Taylor Expansion For Scalar Field

Homework Statement

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.

Homework Equations

Taylor expansion for scalar field:

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

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
$$\rho( \boldsymbol{x} ) + \frac{ r^2 }{ 6 } \nabla^2 \rho = \rho( \boldsymbol{x} ) + \frac{ r^2 }{ 6 } \frac{\partial^2 \rho}{\partial r^2}$$

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: $u = tan(\theta)$.

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

nrqed
Homework Helper
Gold Member

Homework Statement

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.

Homework Equations

Taylor expansion for scalar field:

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

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
$$\rho( \boldsymbol{x} ) + \frac{ r^2 }{ 6 } \nabla^2 \rho = \rho( \boldsymbol{x} ) + \frac{ r^2 }{ 6 } \frac{\partial^2 \rho}{\partial r^2}$$
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.