For 2 same waves traveling parallel to each other:
If one wave travel an extra distant d , as shown in the figure below. the falling-behind waves going to have a delay \tau of arrival. And this delay \tau infers phase shift \phi , vice versa: \boxed{ \text{distant traveled } d \leftrightarrow \text{ delay } \tau \leftrightarrow \text{phase shift } \phi } Give a wave with frequency \omega , wavelength \lambda , distance traveled d , and time to travel d is \tau , then the phase shift \phi is: \phi =& \omega \tau\\ =& 2\pi \nu \tau\\ &(\text{given: } c = \lambda \nu = \frac{d}{\tau} \to \nu = \frac{d}{\tau \lambda} )\\ =& 2\pi \frac{d}{\lambda}
Figure
1.0.1: Time Average of Cosine For Fast Varying Random Phase
Proof:
Because \phi(\psi) \pm \theta(\psi) \mod 2\pi \sim \mathcal{U}([0,2\pi]) so \frac{1}{T} \int_{t-T/2}^{t+T/2} \cos(c_0\psi + c_1 + \phi(\psi)\pm \theta(\psi)) \, d\psi = \frac{1}{T} \int_{t-T/2}^{t+T/2} \cos(c_0\psi + c_1 + \phi(\psi)) \, d\psi
Next, because T\gg\tau_c , we know that \phi(\psi) is fast varying during the underlined integration time T , its like taking lots of samples. So the original time average integral is the same as calculating expectation (ergodicity).
&\frac{1}{T} \int_{t-T/2}^{t+T/2} \cos(c_0\psi + c_1 + \phi(\psi)) \, d\psi \\ \\ &\begin{pmatrix} \because \text{we view each $\psi$ as an independent event} \\ \therefore c_0\psi + c_1 \text{ is just a constant in each event at } \psi \\ \\ \because (c_0\psi + c_1 + \phi(\psi)) \mod 2\pi \sim \mathcal{U}([0,2\pi])\\ \therefore E[\cos( c_0\psi + c_1 +\phi(\psi) )] = E[\cos \phi(\psi) ] \\ \end{pmatrix} \\ \\ &= E [ \cos( \phi)]\\ &= \int_{0}^{2\pi} \cos(\phi)p(\phi) d\phi\\ &= \frac{1}{2\pi} \int_{0}^{2\pi} \cos( \phi) d\phi\\ &= \frac{1}{2\pi} [ \sin( 2\pi) - \sin( 0) ] \\ &= 0