Next, the inverse σ (u) = (τ, x) must be calculated. Use the same considerations as before, but now with U = ( 0 , u ) P = ( τ τ ( u ) , x ( u ) ) . , one gets τ τ = R ( 1 − − λ λ ) λ λ , x = u λ λ , but now with λ depending on u. The condition for P lying in the hyperboloid is − − τ τ 2 + | x | 2 = − − R 2 , or − − R 2 ( 1 − − λ λ ) 2 λ λ 2 + | u | 2 λ λ 2 = − − R 2 , leading to λ λ = R 2 − − | u | 2 2 R 2 .