Introduction to Electrodynamics/Chapter 8/1/5

8.1.5 Boundary Conditions: Reflection and Transmission

Interesting things happen at the boundary between two systems that carry waves if the velocity of the waves are different on either side.

Hi, I'm Jonathan Gardner, and I'm covering section 8.1.5 of Griffith's Introduction to Electrodynamics, 2nd edition. If you have questions, you can ask in the comments below, or make a video response. I'll try to get back with you quickly. Be sure to like and share with your friends.

Suppose we take a string of one mass per unit length and tie it or connect it somehow to a string with a different mass per unit length. When we pull these two connected strings tight, the tension on either side will be the same, but the wave velocities will be different.

(Draw a picture)

So on the left side, we have v1. On the right side, we have v2.

We can write a wave function that describes the motion of waves on both sides of the knot at the same time. There will obviously be one part on the left and another part on the right, because the velocities are different. If we break down the waves into three groups, the incident, the reflected, and the transmitted, then we can write out the wave equation like this. On the left side is the incident and reflected waves, adding together the way waves generally do. On the right is the transmitted wave. Note that the reflected wave is traveling backwards.

Now, let's write down the constraints. The knot in the middle has to be at the same position for both sides. So f+ = f- at x=0.

More subtly, the first derivative of the knot with respect to position has to be the same. The reasoning behind this is that the knot would experience a net force due to the tension if it weren't so. Since we're assuming a massless knot, a net force would be bad. It may take a while to think about and understand why this needs to be true. I don't want to hold you up on it, because we're going to review how electromagnetic fields behave at boundary conditions later and we obviously won't be thinking about knots when we do so.

So, we have Ai + Ar = At at x=0, and k1(Ai + Ar) = k2(At) at x=0.

Doing some simply algebraic manipulations, we can get: Ar = (k1-k2 / k1+k2) Ai. (I encourage you to try this out on your own. It's rather elegant when you see it.) We can also get At = (2k1 / k1+k2) Ai.

We can rewrite the k's in terms of v's. First, we note that the omegas on both sides must be equal. If it weren't so, then there is no way the wave function could meet in the middle. Next, we note that w = kv and that the velocities are not the same, and so now we can relate the two k's with each other via the fact that the omegas are the same.

(equation of Ar, At in terms of Ai in terms of v1 and v2).

Now, we have related the complex amplitudes. I am sure you are interested in the real amplitudes and phase constant. So let's rewrite the complex amplitudes in terms of the real amplitudes and phase constants Ae^id:

(equation of Ar, At and Ai in real and complex parts.)

Now, we have two possibilities. Well, three actually, but we've already excluded the boring case where both sides have the same velocity.

The first case is when the velocity on the right side (v2) is larger than on the left. This is when we have a lighter string on the right side. This implies that the reflected wave is IN PHASE with the incident wave. We can see this since the amplitudes are proportional by some factor that is a positive, real number. This implies that the incident, transmitted, and reflected waves all have the same phase constant. You can imagine that the heavier string is sending a wave that the lighter string simply can't absorb. The extra energy must be reflected back.

Something curious happens when the right side has infinite velocity: The reflected wave basically doubles the incident wave. (The transmitted wave becomes arbitrarily large, which is silly.)

On the other hand, if the velocity on the right side is slower, then you have the case of the tail wagging the dog. You will have the reflected wave out of phase by 180 degrees. The incident and transmitted waves are still in phase. This is caused by the heavier side resisting the incoming wave. It has to slow down when it hits the boundary, and the only way is to cancel some of it with an inverted, reflected wave.

If the other side had a velocity approaching 0, then we will have the case where the reflected wave exactly cancels out the transmitted wave at the boundary --- the knot will not budge an inch. The transmitted wave, of course, is 0. No energy is put into the wall by a string attached to it.

That's pretty much all you care about with boundary conditions and strings. We're going to examine how electromagnetic waves behave in nonconducting media and then look at how waves respond to boundaries.