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Nonrelativistic Lab-frame Velocity as a function of kinetic energy per unit mass

  1. Jul 19, 2005 #1
    Hi everyone--I'm a bit stuck trying to follow a calculation in an article (Nucl. Phys. B360 (1991) p. 145-179) regarding the lab-frame relative velocity of two colliding particles as a function of the kinetic energy per unit mass. (I'll include references to equations in the article, but the article is not required for this particular question.)

    Suppose we have two particles of mass [tex]m[/tex] colliding with one another with energies [tex]E_1, E_2[/tex] and 3-momenta [tex]\mathbf{p_1}, \mathbf{p_2}[/tex]. Define [tex]p_1 = |\mathbf{p_1}|, p_2 = |\mathbf{p_2}|[/tex]

    Define (eq. 3.3)
    [tex]s = 2m^2 + 2E_1E_2 - 2p_1p_2\cos\theta[/tex]

    Which, I believe is the same as the center of mass energy [tex](p_1^\mu+p_2^\mu)^2[/tex].

    Now define the kinetic energy per unit mass in the lab frame, (3.20)
    [tex] \epsilon = \frac{(E_{1,\mathrm{lab}}-m)+(E_{2,\mathrm{lab}}-m)}{2m}[/tex]

    First question: Why can we write

    Second question: Why is is true that the lab velocity is given by
    [tex]v_\mathrm{lab} = \frac{2\epsilon^{1/2}(1+\epsilon)^{1/2}}{1+2\epsilon}[/tex]

    Thanks very much for any assistance!
  2. jcsd
  3. Jul 22, 2005 #2

    Meir Achuz

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    Since s includes cos\theta in the general case, and \epsilon does not, the equation
    \epsilon=(s-4m^2)/4m^2 cannot be true in general. It only holds for the special case of one particle being initialy at rest. That is the meaning of the subscript _lab.
    The equation is easy to derive in this case.
    The equation for v_lab is also for one particle initially at rest.
    If you start with v=p/E, the equation is easily derived.
    The calculation is fully relativistic.
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