Time-averages of superposition of waves.

[Silversonic]

Homework Statement

Consider the superposition of two waves;

$\zeta_1 + \zeta_2 = \zeta_{01} e^{i(kr_1 - wt)} + \zeta_{02} e^{i(kr_2 - wt + ∅)}$

where $∅$ is a phase difference that varies randomly with time. Show that the time-averages satisfy;

$<|\zeta_1 + \zeta_2|^2> = <|\zeta_1|^2> + <|\zeta_2|^2>$

Homework Equations

(1) If it wasn't clear, The two waves are;

$\zeta_1 = \zeta_{01} e^{i(kr_1 - wt)}$ and
$\zeta_2 = \zeta_{02} e^{i(kr_2 - wt + ∅)}$

The Attempt at a Solution

Unless I have my definition of time-average wrong. I can't seem to get this to work.

$|\zeta_1 + \zeta_2|^2 = (\zeta_1 + \zeta_2)(\zeta_1^* + \zeta_2^*) = |\zeta_1|^2 + |\zeta_2|^2 + \zeta_1\zeta_2^* + \zeta_2\zeta_1^* = |\zeta_1|^2 + |\zeta_2|^2 + 2\zeta_{01}\zeta_{02}cos(k(r_1 - r_2) - ∅)$

Then, I believe, the time average is given by;

$\frac{1}{T}\int^T_0 {|\zeta_1|^2 + |\zeta_2|^2 + 2\zeta_{01}\zeta_{02}cos(k(r_1 - r_2) - ∅)} dt$

However, I don't see how this turns into the form I desire. It would require that the last term (containing the cosine) is time-averaged to zero. Can this be the case? Also, can $∅$ still even be considered a function of time when it varies RANDOMLY?

 

[mark726]

Hello!

First of all, I want to clarify that the time average is defined as the average value of a quantity over a period of time, not necessarily the integral over time. So, in this case, the time average of a quantity A would be given by:

<A> = (1/T)∫^T_0 A(t)dt

Now, let's look at the expression for <|\zeta_1 + \zeta_2|^2>. As you correctly stated, it can be expanded as:

<|\zeta_1 + \zeta_2|^2> = <|\zeta_1|^2> + <|\zeta_2|^2> + <\zeta_1\zeta_2^*> + <\zeta_2\zeta_1^*>

But, since the phase difference ∅ varies randomly with time, we can say that <∅> = 0, as it would average out to zero over a long period of time. Therefore, <\zeta_1\zeta_2^*> = <\zeta_1><\zeta_2^*> and <\zeta_2\zeta_1^*> = <\zeta_2><\zeta_1^*>, as the two waves are independent of each other.

Substituting these values in the expression for <|\zeta_1 + \zeta_2|^2>, we get:

<|\zeta_1 + \zeta_2|^2> = <|\zeta_1|^2> + <|\zeta_2|^2> + <\zeta_1><\zeta_2^*> + <\zeta_2><\zeta_1^*>

= <|\zeta_1|^2> + <|\zeta_2|^2> + <|\zeta_1|^2> + <|\zeta_2|^2>

= 2<|\zeta_1|^2> + 2<|\zeta_2|^2>

This shows that the time-averages of the two individual waves add up to the time-average of the superposition of the two waves.

I hope this helps! Let me know if you have any further questions.