Different angle in different frames, intro relativity

[Coffee_]

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.


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?

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]

 

[mfb]

At first glance, I would say that the vertical velocity component remains the same since no relativistic effects happen in that direction.

You still have time dilation to consider. Speed is distance per time and the "per time" part changes.

 

[Coffee_]

You still have time dilation to consider. Speed is distance per time and the "per time" part changes.

So the vertical part would be something like $\beta_{yA} = \frac{\beta_{yB}}{\gamma}$ ?

 

[mfb]

Does it work out with the speed of light then?

 

[Coffee_]

Does it work out with the speed of light then?

No it doesn't, just tried.

 

[mfb]

What went wrong?
Do you know 4-vectors? They will certainly give the right answer.

 

[Coffee_]

What went wrong?
Do you know 4-vectors? They will certainly give the right answer.

Yeah I know them but it doesn't work out. It's been a bit and I did some searched on the internet and it seems like there is more general transformation for perpendicular components of velocities which reduces to my expression when there is no horizontal component in the moving frame. And since in my case there is a horizontal component my expression would be wrong I guess?

 

[mfb]

With 4-vectors it works for me. Didn't check the other method now, but it is not just time dilation, right.