Angle of reflection of a beam light on a moving mirror

[Frostman]

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:

$p'^{\mu} = E'(1, -\cos{ \theta}, -\sin{ \theta}, 0)$
​

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:

$\frac{p^2}{p^1}=\tan{\theta'}=\frac{-E'\sin\theta}{\gamma(-E'\cos\theta+vE')}=\frac{-\sin \theta}{\gamma(v-\cos\theta)}$

$\theta'=-\arctan{\frac{\sin \theta}{\gamma(v-\cos\theta)}}$
​

This is the angle (incident and reflected) saw from $S$.
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?

 

[PeroK]

What is frame $S$ and what is frame $S'$? What direction is the mirror moving in? What Lorentz transformation have you used - a boost in the x-direction?

 

[Frostman]

$S$ is the laboratory system. $S'$ is the comoving system with mirror. The mirror is moving along the x-axis. I used a boost in the x-direction.
The tricky part is the fact that "A mirror moves parallel to its plane", but in the graph drawn the mirror is moving orthogonal to its plane.
I inverted the apex on the angle that I used, it must be $\theta'$ the angle saw in $S'$ and $\theta_i$ and $\theta_r$ in the $S$ system.

 

[PeroK]

The tricky part is the fact that "A mirror moves parallel to its plane", but in the graph drawn the mirror is moving orthogonal to its plane.

That's confusing me. Without doing the full calculations can see what is going to happen in the two cases? 1) Mirror moving in the y-direction (parallel to its plane) and 2) Mirror moving in the x-direction (orthogonal to its plane).

Hint: imagine the mirror is moving at close to $c$.

 

[Frostman]

If the mirror is moving in the y-direction, the angle component parallel to the boost is contracted (y component). Instead the orthogonal component is unchanged (x component).
If the mirror is moving in the x-direction, the angle component parallel to the boost is contracted (x component). Instead the orthogonal component is unchanged (y component).

 

[PeroK]

If the mirror is moving in the y-direction, the angle component parallel to the boost is contracted (y component). Instead the orthogonal component is unchanged (x component).
If the mirror is moving in the x-direction, the angle component parallel to the boost is contracted (x component). Instead the orthogonal component is unchanged (y component).

Is there a conclusion there?

 

[Frostman]

Uhm... I can conclude that in the case that boost is in x-direction the reflected angle is:
$\theta_r=\arctan{\frac{\gamma\sin\theta'}{\cos\theta'}}=\arctan(\gamma\tan\theta')$
Instead, in the y-direction the reflected angle is:
$\theta_r=\arctan{\frac{\sin\theta'}{\gamma\cos\theta'}}=\arctan(\frac{\tan\theta'}{\gamma})$
Is it correct?

 

[PeroK]

Uhm... I can conclude that in the case that boost is in x-direction the reflected angle is:
$\theta_r=\arctan{\frac{\gamma\sin\theta'}{\cos\theta'}}=\arctan(\gamma\tan\theta')$
Instead, in the y-direction the reflected angle is:
$\theta_r=\arctan{\frac{\sin\theta'}{\gamma\cos\theta'}}=\arctan(\frac{\tan\theta'}{\gamma})$
Is it correct?

Those look like the result of detailed calculations. In the first case, assuming the equation is correct, what does that equation tell you?

$\theta_r=\arctan{\frac{\gamma\sin\theta'}{\cos\theta'}}=\arctan(\gamma\tan\theta')$

How does $\theta_r$ relate to $\theta_i$? Why do you have $\theta'$ in your answer?

In the second case, why (physically) do we not have $\theta_r = \theta_i$? In what way is a moving mirror different from a stationary mirror (in frame $S$)? And, again, why does $\theta'$ appear in your answer?

Shouldn't the answer be $\theta_r$ expressed in terms of $\theta$ and $v$ or $\gamma$? I don't interpret the question as asking for $\theta_r$ in terms of $\theta'$.

I suggest you need to think through these problems more. And think what does the answer depend on? What is physically relevant? In frame $S$ what changes when the light reflects?

