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The motion of the earth relative to a star changes the angle at which the star is perceived. Consider to inertial frames of reference I and I'. I' moves with velocity w relative to I along the x-axis. Show that

a) by looking at it in a non-relativistic way (classic) one can show the following correlation

$\tan(\Delta\theta)=\frac{w\sin(\theta)}{c-w\cos(\theta)}$

where $\theta$ is the angle for I and $\theta'$ the angle for I'. $\Delta\theta=\theta'-\theta$ is the perceived angle-difference.

b) by looking at it in a relativistic way that

$\cos(\theta)=\frac{\beta+\cos(\theta')}{1+\beta\cos(\theta')}$ and $\sin(\Delta\theta)=\sin(\theta)\frac{w}{c}$

with $\beta=\frac{w}{c}$.

I've tried getting a visual aid for the problem by drawing all the information I got, but I still can't seem to find an approach to get to the desired equation. I've tried using the law of cosine, but it got me nowhere. Anyone got any ideas?

enter image description here

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  • $\begingroup$ Are you confused on both parts of the problem, or just the relativistic parts? $\endgroup$
    – user854
    Apr 27 '16 at 22:11
  • $\begingroup$ a)Consider that the incoming ray of light is a particle with velocity $\mathbf{u}=-c\left(\cos\theta,\sin\theta\right)$ relatively to the frame $\mathbb{I}$. If $\mathbf{u}'=-u'\left(\cos\theta',\sin\theta'\right)$ is the velocity as seen from frame $\mathbb{I}^{\prime}$ then use the Newtonian addition of velocities $\mathbf{u}^{\prime}=\mathbf{u}-\mathbf{w}$. The magnitude $u^{\prime}$ will be not equal to $c$ since the case is nonrelativistic. $\endgroup$
    – Frobenius
    Apr 27 '16 at 23:00
  • $\begingroup$ By the way I believe that the right equation in (a) must be $\tan\left(\theta'-\theta\right)=-\dfrac{w\sin\theta}{c+w\cos\theta}$. If rain drops fall normal to the ground, $\theta=\pi / 2$, then as you are driving you see the drops falling obliquely with $\theta'\le\pi/2=\theta$ that is $\tan\left(\theta'-\theta\right)\le 0$ while your equation gives $\ge 0$. $\endgroup$
    – Frobenius
    Apr 27 '16 at 23:21
  • $\begingroup$ b)Consider that the incoming ray of light is a particle with velocity $\mathbf{u}=-c\left(\cos\theta,\sin\theta\right)$ relatively to the frame $\mathbb{I}$. If $\mathbf{u}'=-c\left(\cos\theta',\sin\theta'\right)$ is the velocity as seen from frame $\mathbb{I}^{\prime}$ then use relativistic addition of velocities $\mathbf{u},\mathbf{w}$ to find $\mathbf{u}'$. Note that all these describe the phenomenon of aberration. $\endgroup$
    – Frobenius
    Apr 27 '16 at 23:30
  • $\begingroup$ @Frobenius Thanks for the fast help. I've tried doing your approach to get the equation but I can't seem to get $\tan(\theta'-\theta)=...$. I used an addition theorem for the tangent but couldn't get the desired equation. I'm assuming I would get to the same problem doing b). $\endgroup$
    – Rab
    Apr 28 '16 at 5:59
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Hint for (a) :

enter image description here Use the Figure above to prove that :

\begin{equation} \tan\left(\phi^{\prime}-\phi\right)=\dfrac{w\sin\phi}{c-w\cos\phi}=\dfrac{w\sin\theta}{c+w\cos\theta}\;=\;-\tan\left(\theta^{\prime}-\theta\right) \tag{a-01} \end{equation} This explains why I find the last equality instead of the first as I post in one of my comments : we refer to different angles.

