- #1

Frostman

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- 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 four-momentum 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.

The situation is represented in this image. From optical geometry, ##\theta_{incident} = \theta_{reflected}##

The four-momentum 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 four-momentum, 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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