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Clarification on propagation directions of wave equations

  1. Sep 24, 2012 #1
    I would like to ask some questions to confirm if what I have acquired so far are correct, please point out my faults if any.
    If all the below statements are too bulky to read, my question is actually as short as:
    [tex]\psi (x) = A{e^{-j(kx - \omega t)}}[/tex]
    What is the propagation direction of this wave? Positive x or negative x? Why(which terms tell this)?



    Regarding the propagation direction of wave equations, to the best of my understanding, in the following wave equation:
    [tex]\psi (x) = A{e^{ - j(kx - \omega t)}}[/tex], let B = [tex]j(kx - \omega t)[/tex]
    The direction of the wave propagation is towards the positive x direction (+ve), for the following reasons:
    1. when we consider the exponential, we are actually taking only the real part of it, which is [tex]\cos (kx - \omega t)[/tex].
    As [tex]\cos x = \cos ( - x)[/tex], therefore no matter the sign before B is + or - (i.e. [tex]\psi (x) = A{e^{j(kx - \omega t)}}{\rm{ }}or{\rm{ }}\psi (x) = A{e^{ - j(kx - \omega t)}}[/tex]), the propagation direction will not change, remains as positive x direction.

    However, from a lecture note I recently read, I saw something kind of contradicting to the above:
    [tex]\psi (x) = A{e^{j(kx - \omega t)}}{\rm{ + }}B{e^{ - j(kx - \omega t)}}[/tex]
    "Consider a free particle that travels only in the positive direction of x. Let the arbitrary constant B be zero."
    It seems the negative sign on the latter term told us that it's propagating towards negative x direction and that is omitted, which is not consistent with my previous understanding of wave equations.

    I am sorry if I have put too much redundancy here, simply trying to make it as clearly as I can.
    Thanks for reading.
     
  2. jcsd
  3. Sep 24, 2012 #2
    [tex]\psi (x) = A{e^{j(kx - \omega t)}}[/tex]

    The [phase] velocity of the wave given by the equation above is ω/k. You may notice that ω/k = (-ω)/(-k).
     
    Last edited: Sep 24, 2012
  4. Sep 24, 2012 #3

    AlephZero

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    I think there is a typo here. The general solution is of the form f(x-ct) + g(x+ct) for arbitrary functions f and g, representing waves travelling in each directiion. Your notes should be
    [tex]\psi (x) = A{e^{j(kx - \omega t)}}{\rm{ + }}B{e^{ j(kx + \omega t)}}[/tex]
    or something equivalent to that.
     
  5. Sep 24, 2012 #4
    Thanks for the clarification,
    from the phase velocity then it means the propagation direction is determined by the sign difference between k and \omega.

    Also, yes I think that is a typo after reading more, thanks for pointing out:biggrin:
     
    Last edited: Sep 24, 2012
  6. Oct 8, 2012 #5
    Allow me to have one follow up question,
    why it should be kx-wt ending up with a relationship of v = (positive w) / (positive k)?
    The sign before w is a negative sign(-).:confused:

    And in the same way when we consider a wave Acos(kx+wt), we say v = k/-w and the wave propagate in -x direction.

    Would you share more on where does the conclusion [(kx-wt) = positive x direction] stem from?
    Thank you.
     
  7. Oct 8, 2012 #6
    Trying to ans own question:
    for a wave A = Ao cos(kx-wt),
    in case of an increased t, from the relation (kx-wt), the particle possessing same amplitude should have been shifted to the right(in (kx-wt), x is also increased to oppose the change of t making (kx-wt) a constant to keep the amplitude).
    With all particles having the same, the waveform will therefore move in +x direction.

    This is phenomenal explanation I can come up with.
    Please let me know in case there is a concrete mathematics deviation which I will be happier with.:biggrin:
     
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