# Two inertial frames (different angles)

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? • Are you confused on both parts of the problem, or just the relativistic parts?
– user854
Apr 27 '16 at 22:11
• 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. Apr 27 '16 at 23:00
• 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 $\ge 0$. Apr 27 '16 at 23:21
• 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. Apr 27 '16 at 23:30
• @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).
– Rab
Apr 28 '16 at 5:59

Hint for (a) : 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) : 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}

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,

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

• @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.
– Rab
Apr 28 '16 at 20:47
• 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: $\cos\theta = \frac{-\beta + \cos\theta '}{1 - \beta\cos\theta '}$ Apr 28 '16 at 21:04
• Do you mind sharing the source of the question? Apr 28 '16 at 21:09
• @Arjit_Seth It's from a german textbook. I tried translating it as best as I could.
– Rab
Apr 29 '16 at 14:06
• @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. Apr 29 '16 at 18:31