Special relativity - transformation of angle

[Toby_phys]

Homework Statement

Homework Equations

Gamma factor:
$$\gamma = \frac{1}{\sqrt{1-\beta^2}} $$
Lorentz contraction
$$l'=\frac{l}{\gamma}$$
Trig:
$$ cos\theta = \frac{adjacent}{hypotenuse}$$

The Attempt at a Solution

I have all the quantities but the algebra doesn't seem to work out.

Thank you in advance for any help

 

[PeroK]

You've probably approached it the wrong way. Think about energy-momentum transformations for light instead.

 

[Toby_phys]

wow, that was soo easy. Thank you.

However what was wrong with my method? I understand why your way works, but why didn't mine?

 

[PeroK]

wow, that was soo easy. Thank you.

However what was wrong with my method? I understand why your way works, but why didn't mine?

There's a relativity of simultaneity issue that you missed. You need to use a Lorentz Transformation on the x-coordiate, not simply a length contraction.

 

[Toby_phys]

Oh ok, thank you

 

[PeroK]

I found my notes on this. If we do things in the Earth's frame first, with light emitted at $t=0$ with the star at $(0,0)$ and the Earth at $(x,y)$, then $\tan \theta = \frac{y}{x}$

Note that the light reaches Earth at $t = \frac{r}{c}$ where $r^2 = x^2 + y^2$

In the star's frame, the light is emitted from $(0,0)$ at $t'=0$ and reaches the Earth at $(x', y') = (\gamma(x-vt), y)$ at some time $t'$ that isn't important.

So, $\tan \theta' = \frac{y'}{x'} = \frac{y}{\gamma(x-vt)} = \frac{y}{\gamma(x-vr/c)}$

If you work through that and do a bit of trig manipulation, you should get the same answer.