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I Wavevector k in Helmholtz Equation

  1. Dec 16, 2016 #1
    Hello Everyone,

    Helmholtz equations derives from the wave equation by using separation of variables and assuming that the solution is indeed separable ##g(x,y,z,t) = f(x,y,z) T(t)##. The solutions to Helmholtz equations are functions of space, like f(x,y,z), and do not depend on time t.

    the wavevector ##k = \frac{2\pi}{\lambda}## is present in the equation. Is that a constant or a variable? Solutions can span a spectrum of ##k## values so I don't believe the ##k## in the equation can be a constant otherwise the equation would be limited to function having a single and specific k value....

    Thanks
     
  2. jcsd
  3. Dec 16, 2016 #2

    Henryk

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    Gold Member

    Different k values correspond to different waves with respect to wavelength or direction
     
  4. Dec 16, 2016 #3
    True, but in the equation, is the k to be take as a variable?
     
  5. Dec 17, 2016 #4

    Henryk

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    Gold Member

    Oh, that was your question. No, wavevector k is fixed.
     
  6. Dec 17, 2016 #5
    I would say that the wave vector ##\mathbf{k}## is a parameter, since you solve the equation for fixed ##\mathbf{k}## but the value depend on the specific problem you want to solve.
     
  7. Dec 18, 2016 #6
    so are you saying the solutions to this equation can only have a single k since the equation itself has a single k?
     
  8. Dec 18, 2016 #7
    Presumably, this is a matter of convention. But, my personal point of view is the following (maybe other people on this forum will not agree). In your first post you are saying:
    We have therefore a set of solutions ##f_k## labeled by the wave number ##\mathbf{k}##. As, I said before you are then solving the equation for a fixed ##\mathbf{k}## and obtain one ##f_k##. You can then repeat this and solve for arbitrary number of values of ##k## depending on which solutions on the set you are interesting in.
     
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