Circular waveguide wave-equation

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  • #1
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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...
 

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  • #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.
 
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  • #3
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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??
 
  • #4
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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??

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.
 
  • #5
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Thanks Eugene, I was thinking about the problem in too much depth.
 

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