Relating Light Angles in a High-Speed Reference Frame

[bowlbase]

Homework Statement

A space station observes a high-speed rocket passing by at speed β c in the +x direction. The rocket suddenly emits a light ray from a powerful laser. According to the space station, the light ray was emitted at an angle of θ with respect to the +x-axis. However, the technician aboard the rocket ship who fired the light pulse sent it off at an angle of θ′ with respect to the +x′-axis. (As usual, the x′-axis is chosen to be parallel to the x-axis.) Show that the angles are related by the formula
$$cos(\theta^\prime)= \frac {cos(\theta) - \beta}{1- \beta cos(\theta)}$$

Homework Equations

Lorentz Transforms

The Attempt at a Solution

I'm having issues as soon as I start. If the velocity is $\beta c$ then wouldn't

$$cos(\theta)= \frac {\beta c}{c}=\beta$$

Meaning the numerator of the given equation will always be zero?

Clearly this can't be the case but I fail to understand why it's not.

I assume I need to apply the transforms to the problem first and maybe one of the cosines or betas was suppose to be primed or something. My attempts so far are not working at all.

$$v_y=\frac{c sin(\theta)}{1-\frac{\beta c}{c^2}}$$

Trying the same for v_x just gives me c since the betas cancel. Really at a loss for this now.

Thanks for the help.

 

[tiny-tim]

hi bowlbase!

If the velocity is $\beta c$ then wouldn't

$$cos(\theta)= \frac {\beta c}{c}=\beta$$

i don't understand this at all

the ray of light goes from the spaceship to the space station

in one frame it is at angle θ, in the other frame at angle θ'

it goes, of course, at the speed of light

carry on from there ​

 

[bowlbase]

I don't think they necessarily mean that the light is emitted towards the station. So I've placed the ship and station at the origin at the moment the light is emitted. The ship says it sent the beam out at some angle but clearly the station disagrees on the angle.

For the formula I have the light beam as my hypotenuse and the velocity of the ship along the x axis. Given that cosine equals the x component over the hypotenuse I used that to solve for the function.

I guess that doesn't make sense in this case.

I'm really at a loss at how I should go about this.

 

[tiny-tim]

hi bowlbase!

write the equation of the ray in x y and t

use the Lorentz transformation to convert that to x' y' and t'

 

[bowlbase]

Okay, so there's no contraction for y so y'=y. For x':

$$x^\prime = \gamma (x-\beta tc)$$

So for the observer's frame the angle should be:

$$tan(\theta^\prime)=\frac{y^\prime}{x^\prime}$$


$$\gamma = \frac{1}{\sqrt{1-\beta}}$$

$$tan(\theta^\prime) = (\frac{y^\prime \sqrt{1-\beta}}{ (x-\beta tc)})$$

I could square both sides to get the tangent=>cosine identity but at first glance I'd say that denominator would be too messy for that to turn out to be a solution. Am I even on the right track here?

 

[tiny-tim]

hi bowlbase!

(just got up :zzz:)

wouldn't it have helped help if you'd started with the x,y equation of the ray?

 

[bowlbase]

I figured it out! Instead of setting c cos$\theta$=$\beta$c I set it equal to u_x and used the x transform. I started out with the right idea but needed to forget about the beta and think about the x axis. Thanks for the help.