Two space ships fly at relativistic velocities v and u in perpendicular directions?

Question

The student used the textbook formula, and derived their relative velocity w.

After that, the student used the calculator, and found that
cosh(artanh(v/c)) = 7, and
cosh(artanh(u/c)) = 11.

What is the value of
cosh(artanh(w/c))?

Answer 1

Let the first ship moves with velocity v along the x-axis, and the second ship moves with velocity u along the y-axis. All velocities are normalized on the speed of light. In the coordinate system related with the first ship, the second ship has velocity w with components w_x = -v and w_y=u √(1-v^2). The square of the relative velocity w^2 is equal to (w_x)^2+(w_y)^2 = v^2+u^2-v^2 u^2 or

1-w^2 = (1-v^2) (1-u^2).

If we introduce "angles" tanh (ψ_w) = w etc, then cosh (ψ_w)=1/√(1-w^2) etc. Applying the power -1/2 to both sides of the above equation, one has

cosh (ψ_w) = cosh (ψ_v) cosh (ψ_u) = 77.

Answer 2

I don't see how the finite speed of light makes the 5th postulate invalid. But anyway ... if u = 0, then the relative speeds of the ships is just the speed of the 2nd ship. That is, w = v. Therefore, w is NOT greater than v. And if the speeds are slow, the relative speed is w = sqrt ( u^2 + v^2 ) Since the speeds are fast (relativistic) in your problem, it is necessary to specify the observer for the speeds. Is it somebody in ship #1? ship#2? Somebody at the intersection of the perpendicular directions?