Lab frame symmetric/asymmetric energies

[ChrisVer]

I have one question which I need to verify as a thought.
Suppose I have a particle collider for symmetric energies $e^\pm$, that give as a result the $Y(4S)$ resonance which later decays in $B$ mesons. Then the lab-frame is equivalent to the rest frame of the $e^\pm$ system and the $Y(4S)$ is at rest in the lab. In that case I was able to determine the momentum $|p|= \sqrt{\frac{m_Y^2-4m_B^2}{4}}$ and velocities $u=\frac{p}{E}$ of the $B$-mesons and derive their travel length $d_{lab}=\gamma(u) u c \tau$...
If on the other hand the energies of the $e^\pm$ are not equal, say $E_+ \ne E_-$, then the $Y(4S)$ will not be at rest for the lab but have some velocity $\beta$ relative to it.
If I want to derive the length the $B$ mesons travel before decaying, could I boost the result of the symmetric energies ($\beta=0$) to the new lab frame ($\beta \ne 0$) to get the $B$-mesons "new" speed (boost the $d_{lab}$ by $\beta$)?
I am not sure about the directions however...since the B meson result can have any kind of velocity orentation at the first case -with only constraint to be in P-wave - (Y(4S) rest frame= lab frame) , while at the second (Y(4S) boosted relative to the lab) the Y(4S) speed is boosted along the beam's direction alone.

 

[mfb]

Sure.
The B mesons are nearly at rest in the Y(4S) frame, so they will move with the same velocity (this includes the direction!) in the lab frame.

 

[ChrisVer]

Sure.

The B mesons are nearly at rest in the Y(4S) frame, so they will move with the same velocity (this includes the direction!) in the lab frame.

when the Y(4S) was at rest they had some velocity $u$... that velocity would be $u=\sqrt{u_x^2+u_y^2+u_z^2}$, with $u_i$ any number that does not destroy the kinematics.
Let's say that the beam strikes along the $z$-axis. Then the velocity of the Y(4S) in our frame would be $\beta_z$ alone.
So $u$ should be boosted only along the z-component?

 

[ChrisVer]

Or do you mean that $u\approx 0$ (as a number just for completion is u~0.06) and their velocity would be $\beta_Y$ instead?

 

[mfb]

To a good approximation, yes. $\beta_Y \approx 0.4$ for Belle, a bit more for BaBar.

For fully reconstructed decays (of at least one B meson), you can measure the boost of the individual B mesons.