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I am currently studying the textbook Optics, fifth edition, by Hecht. Chapter 2.3 Phase and Phase Velocity says the following:

Examine any one of the harmonic wave functions, such as $$\psi(x, t) = A \sin(kx - \omega t) \tag{2.26}$$ The entire argument of the sine is the phase $\varphi$ of the wave, where $$\varphi = (kx - \omega t) \tag{2.27}$$ At $t = x = 0$, $$ \psi(x,t)|_{\begin{subarray}{l}x=0\\t=0\end{subarray}}=\psi(0,0)=0 $$ which is certainly a special case. More generally, we can write $$\psi(x, t) = A \sin(kx - \omega t + \epsilon) \tag{2.28}$$ where $\epsilon$ is the initial phase. To get a sense of the physical meaning of $\epsilon$, imagine that we wish to produce a progressive harmonic wave on a stretched string, as in Fig. 2.12. In order to generate harmonic waves, the hand holding the string would have to move such that its vertical displacement $y$ was proportional to the negative of its acceleration, that is, in simple harmonic motion (see Problem 2.27). But at $t = 0$ and $x = 0$, the hand certainly need not be on the $x$-axis about to move down-ward, as in Fig. 2.12. It could, of course, begin its motion on an upward swing, in which case $\epsilon = \pi$, as in Fig. 2.13. In this latter case, $$\psi(x, t) = y(x, t) = A \sin(kx - \omega t + \pi)$$ which is equivalent to $$\psi(x, t) = A \sin(\omega t - kx) \tag{2.29}$$ or $$\psi(x, t) = A \cos \left( \omega t - kx - \dfrac{\pi}{2} \right)$$ The initial phase angle is just the constant contribution to the phase arising at the generator and is independent of how far in space, or how long in time, the wave has traveled.
The phase in Eq. (2.26) is $(kx - \omega t)$, whereas in Eq. (2.29) it's $(\omega t - kx)$. Nonetheless, both of these equations describe waves moving in the positive $x$-direction that are otherwise identical except for a relative phase difference of $\pi$. As is often the case, when the initial phase is of no particular significance in a given situation, either Eq. (2.26) or (2.29) or, if you like, a cosine function can be used to represent the wave. Even so, in some situations one expression for the phase may be mathematically more appealing than another; the literature abounds with both, and so we will use both. enter image description here enter image description here

There are so many equivalencies made in this section that it is difficult for a novice to make sense of everything.

How is $\psi(x, t) = y(x, t) = A \sin(kx - \omega t + \pi)$ equivalent to $\psi(x, t) = A \sin(\omega t - kx)$ and $\psi(x, t) = A \cos \left( \omega t - kx - \dfrac{\pi}{2} \right)$?

Furthermore, what precisely is meant by "otherwise identical except for a relative phase difference of $\pi$" with regards to $(kx - \omega t)$ and $(\omega t - kx)$? Is the author saying that $(kx - \omega t)$ and $(\omega t - kx)$ are equivalent? And how is it that both of these equations describe waves moving in the positive $x$-direction?

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  1. I believe they mean to say their wavefunction represents the $y$-coordinate which depends on $x$-coordinate and time, $\psi(x,t) \equiv y(x,t)$. The further equalitites are fairly easy to see by using trigonometric identities.

Hints: $\sin(x + \pi) = -\sin(x) = \sin(-x)$ and $\sin(x)=\cos(x\pm\pi/2)$.

  1. Notice that it is the relative sign of $x$ and $t$ that decides the direction of propagation and hence it does not matter $kx-\omega t$ or $\omega t -kx$. The initial configuration of the hand holding the thread fixes the phase, as you see from figures. If $t$ keeps increasing, in order to keep the phase same, the value of $x$ should increase too -- hence the wave is moving in positive $x$ direction.
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  • $\begingroup$ With regards to 1., it seems to me that the author describes it correctly, since the wave is moving in the positive $x$-direction, no? $\endgroup$ Dec 28 '20 at 14:35
  • $\begingroup$ Yes, that is what I have explained too. $\endgroup$ Dec 28 '20 at 15:44
  • $\begingroup$ Oh, yes, I misinterpreted you. Can you please clarify the "relative sign of $x$ and $t$" part? It isn't clear to me how both result in the wave moving in the positive $x$ direction. $\endgroup$ Dec 28 '20 at 16:26
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    $\begingroup$ Exactly! So since $t$ is always increasing, we can only increase or decrease $x$ to keep the phase same and that determines the direction of propagation. Also, only the relative sign matters because of the properties of sin and cos. Please do read all of it once more. $\endgroup$ Dec 28 '20 at 17:15
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    $\begingroup$ Yes. That's why it's moving in positive direction. $\endgroup$ Dec 28 '20 at 17:24
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I think you would find your answer by noticing that a sine wave is the same as the cosine wave if a half pi shift is implemented. Therefore saying that your wave equation in its sine form is the same as the cosine form with a half pi shift.

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