Relativity - Light At An Angle

[csnsc14320]

1. The problem statement, all variables and given/known data
A light beam is emitted at an angle \theta_o with respect to the x' axis in S'

a) Find the angle \theta the beam makes with respect to the x axis in S.
- Ans. : cos\theta = (cos\theta_o + \frac{v}{c})(1 + \frac{v}{c} cos\theta_o)


3. The attempt at a solution
From an example problem in our book, we know that with a ruler at an angle in the same situation has angle in laboratory frame:

\theta = arctan(tan(\theta_o)\gamma), where \gamma = \frac{1}{\sqrt{1-frac{v^2}{c^2}}}

if you just take cos of both sides, then you have

cos\theta = cos(arctan((tan(\theta_o)\gamma)))

Drawing a triangle with the right leg as tan(\theta_o) and the bottom leg as \frac{1}{\gamma} you get a hypotenuse of \sqrt{(\frac{1}{\gamma})^2 + tan(\theta_o)^2}

from this i narrowed it down to

cos\theta = \frac{\frac{1}{\gamma}}{\sqrt{(\frac{1}{\gamma})^2 + tan(\theta_o)^2}} = \frac{1}{\sqrt{1 + frac{\gamma*2}{cos{\theta_o)^2} - \gamma^2}}

I'm not seeing how this can reduce to my answer I'm suppose to get yet?

[Doc Al]

1. The problem statement, all variables and given/known data
A light beam is emitted at an angle \theta_o with respect to the x' axis in S'

a) Find the angle \theta the beam makes with respect to the x axis in S.
- Ans. : cos\theta = (cos\theta_o + \frac{v}{c})(1 + \frac{v}{c} cos\theta_o)
That answer doesn't look right. It should be:

cos\theta = \frac{(cos\theta_o + \frac{v}{c})}{(1 + \frac{v}{c} cos\theta_o)}

The easy way to derive it is to use the velocity transformations.

[csnsc14320]

That answer doesn't look right. It should be:

cos\theta = \frac{(cos\theta_o + \frac{v}{c})}{(1 + \frac{v}{c} cos\theta_o)}

The easy way to derive it is to use the velocity transformations.

oops I missed the divide sign in my book - your answer is the right one.

I'll give that a shot