Relativistic Light Angles

In summary, the conversation discusses a scenario where a space station observes a high-speed rocket passing by at a speed of βc in the +x direction. The rocket emits a light ray at an angle of θ with respect to the +x-axis, but the technician aboard the rocket ship claims it was emitted at an angle of θ' with respect to the +x'-axis. The formula cos(θ') = (cos(θ) - β) / (1 - β cos(θ)) is used to show the relationship between the two angles. The conversation also mentions the use of Lorentz transforms to solve the problem.
  • #1
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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
[tex] cos(\theta^\prime)= \frac {cos(\theta) - \beta}{1- \beta cos(\theta)}[/tex]


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

[tex] cos(\theta)= \frac {\beta c}{c}=\beta[/tex]

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.

[tex] v_y=\frac{c sin(\theta)}{1-\frac{\beta c}{c^2}} [/tex]

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

Thanks for the help.
 
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  • #2
hi bowlbase! :wink:
bowlbase said:
If the velocity is ##\beta c## then wouldn't

[tex] cos(\theta)= \frac {\beta c}{c}=\beta[/tex]

i don't understand this at all :redface:

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 :smile:
 
  • #3
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.
 
  • #4
hi bowlbase! :smile:

write the equation of the ray in x y and t

use the Lorentz transformation to convert that to x' y' and t' :wink:
 
  • #5
Okay, so there's no contraction for y so y'=y. For x':

[tex]x^\prime = \gamma (x-\beta tc)[/tex]

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

[tex]tan(\theta^\prime)=\frac{y^\prime}{x^\prime}[/tex]


[tex]\gamma = \frac{1}{\sqrt{1-\beta}}[/tex]

[tex]tan(\theta^\prime) = (\frac{y^\prime \sqrt{1-\beta}}{ (x-\beta tc)})[/tex]

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?
 
  • #6
hi bowlbase! :smile:

(just got up :zzz:)

wouldn't it have helped help if you'd started with the x,y equation of the ray? :wink:
 
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  • #7
I figured it out! Instead of setting c cos[itex]\theta[/itex]=[itex]\beta[/itex]c I set it equal to ux 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.
 

1. What is the concept of "relativistic light angles"?

Relativistic light angles refer to the way in which light appears to an observer who is moving at a significant fraction of the speed of light. This phenomenon is described by the principles of special relativity, which state that the laws of physics are the same for all observers moving at a constant velocity.

2. How does the speed of light affect the perception of objects in motion?

The speed of light is a constant in the universe, and according to special relativity, the perception of objects in motion changes as an observer approaches the speed of light. This can result in time dilation, length contraction, and changes in the frequency and direction of light.

3. Can objects appear to be moving faster than the speed of light due to relativistic light angles?

No, according to the principles of special relativity, the speed of light is the maximum speed at which any object can travel. This means that no matter how fast an observer is moving, they will always measure the speed of light to be the same.

4. How do relativistic light angles affect the perception of distances?

As an object moves closer to the speed of light, the distance between that object and an observer appears to decrease. This is due to length contraction, which is a result of time dilation and the constant speed of light.

5. Can relativistic light angles be observed in everyday life?

While the effects of relativistic light angles are typically only observed at very high speeds, they can also be seen in everyday life through technologies such as GPS satellites. The satellites must account for the effects of time dilation in order to accurately calculate positions on Earth.

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