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Is the time average of a sin function = 0?

  1. Aug 8, 2010 #1
    Hello all. I just have a simple question. Suppose a classical charged particle is interacting with a potential of the form V(t) = V Sin(wt). The interaction energy is qV(t). My question is, will the time-average of this interaction be zero? Could someone please show me an equation of how to get that time-average. Thanks a lot.
  2. jcsd
  3. Aug 8, 2010 #2


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    The average of a function on the range [a,b] is

    [tex]\bar{f} = \frac{1}{b-a}\int_a^b dt f(t)[/tex]

    So, to answer this question, you have to specify the time interval over which you want to average the function. Do you want to average it over a full period? Or from t = 0 to infinity?

    For the case of sin(w*t), the [0,infinity) interval doesn't give a well defined average. The full period interval does, and from the definition you should be able to see if it is zero or not.
  4. Aug 8, 2010 #3
    what if the function is cos(w*t) and the time time interval is [- infinity, t]. does it have a well defined average? thanks very much.
  5. Aug 9, 2010 #4


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    Not really. To do an average where the time tends to infinity generally the integrand must tend to some finite limit there, which cos and sine do not. The integral is then defined on a finite range of integration where the "infinite" endpoint is taken to infinity as a limit. For your particular question this would look like

    [tex]\bar{f} = \lim_{T \rightarrow \infty} \frac{1}{t+T}\int_{-T}^t dt \cos(\omega t)[/tex]

    which won't be well defined as [itex]T \rightarrow \infty[/itex]. (However, it may be defined in the sense of a distribution such as the dirac delta function).

    Similar statements hold for sin(wt).
  6. Aug 9, 2010 #5


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    Maybe I'm making a silly mistake, but I'm pretty sure that, as a pointwise limit of bivariate functions, it converges to
    [tex]f(\omega, t) = \begin{cases} 1 & \omega = 0 \\ 0 & \omega \neq 0\end{cases}[/tex]
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