Find the contraction of angles seen by an observer

  • #36
TSny
Homework Helper
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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.
 
  • #37
guyvsdcsniper
237
35
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?

IMG_8283.JPG
 
  • #38
TSny
Homework Helper
Gold Member
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Looks good.
 
  • #39
guyvsdcsniper
237
35
Thank you all for your help. My apologies if I made it more difficult that necessary.
 

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