# Wave reflected by a mirror

1. Dec 20, 2015

### cedricyu803

Hi,

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

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

There is a mirror, $$x=z(t)$$

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

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

For mirror at constant velocity v, this reduces to

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

But how can I prove the general expression?

Thanks

2. Dec 22, 2015

### RUber

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.

3. Dec 22, 2015

Oh right!
Thanks!