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.

Previously, we've examined the wave equation as it relates to strings under tension. We've found that the wave velocity depends on the mass per unit length and the tension. We've also looked at how to write eave equations using complex notation, which makes the job of analyzing wave functions much easier.

For convenience, I've written the formulas on the top here, so that you can quickly refer to them. You really should memorize the different names and formulas for the parts of the wave equation.

Suppose we take a string of one mass per unit length and 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 because the mass per unit length is different. Maybe the left side is heavier and thus slower, or maybe the right side is heavier and slower. We'll examine both cases.

(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.

<math> \tilde{f}(x, t) = \begin{cases} \tilde{A_I}e^{i(\kappa_1x - \omega_1 t)} + \tilde{A_R}e^{i(-\kappa_1x - \omega_1 t)}, & x \le 0 \\ \tilde{A_T}e^{i(\kappa_2x - \omega_2 t)}, & x \ge 0 \end{cases} </math>

Now, let's write down the constraints. The knot in the middle has to be at the same position for both sides.

<math> \tilde{A_I}e^{i(- \omega_1 t)} + \tilde{A_R}e^{i(- \omega_1 t)} = \tilde{A_T}e^{i(- \omega_2 t)} </math>

This implies that the omegas on either side must be equal, which we'll just call omega:

<math> \omega_1 = \omega_2 = \omega </math>

It also implies that the amplitudes must add up to each other:

<math> \tilde{A_I} + \tilde{A_R} = \tilde{A_T} </math>

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 the function didn't smoothly pass through the origin. Since we're assuming a massless knot, a net force would be bad. Don't think about this too hard: It applies to strings. We have a different constraint with electromagnetic waves that gives us the same result. We're going to examine EM waves in the near future.

<math> \begin{align} {\partial \over \partial x}\bigg|_0^- \left (

   \tilde{A_I} e^{i( \kappa_1 x - \omega t)}
 + \tilde{A_R} e^{i(-\kappa_1 x - \omega t)}

\right )

 &=

{\partial \over \partial x}\bigg|_0^+ \left (

   \tilde{A_T} e^{i(\kappa_2 x - \omega t)}

\right ) \\

 i \kappa_1 \tilde{A_I} e^{-i \omega t}

- i \kappa_1 \tilde{A_R} e^{-i \omega t}

 &=
 i \kappa_2 \tilde{A_T} e^{-i \omega t} \\

\kappa_1 (\tilde{A_I} - \tilde{A_R}) &= \kappa_2 \tilde{A_T} \\ \end{align} </math>

These two equations give us pretty much everything we need to calculate the reflected and transmitted waves given the incident wave and the two wave numbers.

<math> \begin{align} \tilde{A_I} + \tilde{A_R} &= \tilde{A_T} \\ \kappa_1(\tilde{A_I} - \tilde{A_R}) &= \kappa_2 \tilde{A_T} \\ \end{align} </math>

Doing some simple algebraic manipulations, we can get:

<math> \begin{align} \tilde{A_R} &= \left({\kappa_1-\kappa_2 \over \kappa_1 + \kappa_2}\right) \tilde{A_I} \\ \tilde{A_T} &= \left({2 \kappa_1 \over \kappa_1 + \kappa_2}\right) \tilde{A_I} \\ \end{align} </math>

Because the omegas match and omega is just kappa v:

<math> {\kappa_1 \over \kappa_2} = {v_2 \over v_1} = {\lambda_2 \over \lambda_1} </math>

So we can rewrite the above as:

<math> \begin{align} \tilde{A_R} &= \left({v_2-v_1 \over v_2 + v_1}\right) \tilde{A_I} \\ \tilde{A_T} &= \left({2 v_2 \over v_2 + v_1}\right) \tilde{A_I} \\ \end{align} </math>

This form is much more elegant and physical than the other form.

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:

<math> \begin{align} A_R e^{i\delta_R} &= \left({v_2-v_1 \over v_2 + v_1}\right) A_I e^{i\delta_I} \\ A_T e^{i\delta_T} &= \left({2 v_2 \over v_2 + v_1}\right) A_I e^{i\delta_I} \\ \end{align} </math>

Just examining the formulas, it is apparent that the phase constant of the incident and the transmitted wave is equal. This isn't surprising.

Depending on the relative velocities, the reflected wave may be in phase or out of phase by 180 degrees.

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.)

When 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 in the following videos.

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