0

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)?

The Pointer
  • 939

2 Answers2

0

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.

dkarkada
  • 421
0

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 $\dot E(t)/E(t)\ll \frac{\nu}{2\pi}=f$.

Newbie
  • 703