# Relativity - Light At An Angle

## Homework Statement

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

## 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
Mentor

## Homework Statement

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.

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