- #1

fliptomato

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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

[tex]\epsilon=\frac{s-4m^2}{4m^2}[/tex]

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!

Best,

Flip