In parallel plates, TM0 mode will become TEM0 mode, but why TE0 mode does not exist? How to explain it with its physical properties?
[Physics] Why TE0 mode does not exist in parallel plates
electromagnetismwaveguide
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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
The electron will accelerate, which, roughly speaking, will create an increasing magnetic field and therefore an electromagnetic wave.
As the electron is moving, its initial potential energy is being converted to the kinetic energy and some of it will be lost to the radiation.
Once the electron hits a positively charged plate (because that's where it would obviously end up), its kinetic energy will be transformed, roughly speaking, to heat.
Presumably, the electron was initially placed between the plates by some external force, so we can say that it got its initial potential energy (eventually lost to the radiation and heat), as it was moved to this position.
The change of the electrostatic energy of the system before the electron was moved to its initial position and after it has landed on the positive plate, depends on where this electron came from.
For instance, if the electron was taken from the positive plate, the potential energy was added to the system and then lost, so the final electrostatic energy of the system would be the same as it was before the whole experiment had started.
If the electron was taken from the negative plate, some energy was lost as the electron was moved into its initial position between the plates and some more energy was lost as the electron completed its trip to the positive plate, moving on its own. In this case the final electrostatic energy of the system would be lower than it was before the experiment. We can say that it was slightly discharged.
If the electron was moved in from the outside, the final electrostatic energy of the system would also be lowered because, at the end of the experiment, the charge on the positive plate will be reduced. The exact sequence of energy changes, in this case, depends on the initial position of the electron.
For illustration purposes, let's look at three basic subcases: A) initial position is in the middle of the gap B) initial position is quarter gap above the neutral line C) initial position is quarter gap below the neutral line
Let's assume that the voltage between the plates is 1V. So, it takes 1eV to move an electron from the positive plate to the negative or 1eV is released (to radiation and heat), when an electron moves from the negative plate to the positive.
The diagrams below show that the introduction of the electron into the system, could leave its energy unchanged (case A), increase it (case B) or decrease it (case C).
In case A, the electron is moved normal to the electric field lines and therefore no work is performed by the system or on the system and, therefore, the energy of the system does not change.
In case B, the electrons is moved into the field, i.e., it has to be pushed in by an external force, which performs the work, and therefore the energy of the system (which now includes the new electron) increases. Based on the position of the electron between the plates, quarter gap above the neutral line, the energy of the system has increased by 0.25eV.
In case C, the electron is moved against the field, i.e., it is pulled by the field, i.e., the system performs some work and, therefore the system loses some energy. Based on the position of the electron between the plates, quarter gap below the neutral line, the energy of the system has decreased by 0.25eV.
As illustrated in the diagrams, the energy loss of the system after the electron has been released depends on the initial position of the electron, but in the end, the total loss of energy by the system, in comparison with its energy before the electron has been introduced, is the same 0.5eV in all three cases and could be calculated based on the fact that the system ended up with one unit of charge less on the positive plate.
NOTE: In the case when the electron was taken from the negative plate and ended up on the positive plate, the charge on both plates was reduced by one unit or, assuming that the voltage in that case was 1V, the energy loss would be 1eV.
Best Answer
The electric field of the TE$_m$ mode in parallel plate waveguides, derived from Maxwell's equations is: $$\mathbf E(\mathbf x) = \hat {\mathbf y} E_0 \sin(\frac{m \pi}{d}x) \ e^{-i\beta z}$$ (See the figure below). You can easily see that for the TE$_0$ case ($m=0$), we have the trivial answer of $\mathbf E=0$, which is why the TE modes start at TE$_1$. This doesn't happen with the TM$_0$ mode because it behaves like a cosine function instead of sine, meaning that it is nonzero at $m=0$.
Check out these lecture notes if you want to see where these equations come from.