View Single Post

 Quote by kev You are right, that I am wrong. You see, I can accept when I have made a mistake :P I would be interested to see how you got to: $$\gamma(v')=\gamma(v) \gamma(u) (1+v_xu/c^2)$$ from: $$v'^2=v'_x^2+v'_y^2$$ $$v'_x=\frac{v_x+u}{1+v_xu/c^2}$$ $$v'_y=\frac{v_y \sqrt{1-u^2/c^2}}{1+v_xu/c^2}$$
Start with $$1-(\frac{v'}{c})^2$$.