# TE and TM modes

1. Apr 17, 2012

### Karthiksrao

I'm a little confused about the analysis of TM and TE modes.

As per definition, TE modes have only H (magnetic field) in the direction of propagation and TM mode has only E in the direction of propagation.

But we also know that when an electromagnetic wave propagates in a medium, E and H fields have to be perpendicular to the direction of propagation (unless the medium is anisotropic (?).

So TE and TM modes exist only in anisotropic media ? If not, what am I missing ?

2. Apr 18, 2012

### Bob S

3. Apr 18, 2012

### Karthiksrao

Thanks. Does it mean that it is completely irrelevant to talk about TE and TM modes when the waves are travelling in any other medium (not a waveguide) ? Do these modes exist only when the waves are travelling in a waveguide ?

4. Apr 18, 2012

### Bob S

Good question.
The important property of a metallic waveguide is total internal reflection. EM waves can also be totally internally reflected inside a dielectric when the direction of propagation exceeds a minimum angle θ to the surface (e.g., total internal reflection and critical angle). So a dielectric can support TE and TM modes. Optical fibers are an example.

"Waveguides can be generally classified as either metal waveguides or
dielectric waveguides. Metal waveguides normally take the form of an
enclosed conducting metal pipe. The waves propagating inside the metal
waveguide may be characterized by reflections from the conducting walls.
The dielectric waveguide consists of dielectrics only and employs
reflections from dielectric interfaces to propagate the electromagnetic wave
along the waveguide
."

See http://www.ece.msstate.edu/~donohoe/ece3323waveguides.pdf

Last edited: Apr 18, 2012
5. Apr 18, 2012

### Karthiksrao

I'm still confused though, is it the total internal reflection that is causing the electric field to have a component along the direction of propagation?

So does it mean when waves are just propagating in any material, without any internal reflection, we just have TEM modes (where both E and H are perpendicular to direction of propagation)?

6. Apr 18, 2012

### Bob S

Yes. Metal and dielectric waveguides are very similar. See http://www.ece.msstate.edu/~donohoe/...waveguides.pdf [Broken]
Yes, like light going through air, lenses, prisms, etc. where refraction is important.

Last edited by a moderator: May 5, 2017
7. Apr 19, 2012

### Born2bwire

When they talk about TE and TM modes, often there is an assumed direction of reference. For example, you will often hear about TE and TM modes in reference to waveguides. But the actual waves generally are still TEM modes in terms of the actual direction of the wave's propagation. The TE and TM is taken in reference to the direction of guidance, typically the direction along the length of the waveguide. What happens is that the wave bounces around inside the waveguide but it does so along the direction of guided propagation. So the direction of guided propagation is not necessarily the same as the direction of the wave's actual propagation dicated by the wavevector. You can think of the case of a parallel plate waveguide where the wave bounces in between the two plates, travelling in a net displacement along the plates. If you look at the actual solutions, you will see that it is a superposition of two plane waves reflecting back and forth inside the waveguide. So while the wave is still technically TEM, we talk about it terms of TE_z and TM_z modes in terms of, in this case, the z-direction.

But I believe true TE and TM modes can exist, like in the regions where sources are present or in special confined cases like a surface wave or evanescant modes.

8. Apr 19, 2012

### Bob S

Actually not. TE and TM stand for transverse electric and transverse magnetic, the field that is perpendicular to the plane of reflection at the boundary (e.g., waveguide wall).

9. Apr 19, 2012

### sunjin09

In a waveguide, the direction of propagation is generally regarded as being along the axis, TE and TM is referencing to the axial direction, so that they do exist in waveguide. Of course, any field can be decomposed into plane waves, which are TEM waves with respect to their own propagation directions. In spherical geometry, even in free space, the referencing direction is generally the radial direction of the coordinate system, so that TE and TM spherical waves exist. But again, if you decompose them into plane waves, each individual component is still TEM w.r.t. its own wave vector.

10. Apr 19, 2012

### Bob S

No. TE and TM refer to the direction orthogonal (transverse) to the axis of the wavegiuide..........
EZ=0 ≡ TE mode,
HZ=0 ≡TM mode

11. Apr 20, 2012

### Born2bwire

This is what both of us previously stated. TE and TM is taken with respect to the direction of guided propagation, which is the axis of the waveguide. This means that the fields are transverse to the direction of guided propagation. This is not what you said in the previous post though. What you said previously,

seems contradictory. It does not make sense to refer to the plane of reflection since, say a rectangulary waveguide has two planes that the waves reflect off of. For the field to be normal to both planes of reflection would require that the only component left must be along the axis of the waveguide.

For the OP, take a look at this set of lecture notes that shows the solution for the rectangular PEC waveguide.

http://www.amanogawa.com/archive/docs/EM15.pdf

The actual field modes look complicated, take a look at slide 272. But as sunjin09 said earlier, you can decompose it into plane waves as seen in the equations on slide 264 (which being the TE mode has no z component to the electric field as z is taken as the direction of guided propagation). The cosine and sine functions can be taken as the superposition of four traveling waves that combine to make two standing waves. Thus,
$$\cos(k_xx)\sin(k_yy)e^{ik_zz} = -.25i (e^{-ik_xx} + e^{ik_xx})(e^{-ik_yy} - e^{ik_yy})e^{ik_zz}$$
So one plane wave travels in the $\mathbf{k} = k_x\hat{x}+k_y\hat{y}+k_z\hat{z}$ direction while another travels in the $\mathbf{k} = -k_x\hat{x}+k_y\hat{y}+k_z\hat{z}$ and so forth. The end result though when you combine all the possible waves is that you have a standing wave in the x and y directions while a traveling component in the z direction. But it is possible to only excite the plane waves that make up the solution individually and watch it bounce back and forth travelling in the z direction as illustrated on slide 243. This would be more apparent in a time-domain simulation where we can watch the wave bounce back and forth. The solutions above are time-harmonic and represent the field distributions in all of space that satisfy the boundary conditions.

12. Apr 20, 2012

### Bob S

Take a look at slide 241 inhttp://www.amanogawa.com/archive/docs/EM15.pdf.
The TE mode is defined by the fact that the E vector is perpendicular to the plane of reflection off the wall of the waveguide. It can be either TEm0 or TEon depending on which wall it reflects off of.

13. Apr 21, 2012

### vanhees71

This is not very helpful. As already stated, TE (TM) waves are such where the electric (magnetic) field modes are perpendicular to the axis of the wave guide. A special mode is the TEM mode, where both fields are perpendicular to the axis. It can only exist in wave guides with multiply connected cross sections (e.g., coaxial cable).

14. Apr 22, 2012

### sunjin09

Honestly, these TE and TM terminologies are somehow arbitrary, they are defined differently. Believe it or not, the FDTD computational EM community calls a TE mode pertaining to the reflection from a half space (E field parallel to the interface) TMZ mode, since they always take the interface as the y-z plane, and look downward from the z-axis, so that a TE mode (as defined in most physics books) with E field in the z direction and H field in the x-y plane (transverse to z), is called "TMZ" ...

15. Apr 23, 2012

### Born2bwire

The electric field vector on that slide in the TE mode is parallel to the reflecting surface. The red vector denotes the direction of the electric field.

16. Apr 23, 2012

### Bob S

Correct. What I mean is that the E vector is perpendicular to the plane made by the incident wave and the reflected wave, not to the reflecting surface.