To give some further help, the vectors in the Figure are defined as follows using the $\:\phi=\pi-\theta\:$ angles

\begin{align} \mathbf{u} & \equiv c\cdot\left(\cos\phi,\sin\phi\right) \tag{a-02}\\ \mathbf{u}^{\prime} & \equiv u^{\prime}\cdot\left(\cos\phi^{\prime},\sin\phi^{\prime}\right) , \quad u^{\prime}=\|\mathbf{u}^{\prime}\| \tag{a-03}\\ \mathbf{w} & \equiv w\cdot\left(1,0\right) , \quad w=\|\mathbf{w}\| \tag{a-04} \end{align}

and of course \begin{equation} \mathbf{u}^{\prime} =\mathbf{u} -\mathbf{w} \tag{a-05} \end{equation}


Hint for (b) :

enter image description here

Use the Figure above to prove that :

\begin{align} \cos\phi & =\dfrac{\beta+\cos\phi^{\prime}}{1+\beta\cos\phi^{\prime}}=\dfrac{\beta-\cos\theta^{\prime}}{1-\beta\cos\theta^{\prime}}=-\cos\theta \qquad \text{or} \tag{b-01}\\ \cos\phi^{\prime} & =\dfrac{-\beta+\cos\phi}{1-\beta\cos\phi}=-\dfrac{\beta+\cos\theta}{1+\beta\cos\theta}=-\cos\theta^{\prime} \tag{b-02} \end{align}

To give some further help, the vectors in the Figure are defined as follows using the $\:\phi=\pi-\theta\:$ angles

\begin{align} \mathbf{u} & \equiv c\cdot\left(\cos\phi,\sin\phi\right) \tag{b-03}\\ \mathbf{u}^{\prime} & \equiv c\cdot\left(\cos\phi^{\prime},\sin\phi^{\prime}\right) \tag{b-04}\\ \mathbf{w} & \equiv w\cdot\left(1,0\right) , \quad w=\|\mathbf{w}\| \tag{b-05} \end{align}

and of course the velocity vector $\mathbf{u}^{\prime}$ is the relativistic sum of velocities $\mathbf{u}$ and $-\mathbf{w}$ :
\begin{align} u_{x}^{\prime} & =\dfrac{u_{x}-w}{1-\dfrac{u_{x}w}{c^{2}}} \tag{b-06x}\\ u_{y}^{\prime} & =\dfrac{u_{y}}{\gamma\left(1-\dfrac{u_{x}w}{c^{2}}\right)} \tag{b-06y}\\ \gamma &=\left(1-\dfrac{w^{2}}{c^{2}}\right)^{-1/2} \tag{b-06$\gamma$} \end{align}

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a) Assume that light has to travel 1 m to reach the origin of the reference frame I, so it'll take $t=1/c $ seconds to reach.

Now, using trigonometry (the sine rule in the first equation), we get:

$$ \frac{\sin{\Delta\theta}}{\omega t} = \frac{\sin{\theta}}{ct} $$

$$ct\cos{\Delta\theta} = 1-{\omega t}\cos{\theta}$$

Dividing the first equation with the second,

$$ \tan{\Delta\theta} = \frac{\omega\sin{\theta}}{c(1-\frac{\omega}{c}\cos{\theta})} = \frac{\omega\sin{\theta}}{c-\omega\cos{\theta}} $$

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  • $\begingroup$ @Arjit_Seth Sorry for my late response. Thanks, now I understand how to solve a). I've tried doing b) in a similar way, but I just can't seem to get it to work. I used the relativistic addition of velocities $d=\frac{a+b}{1+ab/c^2}$, but it got me nowhere. $\endgroup$
    – Rab
    Apr 28 '16 at 20:47
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    $\begingroup$ I like Frobenius' approach to b). The thing about relativistic addition of velocities is that $ {\vec {u'}}$ is the velocity measured with respect to the inertial frame I'. However, as I perform the required calculation, I get: $ \cos\theta = \frac{-\beta + \cos\theta '}{1 - \beta\cos\theta '} $ $\endgroup$ Apr 28 '16 at 21:04
  • $\begingroup$ Do you mind sharing the source of the question? $\endgroup$ Apr 28 '16 at 21:09
  • $\begingroup$ @Arjit_Seth It's from a german textbook. I tried translating it as best as I could. $\endgroup$
    – Rab
    Apr 29 '16 at 14:06
  • $\begingroup$ @Arjit_Seth : I think that the result $\cos\theta = \frac{-\beta + \cos\theta'}{1 - \beta\cos\theta'}$ you refer in your comment above is right and identical to my result , see last equality in equation (b-01) in my answer. $\endgroup$
    – Frobenius
    Apr 29 '16 at 18:31

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