Velocity measurement in relativistic perspectives I understand the precepts of relativistic time dilation, but I'm looking to nail down the local perception of velocity in each frame. My question is threefold:


*

*Given two observers $A$ (rest frame) and $B$, if $A$ observes $B$ traveling
away from starting point $A$ at $V=2c/5$, does $B$ perceive a different
value for $V$ based on dilation, or the same, based on it's local
frame?

*If it is the same (which is my assumption), how does this integrate
with the predicted experience of $B$ from the point of view of $A$
(which I also assume would seem to act as a multiplier on perceived
velocity due to dilation)?

*As a further silly notion, does it change if at some point $B$ stops
and proceeds to return to $A$ at the same previous velocity? My assumption is it does not.

 A: *

*No, $B$ will perceive the same value of $v$ for $A$. According to special relativity, it is impossible for $A$ and $B$ to determine which one of them is moving from their reference frames.

*Both will have the same experience from their perspectives. $A$ will see $B$ contracting along the direction of motion and its clock moving slower, and vice versa.

*If $B$ starts going in the opposite direction with velocity $v$, nothing will change except the perceived direction of motion from both of their reference frames and hence the direction of contraction.
I am not sure if this is all what you were asking. Let me know if I misinterpreted something. 
A: The Special Theory of Relativity gives us no direct transform equation for velocity for strictly two-body situations. There is only the velocity addition formula $s = \frac{u + v}{1 + (vu/c^2)}$, which can be "simplified" by assuming that $s$ equals zero. 
However, the sub-section Principle of relativity in the entry on derivation of Lorentz transforms in Wikipedia states that there is no privileged frame of reference, and explicitly says that you should assume that the apparent velocities perceived by persons A and B have the same value, yet opposing directions. Therefore:$$v = - v'$$
 (which is obviously quite puzzling though, since the variables x and t change under the transforms in inverse proportion - hence time dilatation and length contraction - and therefore it seems the values for v and v' should differ).
And yes, you are right, reversing the direction of the movement changes only the directions of the perceived velocity vectors.
A: The value B sees for V depends on which combination of frames they use for their reference.  If they look entirely in their own frame and are moving at constant velocity, they can't tell what velocity they're moving at.  If they observe A, they will see A moving with a velocity V.  
If however, they look at preset distance markers in A's frame and use the B frame measurement of time, they will see something called the proper velocity.  Brehme referred to this as speedometer velocity[1].  It's the velocity that corresponds to how far B will have travelled in A's frame over the amount of time B experienced.  The expression for proper velocity is $v_\tau = \gamma v = sinh(\phi)$, where $\gamma = \sqrt{1-V^2/c^2}$, and $\phi$ is the rapidity of B's frame which is related to V by $V/c = tanh(phi)$.
This brings us to the third question.  What happens if B stops and returns to A.  Let's say that B slowed to a stop and started back up again in the opposite direction with a constant acceleration.  The change in proper velocity of B while accelerating is $\Delta v_\tau = \gamma v = c sinh(\frac{a\tau}{c})$, where $\tau$ is known as the proper time as measured in B's frame.
References


*

*Flynn, R.W., Spacecraft navigation and relativity, AJP, 53, (1985), 113
http://dx.doi.org/10.1119/1.14092 

*Rindler, W., "Hyperbolic Motion in Curved Space Time", Phys. Rev. 119 2082{2089 (1960).

*Rindler, W, "Special Relativity", Interscience Publishers, 43 (1944)
