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If I have two collinear beams with opposite velocities the cross section is defined by the formula:

$\frac{dN}{dtdV}=\sigma \rho_1 \rho_2 |\underline{v}_1-\underline{v}_2|$ where $\sigma$ is the cross section, $\underline{v}_1$, $\underline{v}_2$ the velocities of the beams and $\rho_1$, $\rho_2$ their densities. I read that if I have two non-collinear beams I must substitute the factor $|\underline{v}_1-\underline{v}_2|$ with $\sqrt{|\underline{v}_1-\underline{v}_2|^2-\frac{|\underline{v}_1 \wedge\underline{v}_2|^2}{c^2}}$. Is that true? How can I derive this?

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Sorry, now it's correct – Andy Bale Apr 26 '11 at 9:56
nobody knows how to help me? – Andy Bale Apr 28 '11 at 22:13

I try to give my answer in case anyone is interested: Weinberg in his book "Quantum theory of fields" Vol 1 in section 3.4 defines the flow factor $v_{\alpha}$ in the formula of the cross section so that it coincides with the velocity of particle 1 in the reference frame in which particle 2 is at rest, in such a way that the cross section remains Loretz invariant in different reference frame. In other word:

$v_{\alpha}= \frac{ ({(p_1 \cdot p_2)^2-(m_1m_2)^2})^{(1/2)} } {E_1E_2}$

if you try to simplify this expression using the equations $\underline{v}=\frac{\underline{p}}{E}$ you find the factor of my question above. To summarize this factor is a definition to make the cross section Lorentz invariant in any reference frame.

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