How to Derive β and Identify Superluminal Motion?

  • Context: Graduate 
  • Thread starter Thread starter Max.Planck
  • Start date Start date
  • Tags Tags
    Motion Superluminal
Click For Summary
SUMMARY

The discussion centers on deriving the parameter β related to superluminal motion, specifically using the equation β = v/c = 1/(cos(θ) + sin(θ)). The user also inquired about plotting β as a function of θ to identify regions of superluminal motion. The original poster, Max.Planck, later resolved the problem independently and offered to share the solution privately.

PREREQUISITES
  • Understanding of kinematics and relativistic motion
  • Familiarity with trigonometric functions and their applications
  • Basic knowledge of plotting functions in mathematical software
  • Concept of superluminal motion in physics
NEXT STEPS
  • Research the implications of superluminal motion in theoretical physics
  • Learn how to plot functions using tools like Python's Matplotlib
  • Study the relationship between velocity, angle, and relativistic effects
  • Explore advanced topics in special relativity and its mathematical foundations
USEFUL FOR

Students and professionals in physics, mathematicians, and anyone interested in the concepts of superluminal motion and its mathematical derivations.

Max.Planck
Messages
128
Reaction score
0
Hi,

I have a question about a problem on superluminal motion. Using the following figure (see attachment) I have to derive that:

a) \beta = \frac{v}{c} = \frac{1}{cos(\theta)+sin(\theta)}

b) Plot \beta as a function of \theta and show where the superluminal motion takes place

Thanks!

Max.Planck
 

Attachments

  • Picture 1.jpg
    Picture 1.jpg
    12.3 KB · Views: 478
Astronomy news on Phys.org


Never mind, i solved it myself, if anybody wants to know the answer please pm me and I will post it here on the forum.

Max.Planck
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
Replies
46
Views
4K
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
1K
Replies
2
Views
2K