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A question about electromagnetic wave function

  1. Dec 31, 2004 #1
    Hi there.
    We always put the time dependent part of the wave functions as e^(iwt).
    Of course there is a reason!!! but I don't know it.
    Can you help me???
    Thanks in advance.
    Somy :smile:
     
  2. jcsd
  3. Dec 31, 2004 #2
    This is due two reasons:

    1) The imaginary exponential is a periodical function

    [tex] e^{iz} = \cos(z) + i \sin(z)[/tex]

    and

    2) It is a very good function to work with it. So physicists, who are very intelligent people, try to use easy functions, and it is more easy to work with and exponential rather than trigonometric ones.

    Maths are the easy part of physics, thats why we try to use the more simplest things to work with.

    Happy new year !!
     
    Last edited: Dec 31, 2004
  4. Dec 31, 2004 #3

    dextercioby

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    To add to what Migui has said,i'm sure you know that the coordinate part of the wavefunctions is put under the form of complex exponentials as well:
    [tex] \exp(i\vec{k}\cdot\vec{r}) [/tex]

    Daniel.

    PS.Migui has given u a simple nontechnical explanation.There's much more behind an explanantion for this fact.To give you a hint:distributions and Fourier transformations.
     
  5. Dec 31, 2004 #4

    Integral

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    Of course the main reason that is used, is because it is the solution to the governing partial differential equation.
     
  6. Dec 31, 2004 #5

    dextercioby

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    Yap,Integral,but think about the 1D wave equation:

    [tex] \frac{\partial^{2}u(x,t)}{\partial x^{2}}=\frac{1}{v^{2}}\frac{\partial^{2}u(x,t)}{\partial t^{2}} [/tex]

    Solve it and tell me whether the result contains complex exponentials.

    Daniel.
     
  7. Dec 31, 2004 #6

    Integral

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    Looks to me like [tex] u(t) = e^{ivt}[/tex] is a solution to the time equation. What am I missing?
     
  8. Dec 31, 2004 #7

    dextercioby

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    The point?

    First of all,your solution is wrong wrt dimensions.Exponential of a meter is kinda hard to swallow.Even complex. :yuck:

    That equation (1D wave eq.posted above) is a typical example of the fact that the method of variable separation reduces the number of solutions,and even affects the physically acceptable solutions.In this specific example,it selects the periodic (sine/cosine) waves propagating along the "x" axis.Both regressive and progressive.
    I'm sure you still remember (and u still have time to show it in 2004 :tongue2: ) 2 facts:
    1)The 1D wave equation admits nonperiodical solutions.
    2)The method of variable separation works best in the case that the solution of the equation is unique,working hand in hand with initial conditions,boundary conditions and Fourier representations.Here,uniqueness suffers from inexistance.

    Daniel.

    PS.Do you remember the solution and how to get it?
     
  9. Dec 31, 2004 #8

    krab

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    In an unbounded domain, the general solution is (let's stick to 1D and use Kurt's notation)
    [tex]u(x,t)=f_{\rm f}(x-vt)+f_{\rm b}(x+vt)[/tex]
    where the f's are arbitrary functions. A fun demonstration is to hit a clothesline sharply with your fist and watch the wave impulse (the f) race away without changing shape, rebound off the end and come back. So why do we always end up using [itex]e^{\pm i\omega t}[/itex]? Or to put it terms of physical objects: why are these arbitrary functions appropriate for the clothesline but not for say a guitar string? This is indeed a very good question; one that bothered me a lot when I learned it. As Kurt said, it is partly to do with the great mathematical tool called Fourier analysis. But the convenience or not of using harmonic functions also has to do with whether the length of the wave shape f is short or long with respect to the length of the 1D domain. A third reason is that if you depart slightly from the wave equation (say v is a function of the wavelength), then the general solution given above is not very helpful.
     
  10. Dec 31, 2004 #9

    Integral

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    So are you trying to say that if it isn't physical it isn't a solution?

    Humm...
     
  11. Jan 1, 2005 #10
    Thanks all,
    But I still didn't get the answer!!!
    Well, I know that they tell us we use this function because of simplicity, but I want to know some deeper mathematical points. dextercioby told me about the fourier transfer, but I really coudn't find the relation between these.
    Can you give me something more???
    Thanks in advance.
    Somy
     
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