# Taylor Expansion For Scalar Field

1. Jun 23, 2016

### ibyea

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):

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

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

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

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

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

2. Relevant equations
Taylor expansion for scalar field:

$$f(\boldsymbol{x'}) = f(\boldsymbol{x}) + \boldsymbol{x} \cdot \nabla f + \frac{ 1 }{ 2 } (\boldsymbol{x} \cdot \nabla)^2 f$$

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

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:

$$\rho + r \frac{ \partial \rho }{ \partial r } + \frac{1}{2} r^2 \frac{ \partial \rho }{ \partial r } + \frac{1}{2} r^2 \frac{ \partial^2 \rho }{ \partial r^2 }$$

Which is not the same as
$$\rho( \boldsymbol{x} ) + \frac{ r^2 }{ 6 } \nabla^2 \rho = \rho( \boldsymbol{x} ) + \frac{ r^2 }{ 6 } \frac{\partial^2 \rho}{\partial r^2}$$

2. Jun 23, 2016

### stevendaryl

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

$\rho(x') = \rho(x) + ...$

the only term that survives, in the limit as $a \rightarrow 0$, is the first term.

What's important is that

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

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

Then I would use a trig substitution: $u = tan(\theta)$.

3. Jun 23, 2016

### ibyea

Thanks! Good to know I wasn't just being too dumb to see the obvious.

4. Jul 11, 2016

### nrqed

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

$$\int_0^\pi d\theta \sin \theta \cos \theta$$
gives zero which kills that term.

The next term introduces a factor $(\cos \theta )^2$. The integral is
$$\int_0^\pi d\theta \sin \theta \cos^2 \theta = \frac{2}{3}$$ 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.