Show that $$\gamma_+ - \gamma_-=\frac{2\beta_0\beta}{\sqrt{(1-\beta_0^2)(1-\beta^2)}}$$ where $$\gamma_+=(1-\beta^2_+)^{-\frac{1}{2}} \ \mbox{and} \ \beta_+=\frac{\beta_0+\beta}{1+\beta_0\beta}$$ $$\gamma_-=(1-\beta^2_-)^{-\frac{1}{2}} \ \mbox{and} \ \beta_-=\frac{\beta_0-\beta}{1-\beta_0\beta}$$
Asked
Active
Viewed 69 times
1
-
I don't know how to go about it. – hubot Jan 07 '16 at 10:12
-
do you know anything about the sign of $\beta$'s? – user26977 Jan 07 '16 at 10:15
-
No, I don't know. – hubot Jan 07 '16 at 10:24
-
Ok then, I will assum that $\beta\beta_0>0$. I think you can try moving the the $\gamma_-$ term to the other side of the equation and squaring both sides afterwards – user26977 Jan 07 '16 at 10:31
-
You just need to substitute the definition of $\beta_{\pm}$ into definition of $\gamma_{\pm}$ and uses the identities: $(1\pm\beta\beta_0)^2 - (\beta\pm\beta_0)^2 = (1-\beta^2)(1-\beta_0^2)$. – achille hui Jan 07 '16 at 10:42
-
What are you think about my answer? – hubot Jan 08 '16 at 16:00
-
Your answer is on the wrong track. you should show $\gamma_{\pm} = \frac{1\pm\beta_0\beta}{\sqrt{(1-\beta^2)(1-\beta_0^2)}}$ instead. – achille hui Jan 09 '16 at 08:11
1 Answers
0
I was moved $\gamma_-$ to other side equation: $$\gamma_+=\frac{2\beta_0\beta}{\sqrt{(1-\beta_0^2)(1-\beta^2)}}+\gamma_-$$ So $$\gamma_+=\frac{2\beta_0\beta}{\sqrt{(1-\beta_0^2)(1-\beta^2)}}+\frac{1}{\sqrt{1-\beta^2_-}}$$ Therefore $$\gamma_-=\frac{2\beta_0\beta\sqrt{1-\beta^2}+\sqrt{(1-\beta^2_0)(1-\beta^2)}}{\sqrt{(1-\beta_0^2)(1-\beta^2)}\sqrt{1-\beta^2_-}}$$ So we come to the conclusion that $$\gamma_-^2=\left(\frac{2\beta_0\beta\sqrt{1-\beta^2}+\sqrt{(1-\beta^2_0)(1-\beta^2)}}{\sqrt{(1-\beta_0^2)(1-\beta^2)}\sqrt{1-\beta^2_-}}\right)^2$$ $$\gamma_-^2=\frac{\left(2\beta_0\beta\sqrt{1-\beta^2}+\sqrt{(1-\beta^2_0)(1-\beta^2)}\right)^2}{\left(\sqrt{(1-\beta_0^2)(1-\beta^2)}\sqrt{1-\beta^2_-}\right)^2}$$ Well I think?
hubot
- 129