 

[PeroK]

Hint: what about velocity components?

 

[Frostman]

Those look like the result of detailed calculations. In the first case, assuming the equation is correct, what does that equation tell you?

$\theta_r=\arctan{\frac{\gamma\sin\theta'}{\cos\theta'}}=\arctan(\gamma\tan\theta')$

Because $\gamma$ is in the numerator we have an dilatation, so the angle I will get will be $\theta_r \rightarrow \frac{\pi}{2}$ for $\gamma \rightarrow \infty$

How does $\theta_r$ relate to $\theta_i$? Why do you have $\theta'$ in your answer?

I have $\theta'$, because I used the Lorentz transformation. You're right to tell me how $\theta_r$ relates to $\theta_i$.

In the second case, why (physically) do we not have $\theta_r = \theta_i$? In what way is a moving mirror different from a stationary mirror (in frame $S$)? And, again, why does $\theta'$ appear in your answer?

Because the orthogonal component to the boost is not affected instead the parallel one get a contraction.

Shouldn't the answer be $\theta_r$ expressed in terms of $\theta$ and $v$ or $\gamma$? I don't interpret the question as asking for $\theta_r$ in terms of $\theta'$.

Absolutely yes.

Hint: what about velocity components?

Let's start over by doing a bit of order.

$S'$: it is the rest frame, where the mirror has $\vec{v}=0$
$S$: it is the frame where we see that the mirror has $\vec{v} \neq 0$. More specifically $\vec{v}=(v_x, 0, 0)$.

The components of an incident photon are the following:
$$
\vec{u_i'}=\begin{pmatrix}
-c \cos \theta_i' \\
-c \sin \theta_i' \\
0 \\
\end{pmatrix}
$$$$
\vec{u_i}=\begin{pmatrix}
-c \cos \theta_i \\
-c \sin \theta_i \\
0 \\
\end{pmatrix}
$$
Now we can use velocity-addition formula:
$$
u_x = \frac{u_x'+v}{1+\frac{vu_x'}{c^2}}
$$
The same we can do with $\theta_r$ and $\theta_r'$. We can find a relation that relates $\theta_i$ with $\theta_i'$ and $\theta_r$ with $\theta_r'$ and remember that $\theta_i' = \theta_r'$ in $S'$ and getting $\theta_i = C \theta_r$ for $S$

 

[PeroK]

Let me tell you how I think about it. In the rest frame of the mirror, we have angle of incidence equals angle of reflection: $\theta'_i = \theta'_r$.

That means that $u'_y$ is unchanged after reflection and $u'_x$ changes sign.

Note that if $u'_y$ is constant in this frame, then $u_y$ is constant in every frame (as long as the frames agree on the y-direction).

First, let's consider the case where the mirror moves in the y-direction. In this case, the sign of $u_x$ must change on reflection. You can check this by looking at the velocity transformation formula, but it must be true by thinking about the symmetry of motion in the $\pm x$ directions in this case.

Therefore, in this case, we have $\theta_i = \theta_r$. And that makes sense physically as the mirror moving in the y-direction does not affect the reflection process.

Now, can you analyse the case where the mirror is moving in the x-direction - towards the light source?

 

[Frostman]

Now, can you analyse the case where the mirror is moving in the x-direction - towards the light source?

$u_x'$ changes the sign, $u_y'$ doesn't change the sign in $S'$ after reflection. Passing in $S$, $u_x$ and $u_y$ change their value on incident and reflection because we have these laws: $u_x = \frac{u_x'+v}{1+vu_x'}$ and $u_y = \frac{\gamma u_y'}{1+vu_x'}$

 

[PeroK]

$u_x'$ changes the sign, $u_y'$ doesn't change the sign in $S'$ after reflection. Passing in $S$, $u_x$ and $u_y$ change their value on incident and reflection because we have these laws: $u_x = \frac{u_x'+v}{1+vu_x'}$ and $u_y = \frac{\gamma u_y'}{1+vu_x'}$

That's not an analysis. That's quoting two formulas that I already know.

