Why are phase constants of incident, reflected and transmitted simple waves equal in absolute value? I was reading Griffiths's book of electrodynamics and i got stuck on the ninth chapter, where he analyses the propagation of a simple wave - fixed form and constant velocity - using travelling pulses through ropes as example. The situation is about the transmission and reflection of the incident pulse coming from rope 1 and reaching rope 2 at a knot in the origin, with both ropes in the z axis. The incident wave is given by:
$\tilde{f}_{I}(z,t) = \tilde{A}_{I}e^{i(k_{1}z-\omega t)},\quad (z < 0)$.
Similarly to the incident, the reflected wave is described below:
$\tilde{f}_{R}(z,t) = \tilde{A}_{R}e^{i(-k_{1}z-\omega t)},\quad (z < 0)$.
And finally the transmitted wave:
$\tilde{f}_{T}(z,t) = \tilde{A}_{T}e^{i(k_{2}z-\omega t)},\quad (z > 0)$.
There is a person shaking the rope at $-\infty$, generating the wave. We work with those expressions trying to find the amplitude of the reflected and transmitted waves around the knot. Considering $\tilde{A}_{R} = A_{R}e^{i\delta_{R}}$ and $\tilde{A}_{T} = A_{T}e^{i\delta_{T}}$, we come to  two expressions:  
$A_{R}e^{i\delta_{R}}=(\frac{v_{2}-v_{1}}{v_{2}+v_{1}})A_{I}e^{i\delta_{I}} \quad (1) \quad $$A_{T}e^{i\delta_{T}}=(\frac{2v_{2}}{v_{2}+v_{1}})A_{I}e^{i\delta_{I}}\quad (2)$
Now, the real question is: why if the velocity $v_{2}$ is greater than $v_{1}$ all the phase constants, i.e., $\delta_{I}$, $\delta_{R}$ and  $\delta_{T}$, are going to be equal? And why, if $v_{1}$ is greater than $v_{2}$, $\delta_{R}$ is going to be $\pi$ radians out of phase from $\delta_{I}$ and $\delta_{T}$?. It seems easy to understand that, in the first case, both the constants are going to have the same sign, and in the second they're going to have oposite signs, but it seems not so obvious why they are going to have the exact same absolute value. I'd really appreciate any help (sorry for the extremely long post, i'm a newbie here).
 A: The equations you have here relate the amplitudes to the left on the interface with the amplitudes to the right of the interface. The amplitudes (of the rope or electric fiel) are given as complex numbers, $\bar A=A\exp(i\delta)$, with it's absolute value $A$ (a real number) and the complex phase $\exp~i\delta$.
(case $v_2>v_1$)
Using Eulers formula you can rewrite your first formula as
$$ A_{R}(cos(\delta_R)+i\cdot sin(\delta_R))=+\left|\frac{v_{2}-v_{1}}{v_{2}+v_{1}}\right|A_{I}(cos(\delta_I)+i\cdot sin(\delta_I)).$$
The point is that we can't measure complex numbers. (What's "$i\cdot 0.3m$ of rope"?) So we want to cancel the terms proportional to $i$, i.e. the sines. Since the cosine has the same argument we cancel the whole bracket, i.e. the whole phase, this way. (I could also have left the $\exp(i\delta)$'s as it is... but I thought that the point becomes more apparent this way.) We can cancel the terms in the brackets by setting
$$\delta_R=\delta_I, $$
or: the phases are the same. Similarly for your second equation. This yields $\delta_R=\delta_I=\delta_T $.
(case $v_2<v_1$)
In this case you get a minus sign in your equation number 1:
$$ A_{R}e^{i\delta_{R}}=-\left|\frac{v_{2}-v_{1}}{v_{2}+v_{1}}\right|A_{I}e^{i\delta_{I}}$$
Writing -1 as $\exp(-\pi~i)$:
$$ A_{R}e^{i\delta_{R}}=e^{-\pi~i} \left|\frac{v_{2}-v_{1}}{v_{2}+v_{1}}\right|A_{I}e^{i\delta_{I}}=\left|\frac{v_{2}-v_{1}}{v_{2}+v_{1}}\right|A_{I}e^{i(\delta_{I}-\pi)}$$
Again using Eulers formula:
$$ A_{R}(cos(\delta_R)+i\cdot sin(\delta_R))=\left|\frac{v_{2}-v_{1}}{v_{2}+v_{1}}\right|A_{I}(cos(\delta_I-\pi)+i\cdot sin(\delta_I-\pi)).$$
Therefore $\delta_R=\delta_I-\pi$. The second equation still yields $\delta_T=\delta_I$, no phase shift here. To summarize: 
For $v_2>v_1$: All phases are equal.
For $v_1<v_2$: The phase of the reflected wave is shifted by $\pi$, while the phases for transmitted and incident wave are the same.
A: Although this question was asked some time ago, I still decided to give my own answer to it. The thing is, I was reading Griffiths's Introduction to Electrodynamics recently and I also got stuck at this very point. My own explanation goes as follows.
