Prove the relation between relative velocity ,momentum and energy In Chapter 8 of F.Mandl's book Quantum field theory, during the derivation of the differential cross section, the following relation is used:
$$E_1E_2v_{rel}=\sqrt{(p_1p_2)^2-m_1^2m_2^2} \,\, ,$$
where $v_{rel}$ is the relative velocity between two colliding particles, $m_1$ and $m_2$ are the rest masses.
How is this equation derived?
 A: Consider the following results:


*

*From the definition of scalar product of four vectors,
$$ \tag{1}(p_1 p_2)^2 \equiv (p_{1\mu}p_2^\mu )^2 = (E_1E_2 - \textbf{p}_1 \cdot \textbf{p}_2 )^2.$$

*The usual dispersion relations:
$$ \tag{2} E_i = \sqrt{ | \textbf{p}_i |^2 + m_i^2}.$$

*The velocity $\textbf{v}_i$ in terms of momentum and energy:
$$ \tag{3} \textbf{v}_i = \textbf{p}_i / E_i .$$

*From the collinearity of the particles:
$$ \tag{4} \textbf{p}_1 \cdot \textbf{p}_2 = |\textbf{p}_1| \, | \textbf{p}_2|. $$

*The relative velocity $v_{rel} \equiv |\textbf{v}_1-\textbf{v}_2|$ can be written as:
$$ \tag{5} v_{rel} E_1 E_2 = \left| \textbf{p}_1 E_2 - \textbf{p}_2 E_1 \right|.$$
To see this, we use (3):
$$ v_{rel} \equiv |\textbf{v}_1-\textbf{v}_2|
= \left| \frac{\textbf{p}_1}{E_1} - \frac{\textbf{p}_2}{E_2} \right|
= \frac{1}{E_1 E_2} \left| \textbf{p}_1 E_2 - \textbf{p}_2 E_1 \right|
$$


Now, starting from (1), expanding the energy squared with (2) and using (4), we have:
$$ \tag{6} (p_1 p_2)^2 - m_1^2 m_2^2 = 2 (\textbf{p}_1 \cdot \textbf{p}_2)^2 + m_1^2 | \textbf{p}_2 |^2 + m_2^2 | \textbf{p}_1|^2 - 2 E_1 E_2 \textbf{p}_1 \cdot \textbf{p}_2.$$
Now you can just square (5) again using (2) and (4) and check that what you get is equal to the RHS of (6).
