- #1
cedricyu803
- 20
- 0
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
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