We all agree that $$
\tilde{A}_R=\frac{v_2-v_1}{v_1+v_2} \tilde{A}_I.
$$ 
Now, let's rewrite that to a following form, using Euler's formula and defining $\alpha$ as $\alpha=\frac{v_2-v_1}{v_1+v_2}$ (that'll simplify the calculations a little bit). From the definition of $\tilde{A}_R$ and $\tilde{A}_I$ we get: $A_R e^{i\delta_R}=\alpha A_I e^{i \delta_I}$. Using Euler's formula we get:$$
A_R (i \sin{\delta_R}+\cos{\delta_R} )= \alpha A_I (i \sin{\delta_I} + \cos{\delta_I}).
$$
We can rearrange this to a following form:$$
A_R = \alpha A_I \frac{i \sin{\delta_I} + \cos{\delta_I}}{i \sin{\delta_R}+\cos{\delta_R}}
$$
Now we clearly see the "problems" with our equations: $A_R$ and $A_I$ are both real numbers, as well as $\alpha$; $A_R$, according to our equation, is, however, clearly a complex number (unless we somehow manage to make the imaginary part disappear). But that ins't the most problematic part. The equality suggests that $A_R$ depends on $A_I$ (OK, THAT is completely fine - after all, we expected that, because $A_I=0$ would give us $A_R=0$, and the bigger $A_I$ is, the bigger value of $A_R$ we should get), but the weird part is that it also tells us that $A_R$ is depending on $\delta_I$ and $\delta_R$! Think about it - these two are phase constants - numbers that appear in the equation of sinusoidal wave $f(z,t)=A \cos{(kz-\omega t+\delta)}$ as the result of choosing $t=0$ at some "inappropriate moment" - we can always start our "clock" at such a moment that $\delta$ wil be equal to zero. And of course, (one could say that it's trivial consequence of the previous fact) $A$ is totally independent of $\delta$ and vice versa. This implies that $A_R$ cannot be a function of either $\delta_I$ or $\delta_R$, and hence, the values of these to phase constants should behave in a way which would enable us to reduce the whole fraction $\frac{i \sin{\delta_I} + \cos{\delta_I}}{i \sin{\delta_R}+\cos{\delta_R}}$ to some constant value (independent of $\delta_I$ and $\delta_R$). Let's call that constant $1/C$ (it can be any real number you want, e.g. 5, 3.14, etc.). That means:$$
\frac{i \sin{\delta_I} + \cos{\delta_I}}{i \sin{\delta_R}+\cos{\delta_R}} = \frac{1}{C}
$$
or, in one line:$$
i \sin{\delta_R}+\cos{\delta_R}=C(i \sin{\delta_I} + \cos{\delta_I})
$$
In order for such an equation to be true we need the imaginary part and the real part to be true at the same time. So, $\sin{\delta_R}=C\sin{\delta_I} \wedge \cos{\delta_R}=C\cos{\delta_I}$, from here we get $\delta_R=\arcsin{(C\sin{\delta_I})}$, and similarly, $\delta_R=\arccos{(C\cos{\delta_I})}$. Let's try putting one of these to our equation; we get:$$
A_R(iC \sin{\delta_I}+\sqrt{1-C^2\sin^2{\delta_I}})=\alpha A_I (i\sin{\delta_I}+\cos{\delta_I})
$$
The same happens when we put $\delta_R$ from our second equation. This means that we will be able to find phase constants which make $\frac{i \sin{\delta_I} + \cos{\delta_I}}{i \sin{\delta_R}+\cos{\delta_R}}$ constant only if this constant value will be equal to $\pm 1$. 
Great! Now it's time to actually look for certain relation between $\delta_R$ and $\delta_I$ which will enable us to cancel the whole fraction to either 1 or -1. 
What happens if $v_2>v_1$? Well, $\alpha$ is positive, so apparently we should look for phase constants such that $\sin{\delta_R}=\sin{\delta_I} \land \cos{\delta_R}=\cos{\delta_I}$ ($C=1$). From here we get:$$
(\delta_R=\delta_I+2k\pi \lor \delta_R=\pi-\delta_I+2k\pi) \land (\delta_R=\delta_I+2l\pi \lor \delta_R=-\delta_I+2l\pi) \land l, k \in Z
$$
We choose the common solution of these: $\delta_R=\delta_I+2k\pi$; because adding $2\pi$ isn't changing anything we can write: $$\delta_R=\delta_I.$$
We follow the same method when $v_1>v_2$ (and $\alpha<0$), however now we see that right-hand side of $A_R (i \sin{\delta_R}+\cos{\delta_R} )= \alpha A_I (i \sin{\delta_I} + \cos{\delta_I})$ is negative, while the left-hand side is positive, hence now we should choose $-1$ as $1/C$ (or $C$ - it doesn't really matter). So, $\sin{\delta_R}=-\sin{\delta_I} \land \cos{\delta_R}=\cos{(\pi-\delta_I)}$ From that we get $$
(\delta_R=-\delta_I+2k\pi \lor \delta_R=\pi+\delta_I+2k\pi) \land (\delta_R=\pi-\delta_I+2l\pi \lor \delta_R=\delta_I-\pi+2l\pi) \land k,l \in Z.
$$
The common part of the solution is $\delta_R=\delta_I-\pi$ or $\delta_R=\delta_I+\pi$.