 

[Frostman]

If the mirror is moving in the x-direction, $u_x'$ changes its sign (from $-u_x'$ to $u_x'$) because the mirror plane is orthogonal to x-axis while $u_y'$ doesn't change direction, it keeps going along $-y$.
In $S$ $u_y$ doesn't change its sign, keep maintain the sign of $u_y'$, while for $u_x$ the sign can change if $|u_x'|<v$ , if not we have the same sign as well, it depends from numerator of $u_x$.
I don't know if now is better as analysis.

 

[PeroK]

If the mirror is moving in the x-direction, $u_x'$ changes its sign (from $-u_x'$ to $u_x'$) because the mirror plane is orthogonal to x-axis while $u_y'$ doesn't change direction, it keeps going along $-y$.
In $S$ $u_y$ doesn't change its sign, keep maintain the sign of $u_y'$, while for $u_x$ the sign can change if $|u_x'|<v$ , if not we have the same sign as well, it depends from numerator of $u_x$.
I don't know if now is better as analysis.

What are your conclusions? I want to know whether $\theta_i = \theta_r$ or not. I want a straight answer: yes or no, and why.

 

[Frostman]

No, it's not equal.

In $S'$ we have:

$u_{ix}' = -\cos \theta_i'$
$u_{rx}' = \cos \theta_r'$

In $S$ we have:

$u_{ix} = -\cos \theta_i = \frac{-\cos \theta_i' + v}{1-v\cos \theta_i'}$
$u_{rx} = \cos \theta_r = \frac{\cos \theta_r' + v}{1+v\cos \theta_r'} = \frac{\cos \theta_i' + v}{1+v\cos \theta_i'}$

The last step come from the fact that in $S'$ we have $\theta_r' = \theta_i'$.

The value of $|u_{ix}|\neq |u_{rx}|$ as it was with $|u_{ix}'|= |u_{rx}'|$

 

[PeroK]

No, it's not equal.

In $S'$ we have:

$u_{ix}' = -\cos \theta_i'$
$u_{rx}' = \cos \theta_r'$

In $S$ we have:

$u_{ix} = -\cos \theta_i = \frac{-\cos \theta_i' + v}{1-v\cos \theta_i'}$
$u_{rx} = \cos \theta_r = \frac{\cos \theta_r' + v}{1+v\cos \theta_r'} = \frac{\cos \theta_i' + v}{1+v\cos \theta_i'}$

The last step come from the fact that in $S'$ we have $\theta_r' = \theta_i'$.

The value of $|u_{ix}|\neq |u_{rx}|$ as it was with $|u_{ix}'|= |u_{rx}'|$

I don't see the answer there. How can I see that $\theta_i \ne \theta_r$?

I don't see any calculation of $\theta_r$.

 

[Frostman]

I wrote all the passages that I have done:
In $S'$, we have $\theta_i'=\theta_r'$

For the beam incident we have:
$u_{ix}'=-\cos \theta_i'$
$u_{iy}'=-\sin \theta_i'$

For the beam reflected we have:
$u_{rx}'=\cos \theta_r'$
$u_{ry}'=-\sin \theta_r'$

In $S$ we have:

For the beam incident we have:
$u_{ix}=-\cos \theta_i = \frac{-\cos \theta_i'+v}{1-v\cos\theta_i'}$
$u_{iy}=-\sin \theta_i =\frac{-\gamma \sin \theta_i'}{1-v\cos\theta_i'}$

For the beam reflected we have:
$u_{rx}=\cos \theta_r = \frac{\cos \theta_r'+v}{1+v\cos\theta_r'}$
$u_{ry}=-\sin \theta_r =\frac{-\gamma \sin \theta_r'}{1+v\cos\theta_r'}$

Now let's calculate:
$\frac{u_{iy}}{u_{ix}}=\tan\theta_i=\frac{\gamma \sin \theta_i'}{\cos \theta_i'-v}$
$\frac{u_{ry}}{u_{rx}}=\tan\theta_r=\frac{\gamma \sin \theta_r'}{\cos \theta_r'+v}$

