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Waves: When do we use $e^{i(-kx + wt)}$ as opposed to $e^{i(kx - wt)}$

  1. May 24, 2014 #1
    Both of these represent waves moving in the $+x$ direction. I have seen both used in Howard Georgi's book on waves and oscillations, but it has not been explained which is used in which circumstances. What is the difference between the two and when do we use which?

    Thanks in advance.
     
  2. jcsd
  3. May 24, 2014 #2
    Is this the solution that the two are exactly equivalent?

    I think i might have found the solution. That the two are exactly equivalent: Because $e^{i(-kz + wt)} = cos(-k(z- (w/k)t) = cos(k(z-vt)), and $e^{i(kz - wt)}= cos(k(z- (w/k)t) = cos(k(z-vt)).$

    This is rather interesting, because in some displacement solutions, one form is easier to work with than the other as terms cancel easily. Is there a general rule as to when to use which [just for neatness and appropriateness?]
     
  4. May 24, 2014 #3
    Yes the two waves are indeed entirely equivalent because one is the complex conjugate of the other. I'm not a big fan of the statement $$e^{i(-kz + wt)} = cos(-k(z- (w/k)t) = cos(k(z-vt))$$ because the right side is just the real part of the left side. That means they are NOT equal unless the imaginary part vanishes which is not the case. I doubt very much that one of them is easier to work with in any situation. Math is totally symmetric under the replacement i --> (-i).
     
  5. May 25, 2014 #4
    Thanks very much. When we consider the movement of peaks is fits and spurts when two waves of opposite directions are coming with some phase difference, we differentiate to find a maximum by taking the derivative with respect to x and then use implicit differentiation to find the derivative with respect to $t$. If we use the $$e^{i(-kx + wt)}$$ and $$e^{(i(kx+wt))}$ then if we take the x derivative and set it equal to zero the $$e^{iwt})$$ will get cancelled out which is undesirable since you want the time derivative. If we use the other form however, the mathematics is smooth (I.e if we use $$e^{i(kx-wt)}$$ and $$e^{i(kx+wt)}$$. This is an example of a problem where one notation presents the problem neatly as compared with the other.
     
  6. May 25, 2014 #5

    vanhees71

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    Whether you use the one or the other solution, representing a "right-moving wave" is indeed just a matter of convention. They are completely equivalent. The bad thing is that different communities use the one and others the other convention. Usually, physicists use the version with [itex]\exp(-\mathrm{i} \omega t)[/itex], while in electrical engineering they write [itex]\exp(+\mathrm{j} \omega t)[/itex], using also j instead of i for the imaginary unit ;-)). This is only a problem when comparing solutions of the wave equation in different textbooks/papers using different conventions.
     
  7. May 25, 2014 #6
    Your latex is messed up. Try using ##e^{-i(kx+wt)}## instead of ##e^{i(kx+wt)}## and you will see what I mean.
     
  8. May 26, 2014 #7
    Sorry - corrected that: Here is the corrected version: Thanks very much. When we consider the movement of peaks is fits and spurts when two waves of opposite directions are coming with some phase difference, we differentiate to find a maximum by taking the derivative with respect to x and then use implicit differentiation to find the derivative with respect to $t$. If we use the $$e^{i(-kx + wt)}$$ and $$e^{(i(kx+wt))}$$ then if we take the x derivative and set it equal to zero the $e^{iwt})$ will get cancelled out which is undesirable since you want the time derivative. If we use the other form however, the mathematics is smooth (I.e if we use $e^{i(kx-wt)}$ and $e^{i(kx+wt)}$). This is an example of a problem where one notation presents the problem neatly as compared with the other.
     
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