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I am reading a book called "Special Relativity" by Anthony Philip French.

There is a small section that discusses the situation of trying to observe a distant star using a telescope from earth while considering the orbit of the earth around the sun.

Here is a depiction from the book

enter image description here

The book says

A telescope on a stationary earth (a) would have to be pointed at the true altitude $\theta_0$ in order that the rays of light from the star should travel along the axis of the instrument and form an image at the center of the field of view. But on a moving earth (b) the telescope would have to be tilted at a slightly different angle, $\theta$. The difference of angles is the aberration, $\alpha$.

At positions 1 and 3 we have a situation like that depicted in Fig. 2-2, in which the aberration angle is only of magnitude $v\sin{\theta_0/c}$.

Note that it is not clear to me if it is written $v\sin{(\theta_0/c)}$ or $\frac{v\sin{\theta_0}}{c}$, though I think it must be the latter (because then it agrees with a result about raindrops that I mention below).

My question is how to arrive at this aberration angle?

Here is Fig. 2-2

enter image description here

The book mentions that the situation is analogous to that of a raindrop falling vertically at speed $w$ (relative to the earth), but relative to a vehicle with speed $v$ the raindrop has a velocity that forms an angle of $\tan{\frac{v}{w}}$ with the vertical.

I believe the calculation is the following

$$\vec{v}_{rd,S}=-w\hat{j}$$ $$\vec{v}_{rd,S'}=\vec{v}_{S,S'}+\vec{v}_{rd,S}=-v\hat{i}-w\hat{j}$$

where the subscript has form $x,y$ where $x$ is the object tracked and $y$ is the reference frame. $S$ is earth and $S'$ is the vehicle. Thus, $\vec{v}_{S,S'}$ is the velocity of the earth relative to the vehicle, which is $-v\hat{i}$.

$\vec{v}_{rd,S'}$ is the velocity of the raindrop relative to the vehicle, and we see that this vector forms an angle $\alpha$ with the vertical axis, such that $\tan{\alpha}=\frac{v}{w}$.

Now, let's try to apply similar thinking to the case of Earth at position 1 in the first picture I posted above.

enter image description here

$$\vec{v}_{l,S}=-c(\cos{\theta}\hat{j}+\sin{\theta}\hat{k})$$

is the velocity of the ray of light from the star relative to a frame that is at rest relative to the star (perhaps approximately the sun?).

We would like the velocity of the ray of light relative to the Earth because then we can find out the angle of this vector relative to the y-axis (so we can compare with the $\theta_0$ we observe from the $S$ frame).

$$\vec{v}_{l,S'}=\vec{v}_{S,S'}+\vec{v}_{l,S}$$

Now, I already sense a potential problem because I don't think this Galilean transformation can be used with the speed of light which is constant in every frame.

Indeed, if we plug in values we get

$$\vec{v}_{l,S'}=(-v_{S',S}-c\cos{\theta})\hat{j}-c\sin{\theta}\hat{k}$$

which has a length larger than $c$.

Despite this, if we continue down this path we have

enter image description here

and so

$$\sin{\alpha}=\frac{c}{\lVert v_{l, S'} \rVert}\sin{\theta}<\sin{\theta} \implies \alpha<\theta$$

which seems to be qualitatively the result we are looking for.

The aberration angle would be

$$\theta - \alpha = \theta - \sin^{-1}\left ( \frac{c}{\lVert v_{l, S'} \rVert}\sin{\theta} \right )$$

I also saw the following calculation in Wikipedia

$$\tan{\alpha}=\frac{c\sin{\theta}}{v_{S',S}+c\cos{\theta}}$$

$$=\frac{\sin{\theta}}{\frac{v_{S',S}}{c}+\cos{\theta}}$$

and for $\theta=\pi/2$ we have

$$\tan{\alpha}=\frac{c}{v_{S',S}}$$

Then they say that this reduces to

$$\tan(\theta-\alpha)=v_{S',S}/c$$

which I am not sure how they obtained. If we accept that this is true, then when $v_{S',S}/c << 1$ we can approximate $\tan{\theta-\alpha}$ by $\theta-\alpha$ and so we have

$$\theta - \alpha = v_{S',S}/c$$

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The calculations in the original post are correct (at least as far as the classical approach to the problem goes) but incomplete.

Starting from

$$\tan{\alpha}=\frac{c\sin{\theta}}{v_{S',S}+c\cos{\theta}}\tag{1}$$

$$\tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}}\tag{2}$$

we use the formula for $\tan{(\theta-\alpha)}$, where $\theta-\alpha$ is the aberration angle we are looking for.

$$\tan{(\theta-\alpha)}=\frac{\tan{\theta}-\tan{\alpha}}{1+\tan{\theta}\tan{\alpha}}\tag{3}$$

I won't go through the steps, but eventually we reach

$$\tan{(\theta-\alpha)}=\frac{\frac{v_{S',S}}{c}\sin{\theta}}{1+\frac{v_{S',S}}{c}\cos{\theta}}\tag{4}$$

At this point it seems that approximations are used.

Because $\frac{v_{S',S}}{c}$ is much smaller than 1, then (4) is approximately $\frac{v_{S',S}}{c}\sin{\theta}$.

Furthermore, because $\theta-\alpha$ is very close to zero and $\tan{x}\approx x$ then we can say that

$$\theta-\alpha \approx \tan{(\theta-\alpha)}=\frac{v_{S',S}}{c}\sin{\theta}\tag{5}$$

which is aberration angle.

A few observations

  • Considering the first picture in the original post, the discussion so far has been about the earth being at a position such as 1 or 3. When $\theta=\frac{\pi}{2}$, we have the case of the star being "directly above" the earth, meaning the rays of light arrive perpendicular to the velocity of the earth (as viewed from the sun's frame, S).

  • However, when the earth is at positions 2 or 4 in its orbit, rays of light also arrive perpendicular to the velocity of the earth

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