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Wave reflected by a mirror

  1. Dec 20, 2015 #1
    Hi,

    In (1+1)D Minkowski spacetime, with coordinates (t,x),

    let's say there is an incoming plane wave of frequency [tex]\omega[/tex],
    [tex]\phi_{in}(t,x)=e^{-i\omega (t+x)}[/tex].

    There is a mirror, [tex]x=z(t)[/tex]

    It reflects the incoming plane wave and emits an outgoing plane wave.

    Question:
    why is the outgoing wave
    [tex]\phi_{out}=e^{-i\omega (2\tau_u-u)}[/tex],
    where
    [tex]u=t-x[/tex],
    [tex]\tau_u-z(\tau_u)=u[/tex],
    i.e. it is the retarded time.
    ??

    For mirror at constant velocity v, this reduces to

    [tex]\phi_{out}=e^{-i\omega\frac{1+v}{1-v}\cdot u}[/tex],
    the two Doppler shifts are obvious.

    But how can I prove the general expression?

    Thanks
     
  2. jcsd
  3. Dec 22, 2015 #2

    RUber

    User Avatar
    Homework Helper

    I look at this as a 2-step problem.
    Step 1. Ensure proper directionality
    ##\phi_{in} = e^{-i\omega(t+x)}## so the reflection will be headed in the opposite direction
    ##\phi_{out} = Ae^{-i\omega(t-x)}##.
    Step 2. Match function value at ##x=z(t)## i.e. enforce continuity at the mirror.
    ##\phi_{in}(\tau,z(\tau)) = e^{-i\omega(\tau+z(\tau))} = Ae^{-i\omega(\tau-z(\tau))}=\phi_{out}(\tau,z(\tau))##
    Solving for A gives:
    ##A = \frac{e^{-i\omega(\tau+z(\tau))}}{e^{-i\omega(\tau-z(\tau))}} = e^{-i\omega (2 z(\tau) )}##

    For the constant velocity term, it looks like ##\tau## and ##z(\tau)## can be determined by v, and some simplification is applied.
     
  4. Dec 22, 2015 #3
    Oh right!
    Thanks!
     
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