Thread about Jackson's Classical Electrodynamics 3rd edition

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I am reading now the solution of problem 3.19 from Jackson (if you know of another more complete solution let me know);

here:
http://homerreid.dyndns.org/physics/jackson/jack3c.pdf

And it seems he's stuck at calculating the integral ##\int_0^\infty \frac{\sinh(kz_0)}{\sinh(kL)}kdk##, can you help both of us?

Thanks in advance!
Cheers!
 
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I read someone's remark on solving Jackson's problems:
"Getting any single Jackson problem completely correct could be a life’s work!"

LOL! wish I had the time to solve completely correctly the 350 problems in the third edition, but I have exams... :-D
 
Mathematica gives as a result (of course only valid for ##L>z_0##)
$$\frac{1}{4L^2} \left [\psi'\left (\frac{L-z_0}{2L} \right )-\psi' \left(\frac{L+z_0}{2L} \right ) \right ],$$
where
$$\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}$$
is the socalled "digamma" function.
 
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vanhees71 said:
Mathematica gives as a result (of course only valid for ##L>z_0##)
$$\frac{1}{4L^2} \left [\psi'\left (\frac{L-z_0}{2L} \right )-\psi' \left(\frac{L+z_0}{2L} \right ) \right ],$$
where
$$\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}$$
is the socalled "digamma" function.
Does this expression simplifies to the expression in problem 3.19 in Jackson's? i.e with the ##\sec^2##?
 
I have a question regarding problem 5.5 from Jackson's third edition.

Do I actually need to post it (I mean I assume that you have a legal or illegal copy of the book).

I've found a solution to problem 5.5 from here:
http://pages.uoregon.edu/gbarello/Resources/Papers/Homework/Electrodynamics/HW4.pdf
on page 8 the solution to assignment 5.5b, the last solution has the wrong coefficients, so I wonder where did he go wrong there?

Thanks, jesus and out... :-)
 
Does someone know if some of the problems in Jackson's third edition can be found in reference books?

There are some problems about superconducting material which I thought may be found in Tinkham's reference.
 
vanhees71 said:
Admittedly the calculation is a bit confusing, because of the shorthand notation used by Jackson. Calculated is the volume integral

For ##n \geq 2## I cannot verify generally Jackson's result, but all you need is to show that the integral is 0 for ##a \rightarrow 0##. Now introduce again spherical coordinates, and you see that all these integrals are proportional to
$$3 a^2 \int_0^R \mathrm{d} r \frac{r^{n+2}}{(r^2+a^2)^{5/2}}=3a^2 I_n.$$
[corrected in view of the hint in #7]
For ##n=2## we get
$$I_2=\int_0^R \mathrm{d} r \frac{r^{4}}{(r^2+a^2)^{5/2}}=\mathrm{arsinh} \left ( \frac{R}{a} \right )-\frac{4R^3+3ra^3}{4(R^2+a^2)^{3/2}}.$$
Obviously one has
$$\lim_{a \rightarrow 0} a^2 I_2=0.$$
I'm not very good in curvilinear coordinates, let me try to expand it in cartesian coordinates instead. I'm not sure whether the reasoning is correct though (note that odd components of ##\mathbf{h} = |\mathbf{x^{'}} - \mathbf{x}|## cancel)...
\begin{align*}
\rho(\mathbf{x}^{'}) & =\rho(\mathbf{x}) + \frac{1}{1!}\frac{\partial \rho}{\partial \mathbf{h}} +\frac{1}{2!}\frac{\partial^2 \rho}{\partial \mathbf{h}^2} + \cdots\\
&=\rho(\mathbf{x}) + (\mathbf{x^{'}} - \mathbf{x})\cdot\nabla \rho + (\sum_{i} (x^{'}_{i} - x_{i} \frac{\partial}{\partial e_{i}}))(\sum_{i} (x^{'}_{i} - x_{i} \frac{\partial}{\partial e_{i}}))\rho + \cdots\\
&=\rho(\mathbf{x}) + \frac{1}{2} ( (x^{'} -x)^2 \frac{\partial^2 \rho}{\partial x^2} + (y^{'} -y)^2 \frac{\partial^2 \rho}{\partial y ^2}+ (z^{'} - z)^2 \frac{\partial^2 \rho}{\partial z^2})+ \cdots\\
&=\rho(\mathbf{x}) + \frac{|\mathbf{x^{'}} - \mathbf{x}|^2}{6} ( \frac{\partial^2 \rho}{\partial x^2} + \frac{\partial^2 \rho}{\partial y ^2} + \frac{\partial^2 \rho}{\partial z^2})+ \cdots\\
&=\rho(\mathbf{x}) + \frac{r^2}{6}\nabla^2 \rho + \mathcal{O}(r^3)
\end{align*}
Since ##\rho## does not vary much in the volume of interest we may assume the function is spherically symmetric, we see that $$ (x^{'} -x)^2 \frac{\partial^2 \rho}{\partial x^2} \approx (y^{'} -y)^2 \frac{\partial^2 \rho}{\partial x^2}$$

and $$ 3 (x^{'} -x)^2 \frac{\partial^2 \rho}{\partial x^2} \approx |\mathbf{x^{'}} - \mathbf{x}|^2 \frac{\partial^2 \rho}{\partial x^2}$$

Right?...
 
By the way, judging by how seasoned the advice Jackson gives, how comprehensive the treatment is and the copious amounts of references given, I wonder, did Jackson actually read through all the books he cite? (or at least read a few chapter)