# Lab frame symmetric/asymmetric energies

1. Jun 13, 2015

### 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.

Last edited: Jun 13, 2015
2. Jun 13, 2015

### Staff: Mentor

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.

3. Jun 13, 2015

### ChrisVer

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?

4. Jun 13, 2015

### 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?

5. Jun 14, 2015

### Staff: Mentor

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.