Relativity Lite

100 | Appendix or v c c V c A x x c c x ≡ − = − + + −         ≅ − + 1 1 1 1 1 1 1 1 2 0 2 2 0 2 2 γ ( ) A x x c V c x ( ) − +       0 2 0 2 2 2 1 2 . (A22) Then we can rewrite (A12a) as At c V c x +       = − 0 2 2 1 γ , (A23a) or t c A V c x = − −     γ 2 0 1 . (A23b) We can obtain the corresponding relation between distance and γ by solving (A21) for x − x 0 , which we also called r in chapter 3. r x x c A V c x ≡ − = − +         ( ) 0 2 0 2 2 1 γ . (A24) INITIAL ACCELERATION TOWARDS ALPHA CENTAURI Let us use these to find how long it takes to boost a rocket to v = 0.6 c ( γ = 1.25 ) at an acceleration of 1 g . * If the initial velocity with V x0 = 0 , we have from (A23b) t = 0.726536 yr . Likewise (A24) gives the distance traveled when this speed is reached, r = 0.242179 c yr . Inverting (A9a), τ =     +         c A V c v c x -arcsinh arctanh 0 (A9c) gives τ = 0.671462 yr , as does inverting (A11a), τ = +     −         c A At V c V c x x arcsinh arcsinh 0 0  . (A11d) We can use this to confirm the caution after equation (A11) that for extended periods of acceleration, the ratio t/τ = 1.08 is indeed not γ = 1.25 even though we still have v c = − 1 1 2 γ if γ is defined as in equation (A22). * A 1 g acceleration can be written in the useful form A/c = 1/(0.968715 yr) .