Find the contraction of angles seen by an observer

[guyvsdcsniper]

Im just confused. I understand that we need to get C'D into the mix. I want this because it runs parallel to the direction of motion. Anything perpendicular will not get contracted, only going in the direction of motion will result in contracts so that's why we need the parallel components.

I know that $\tan \alpha' = \frac{|C'D|}{|AD|}$ and $\tan \beta' = \frac{|C'D|}{|DB|}$ . Tan allows me to relate something contracted in one frame to something that remains the same in both frames

I also know that C'D = ϒCD, which is the contracted length is equal to the uncontracted length times gamma. I know that gamma is equal to √(¾) is is √(3)/2 for this problem.

So this tells me that the contracted length is √(3)/2 CD.

So then let me say $\tan \alpha' = √(3)/2* \frac{|CD|}{|AD|}$ And I know that $\frac{|CD|}{|AD|}$ is tan60 degrees, which is equivalent to √3.

so √(3)/2 * √(3) = 1/2. therefore tan^-1(1/2) = 26.6 degrees?

Is this correct?

 

[guyvsdcsniper]

Okay, but we've done all the work except solve it. There's not much left to do except a final step which gives the answer.

You're honestly saying that you cannot finish the job from here:Note that $C' \rightarrow C$ is the only point on the triangle that changes from one frame to the the other. The line $ADB$ is unaffected by length contraction, as it is perpendicular to the direction of motion.

Im sorry if its coming off that way . I genuinely want to learn how to approach this problem and I am not consciously trying to ask for the answer. I understand you all have put in the effort to help set the problem up and help me understand, but its just hard for me to really grasp it.

 

[PeroK]

So then let me say $\tan \alpha' = √(3)/2* \frac{|CD|}{|AD|}$ And I know that $\frac{|CD|}{|AD|}$ is tan60 degrees, which is equivalent to √3.

so √(3)/2 * √(3) = 1/2. therefore tan^-1(1/2) = 26.6 degrees?

Is this correct?

$\alpha' = 60$ degrees. The primed frame is the rocket frame.

 

[guyvsdcsniper]

$\alpha' = 60$ degrees. The primed frame is the rocket frame.

Your right. So then I should get this as my answer. And then I can just repeat the same for beta.

Then add those two up and subtract by 180 to get the 3rd angle?

 

[PeroK]

$\alpha$ can't be greater than $60$ degrees.

You need to be more careful in your calculations.

 

[TSny]

You are getting very close. But you are making some careless errors regarding which length is the contracted length and the definition of gamma.

I also know that C'D = ϒCD, which is the contracted length is equal to the uncontracted length times gamma.
I know that gamma is equal to √(¾) is is √(3)/2 for this problem.

The relation C'D = ϒCD is correct. But C'D is the uncontracted length and CD is the contracted length.

Gamma is not equal to $\sqrt{1-v^2/c^2}$. Gamma is $\large \frac{1}{\sqrt{1-v^2/c^2}}$

Please try again.

 

[guyvsdcsniper]

You are getting very close. But you are making some careless errors regarding which length is the contracted length and the definition of gamma.The relation C'D = ϒCD is correct. But C'D is the uncontracted length and CD is the contracted length.

Gamma is not equal to $\sqrt{1-v^2/c^2}$. Gamma is $\large \frac{1}{\sqrt{1-v^2/c^2}}$

Please try again.

So it should be this? Which I then just repeat to β and then sum the two and subtract from 180?

 

[TSny]

Looks good.

 

[guyvsdcsniper]

Thank you all for your help. My apologies if I made it more difficult that necessary.