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Circular waveguide wave-equation

  1. Aug 27, 2007 #1
    I am solving the wave-equation (more specifically Helmholtz Eq.) in cylindrical coordinates.
    I've seperated the equation into 3 ODE's.
    - The radial differential equation
    - The phase differential equation
    - The z differential equation (direction of which the EM wave propagates)

    My issue is the solution to the phase's differential equation. It has the simple solution:
    [tex] Ae^{im\phi}+c.c. [/tex] (easy to prove).

    Why is 'm' an integer?
    Are the phases 'quantised'?

    I've read in many books that m must be an integer to allow continuity at [tex] 2\pi [/tex], but thats as far as they go...I'm very confused...
     
  2. jcsd
  3. Aug 27, 2007 #2

    Dick

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    [tex] Ae^{im\phi}+c.c. [/tex] is just 2Acos(m*phi) (at least if A is real). As phi is an angular coordinate then the value of the solution at phi must equal to the value of the solution at phi+2pi because they represent the same point. Hence, m is an integer. The phase isn't quantized, 'm' is quantized by the requirement that solution be single valued.
     
    Last edited: Aug 27, 2007
  4. Aug 28, 2007 #3
    Your exact explanation to 'm' is whats confusing me...
    I understand that phi must equal phi+2pi, but how do you go about saying that m*phi allows the relation (or phase continuity) phi+2pi to be satisfied??
     
  5. Aug 28, 2007 #4
    The condition

    [tex] \cos[m (\phi + 2 \pi)] \equiv \cos[m \phi + 2 m \pi] = \cos[m \phi] [/tex]

    can be satisfied only if m is integer.

    Eugene.
     
  6. Aug 28, 2007 #5
    Thanks Eugene, I was thinking about the problem in too much depth.
     
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