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Divergence and Radially Symmetric Fields

  1. Mar 20, 2013 #1
    Is it possible for a spherically symmetric field, on all of R^3, to have a divergence of 0? (assuming the field is nonzero)

    Relevant equation:

    F=f(ρ)a (a is a unit vector of <x,y,z>) and f(ρ) is scalar fxn, and ρ = lal
    Last edited: Mar 20, 2013
  2. jcsd
  3. Mar 21, 2013 #2
    You have almost answered your own question! Do you know what the expression for the divergence of your function of ρ only would look like? What happens if that expression is 0?
  4. Mar 21, 2013 #3
    Well after some calculations I came with the conclusion that it is possible with the following field:

    F= r/(p^3) where r = <x,y,z> and p^3 = (x^2+y^2+z^2)^(3/2). The divergence is 0.

    But does this exist on all of R^3?, since the denominator could be 0 at x,y,z=0?
    Last edited: Mar 21, 2013
  5. Mar 21, 2013 #4
    That is indeed not defined at the origin, and we can show that the same holds true of any function with that property.
    If we take the divergence of the most general type of spherically symmetric function: [itex]F(r) = f(r)\hat{r}[/itex], we get [itex](\nabla\cdot F)(r) = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2 f(r))[/itex]. If we set this equal to 0, we see that r2*f(r) must be a constant if f(r) is to be only a function of r, which means [itex]f(r) = \frac{k}{r^2}[/itex] for some constant k. This is necessarily not defined when r = 0.
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