How Does the Phase Relation of the Electric Field Vector Change with Time?

[yungman]

Homework Statement

This is part of the derivation of the direction of rotation of an ellipse in EM wave polarization. I need to find the direction of the change of phase $\psi$ of the electric field vector with increase of time t. To make the long story short, for example:

$$\psi\;=\; \tan^{-1}\left(\frac{\cos(\omega t+\frac{\pi}{2})}{\cos \omega t}\right)\;=\; \tan^{-1}\left(\frac{-\sin(\omega t)}{\cos \omega t}\right)\;=\;-\omega t$$
From this we can conclude $\psi$ DECREASE with INCREASE of t.

But if the constant is $\delta$ where $0<\delta<\pi$, how do I find the relation of $\psi$ with time t?

Homework Equations

$$\psi\;=\; \tan^{-1}\left(\frac{\cos(\omega t+\delta)}{\cos \omega t}\right)$$

The Attempt at a Solution

$$\psi\;=\; \tan^{-1}\left(\frac{\cos(\omega t+\delta)}{\cos \omega t}\right)\;=\; \tan^{-1}\left(\frac{\sin(\omega t+\frac{\pi}{2}+\delta)}{\cos \omega t}\right)\;=\;\tan^{-1}\left(\frac {\cos\omega t \cos \delta\;-\; \sin\omega t \sin \delta}{\cos \omega t}\right)$$I don't know how to go beyond this to find the relation of $\psi$ with t. Please help.

Thanks

Alan

 

[mfb]

Are you sure that there is an easy way to write this relation?
You can simplify the expression a bit:

$$ \psi = \tan^{-1} \left( \cos(\delta ) - \sin(\delta ) \tan(\omega t) \right)$$

 

[yungman]

Are you sure that there is an easy way to write this relation?
You can simplify the expression a bit:

$$ \psi = \tan^{-1} \left( \cos(\delta ) - \sin(\delta ) \tan(\omega t) \right)$$

Thanks for your reply. That is the next step. I am not sure there is an easier way, the goal is to determine the direction of rotation with increase time t. My main question is how to go any further. How do I find out whether $\psi$ increase or decrease with time.

Do you think taking the derivative and look at whether it's +ve or -ve will work? I don't know the answer. This is not a home work, it's been a while since I studied calculus and I am very rusty in it.

 

[mfb]

Do you think taking the derivative and look at whether it's +ve or -ve will work? I don't know the answer. This is not a home work, it's been a while since I studied calculus and I am very rusty in it.

That should work, but you can do this a bit quicker: tan^-1 and tan are strictly increasing. If sin(δ) is positive, ψ is going down if t increases, otherwise ψ and t increase together.

 

[yungman]

That should work, but you can do this a bit quicker: tan^-1 and tan are strictly increasing. If sin(δ) is positive, ψ is going down if t increases, otherwise ψ and t increase together.

Thanks. I am just confused when there is some constant inside. I just digged up my old notes on inverse tangent and need to read through it. Have not deal with this for years!

 

[yungman]

I worked on this more:

$$\psi = \tan^{-1} \left( \cos(\delta ) - \sin(\delta ) \tan(\omega t) \right)\;=\;\tan^{-1} \left( K_1 - K_2 \tan(\omega t) \right) \;\hbox {where }\;K-1=\cos \delta\;,\;K_2=\sin \delta$$

$$\Rightarrow \frac { d[tan^{-1}(K_1+K_2\tan\omega t)]}{d[K_1+K_2\tan\omega t]}\;=\;\frac{1}{1+(K_1+K_2 \tan\omega t)}\;\Rightarrow\; \psi'\;=\;\frac { d[tan^{-1}(K_1+K_2\tan\omega t)]}{d t}\;=\;\frac{\omega K_2 \sec^2\omega t}{1+(K_1+K_2\tan\omega t)^2}$$

$$\psi'\;=\;\frac{\omega K_1 (1+\tan^2\omega t)}{(1+K^2_1 + 2K_1K_2\tan\omega t +K^2_2 \tan^2\omega t)}\;=\;\frac {(K_3+K_3 tan\omega t)}{(K_4+K_5\tan\omega t+ K_6 \tan^2\omega t)}$$

$K_1$ to $K_6$ are constants. With this relationship, how can I predict the change of direction of phase with time t?

I don't need to find the value of $\psi$, I just need to know whether $\psi$ increase or decrease with t.

 

[mfb]

Well, see my approach above, it avoids those messy expressions.
Alternatively, consider
$$\psi'\;=\;\frac{\omega K_2 \sec^2\omega t}{1+(K_1+K_2\tan\omega t)^2}$$
The denominator is always positive, and ω and sec² are positive as well. Therefore, the derivative has the same sign as $K_2=-\sin(\delta)$ (you swapped the sign somewhere).

 

[yungman]

Thanks