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**1.( Everything is in natural units, hbar= 1 , c =1 )**

Situation: Frame A is the lab frame. Frame B is moving at velocity ##\beta## away from frame A along the x-axis. At some point a photon is released from the origin in the B frame. The photon makes an angle ##\theta_{B}## with the x'-axis.

Situation: Frame A is the lab frame. Frame B is moving at velocity ##\beta## away from frame A along the x-axis. At some point a photon is released from the origin in the B frame. The photon makes an angle ##\theta_{B}## with the x'-axis.

**2. Relevant reasoning: The horizontal and vertical velocity components in the B frame are just equal to ##cos(\theta_{B})## and ##sin(\theta_{B})##. From the relativistic addition formula, the horizontal velocity component can be easily found being:**

##\beta_{xA}=\frac{\beta +cos(\theta_{B})}{1+\beta cos(\theta_{B})}##

My question is, what happens to the vertical component? How does the vertical component in frame B relate to frame A?

##\beta_{xA}=\frac{\beta +cos(\theta_{B})}{1+\beta cos(\theta_{B})}##

My question is, what happens to the vertical component? How does the vertical component in frame B relate to frame A?

## The Attempt at a Solution

:At first glance, I would say that the vertical velocity component remains the same since no relativistic effects happen in that direction. This makes a lot of sense except that this leads to a paradoxical situation. In frame A:

##sin(\theta_{A})=\frac{\beta_{yA}}{\beta_{A}}##

Since ##\beta_{A}=1## , the velocity of a photon doesn't change, and ##\beta_{yA}## is the same as in frame B, this would mean the angle doesn't change at all.

QUESTION: Where is my reasoning mistake? I assume in the last part. Could someone elaborate.[/B]