Fractions in Einstein Relativity Theory

AI Thread Summary
The discussion centers on simplifying the equation t = (L/(v+c)) + (L/(v-c)) into the form t = (2L/c)(1/(1 - (v^2/c^2))). The main problem is understanding how to transform (1/(1 - (v^2/c^2))) into the equivalent form (1 + (v^2/c^2))/(1 - (v^2/c^2)^2). Participants suggest factoring the denominator as a difference of squares to facilitate the transformation. The conversation highlights the importance of recognizing algebraic identities in relativity calculations. Overall, the thread emphasizes the need for clarity in manipulating fractions within the context of Einstein's theory.
Norway
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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.
 
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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
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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