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Lab frame symmetric/asymmetric energies

  1. Jun 13, 2015 #1

    ChrisVer

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    I have one question which I need to verify as a thought.
    Suppose I have a particle collider for symmetric energies [itex]e^\pm[/itex], that give as a result the [itex]Y(4S)[/itex] resonance which later decays in [itex]B[/itex] mesons. Then the lab-frame is equivalent to the rest frame of the [itex]e^\pm[/itex] system and the [itex]Y(4S)[/itex] is at rest in the lab. In that case I was able to determine the momentum [itex] |p|= \sqrt{\frac{m_Y^2-4m_B^2}{4}}[/itex] and velocities [itex]u=\frac{p}{E}[/itex] of the [itex]B[/itex]-mesons and derive their travel length [itex]d_{lab}=\gamma(u) u c \tau[/itex]...
    If on the other hand the energies of the [itex]e^\pm[/itex] are not equal, say [itex]E_+ \ne E_-[/itex], then the [itex]Y(4S)[/itex] will not be at rest for the lab but have some velocity [itex]\beta[/itex] relative to it.
    If I want to derive the length the [itex]B[/itex] mesons travel before decaying, could I boost the result of the symmetric energies ([itex]\beta=0[/itex]) to the new lab frame ([itex]\beta \ne 0[/itex]) to get the [itex]B[/itex]-mesons "new" speed (boost the [itex]d_{lab}[/itex] by [itex]\beta[/itex])?
    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
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  3. Jun 13, 2015 #2

    mfb

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    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.
     
  4. Jun 13, 2015 #3

    ChrisVer

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    when the Y(4S) was at rest they had some velocity [itex]u[/itex]... that velocity would be [itex]u=\sqrt{u_x^2+u_y^2+u_z^2}[/itex], with [itex]u_i[/itex] any number that does not destroy the kinematics.
    Let's say that the beam strikes along the [itex]z[/itex]-axis. Then the velocity of the Y(4S) in our frame would be [itex]\beta_z[/itex] alone.
    So [itex]u[/itex] should be boosted only along the z-component?
     
  5. Jun 13, 2015 #4

    ChrisVer

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    Or do you mean that [itex]u\approx 0[/itex] (as a number just for completion is u~0.06) and their velocity would be [itex]\beta_Y[/itex] instead?
     
  6. Jun 14, 2015 #5

    mfb

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    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.
     
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