That is correct, but irrelevant.Lorentz operator does not have the property: f[sub](a+b)[/sub]=f[sub](a)[/sub]+f[sub](b)[/sub]

Specifically, the Lorentz transform operates as a linear transform on vectors, thus it is an analog of multiplication, and we have in one dimension (for velocities parallel or anti-parallel): $$\Lambda( a \hat{x} )\Lambda( b \hat{x} ) = \Lambda( \frac{a + b}{1 + \frac{a b}{c^2}} \hat{x} )$$.

And for arbitrary directions:

$$\Lambda( \vec{a} )\Lambda( \vec{b} ) = { \huge \Lambda } \left( \frac{ \left( 1 + \frac{\vec{a} \cdot \vec{b}}{c^2 + c \sqrt{c^2 - \vec{a}^2}} \right) \vec{a} + \sqrt{ 1 - \frac{\vec{a}^2}{c^2}} \vec{b}}{1 + \frac{ \vec{a} \cdot \vec{b} }{c^2}} \right) {\huge R } \left( \frac{ \left( \left( 1 + \frac{\vec{a} \cdot \vec{b}}{c^2 + c \sqrt{c^2 - \vec{a}^2}} \right) \vec{a} + \sqrt{ 1 - \frac{\vec{a}^2}{c^2}} \vec{b} \right) \times \left( \left( 1 + \frac{\vec{a} \cdot \vec{b}}{c^2 + c \sqrt{c^2 - \vec{b}^2}} \right) \vec{b} + \sqrt{ 1 - \frac{\vec{b}^2}{c^2}} \vec{a} \right) }{ \left(\vec{a} + \vec{b}\right)^2 - \left(\frac{ \vec{a} \times \vec{b} }{c}\right)^2 } \right) $$

which is just one of the ways the Lorentz transformation is famously related to rotation.

Need to double check the above for sign errors.

Since only the square of the velocity enters into proper time calculations in an inertial frame, the direction is irrelevant.Tangential speed is not linear speed.(need to figure out the velocity)

You have assumed that either a or b may be taken as motionless in some inertial frame, and that contradicts your assumption that both a and b describe curved trajectories in some inertial Cartesian coordinate system. Thus neither of your cases applies.You have two cases.

-a is stationary it has the proper time and Lorentz operator is applied to time b

-b is stationary it has the proper time and Lorentz operator is applied to time a.

You need the speed between a and b, which is a function of time, V(a,b)[sub](t)[/sub].

But, because the speed is the same from a-b and b-a, you can use the general form V(a,b)[sub](t)[/sub]=V(b,a)[sub](t)[/sub] without the exact calculation

General relativity does not save you since due to the flatness of space time assumed when you specified that you were going to describe a special relativity "paradox" means that in no coordinate system may the curved trajectories you depict can fairly represent geodesics of space time

Again you run afoul of formal debate rules. Because you had only a vague idea as to what physical situation you wish to describe as a "paradox" we still don't have a topic of debate many iterations in. This is not the thread to debate in -- this is the thread in which to propose aNow I realize that not enough one post of mine in debate.

I'll need to give a reply. There are too many mistakes.

*formal*debate where I would be motivated to get my signs right in the above rotation matrix.