As I understand, in an ideal TEM transmission line we can set up the telegrapher equations and solve to show that the line propagates voltage and current waves.

My confusion arises when we recognize that the boundary conditions we have enforced for this ideal situation imply the electric field in the system is always perpendicular to the surfaces of the conductors, and the magnetic field is always parallel to the surfaces but perpendicular to the direction of propagation (direction of the line).

If there is AC current in the line, then locally charges are being accelerated in the direction of propagation.

If charges are being accelerated, locally, along the line in one direction or the other, then doesn't there need to be, locally, either some electric field that is parallel to the direction of propagation, or (more weirdly) some velocity of charge carriers in the line that is perpendicular both to the magnetic field and the direction of propagation?

Furthermore, doesn't the mere fact that we have modulated charge density along the line (before or because we have modulated current density) imply that there must be electric field pointing between the peaks and the valleys of this modulated charge density, and hence along surface of the line, not perpendicular to it?

I believe that in situations with nonideal conductors electric field along the direction of propagation is not forbidden roughly up to the skin depth of the conductor, but as I understand it this is not the case for the ideal situation of lossless conductors, which is the system confusing me right now.

Am I wrong about the assumption that there is no electric field along the conductor? Is there some sense in which the lumped element approximation of the transmission line buries this fact? Can, in classical electrodynamics, charges be accelerated without any local electric or magnetic fields (something that would be surprising to me and indicate I really have misunderstood something)?

Fundamentally my question is 'where, explicitly, is the force coming from that is accelerating the charges along the line?' I don't know if this is possible, but I would appreciate the answer being in terms of electric and magnetic fields, not scalar or vector potentials.

As always, thanks for your attention.

  • $\begingroup$ In my opinion, under the assumption (or mathematical model) of "perfect" conductor, we don't need any electric field to drive or even accelerate the charges inside the conductor. They can just move freely, at no cost. $\endgroup$
    – George C
    Jan 16, 2021 at 7:50
  • $\begingroup$ Think about how we introduce the concept of "perfect conductor": we let the conductivity equal ∞. Then Ohm's law J=σE implies that we can get arbitrary J with zero E inside the conductor. $\endgroup$
    – George C
    Jan 16, 2021 at 7:50

1 Answer 1


There is a longitudinal component of electric field! As you suspect, this component must be present to drive the longitudinal current flow, but it is far smaller than the transverse component.

Consider a parallel plate, metallic waveguide. You can (rigorously) think of the modes of this structure as $p$-polarized waves bouncing back and forth between the metal plates. Therefore there’s a longitudinal component of electric field! The angle of incidence to the plates is set by the wavelength and plate separation in order to provide constructive interference (that’s how you solve for the mode).

For the lowest order mode of a structure with width larger than the wavelength, the angle of incidence (to the metal) is close to 90$^\circ$, which explains why the longitudinal component of electric field is small compared to the transverse component. Such a mode will have an “effective refractive index” of close to 1 (or whatever the index of the core dielectric is).

  • $\begingroup$ My understanding is that the bouncing wave picture is only applicable to TE and TM modes of a waveguide in which you explicitly allow this longitudinal component of either the electric or magnetic field. In the derivation of TEM waves you set $E_z = H_z = 0$ a priori, which would preclude any bouncing wave interpretation of a solution, but then you still deduce currents in the conductors in the ideal lossless case (see Pozar Ch3 for example). My current mental model is actually that, assuming there is already some current in the transmission line, the magnetic field produced by... $\endgroup$ Jan 1, 2021 at 19:39
  • $\begingroup$ each conductor will induce cyclotronic orbits in the current carriers on the surface of the other conductor, which will in turn propagate the magnetic field further along the waveguide, and this process will continue freely with no loss. The longitudinal electric field induced by the modulated charge density will then be cancelled perfectly by the longitudinal electric field induced by the propagating magnetic flux, but the induced transverse electric field will be left in tact to generate the voltage difference along the line... $\endgroup$ Jan 1, 2021 at 19:44
  • $\begingroup$ This is completely hand-wavey and unverified though, and seems suspiciously complicated to me. I'd like to be able to confirm or deny it by a calculation but I don't really know how. $\endgroup$ Jan 1, 2021 at 19:46
  • $\begingroup$ @AndreasButler As your cognitive dissonance can attest, the $E_z = 0$ assumption is an approximation. It works well, though, in the right conditions! Think of it this way: If there is a positive charge accumulation in the conductor adjacent to a negative charge accumulation (one half wavelength away), then there is an electric field pointing from from the positive charges to the negative charges, just by Gauss’ Law. Does this help? $\endgroup$
    – Gilbert
    Jan 2, 2021 at 0:20
  • $\begingroup$ Hm, sorry, no, but I appreciate you sticking with me. When someone derives the expressions for TEM modes I do not see where we they are ever making any approximations $E_z \sim 0$. Pretty explicitly it is always taken a priori that $E_z = H_z = 0$. I understand that in reality we never encounter true TEM modes because of loss, but the point of my confusion is that in the mathematics of the ideal lossless TEM waveguide we stipulate $E_z = 0$ and then Maxwell's equations give longitudinal current. There isn't really leeway for that condition to be relaxed after the fact, as far as I can see. $\endgroup$ Jan 2, 2021 at 1:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy