Solve equation on mathematical physics Show that
$$\gamma_+ - \gamma_-=\frac{2\beta_0\beta}{\sqrt{(1-\beta_0^2)(1-\beta^2)}}$$
where
$$\gamma_+=(1-\beta^2_+)^{-\frac{1}{2}} \ \mbox{and} \ \beta_+=\frac{\beta_0+\beta}{1+\beta_0\beta}$$
$$\gamma_-=(1-\beta^2_-)^{-\frac{1}{2}} \ \mbox{and} \ \beta_-=\frac{\beta_0-\beta}{1-\beta_0\beta}$$

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?
