Thread about Jackson's Classical Electrodynamics 3rd edition

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SUMMARY

This discussion focuses on the derivations and solutions to exercises in "Classical Electrodynamics" (3rd edition) by J.D. Jackson, specifically addressing the Taylor expansion involving the Laplacian. The participants clarify why the first-order derivative terms vanish in the expansion, attributing this to symmetry in the integrals over spherical domains. Key calculations are provided, including the evaluation of integrals and the behavior of charge distributions under the Poisson equation. The discussion also references external resources, such as Math Stack Exchange, for additional insights.

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  • Understanding of multivariable calculus, particularly Taylor expansions.
  • Familiarity with vector calculus, including Laplacians and gradients.
  • Knowledge of electrostatics and the Poisson equation.
  • Experience with integral calculus, especially in spherical coordinates.
NEXT STEPS
  • Study the derivation of Taylor expansions in multiple dimensions, focusing on symmetry properties.
  • Learn about the application of the Poisson equation in electrostatics, particularly in spherical coordinates.
  • Explore advanced vector calculus techniques, including the evaluation of integrals over symmetric domains.
  • Review solutions to exercises in "Classical Electrodynamics" to deepen understanding of the material.
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Students and professionals in physics, particularly those studying electrodynamics, as well as educators and researchers seeking to clarify complex derivations in Jackson's textbook.

  • #31
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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  • #32
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
 
  • #33
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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  • #34
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##?
 
  • #36
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... :-)
 
  • #37
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.
 
  • #43
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?...
 
  • #44
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)
 

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