# Energy propagation in a circular waveguide

1. Oct 15, 2008

### bdforbes

I have a perfectly conducting circular waveguide. I want to calculate the time-averaged Poynting vector of a circularly polarised TE mode, ie:

$$H_z = H_0 J_{n}(\rho \chi)e^{in\phi}$$

Where $$\chi$$ is the appropriate eigenvalue.
My result for <S> implies helical energy flow; it has a z component and a phi component. Does this make sense? Instinctively I would have thought the energy would just flow in the z direction, but since it's circularly polarised I'm not so sure.

The circularly polarised mode is simply two orthogonal TE modes of the same wavenumber out of phase by pi/2. Couldn't we consider the total energy propagation to be the superposition of the propagation of each component? In that case it seems like the energy should only go in the z direction.

2. Oct 15, 2008

### BerryBoy

Since the time-avergaged poynting vector always points in the direction of propagation, it certainly would point in the z direction for a single mode. This usually always happens in real life fibre optic cables. Even when linearly polarized light is focussed into a fibre waveguide, very slight anisotropies in the manufacturing will cause the fibre core to be slightly birefrengent. This will cause component axis of the E field to become out of phase and result in partial circular polarized light at the end of it and a dispersion known as Polarisation Mode Dispersion; not a desired thing.

Sam

3. Oct 15, 2008

### bdforbes

That's interesting but I think my situation is a bit different. I'm using a conductor-walled waveguide not a dielectric waveguide, and I'm assuming there is always circularly polarised light with no dispersion.

I've checked and rechecked my working, I can't see an error. My logic in the previous post was probably wrong; the amplitudes of the two modes will superimpose, but the intensities won't, so there's no reason to assume the Poynting vector will only be in the z-direction.

The problem notes that circularly polarised light has angular momentum; I don't know how to deal with this mathematically but I assume that implies helical energy flow. Could I show this somehow? What does angular momentum mean when applied to light?

4. Oct 16, 2008

### BerryBoy

Hey there, so the difference here is that your electric field must be zero at the conducting wall instead of a value in the case of a dielectric (as the return signal will be reflected at the same amplitude exactly pi out of phase).

The fact that the two polarization modes are orthogonal means that they can always be resolved and that what you do to one does absolutely nothing to the other. Thus you can model the energy flow as the sum of the energies of the different polarizations (I'm not going to use the word mode to avoid confusion with TE modes):

$$S_t = S_\| + S_\bot$$

Since the time-averaged poynting vector doesn't oscillate in time or space. The poynting vector will always be pointing in that direction. The only case you get helical energy flow is in skew modes where the electric field direction is circular across the cross-section of the guide. In circularly polarized TE modes, at any given time the electric fields across a cross-section of the waveguide point in the same direction (they rotate with time not space).

Could I possibly see what you've calculated?

Sam

5. Oct 16, 2008

### bdforbes

Hmm what you're saying makes sense.

I started with the H_z component from my first post, then applied these formulae:

$$E_t = (-i\omega/\gamma^2)\widehat{z}\times\nabla_t{B_z}$$
$$H_t = (ik/\gamma^2)\nabla_t{H_z}$$

From which I got:

$$H_t = \frac{ikH_0e^{in\phi}}{\gamma^2}(\chi J_n'(\rho \chi)\widehat{\rho}+in J_n(\rho \chi)\widehat{\phi}/\rho)$$
$$E_t = \frac{ikZH_0e^{in\phi}}{\gamma^2}(inJ_n(\rho \chi)\widehat{\rho}/\rho-\chi J_n'(\rho \chi)\widehat{\phi})$$

Next I used:

$$<S>=\frac{1}{2}Re(E\times H^*)$$

The component that I get in the phi direction is

$$-E_\rho H_z^*\widehat{\phi}$$

This appears to come out real. Can you see any errors there?

Thanks for the help. To be honest, I didn't expect anyone to answer this question.

6. Oct 16, 2008

### BerryBoy

Ouch, I'm not used to seeing these equations in this form. Far too analytical for me; going to go and revise some stuff. But shouldn't $E_t$ be:

$$E_t = (-i\omega/\gamma^2)\widehat{z}\nabla_t \times {B_z}$$

Based on memory of the look of the equation (not a very technical appraoch I'm aware :tongue2:).

Haha. I answered because I need to learn how to model on a computer multiple TEM modes propagating down a fibre optic cable with all kinds of dispersion and what have you. I noticed your thread while reading about circular polarizations down a fibre optics cable and why polarization is not preserved down the cable.

Sam

7. Oct 16, 2008

### BerryBoy

Nope, I'm talking absolute rubbish.... scratch that. Your equation stands

8. Oct 16, 2008

### bdforbes

I was thinking about how you superimposed the intensities. Suppose we consider the superposition:

$$<S_1>=\frac{1}{2}Re(E_1\times H_1^*)$$
$$<S_2>=\frac{1}{2}Re(E_2\times H_2^*)$$

The total intensity is:

$$<S>=\frac{1}{2}Re((E_1+E_2)\times (H_1^*+H_2^*))$$
$$= <S_1> + <S_2> + \frac{1}{2}Re(E_1\times H_2^* + E_2\times H_1^*)$$

These interaction terms might be purely imaginary, in which case you would be correct, but I'm not so sure that they are.

EDIT: I see you mentioned TEM modes above. In a TEM mode the Poynting vector would be without a doubt purely in the z direction. But I have magnetic induction in the z-direction too, are you considering that?

9. Oct 16, 2008

### bdforbes

I worked through the problem as above with each component separate, and I still got the same answer. The radial intensity canceled out neatly, but the angular intensity remained. I'm still not convinced that I'm wrong.

10. Oct 16, 2008

### BerryBoy

E1 oscillates in the same direction as H2 and H1 osciallates in the same direction as E2, so I'd expect their vector product to be zero.

11. Oct 16, 2008

### bdforbes

I don't think that's true, because H1 and H2 have components in the z-direction, whereas E1 and E2 are transverse.

12. Oct 16, 2008

### BerryBoy

How very true.

Again I've gone off into dielectric mode written about a TEM mode. I've been desperately trying (as you may have noticed), to avoid using any maths in this treatment; a futile effort it seems. I'm going to brush up on my maxwell maths and try an analytical approach; I wonder if the sum of the vector products is zero (could this be reasonable with a physical description of the fact that we chose them to be identical except for polariztion so a cross product of one is the negative of the other)?

I will keep an eye on this thread; since I am curious and would like to know what is going on. Please update this if you discover what the poynting vector is doing.

Sam