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I am following a training course and came across this proof, from my colleague, that the ordinary law of reflection $\theta_i = \theta_r$ does not hold in relativity:

Let $S$ be a perfectly reflecting mirrror.

Obviously if it is at rest the canonical law of reflection $\theta_i=\theta_r$ holds. Now suppose the mirror is moving with velocity $\overline v$ in the frame $\Sigma$, and let the frame of the mirror be $\Sigma'$. Then in $\Sigma'$,

$$\theta'_i\equiv\theta'_r$$

which means that

$$ \cos \theta =\frac{\cos \theta'_i+\beta}{1+\beta \cos \theta'_i}. $$

For the reflected ray changes $\theta$ before was $\theta_i $ is now $\theta =\pi-\theta_r $, similarly $\theta'= \pi- \theta'_r $ waves $ \cos \theta = - \cos (\theta_r)$ which is replaced provides:

$$ -\cos \theta_r =\frac{-\cos \theta'_r+\beta}{1-\beta \cos \theta'_r}, \quad \cos \theta_r =\frac{\cos \theta'_r-\beta}{1-\beta \cos \theta'_r} $$

$$\tan \left(\frac{\theta_i}{2}\right) =\sqrt{\frac{1-\beta}{1+\beta}}\tan \left(\frac{\theta'_i}{2}\right), \qquad \tan \left(\frac{\theta_r}{2}\right) =\sqrt{\frac{1+\beta}{1-\beta}}\tan \left(\frac{\theta'_r}{2}\right) $$

hence

$$ \tan\left( \frac{\theta_i}{2}\right)=\left(\frac{1-\beta}{1+\beta}\right)\tan \left( \frac{\theta_r}{2} \right) $$ and the law of reflection is no longer valid.

I don't think this derivation is very clear. Is there anyone who can help me to understand the steps in this unclear derivation, or recommend a book where I can find a proof of the final result?

If more information is needed, a picture of the derivation can be found here and here.

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    $\begingroup$ I'm a bit unclear about the question; are you asking for a step-by-step explanation of the quoted notes? $\endgroup$
    – Nat
    May 14 '18 at 21:11
  • $\begingroup$ With lot of sincerity: yes. I would understand the procedure to have at the end $\theta_i<\theta_r$. I do not understand the link of the formulas and I have seen at least twenty books in Italian and English but I have not found anything that refers to what I have reported. If you know some text or you could help me, I'm grateful. $\endgroup$
    – Sebastiano
    May 14 '18 at 21:27
  • $\begingroup$ Hi Sebastiano, the figure legends are in Italian, but the text is in English. I suppose you made the translation yourself. Can you post the original text in Italian too? $\endgroup$
    – magma
    May 16 '18 at 7:37
  • $\begingroup$ @magma Done! I have done a translation because here we must write in English language. Best regards. $\endgroup$
    – Sebastiano
    May 16 '18 at 20:07
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    $\begingroup$ This is too hard to answer without knowing exactly what you're confused about. Can you point to a specific step you don't understand and explain why it's confusing? $\endgroup$
    – knzhou
    May 16 '18 at 20:27
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There are two inertial frames of reference involved; a "primed" frame in which the mirror is at rest, and an "unprimed" frame in which the mirror is moving in the direction opposite the mirror's surface normal. Label the coordinates such that the direction the mirror is moving in (i.e. the direction opposite the surface normal) is the $x$ and $x'$ direction, the other direction in the plane containing both the incident and reflected beams is the $y$ and $y'$ direction, and the direction that's unimportant to this problem is the $z$ and $z'$ direction.

In either frame, the components of an incident photon's three-velocity are given by the projections of the three-velocity onto the coordinate axes. I.e., the three-velocity of an incident photon, as measured in each of the two frames of reference, is

$$\mathbf{u}_i=\left(\begin{array}{c}c\cos\theta_i \\ c\sin\theta_i \\ 0\end{array}\right)$$

and

$$\mathbf{u}'_i=\left(\begin{array}{c}c\cos\theta'_i \\ c\sin\theta'_i \\ 0\end{array}\right)\ \ ,$$

where $\theta_i$ and $\theta'_i$ are the angle of incidence as measured in each of the two frames of reference, and $c$ of course is the speed of light.

The derivation in the question uses a version of the relativistic velocity addition formula, a proof of which can be found in the linked-to Wikipedia article. That formula states that if a primed frame is moving with speed $v$ in the $x$ direction as measured in an unprimed frame, speeds along the $x$ and $x'$ direction, as measured in each of the two frames, are related as

$$u_x=\frac{u'_x+v}{1+v u'_x/c^2}\ \ .$$

Plugging in the $x$ and $x'$ components of $\mathbf{u}_i$ and $\mathbf{u}'_i$ into the relativistic velocity addition equation, i.e. setting

$$u_x=c \cos\theta_i$$

and

$$u'_x = c \cos\theta'_i$$

gives

$$c \cos\theta_i=\frac{c \cos\theta'_i+v}{1+v c \cos\theta'_i/c^2}\ \ .$$

Dividing both sides of that equation by $c$ gives

$$\cos\theta_i=\frac{\cos\theta'_i+\beta}{1+\beta \cos\theta'_i}\ \ ,$$

where $\beta = v/c$.

