# Why is the phase velocity in a transmission line not affected by its geometry?

When deriving the speed of light in vacuum, one usually starts from Maxwell's equations, does some calculus and finds a wave solution with the phase velocity $$c = 1/\sqrt{\mu_0 \varepsilon_0}$$. This is clear to me.

When analyzing transmission lines, the approach is not quite the same. One starts with a model of distributed inductance, capacitance, resistance and admittance along the transmission line from which one obtains the telegrapher's equations.

Solving these equations (with a sinusoidal ansatz) again leads to a wave solution of the form $$\exp(\mathrm i \omega t - \gamma x)$$ where $$\gamma = \sqrt{(R + \mathrm i \omega L) (G + \mathrm i \omega C)} .$$ For a lossless transmission line ($$R = 0 = G$$) we thus find $$v_{\mathrm{ph}} = \omega / \operatorname{Im}(\gamma) = 1 / \sqrt{L C}$$.

Now, in the literature I've been looking at (on calculating $$v_{\mathrm{ph}}$$ for coplanar waveguides, but this is unimportant, I suspect) it is stated as an obvious fact that in the absence of any dielectric except vacuum $$v_{\mathrm{ph}} = c$$ in the transmission line. This is not obvious to me at all. Shouldn't the waveguide geometry (which determines $$L$$ and $$C$$, after all) be able to affect this value? The idea seems to be that only the space in which the fields propagate determine this velocity and that the conductors, carrying merely currents and charges, can't affect it.

Can you justify this fact and make it intuitive for me?

Does this still hold with a lossy transmission line? That is, if I "switch on" the resistance $$R > 0$$ without changing the geometry of an existing transmission line or any nearby dielectrics, will $$L$$ and $$C$$ "magically" adjust in order to keep $$\operatorname{Im}(\gamma)$$ (and thus $$v_{\mathrm{ph}}$$) constant?

• The geometry matters. Look up "slow wave structures". Commented Jan 11, 2023 at 2:01

If you have that then the fundamental mode of propagation is essentially the static $$E$$ and $$H$$ fields along the line and because of the assumed homogeneity that field has no longitudinal component, that is a TEM mode. The fields can be derived from a scalar potential $$\Phi$$ so that with $$\kappa = \omega \sqrt{\epsilon \mu}$$: $$\mathbf E_t =\nabla_t \Phi e^{-\mathfrak j \kappa z} \qquad E_z=0\\ \mathbf H_t = \pm \sqrt{\frac{\epsilon}{\mu}} \mathbf {\hat z} \times \mathbf E_t \qquad H_z=0 \\ \nabla^2_t \Phi = 0 \\ \Phi({\mathcal{C}_1}) =\Phi_1 \qquad \Phi({\mathcal{C}_2}) =\Phi_2$$ where $$\mathcal{C_1}$$ and $$\mathcal{C_2}$$ are the contours of the conductors at which the scalar potential $$\Phi$$ are given constants. If the medium is lossy then the $$\epsilon$$ and $$\mu$$ are complex quantities.
• Ah, I see, that makes a lot of sense! A lossy medium (i.e. complex $\epsilon$ and $\mu$) correspond only to a nonzero $G$ though, wouldn't it? A nonzero $R$ means that the losses happen in the conductor, after all, not in the dielectric. Or am I missing something? Commented Jan 11, 2023 at 10:48
• Exactly, but note that while $R$ usually models the losses in the conductor that I completely neglect in the above one can have both $\epsilon$ and $\mu$ be lossy and then result in a complex $\kappa$ and wave impedance $\sqrt{\frac{\mu}{\epsilon}}$. Thus media contribute to $R$ as well as to $G$, but transmission lines are rarely used on magnetic material (except in a few ferrite-based applications) so only then only $\Im[ {\epsilon}]$ would show up in modeling $G$. To include conductor losses formally you would need to use the so-called "impedance boundary" approximation with the TEM waves. Commented Jan 11, 2023 at 14:34
• neglect $G$, keep $R<<\omega L$, then $\gamma = \sqrt{-\omega^2 LC} \sqrt{1-\mathfrak j R/(\omega L)} \approx \mathfrak j \omega \sqrt{LC} (1-\mathfrak j R/(2\omega L)$, so in the propagation term $e^{-\gamma z} = e^{-\mathfrak j \kappa z} e^{-\alpha z}$ where $\alpha = R/(2Z_0)$, $Z_0=\sqrt{L/C}$ showing that $R$ contributes only to the exponential amplitude decay and not to the phase velocity. Commented Jan 11, 2023 at 15:46