 #1
Frostman
 115
 17
 Homework Statement:
 A mirror moves parallel to its plane. A beam of light hits this mirror at an angle θ. At what angle will the beam be reflected?
 Relevant Equations:

\begin{cases}
p^0 = \gamma(p'^0+vp'^1)\\
p^1 = \gamma(p'^1+vp'^2)\\
p^2=p'^2\\
p^3=p'^3\\
\end{cases}
Before to open this topic, I found this there. It's quite similar, if not the same, but I'm a little confused, so I'm here.
The situation is represented in this image. From optical geometry, ##\theta_{incident} = \theta_{reflected}##
The fourmomentum in ##S'## is the following one:
I'm not sure is it ok or I missed some considerations. And if the angle is the most simplified one.
EDIT: I'm wrong, what I find now is the incident angle, that is different from reflected angle.
The situation is represented in this image. From optical geometry, ##\theta_{incident} = \theta_{reflected}##
The fourmomentum in ##S'## is the following one:
##p'^{\mu} = E'(1, \cos{ \theta}, \sin{ \theta}, 0)##
Now we can use Lorentz transformation in order to pass from ##S'## to ##S##.
##p^{\mu}=\Lambda^{\mu}{}_{\nu} p'^{\mu}##
Obtaining this system of equations\begin{cases}
p^0 = \gamma(E'vE'\cos\theta)\\
p^1 = \gamma(E'\cos\theta+vE')\\
p^2=E'\sin\theta\\
p^3=p'^3=0\\
\end{cases}
We can find the new angle by doing this:p^0 = \gamma(E'vE'\cos\theta)\\
p^1 = \gamma(E'\cos\theta+vE')\\
p^2=E'\sin\theta\\
p^3=p'^3=0\\
\end{cases}
##\frac{p^2}{p^1}=\tan{\theta'}=\frac{E'\sin\theta}{\gamma(E'\cos\theta+vE')}=\frac{\sin \theta}{\gamma(v\cos\theta)}##
##\theta'=\arctan{\frac{\sin \theta}{\gamma(v\cos\theta)}}##
This is the angle (incident and reflected) saw from ##S##.##\theta'=\arctan{\frac{\sin \theta}{\gamma(v\cos\theta)}}##
I'm not sure is it ok or I missed some considerations. And if the angle is the most simplified one.
EDIT: I'm wrong, what I find now is the incident angle, that is different from reflected angle.
##p'^{\mu}_i = E'(1, \cos{ \theta}, \sin{ \theta}, 0)##
It is the incident fourmomentum, the reflected one is the following:
##p'^{\mu}_r = E'(1, \cos{ \theta}, \sin{ \theta}, 0)##
What I have to do is now apply the same transformation and find out that:
##\theta'_r=\arctan{\frac{\sin \theta}{\gamma(v+\cos\theta)}}##
Instead, before I found the
##\theta'_i=\arctan{\frac{\sin \theta}{\gamma(v\cos\theta)}}##
So, conclusion the angle in ##S## is not the same as it was in ##S'##, am I wrong?
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