How did Jackson simplify the Taylor expansion for charge density in 3-D?

[techmologist]

On p35 of Jackson's Classical Electrodynamics 3rd Edition, the author gives the expansion of the charge density $\rho(\mathbf{x'})$ around $\mathbf{x'}=\mathbf{x}$ as

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

where $$r = |\mathbf{x} - \mathbf{x'}|$$

My question is, where did the first order terms go, and why is the second order term not in the form of a Hessian matrix? For instance:

$$\rho(\mathbf{x'}) = \rho(\mathbf{x}) + \mathbf{r} \cdot \nabla \rho + \frac{1}{2}\mathbf{r^T}H_{\rho}\mathbf{r} + ...$$.

He says the charge density doesn't change much in the small spherical volume under consideration, so maybe that explains the absense of first order terms, but I still don't see how to get his version of the second order term. Thank you.

 

[techmologist]

Maybe it would help to give some context. Jackson is demonstrating that

$$\Phi(\mathbf{x}) =\frac{1}{4\pi\epsilon_0} \int \frac{\rho(\mathbf{x'})}{|\mathbf{x} -\mathbf{x'}|}d^3x'$$ (over all space) is a solution to the Poisson equation:

$$\nabla^2 \Phi = -\frac{\rho(\mathbf{x})}{\epsilon_0}$$.

He does this by first showing that

$$\Phi_a(\mathbf{x}) = \frac{1}{4\pi\epsilon_0}\int \frac{\rho(\mathbf{x'})}{(|\mathbf{x} -\mathbf{x'}|^2+a^2)}d^3x'$$

satisfies Poisson's equation and then letting a->0. After taking the Laplacian operator inside the integral, he then argues that for reasonable charge distributions, any contribution from outside a small sphere of radius R centered at x goes to zero like a^2. He then Taylor expands $\rho(\mathbf{x'})$ about $\mathbf{x'} = \mathbf{x}$ in the strange way shown above, and computes resulting integral (over the small ball of radius R) to get $-\rho/\epsilon_0$. I just don't understand the way he did the taylor expansion.

 

[techmologist]

Maybe it would help to give some context. Jackson is demonstrating that

$$\Phi(\mathbf{x}) =\frac{1}{4\pi\epsilon_0} \int \frac{\rho(\mathbf{x'})}{|\mathbf{x} -\mathbf{x'}|}d^3x'$$ (over all space) is a solution to the Poisson equation:

$$\nabla^2 \Phi = -\frac{\rho(\mathbf{x})}{\epsilon_0}$$.

He does this by first showing that

$$\Phi_a(\mathbf{x}) = \frac{1}{4\pi\epsilon_0}\int \frac{\rho(\mathbf{x'})}{(|\mathbf{x} -\mathbf{x'}|^2+a^2)}d^3x'$$

satisfies Poisson's equation and then letting a->0. After taking the Laplacian operator inside the integral, he then argues that for reasonable charge distributions, any contribution from outside a small sphere of radius R centered at x goes to zero like a^2. He then Taylor expands $\rho(\mathbf{x'})$ about $\mathbf{x'} = \mathbf{x}$ in the strange way shown above, and computes resulting integral (over the small ball of radius R) to get $-\rho/\epsilon_0$. I just don't understand the way he did the taylor expansion.

I see what Jackson is doing now. He must have omitted a few steps of reasoning out of respect for the reader's intelligence. Apparently, Jackson has a very deep respect for the reader's intelligence.

The reason he can write the Taylor expansion that way is because it's being integrated over the small ball of radius R. The first order term integrates out to zero by spherical symmetry. The second order term has the form he gave it because you are free to choose coordinates that diagonalize the Hessian matrix. The other factor in the integrand depends on r = |x-x'| only, so when you integrate over the interior of the spere, the Laplacian term of the charge density is all that's left of second order terms.

Also, the integral I gave in the previous post should of course be

$$\Phi_a(\mathbf{x}) = \frac{1}{4\pi\epsilon_0}\int \frac{\rho(\mathbf{x'})}{\sqrt{(|\mathbf{x} -\mathbf{x'}|^2+a^2)}}d^3x'$$