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Fourier Transform and zero wave-vector

  1. Jul 19, 2011 #1
    I am not sure if I am crazy, I am not a mathematician or physicist by training, but I recall doing some work where if I was interested in the limit of a function in "r space" as r-> Infinity I could just use the value of the function in "k-space" at 0 to get the value I was interested in. Is this a real thing or did I make this up(I have been known to be "creative" sometimes with my mathematics)? Perhaps more simply I could write:
    Limit r-> Infinity of C(r) is equal to C_hat(k) for k = 0
    Thank you in advance
     
  2. jcsd
  3. Jul 20, 2011 #2
    You probably didn't make it up, but it's not a general rule. It's a statement that could be true in certain contexts.
     
  4. Jul 20, 2011 #3

    jasonRF

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    Do you have any more details about exactly what you were doing? This sounds a little like the final value theorem for Laplace transforms (see wikipedia), but without more information it isn't obvious to me what applies.

    By the way, to me your limit "theorem" (at least as written) doesn't seem to make a lot of sense. If lim C(r) = 0, then you are saying that C_hat(k) = 0 at k=0. As a counter example, let C(r) = exp(-0.5 r^2); clearly lim C(r) = 0. The Fourier transform is proportional to exp(-0.5 k^2), which is not zero at k=0. Now if lim C(r) isn't zero, then its Fourier transform does not exist if k is real. Thus k is complex, but then the Fourier transform doesn't converge for k on the real axis, which includes zero of course.

    cheers,

    jason
     
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