We can perform a similar procedure with a reflected photon. The three-velocity of a reflected photon, as measured in each of the two frames of reference, is

$$\mathbf{u}_r=\left(\begin{array}{c}-c\cos\theta_r \\ c\sin\theta_r \\ 0\end{array}\right)$$

and

$$\mathbf{u}'_r=\left(\begin{array}{c}-c\cos\theta'_r \\ c\sin\theta'_r \\ 0\end{array}\right)\ \ ,$$

where $\theta_r$ and $\theta'_r$ are the angle of reflection as measured in each of the two frames of reference. In the case of the reflected photon, the values we plug in to the relativistic velocity addition formula are

$$u_x=-c \cos\theta_r$$

and

$$u'_x = -c \cos\theta'_r\ \ ,$$

giving

$$- c \cos\theta_r=\frac{-c \cos\theta'_r+v}{1+v (-c \cos\theta'_r)/c^2}\ \ .$$

Dividing both sides of that equation by $-c$ gives

$$\cos\theta_r=\frac{\cos\theta'_r-\beta}{1-\beta \cos\theta'_r}\ \ .$$

The above formulas for $\cos\theta_i$ and $\cos\theta_r$ are inconvenient for the purposes of comparing $\theta_i$ and $\theta_r$, largely due to $\theta'_i$ and $\theta'_r$ each appearing twice in the equations. We can arrive at simpler equations by using the tangent half-angle formula

$$\tan\left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\ \ .$$

Applying the tangent half-angle formula to $\theta_i$ gives

$$ \begin{align} \tan\left(\frac{\theta_i}{2}\right)&=\sqrt{\frac{1-\cos\theta_i}{1+\cos\theta_i}}\\ &=\sqrt{\frac{1-\frac{\cos\theta'_i+\beta}{1+\beta \cos\theta'_i}}{1+\frac{\cos\theta'_i+\beta}{1+\beta \cos\theta'_i}}}\\ &=\sqrt{\frac{1+\beta \cos\theta'_i-(\cos\theta'_i+\beta)}{1+\beta \cos\theta'_i+\cos\theta'_i+\beta}}\\ &=\sqrt{\frac{(1-\beta)(1-\cos\theta'_i)}{(1+\beta)(1+\cos\theta'_i)}}\\ &=\sqrt{\frac{1-\beta}{1+\beta}}\sqrt{\frac{1-\cos\theta'_i}{1+\cos\theta'_i)}}\\ &=\sqrt{\frac{1-\beta}{1+\beta}}\tan\left(\frac{\theta'_i}{2}\right)\ \ . \end{align} $$

Applying the tangent half-angle formula to $\theta_r$ would proceed similarly, except that $\theta'_i$ is everywhere replaced by $\theta'_r$, and $\beta$ is everywhere replaced by $-\beta$. The result is thus

$$\tan\left(\frac{\theta_r}{2}\right)=\sqrt{\frac{1+\beta}{1-\beta}}\tan\left(\frac{\theta'_r}{2}\right)$$

or

$$\tan\left(\frac{\theta'_r}{2}\right)=\sqrt{\frac{1-\beta}{1+\beta}}\tan\left(\frac{\theta_r}{2}\right)\ \ .$$

But since the mirror isn't moving in the primed frame, the normal law of reflection is valid in the primed frame, $\theta'_i=\theta'_r$, and so we have

$$ \begin{align} \tan\left(\frac{\theta_i}{2}\right)&=\sqrt{\frac{1-\beta}{1+\beta}}\tan\left(\frac{\theta'_i}{2}\right)\\ &=\sqrt{\frac{1-\beta}{1+\beta}}\tan\left(\frac{\theta'_r}{2}\right)\\ &=\sqrt{\frac{1-\beta}{1+\beta}}\left(\sqrt{\frac{1-\beta}{1+\beta}}\tan\left(\frac{\theta_r}{2}\right)\right)\\ &=\left(\frac{1-\beta}{1+\beta}\right)\tan\left(\frac{\theta_r}{2}\right)\ \ , \end{align} $$

which was the result to be shown.

Note that for $0<\beta<1$,

$$0<\left(\frac{1-\beta}{1+\beta}\right)< 1\ \ ,$$

so

$$\tan\left(\frac{\theta_i}{2}\right)<\tan\left(\frac{\theta_r}{2}\right)$$

and $\theta_i<\theta_r$.

