View Single Post
Silversonic
#1
Mar8-12, 09:45 PM
P: 129
1. The problem statement, all variables and given/known data

Consider the superposition of two waves;

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

where [itex] ∅ [/itex] is a phase difference that varies randomly with time. Show that the time-averages satisfy;

[itex]<|\zeta_1 + \zeta_2|^2> = <|\zeta_1|^2> + <|\zeta_2|^2> [/itex]

2. Relevant equations

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

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


3. The attempt at a solution

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

[itex]|\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) - ∅)[/itex]

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

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

However, I don't see how this turns in to 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 [itex] ∅ [/itex] still even be considered a function of time when it varies RANDOMLY?
Phys.Org News Partner Science news on Phys.org
Bees able to spot which flowers offer best rewards before landing
Classic Lewis Carroll character inspires new ecological model
When cooperation counts: Researchers find sperm benefit from grouping together in mice