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Taylor Expansion For Scalar Field

  1. Jun 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Page 35 of Jackson's Electrodynamics (3rd ed), it gives the following equation (basically trying to prove your standard 1/r potential is a solution to Poisson equation):

    [tex] \nabla^2 \Phi_a = \frac{ -1 }{ \epsilon_0 } \int \frac{ a^2 }{( r^2 + a^2)^{5/2} } \rho( \boldsymbol{x'} ) d^3 x' [/tex]

    Where r is distance from center of a sphere with radius R, centered at [itex] \boldsymbol{x} [/itex] and [itex] a [/itex] is a parameter much smaller than R whose limit approaches to 0. And note that [itex] r = \sqrt{ (x' - x)^2 + (y' - y)^2 + (z' - z)^2} [/itex]. R is chosen such that [itex] \rho( \boldsymbol{x'} ) [/itex] changes little inside the sphere. The integral is zero outside the sphere as [itex] a [/itex] approaches zero, so only the inside of the sphere is considered.

    Then the book says, "with a Taylor series expansion of the well behaved [itex] \rho( \boldsymbol{x'} ) [/itex] around [itex] \boldsymbol{x'} = \boldsymbol{x} [/itex], one finds":

    [tex] \nabla^2 \Phi_a = \frac{ -1 }{ \epsilon_0 } \int_0^R \frac{ a^2 }{( r^2 + a^2)^{5/2} } [ \rho( \boldsymbol{x} ) + \frac{ r^2 }{ 6 } \nabla^2 \rho + \ldots ] r^2 dr + O( a^2 ) [/tex]

    Essentially I tried to fill in the gaps between those two steps.

    2. Relevant equations
    Taylor expansion for scalar field:

    [tex] f(\boldsymbol{x'}) = f(\boldsymbol{x}) + \boldsymbol{x} \cdot \nabla f + \frac{ 1 }{ 2 } (\boldsymbol{x} \cdot \nabla)^2 f [/tex]


    3. The attempt at a solution
    Well, essentially I attempted to solve the problem by using the relevant equations. First of all, I changed [itex] d^3 x' [/itex] into spherical polar and the angular parts integrate into [itex] 4 \pi [/itex].

    This is where I get lost because when I did the Taylor expansion, the first order terms did not go away. Secondly the second order term is a problem because I never managed to get a six at the bottom. In fact, what I get is:

    [tex] \rho + r \frac{ \partial \rho }{ \partial r } + \frac{1}{2} r^2 \frac{ \partial \rho }{ \partial r } + \frac{1}{2} r^2 \frac{ \partial^2 \rho }{ \partial r^2 } [/tex]

    Which is not the same as
    [tex] \rho( \boldsymbol{x} ) + \frac{ r^2 }{ 6 } \nabla^2 \rho = \rho( \boldsymbol{x} ) + \frac{ r^2 }{ 6 } \frac{\partial^2 \rho}{\partial r^2} [/tex]
     
  2. jcsd
  3. Jun 23, 2016 #2

    stevendaryl

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    I really don't understand that expansion, either, but it seems pretty much irrelevant, because when you write:

    [itex]\rho(x') = \rho(x) + ...[/itex]

    the only term that survives, in the limit as [itex]a \rightarrow 0[/itex], is the first term.

    What's important is that

    [itex]lim_{a \rightarrow 0} \int_0^R \frac{-3a^2}{(r^2 + a^2)^{\frac{5}{2}}} r^2 dr = -1[/itex]

    Jackson claims that this can be done by "direct integration", but I don't see that, at all. I would first let [itex]r = a u[/itex], to get:
    [itex]lim_{a \rightarrow 0} \int_0^\frac{R}{a} \frac{-3}{(u^2 + 1)^{\frac{5}{2}}} u^2 du[/itex]
    [itex]= \int_0^\infty \frac{-3}{(u^2 + 1)^{\frac{5}{2}}} u^2 du[/itex]

    Then I would use a trig substitution: [itex]u = tan(\theta)[/itex].
     
  4. Jun 23, 2016 #3
    Thanks! Good to know I wasn't just being too dumb to see the obvious.
     
  5. Jul 11, 2016 #4

    nrqed

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    If we align the z axis with the vector ##\nabla f##, then the ## \boldsymbol{x} \cdot \nabla f## term introduces an extra factor ##\cos \theta##. Integrating

    [tex]\int_0^\pi d\theta \sin \theta \cos \theta [/tex]
    gives zero which kills that term.

    The next term introduces a factor ##(\cos \theta )^2##. The integral is
    [tex]\int_0^\pi d\theta \sin \theta \cos^2 \theta = \frac{2}{3} [/tex] which is 1/3 of the result with no factor of ##\cos \theta##. Taking into account the factor of 1/2 from the Taylor expansion, we get the factor of 1/6 of Jackson.
     
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