# Homework Help: Special Relativity Angles

1. Feb 15, 2012

### icantsee99

1. The problem statement, all variables and given/known data
A meter stick is at rest in the laboratory frame. It lies in the x-y plane, making an angle of 30° with the x-axis. What angle does this meter stick make with the x'-axis of a reference frame moving at V=0.8c in the x-direction.

Laboratory frame is the rest frame.
Variables: θ=30°, θ', V=0.8c

2. Relevant equations
Length Contraction
x = x'*√1-V2/c2

3. The attempt at a solution
I tried treating the meter stick as a unit vector so that x=(1)cosθ and x'=L'cos(θ'). y=y' obviously.
cos(θ') = cos(30)/√1-0.82
= 0.866/0.6
= 1.44
θ'=cos-1(1.44)=domain error

I feel like I'm making a stupid mistake and I just can't see it

2. Feb 16, 2012

### vela

Staff Emeritus
In the moving frame, you expect x' < x because of length contraction, so it must be that x' = x/γ. You're using x' = γx.

3. Feb 16, 2012

### icantsee99

ahh so it would be tangent instead of cosine, right?

In that case,
tan(θ')=tan(30)/.6
θ'= 43.9°

Thanks!

4. Feb 16, 2012

### vela

Staff Emeritus
No, I don't think that's right. How did you get tangent? That's a gamma, not a y. Also, remember the meter stick is no longer 1 meter long in the moving frame.

5. Feb 16, 2012

### icantsee99

Oh, I thought it was a y. Regardless, you can just say

tanθ=y/x tanθ'=y'/x'

(tanθ)/(tanθ')=yx'/xy'

but since y=y'

(tanθ)/(tanθ')=x'/x

xtanθ=x'tanθ'

so then use

tanθ'=tanθ/√1-V2/c2

plugging in the numbers gives me θ'=43.9, which is the answer in the book.

please let me know if my reasoning is wrong, I want a better understanding of this.

6. Feb 16, 2012

### vela

Staff Emeritus
That's right. Just wanted to make sure your reasoning for going from cos to tan was correct.

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