# Rectangular Optical Wavguide - EM Wave

1. Jul 28, 2007

### NotGreatAtPhys

Hello All, this is my first post, and while I'm a newb to the forums I'm not a newb to Optics. However, I have a very light background in low level electromagnetics, I'm more of a network design person.

I'm working on a very very long term personal project to understand optics better and I've come to the point where I would like to understand how to calculate how an EM-Wave propagates down a Rectangular Optical Waveguide. This waveguide is made of one glass material "n=1.47", it has nothing but air surrounding it. "See the attached image"

If I take the standard wave equation of:
$${ \partial^2 u \over \partial t^2 } = c^2 \nabla^2 u$$

and I calculate that the speed of light in the medium "with a refractive index or n=1.47" is
$$c = { c_o \over n } = { 1 \over \sqrt{ \mu \varepsilon } }$$
which equals: 203945578.23

What's my next step?
Am I even going in the right direction?

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2. Jul 29, 2007

### olgranpappy

both Jackson and Griffiths (and probably every EM book ever written) have sections on waveguides, and in particular on rectangular waveguides like the one you are interested in... I would probably start with Griffiths if I were you:

Griffiths, "Introduction to Electrodynamics", section 9.5 (Guided Waves), especially section 9.5.2 (TE waves in a rectangular Wave Guide)

3. Jul 29, 2007

### NotGreatAtPhys

Thanks, I'll start there.

4. Jul 29, 2007

### NotGreatAtPhys

I got a hold of a copy of: Griffiths, "Introduction to Electrodynamics", section 9.5 (Guided Waves) .. and read through chapter 9...

I guess my problem is the level of math is above what I'm used to.

Can anyone explain the standard wave equation?

$${ \partial^2 u \over \partial t^2 } = c^2 \nabla^2 u$$

What do each part represent, I can't seem to visualize what it stands for.

5. Jul 29, 2007

### olgranpappy

It's easier to start off with just one space dimension, in which case we have:

$$\frac{1}{c^2}\frac{\partial^2}{\partial t^2}u(x,t)=\frac{\partial^2}{\partial x^2}u(x,t)$$

Now, consider a function $$u(x,t)$$ that depends on space and time only through the single variable $$x-ct$$. That is, for any function of one variable $$f(w)$$ define
$$u(x,t)=f(x-ct)$$

then
$$\frac{\partial u}{\partial x}=f'(x-ct)$$

and
$$\frac{\partial u}{\partial t}=-cf'(x-ct)$$

thus
$$\frac{\partial^2 u}{\partial x^2}=\frac{1}{(-c)^2}\frac{\partial^2 u}{\partial t^2}$$
I.e., u satisfies the wave equation... so how to interpret this form of u?

well, for example, if the function f has a pronounced peak somewhere, then as time goes on the location of the peak moves along with a velocity c. If at t=0 f has a peak at x=x_0 then at t=t' f has a peak at x_0+ct'. Etc.

c is the velocity at which the wave (who's amplitude is described by u(x,t) at time t and position x) travels.

in the context of EM we usually arrive at wave equations where the roll of u is played by the amplitudes of the fields.

Last edited: Jul 29, 2007
6. Jul 29, 2007

### NotGreatAtPhys

Ok, I've got to go back to the basics again:

Can you explain curl?

as in how do I do it, I've read many portions of many books that say by taking the curl of Maxwell's equations 2 and 4, we can derive the Wave equation for E. However I don't understand how they get from this: "See the attached"

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7. Jul 29, 2007

### olgranpappy

"Curl" can be thought of as a differential operator that acts on vectors it's also written with the nabla $$\nabla$$ and cross-product $$\times$$ symbols as
$$curl\vec V=\vec \nabla\times \vec V$$
I.e., curl V is a vector whose cartesian components are given by
$$(curl V)_x=\frac{\partial V_z}{\partial y}-\frac{\partial V_y}{\partial z}$$

and

$$(curl V)_y=\frac{\partial V_x}{\partial z}-\frac{\partial V_z}{\partial x}$$

and

$$(curl V)_z=\frac{\partial V_y}{\partial x}-\frac{\partial V_x}{\partial y}$$

8. Jul 30, 2007

### olgranpappy

...now, you get to start memorizing properties of the curl. For example, since the curl of a vector is also a vector you can take the curl twice. For example:
$$\vec \nabla \times (\vec \nabla \times \vec V)\;.$$

It turns out that that is equal to
$$-\nabla^2\vec V+\vec\nabla(\vec \nabla\cdot\vec V)$$

So, for example, it you have $$\vec \nabla\cdot \vec V=0$$ (like if V was a magnetic field) then you would have
$$\nabla\times\nabla\times\vec B=-\nabla^2\vec B\;.$$

Low and behold there is the spatial part of a wave equation!

9. Jul 30, 2007

### olgranpappy

... but from maxwell's equation we also know that (in the absence of sources)
$$\nabla\times B=\frac{\partial E}{c\partial t}$$

and $$\nabla\times E=\frac{\partial B}{-c\partial t}$$so that
$$\nabla\times\nabla\times B=\nabla\times\frac{\partial E}{c\partial t}=-\frac{\partial^2 B}{c^2\partial t^2}$$

Low and behold, the other half of your wave-equation. Thus all together we have:
$$-\nabla^2\vec B=\frac{\partial^2}{-c^2\partial t^2}\vec B$$

...we can look for solutions to maxwell's equation that are travelling waves whose velocity is c.