T_z=\int _{0}^{ \pi {\lambda \over v}} }\! \sqrt{({dt_z \over dq})^2-{({dx_z \over dq})^2 \over c^2}} {dq}
=
\int _{0}^{ \pi {\lambda \over v}} }\! \sqrt{(1-{(v \cos({v \over \lambda} q))^2 \over c^2}} {dq} < T_p