$\frac{\tan\theta_i}{\tan\theta_r}=\frac{\gamma \sin \theta_i'}{\cos \theta_i'-v}\frac{\cos \theta_r'+v}{\gamma \sin \theta_r'}$

Remembering that $\theta_i'=\theta_r'$ we have:

$\frac{\tan\theta_i}{\tan\theta_r}=\frac{\cos \theta_r'+v}{\cos \theta_r'-v}$

Using the inverse Lorentz transformation we have:

$\cos \theta_r' = \frac{\cos \theta_r - v}{1-v\cos\theta_r}$

Substituting this in the relationship between the tangents we have:

$\frac{\tan\theta_i}{\tan\theta_r}=\frac{\cos \theta_r (1-v^2)}{\cos\theta_r(1+v^2)-2v} \implies \tan\theta_i = \frac{\cos \theta_r (1-v^2)}{\cos\theta_r(1+v^2)-2v}\tan\theta_r$

This shows that $\theta_r \neq \theta_i$ in $S$.

Could it be ok now? Or I missed again something? Or have I made a mistake in the starting hypotheses?

 

[PeroK]

I'll look at it later if I have time. You can simplify that last expression to $\frac{1 + v}{1-v}$.

 

[Frostman]

Thank you for your time.

 

[PeroK]

I wrote all the passages that I have done:
In $S'$, we have $\theta_i'=\theta_r'$

For the beam incident we have:
$u_{ix}'=-\cos \theta_i'$
$u_{iy}'=-\sin \theta_i'$

For the beam reflected we have:
$u_{rx}'=\cos \theta_r'$
$u_{ry}'=-\sin \theta_r'$

In $S$ we have:

For the beam incident we have:
$u_{ix}=-\cos \theta_i = \frac{-\cos \theta_i'+v}{1-v\cos\theta_i'}$
$u_{iy}=-\sin \theta_i =\frac{-\gamma \sin \theta_i'}{1-v\cos\theta_i'}$

For the beam reflected we have:
$u_{rx}=\cos \theta_r = \frac{\cos \theta_r'+v}{1+v\cos\theta_r'}$
$u_{ry}=-\sin \theta_r =\frac{-\gamma \sin \theta_r'}{1+v\cos\theta_r'}$

Now let's calculate:
$\frac{u_{iy}}{u_{ix}}=\tan\theta_i=\frac{\gamma \sin \theta_i'}{\cos \theta_i'-v}$
$\frac{u_{ry}}{u_{rx}}=\tan\theta_r=\frac{\gamma \sin \theta_r'}{\cos \theta_r'+v}$

$\frac{\tan\theta_i}{\tan\theta_r}=\frac{\gamma \sin \theta_i'}{\cos \theta_i'-v}\frac{\cos \theta_r'+v}{\gamma \sin \theta_r'}$

Remembering that $\theta_i'=\theta_r'$ we have:

$\frac{\tan\theta_i}{\tan\theta_r}=\frac{\cos \theta_r'+v}{\cos \theta_r'-v}$

Using the inverse Lorentz transformation we have:

$\cos \theta_r' = \frac{\cos \theta_r - v}{1-v\cos\theta_r}$

Substituting this in the relationship between the tangents we have:

$\frac{\tan\theta_i}{\tan\theta_r}=\frac{\cos \theta_r (1-v^2)}{\cos\theta_r(1+v^2)-2v} \implies \tan\theta_i = \frac{\cos \theta_r (1-v^2)}{\cos\theta_r(1+v^2)-2v}\tan\theta_r$

This shows that $\theta_r \neq \theta_i$ in $S$.

Could it be ok now? Or I missed again something? Or have I made a mistake in the starting hypotheses?

This doesn't look quite right. I would have expected the angle of the incident light to be relevant. You can express this as either $\cos \theta'$ or $|u'_x|$, being the component of velocity in the x-direction in the mirror frame.

If we assume for simplicity that $u_x > v$, then I get:
$$\tan \theta_r = \frac{1 - v\cos \theta'}{1 + v\cos \theta'} \tan \theta_i$$
Hence $$\theta_r < \theta_i$$