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Lorentz transformation lab to CM system

  1. Feb 17, 2008 #1
    Which Lorentz transformation takes the lab system to the CM system?

    Lab system: [tex] p_a = (E^{lab}_a, \vec{p}_a) [/tex] and [tex] p_b = (m_b, \vec{0}) [/tex]
    CM system: [tex] p_a = (E^{CM}_a, \vec{p}) [/tex] and [tex] p_b = (E^{CM}_b, -\vec{p}) [/tex]

    For a binary reaction a+b->c+d, the textbooks I have say quite a lot about the kinematics of such reactions (Mandelstam variables and all that) but how do I go about doing a Lorentz transformation between the lab and CM system? That must be possible to do.
     
  2. jcsd
  3. Feb 18, 2008 #2
    Anyone?
     
  4. Feb 18, 2008 #3
    The velocity of CM in the scattering of two particles with rest mass [itex]m_1[/itex] and [itex]m_2[/itex] is given by

    [tex]\vec{u}_{CM}=\frac{\vec{p}_1+\vec{p}_2}{E_1+E_2}\,c^2[/tex]

    from which you can find the Lorentz transformation.
     
  5. Feb 19, 2008 #4
    So a lorentz transformation from the lab frame to the CM frame would be
    [tex]
    \left[ \begin{array}{cccc} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right]
    [/tex]
    with [tex] \beta = ||\vec{u_{CM}}|| [/tex] and [tex] \gamma = (1-\beta^2)^{-1/2} [/tex]?
    It doesn't make any sense to me.
     
  6. Feb 19, 2008 #5
    The crusial thing is to make sense in nature, not to any of us.
    First system of center of mass for two particles is defined by the requirement, that [itex](\vec{p}_1+\vec{p}_2)_{CM}=0[/itex]. The general Lorentz transormation [itex](c=1)[/itex] is

    [tex]\Lambda=\begin{pmatrix} \gamma& \gamma\,\vec{\beta} \cr \gamma\,\vec{\beta} & \delta_{\alpha\beta}+\frac{\gamma-1}{\beta^2}\,\beta_\alpha\,\beta_\beta \end{pmatrix}[/tex]

    Take now the 4-vector [itex]p^\alpha_1=(E_1,\vec{p}_1)[/itex] and apply [itex]\Lambda[/itex] with [itex]\beta=\frac{\vec{p}_1+\vec{p}_2}{E_1+E_2}[/itex], to get for the 3-momentum [itex]\vec{P}_1,\,\vec{P}_2[/itex] to the CM system:

    [tex]\vec{P}_1=\Lambda\,\vec{p}_1\Rightarrow \vec{P}_1=\gamma\,\left(E_1+\frac{\gamma}{\gamma+1}\,\vec{\beta}\,\vec{p}_1\right)\,\vec{\beta}+\vec{p}_1[/itex]
    [tex]\vec{P}_2=\Lambda\,\vec{p}_2\Rightarrow \vec{P}_2=\gamma\,\left(E_2+\frac{\gamma}{\gamma+1}\,\vec{\beta}\,\vec{p}_2\right)\,\vec{\beta}+\vec{p}_2[/itex]

    Adding now the two equations we get

    [tex]\vec{P}_1+\vec{P}_2=\gamma\,\left(E_1+E_2+\frac{\gamma}{\gamma+1}\,\vec{\beta}\,(\vec{p}_1+\vec{p}_2)\right)\,\vec{\beta}+\left(\vec{p}_1+\vec{p}_2\right)\Rightarrow \vec{P}_1+\vec{P}_2=\gamma\,\left(E_1+E_2+\frac{\gamma}{\gamma+1}\,\vec{\beta}\,(E_1+E_2)\,\vec{\beta}\right)\,\vec{\beta}+\left(E_1+E_2\right)\,\vec{\beta}\Rightarrow[/tex]
    [tex]\vec{P}_1+\vec{P}_2=\gamma\,(E_1+E_2)\,\left(1+\frac{\gamma}{\gamma+1}\,\beta^2\right)\,\vec{\beta}+\left(E_1+E_2\right)\,\vec{\beta}\Rightarrow \vec{P}_1+\vec{P}_2=0[/tex]

    Now does it makes sense?
     
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