ADDENDUM:

As a response to the comments on this answer, the following provides additional clarification of the confusing statement in the question that "For the reflected ray changes $\theta$ before was $\theta_i $ is now $\theta =\pi-\theta_r $, similarly $\theta'= \pi- \theta'_r $ waves $ \cos \theta = - \cos (\theta_r)$":

For a three-velocity $\mathbf{u}$ in general, the $x$ component of $\mathbf{u}$, $u_x$, is given by the scalar projection of $\mathbf{u}$ onto $\hat{\mathbf{x}}$, the unit vector in the $x$ direction. As per the scalar projection Wikipedia article, the scalar projection can be expressed as

$$u_x=\left|\mathbf{u}\right|\cos\theta\ \ ,$$

where $\left|\mathbf{u}\right|$ is the length of $\mathbf{u}$, and $\theta$ is the angle between $\mathbf{u}$ and $\hat{\mathbf{x}}$.

The speed of either an incident or a reflected photon is $c$, i.e.,

$$\left|\mathbf{u}_i\right|=\left|\mathbf{u}_r\right|=c\ \ ,$$

so in either case we use $\left|\mathbf{u}\right|=c$ in the scalar projection equation for $u_x$.

The angle of incidence $\theta_i$ is defined as the angle between $-\mathbf{u}_i$ and $\hat{\mathbf{n}}$, where $\hat{\mathbf{n}}$ is the mirror surface's unit normal. But that angle is the same as the angle between $\mathbf{u}_i$ and $-\hat{\mathbf{n}}$, which is the same as the angle between $\mathbf{u}_i$ and $\hat{\mathbf{x}}$, since we've defined our coordinate system such that $\hat{\mathbf{x}}=-\hat{\mathbf{n}}$. Thus, when dealing with $\mathbf{u}_i$, we just use $\theta=\theta_i$ in the scalar projection equation for $u_x$, i.e.

$$u_x=\left|\mathbf{u}\right|\cos\theta=c\cos \theta_i\ \ ,$$

as in the original answer above.

On the other hand, the angle of reflection $\theta_r$ is defined as the angle between $\mathbf{u}_r$ and $\hat{\mathbf{n}}$. The angle between $\mathbf{u}_r$ and $\hat{\mathbf{x}}$ differs from $\theta_r$ by $\pi$ radians, because the angle between $\hat{\mathbf{n}}$ and $\hat{\mathbf{x}}$ is $\pi$ radians. Thus, when dealing with a reflected photon, we use $\theta=\pi-\theta_r$ instead of $\theta=\theta_i$ in the scalar projection equation for $u_x$,

$$u_x=\left|\mathbf{u}\right|\cos\theta=c\cos (\pi-\theta_r)=-c\cos\theta_r\ \ ,$$

as in the original answer.

The scalar projection equations for $u'_x$ are the same as the scalar projection equations for $u_x$, for the same reasons, except that they use $\theta'_i$ and $\theta'_r$ instead of $\theta_i$ and $\theta_r$.

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  • $\begingroup$ Thank you so much for your patience and your response. I ask you if it's possible, that you add a figure that can better understand some of your steps. In particular I would like a addition clarification on the components of the vectors $\mathbf {u} _r $ ed $\mathbf {u} '_ r $: how hane you find? After can you explain me the role of this concept with a drawing: For the reflected ray changes $\theta$ before was $\theta_i $ is now $\theta =\pi-\theta_r $, similarly $\theta'= \pi- \theta'_r $ waves $ \cos \theta = - \cos (\theta_r)$. Thank you. I have upvoted +1 your answer. $\endgroup$
    – Sebastiano
    May 21 '18 at 20:10
  • $\begingroup$ Can you add the details of the my previous comment, please? It is very important for me. Thanks. $\endgroup$
    – Sebastiano
    May 23 '18 at 21:35
  • $\begingroup$ Thank you very much for the answer and the addendum! I have checked and "mille grazie". $\endgroup$
    – Sebastiano
    May 25 '18 at 19:29
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I would post this video as an answer: https://www.youtube.com/watch?v=v8lzsYh3JRY&list=PLj6DWzIvBi4PFDXCCV1bNhVUgDLTwVbFc&index=13

A variation would be to shine a horizontal light beam at a 45 degree mounted mirror in a moving ship. Either way (light clock or mirror), the resultant beam would be vertical. The video shows how to calculate the angle from vertical seen from a stationary perspective of a moving light clock or mirror. Instead of using time dilation, the same problem can be solved by length contraction of the 45 degree mirror's base although this is a far more complex way to tackle the problem. That solution can be seen here:

https://www.physicsforums.com/threads/45-dergree-mirrors-in-special-relativity.352427/

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