Fractions in Einstein Relativity Theory

Click For Summary
SUMMARY

The discussion focuses on the manipulation of the formula for time dilation in Einstein's relativity, specifically transforming the equation t = \frac{L}{v+c} + \frac{L}{v-c} into t = \frac{2L}{c}\left(\frac{1}{1 - \frac{v^2}{c^2}}\right). Participants explore how to equate \left(\frac{1}{1 - \frac{v^2}{c^2}}\right) with \left(\frac{1 + \frac{v^2}{c^2}}{1 - \left(\frac{v^2}{c^2}\right)^2}\right) through algebraic manipulation, particularly by factoring the difference of squares in the denominator. The conversation emphasizes the importance of understanding algebraic identities in the context of physics equations.

PREREQUISITES
  • Understanding of Einstein's theory of relativity
  • Familiarity with algebraic manipulation, specifically factoring
  • Knowledge of the concepts of time dilation and Lorentz transformations
  • Basic proficiency in working with fractions and rational expressions
NEXT STEPS
  • Study the derivation of Lorentz transformations in detail
  • Practice algebraic manipulation techniques, focusing on factoring expressions
  • Explore the implications of time dilation in various relativistic scenarios
  • Learn about the significance of the speed of light (c) in relativity
USEFUL FOR

Students of physics, particularly those studying relativity, educators teaching advanced algebra, and anyone interested in the mathematical foundations of Einstein's theories.

Norway
Messages
50
Reaction score
3

Homework Statement


From the formula t = \frac{L}{v+c} + \frac{L}{v-c} I've made t = \frac{2L}{c}\left(\frac{1}{1 - \frac{v^2}{c^2}}\right). This is the problem:
\left(\frac{1}{1 - \frac{v^2}{c^2}}\right) = \left(\frac{1 + \frac{v^2}{c^2}}{1 - \left(\frac{v^2}{c^2}\right)^2}\right)
How?

Homework Equations


That's kinda what I'm asking for. :-b


The Attempt at a Solution


I've gotten this far, but no more. I've tried to see how they relate, but I can't figure anything out.
 
Physics news on Phys.org
Norway said:
\left(\frac{1 + \frac{v^2}{c^2}}{1 - \left(\frac{v^2}{c^2}\right)^2}\right)

Factor the difference of squares in the denominator.
 
You remember what x^2 - y^2 equals to?
 
I get it. Thanks so much! Kinda embarrassing. :-b
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
1
Views
1K
  • · Replies 27 ·
Replies
27
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
25
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K