I am currently working on [Introduction to Electrodynamics] of David J.Griffiths, chapter 9.5.3.

The book says: In the coaxial transmission line of inner radius a and outer radius b, TEM modes are admitted so that we have $$\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}=0\\ \frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}=0\\ \frac{\partial B_x}{\partial x}+\frac{\partial B_y}{\partial y}=0\\ \frac{\partial B_y}{\partial x}-\frac{\partial B_x}{\partial y}=0 $$ from Maxwell's equations together with $\rho = \mathbf{J} = 0$. The equations are those of electrostatics and magnetostatics in two dimensions.

I understood so far. But here's what I don't understand :

Since the equations are those of electrostatics and magnetostatics in two dimensions, the solution can be borrowed from the case of an infinite line charge and an infinite straight current, so

$$\mathbf{E_0} = \frac{A}{s} \mathbf{\hat{s}}, \mathbf{B_0}=\frac{A}{cs}\mathbf{\hat{\phi}} $$.

I partially understood that if there are cylindrical symmetry in electrostatics then the only situation that can occur is infinite line charge, but there can not only be a infinite line, but also a infinite cylindrical surface charge, etc. How can we just put $\mathbf{E_0} = \frac{A}{s} \mathbf{\hat{s}}$ given that the coaxial line is not a single line at s=0?


1 Answer 1


I think cylindrical symmetry does answer your question. And... there's some kind of "uniqueness" logic working on here.

The number of "independent" TEM modes is related to the homology group of the cross-section (while this fact is not stated in common textbooks). In Griffiths' book it is even "proved" that TEM modes are impossible, but that holds only for waveguides with simply connected cross-section. When there are several "holes" in the cross section, various modes that "encloses" those holes can occur.

And since your problem, the coaxial waveguide, has only one hole, the $\frac{A}{s} \hat{\mathrm{s}}$ solution is the solution!

I hope this does answer your question or at least provides some insights.^^

  • $\begingroup$ I found it difficult for me to understand what homology group is, because of my lack of knowledge in Algebra. Could you explain more about that? I kind of feel that the connectedness of cross-section is something that matters, and there is a hole in there, but I am not understanding why there is a single hole and why is it s=0. Shouldn't there be two holes since there are two cylindrical boundaries; s=a and s=b ? $\endgroup$ May 5, 2021 at 11:54
  • 1
    $\begingroup$ Hi 승우, I'm afraid that I am not understanding your question very well. But what I meant by "one hole" is that, in terms of the violation of simply-connected-ness, the inner boundary s=a makes one "puncture" in the middle of the disk s<=b. $\endgroup$
    – L-C
    May 6, 2021 at 16:36
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    $\begingroup$ And sorry for making you confused by using some highbrow terminology "homology". It is more about geometry and topology rather than algebra, though. Homology is just the study of counting holes of various shapes. So what I wanted to say is, the number of independent TEM modes depends on the number of independent holes in the cross-section! $\endgroup$
    – L-C
    May 6, 2021 at 16:41
  • $\begingroup$ Ahh I understand why is there one hole. I got to know about homology too, though naively. Thank you !! $\endgroup$ May 8, 2021 at 1:45

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