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I have the time dependence of the electric field $A(t)$ written in terms of the amplitude $E(t)$ and phase $\phi(t)$ as

$$A(t) = E(t) \cos[\nu t + \phi(t)].$$

I am told that this "varies slowly in an optical period $2\pi/\nu$."

I'm a bit confused as to what this $2\pi/\nu$ is supposed to be. Is this the angular frequency? Angular frequency is defined as

$${\displaystyle \omega ={\frac {2\pi }{T}}={2\pi f},}$$

where $T$ is the period and $f$ is the ordinary frequency.

Furthermore, how can $\cos[\nu t + \phi(t)]$ be written so that the so-called "optical period" of $2\pi/\nu$ is clearer (to a novice)?

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  • $\begingroup$ Can you clarify what you mean by “I’m a bit confused…”. Also what is the significance of introducing angular frequency in the first place? $\endgroup$
    – Newbie
    Jan 13, 2022 at 5:42
  • $\begingroup$ @Newbie What I mean is that $2\pi/\nu$ is described as being the "optical period," but "period" is conventionally known to be the $T$ in the denominator of the angular frequency, so it isn't clear to me (as a novice) what $2\pi/\nu$ is supposed to be. I was hoping that someone more knowledgeable would be able to lay this all out in a clearer way so that I may understand. $\endgroup$ Jan 13, 2022 at 5:44
  • $\begingroup$ Just considering the fact that the argument of the $\cos$ function should be dimensionless (assuming with units of radians) and time $t$ is e.g., in units of seconds, can you obtain the unit for $\nu$? $\endgroup$
    – Newbie
    Jan 13, 2022 at 5:47
  • $\begingroup$ Lets call the optical period $T_{o}$. Based on the definition in the problem $T_{o}=2\pi/\nu$. Thus $\nu T_{o}=2\pi$ and yes $\nu$ does in fact look like the angular frequency here. $\endgroup$
    – Newbie
    Jan 13, 2022 at 5:49
  • $\begingroup$ @Newbie Ohh, ok, I see. The notation was confusing me. So this is just the usual spectral decomposition of the 1-D electric field en.wikipedia.org/wiki/… $\endgroup$ Jan 13, 2022 at 5:55

2 Answers 2

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As mentioned in my comments considering the fact that the period of a $\cos$ function is $2\pi$ one can write $$\cos[\nu t+\phi(t)]=\cos[\nu t+2\pi+\phi(t)]=\cos[\nu(t+\frac{2\pi}{\nu})+\phi(t)]$$ Hence $T_{\rm o}=\frac{2\pi}{\nu}$ is the period of the $\cos$ function only if $\phi(t)$ is constant since in that case $$\frac{d}{dt}[\nu t+\phi(t)]=\nu+\dot{\phi}(t)=\nu$$ However, in general $\phi(t)$ changes as a function of time and therefore, the angular frequency is $$\omega=\nu+\dot{\phi}(t)\rightarrow T=\frac{2\pi}{\omega}=\frac{2\pi}{\nu+\dot{\phi}(t)}$$

In your question it is mentioned that the variation of the electric field over the time period $T_{\rm o}$ is negligible. This implies that the variations of $E(t)$ and $\phi(t)$ over the timescale $T_{\rm o}$ are not significant. For $\phi(t)$ this means $\dot\phi(t)\ll\nu$. For $E(t)$ this means $\dot E(t)/E(t)\ll \frac{\nu}{2\pi}=f$.

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  • $\begingroup$ I have one additional question: This article en.wikipedia.org/wiki/… seems to claim that $\phi$ is usually a constant, so why is it a function of $t$ in my case? Isn't "phase" usually defined to be everything within the $\cos$ function? So why is $\phi(t)$ said to be the "phase"? $\endgroup$ Jan 13, 2022 at 6:04
  • $\begingroup$ I'm addressing this in my answer. $\endgroup$
    – Newbie
    Jan 13, 2022 at 6:05
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    $\begingroup$ They're approximately constant over one optical period. But if you're doing an interference or diffraction problem for example, the differences can add up and you'll have to take that into account. $\endgroup$
    – dkarkada
    Jan 13, 2022 at 6:09
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    $\begingroup$ @ThePointer Definitely go through the answer again. I have added important edits. $\endgroup$
    – Newbie
    Jan 13, 2022 at 6:15
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    $\begingroup$ @ThePointer Response revised significantly. $\endgroup$
    – Newbie
    Jan 13, 2022 at 6:29
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Normally, a wave is given by $A(t)=E\cos(\nu t+\phi)$, where $E$ and $\phi$ are constants. (The notation is a little weird, as you suspected; we usually denote angular frequency by $\omega$ rather than $\nu$.) Anyways, the period of this wave is $T=2\pi/\nu$.

In your situation, the optical period is the period I just mentioned (the period of the wave if the amplitude and phase were constants). But, the amplitude and phase aren't constants in your case. This means that the "true" period will be have to be corrected, depending on how quickly $\phi(t)$ and $E(t)$ change.

However, the statement that $E(t)$ and $\phi(t)$ "vary slowly in an optical period" basically tells you that the correction will be small. In other words, the true period won't be very different from the optical period, since the amplitude and phase will only change a little bit over the course of one optical period.

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