Why Is the Time Average of the Cosine Squared Term in the Poynting Vector 1/2?

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Discussion Overview

The discussion revolves around understanding the time average of the cosine squared term in the Poynting vector for monochromatic plane waves. Participants explore the mathematical reasoning behind the claim that this average is 1/2, including potential methods for calculating it.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant seeks clarification on why the time average of cos²(kz - ωt + φ) is stated to be 1/2.
  • Another participant suggests using the trigonometric identity cos²x = 1/2(1 + cos2x) to aid in understanding.
  • A participant mentions that finding the time average likely involves integrating the cos² term over a period.
  • Another participant provides the formula for the time average of a periodic function, indicating how to apply it to the cos² term.
  • One participant questions the necessity of integration, suggesting that the average can be inferred from the graph of the function.
  • The original poster expresses gratitude and indicates they will explore both graphical and integration methods for understanding.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to determine the time average, with some favoring graphical interpretation and others advocating for integration. The discussion remains open-ended regarding the most effective approach.

Contextual Notes

There are varying levels of familiarity with mathematical concepts among participants, which may affect their interpretations and suggestions. The discussion does not resolve the specific steps needed to arrive at the time average.

ronaldoshaky
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Hello.

I am reading in my book about the Poynting vector for monochromatic plane waves. It includes a cosine term: cos^2 (kz - omega t + phi). My book states that the time average of this term is 1/2. Can anyone explain this? I don't understand how they work that out.

Thank you
 
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Hello ronaldoshaky! :smile:

(have an omega: ω and a phi: φ and try using the X2 tag just above the Reply box :wink:)

Use one of the standard trigonometric identities …

cos2x = 1/2 (1 + cos2x) :wink:
 
Hi tiny-tim.

Does finding the time average have something to do with integrating the cos^2 term?

Thanks again
 
The time average of any periodic function, f(t), with period T is given by

\langle f\rangle_t=\frac{\int_{t_0}^{t_0+T}f(t)dt}{\int_{t_0}^{t_0+T}dt}=\frac{1}{T}\int_{t_0}^{t_0+T}f(t)dt

Apply that to your \cos^2 term
 
Hi ronaldoshaky! :smile:
ronaldoshaky said:
Does finding the time average have something to do with integrating the cos^2 term?

"integrating" is a very technical word to use …

can't you tell the average of cosx (or of cos2x = (1 + cos2x)/2) just by looking at the graph?! :smile:
 
Thanks to all who replied. I will do both the graph and the integration. This has helped me a lot!
 

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