By "modes" I here refer to TE (transverse electrical; s-polarized) and TM (transverse magnetic; p-polarized) modes.

After going through these notes(p 7-8) I somewhat understand the TE and TM modes: The idea is that any vector can be resolved into two components, one parallel and one perpendicular with respect to the chosen resolution axis. Along the same lines, we can think that there is a plane wave incident on a plane or a slab.
$$\vec{E} = \vec{E}_{0}(x,y) e^{-ik_{z}z} $$ Our $yz$-plane or slab is chosen (by convention) as $\hat{y}$. We resolve $\vec{E}$ into two vectors, one along $\hat{y}$ (TE mode) and another perpendicular to it (TM mode). At the end we just add the contribution from both to get the total $\vec{E}$.

My question is: How does the TEM (transverse electromagnetic) mode fit into this? How do we see the following:

We have that $E_{z} = 0$, $H_{z}\ne 0$ for a TE mode and $H_{z} = 0$, $E_{z}\ne 0$ for a TM mode.

Physical significance is what I am looking for. I checked a lot of references(using normal and transverse as seperate modes!) but these only add to my confusion regarding these modes.


1 Answer 1


You're confused because there are two separate concepts here: the plane interface problem, and waveguide modes. The definition for what is TE or TM is completely different between the two cases. Confusing, I agree.

In waveguide modes, the definition is based on what is transverse to the direction of propagation in the waveguide. So TE modes have an electric field completely transverse to the direction of propagation, with a non-zero transverse magnetic field. TM modes have a magnetic field completely transverse to the direction of propagation, with a non-zero transverse electric field. In TEM modes both the electric and magnetic field have zero components in the transverse direction.

For the plane interface problem, the meaning of transverse is different. Now we're concerned with transverse to the plane of incidence, not the direction of propagation. It's kind of an opposite definition. So TE modes (s-polarized) have electric field completely transverse to this plane. This plane is the xz-plane in your notes, so the electric field is y-polarized, and the magnetic field has no polarization in the y-direction. TM modes (p-polarized) have magnetic fields completely transverse to this plane (y-polarized) and electric fields have no components in the y-direction.

This definition of TE/TM for the plane interface problem can vary. I believe I have seen the definition of TE and TM switched so that they describe quantities that are transverse to the direction of propagation. This makes the definition more consistent with the definition used in waveguide modes.

As per usual, keeping track of conventions is half the battle.

Some references:

  • Chapter 8 of Classical Electrodynamics by Jackson. That's based on my super old 2nd edition. I'm guessing it's true for 3rd edition as well.
  • A good writeup from Rutgers
  • Some slides from MIT.
  • Elements of Electromagnetics by Sadiku, chapter 12 in the second edition.
  • Advanced Engineering Electromagnetics by Balanis, chapter 8
  • $\begingroup$ Nice and helping one. To clarify something: you said for the direction of propagation, TE modes have an electric field completely transverse to the direction of propagation, with a non-zero transverse magnetic field? This last part(in italics) is bit unclear to me. Confusion is same for TM also. $\endgroup$
    – Shamina
    Mar 16, 2017 at 8:29
  • $\begingroup$ @Shamina For waveguides, if $z$ is the direction of propagation (i.e. transverse), so that the $xy$ cross section is the same for any $z$ value, then for TE modes, the electric field has vector components in the $x$ and $y$-direction, but no $z$ components. However, the magnetic fields can have vector components in all three directions. So we can write $\mathbf{E}_\text{TE} (x,y,z) = ( \mathbf{\hat{x}} E_x(x,y) + \mathbf{\hat{y}} E_y(x,y) ) e^{ik_z z}$ and $\mathbf{H}_\text{TE} (x,y,z) = ( \mathbf{\hat{x}} H_x(x,y) + \mathbf{\hat{y}} H_y(x,y) + \mathbf{\hat{z}} H_z(x,y) ) e^{ik_z z}$. $\endgroup$
    – LedHead
    Mar 16, 2017 at 11:21
  • $\begingroup$ But if I am not mistaken, for EM waves(isotropic) $E⃗ ⊥k⃗ ⊥H⃗ $ . This should hold? if $k⃗$ is in $\hat{z}$ then $E⃗$ and $H⃗$ are in $x−y$ plane? Then what is TE and TM about it? So I am really confused now. $\endgroup$
    – Shamina
    Mar 16, 2017 at 12:20
  • $\begingroup$ $\mathbf{E} \perp \mathbf{k} \perp \mathbf{H}$ is true for plane waves propagating in an infinite homogeneous medium. It is not in general true for waveguide modes. For more details, look at Chapter 8 in Classical Electrodynamics by Jackson. (I have a really old 2nd edition, but I'd guess 3rd ed. is the same.) This is a good write up, too. $\endgroup$
    – LedHead
    Mar 16, 2017 at 13:12
  • $\begingroup$ Thanks for this. The reason I thought it to be $E ⊥ k ⊥ H$. Because, $∇⋅ E$ =0 and $∇⋅ H =0$(Transverse conditions). I always thought this is in general true the mutually perpendicular. I never thought they are not necessarily $\perp$ in general, atleast inside the waveguide, reason I don't know. But thanks to you to explain it me today, the illusion I was taking with me. $\endgroup$
    – Shamina
    Mar 16, 2017 at 14:16

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