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?
Best Answer
This is an excellent question but you must restrict it for the case of a homogeneous transmission line, otherwise it is not true. The line must consist of a pair of parallel conductors that individually may have different but otherwise arbitrary cross sections, one may contain other as in a coaxial line, and the lines are embedded in propagating medium that is homogeneous both in cross-section and along the axis.
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.
The intuition you are asking for is in the static nature of this propagating field it being a slight modification, with restriction to the conductors, of a plane wave whose phase velocity then depends only on the medium. This does not mean that a different, not TEM, mode may not propagate at a geometry dependent speed, and, in fact, the phase velocity of all the other modes, be it TE or TM, does depend on the shape of the contours and their separation.