Why Don't First-Order Terms Disappear in the Taylor Expansion for Scalar Fields?

[ibyea]

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]

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).

 

[ibyea]

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

 

[nrqed]

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.