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Separable PDE for electric field in a cavity

  1. Sep 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Here is a photo of a page in Laser Physics by Hooker:
    https://www.evernote.com/shard/s245/sh/2172a4e7-63c7-41a0-a0e7-b1d68ac739fc/7ba12c241f76a317a6dc3f2d6220027a/res/642710b5-9610-4b5b-aef4-c7958297e34d/Snapshot_1.jpg?resizeSmall&width=832

    I have 3 questions (I'm a bit rusty after a long summer):
    1. Is the "c.c." bascially [itex]\textbf{U}(r) e^{i \omega t}[/itex] or is the U actually U*? (Just checking)

    2. Why can't we say [itex] \textbf{U}(r) = X(x)Y(y)Z(z) [/itex] instead of separating it into Ux Uy and Uz?

    3. Finally: Having gone through the PDE and got an answer with boundary solutions [itex]\textbf{U}(r)_{x} = 0[/itex] at edges, I don't understand where the [itex]\frac{\pi}{L}[/itex] comes from. The π I know is just put in to make the mode numbers simpler later, but how does the L get there?
    And how is the x component a cos function? I get a sin!

    Thanks for the help!! I'm just a bit rusty!
     
  2. jcsd
  3. Sep 3, 2013 #2
    Link does not work.
     
  4. Sep 4, 2013 #3
    Oh, ok here you go (attached)
     

    Attached Files:

  5. Sep 4, 2013 #4

    vanhees71

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    2016 Award

    the +c.c. simply means "plus the conjugate complex expression", i.e.,
    [tex]a+ \text{c.c.}=a+a^*.[/tex]
     
  6. Sep 4, 2013 #5
    To answer your questions:

    1: Yes
    2: Note: U (a 3D vector) is a function of r (which contains x, y, and z.)
    3: L is the cavity size (some times you will see 2*Pi/L) depending on boundary conditions and such.

    Hope this helps!

    As for the cos, I am unsure about this as well. Perhaps someone else can shed some light on this matter.
     
    Last edited: Sep 4, 2013
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