + y^2 + z^2)}^{n/2}\sqrt{x^2 + y^2 + z^2}[/tex]

    How can I extract

    [tex]\underline{F_x}, \underline{F_y}, \underline{F_z}[/tex]?

    It's probably really simple but I can't see it! Thanks in advance.
     
  2. jcsd
  3. Nov 29, 2004 #2
    Yeah I was being daft, after sleeping on it I came up with:

    [tex] \underline{F} = [{(x^2 + y^2 + z^2)}^{n/2}x, {(x^2 + y^2 + z^2)}^{n/2}y, {(x^2 + y^2 + z^2)}^{n/2}z] [/tex]

    That's better, yeah?

    so
    [tex] \underline{F}_x = {(x^2 + y^2 + z^2)}^{n/2}x[/tex]
    [tex] \underline{F}_y = {(x^2 + y^2 + z^2)}^{n/2}y[/tex]
    [tex] \underline{F}_z = {(x^2 + y^2 + z^2)}^{n/2}z[/tex]

    now
    [tex]\frac{\partial\underline{F}_x}{\partial x} = nx^2{(x^2 + y^2 + z^2)}^{(n/2)-1} + {(x^2 + y^2 + z^2)}^{n/2} [/tex]

    Hmm if n = 0 the first term is zero as required but the second term would become 1. So how can I find the values of n for which this is zero?
     
    Last edited: Nov 29, 2004
  4. Nov 29, 2004 #3
    Calculating [itex]\nabla \cdot \vec{F}[/itex] should give you:

    [tex]3(x^2+y^2+z^2)^{n/2} + n(x^2+y^2+z^2)^{n/2}[/tex]

    Set that equal to zero and solving for n whould give you n = -3 as long as [itex](x^2+y^2+z^2)^{n/2} \ne 0[/itex].

    I think...
     
    Last edited: Nov 29, 2004
  5. Nov 30, 2004 #4
    Thanks, I was having trouble simplifying my expression for divF, knowing what I was aiming for gave me the confidence to proceed lol! Cheers,

    Matt.
     
  6. Nov 30, 2004 #5
    Following on i'm trying to find the value of [tex]\lambda[/tex] which makes

    [tex] \frac{\lambda\underline{a}}{|\underline{r}|^3} - \frac{(\underline{a}.\underline{r})\underline{r}}{|\underline{r}|^5} [/tex]

    solenoidal. Where a is uniform.

    I think I have to use div(PF) = PdivF + F.gradP (where P is a scalar field and F a vector field)

    and grad(a.r) = a for fixed a.

    So when calculating Div of the above, there should the a scalar field in there somewhere that I can seperate out?!

    I need some pointers please!
     
  7. Nov 30, 2004 #6
    I don't understand what you mean when you say that 'a is uniform'. Fiddle around with the algebra and show me how far you get.
     
  8. Nov 30, 2004 #7
    If this is wrong I can post my working (but its tedious to keep latexing my results!)

    I get the expression down to [tex] \frac{(\lambda -1)a}{x^2+y^2+z^2}[/tex]

    so could I just say that if [tex] \lambda = 1 [/tex] this would give "[tex] \underline{F}[/tex]" say to be zero which implies dF/dx, dF/dy, dF/dz are all zero so divF = 0 which means it is solenoidal?

    Thanks for being patient.

    edit** is the (lambda -1)a the scalar field P to plug into that formula?
     
    Last edited: Nov 30, 2004
  9. Nov 30, 2004 #8
    Where did you get that formula from? It makes no sense to me (i.e. you can't distribute an operator acting on different objects, scalars and vectors in this case). I know it's tedious to show your work but I can't really tell what your doing. For example, you obtained an expression which isn't even a vector!
     
  10. Dec 1, 2004 #9
    sorry it was meant to be..

    [tex] \frac{(\lambda -1)a}{(x^2 + y^2 + z^2)^{3/2}} = \frac{(\lambda -1)a}{|\underline{r}|^3}[/tex]

    I forgot the 3/2 power which didn't make it a vector as you pointed out!
     
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