Prove that relative velocity, momentum, and energy, are related by $E_1E_2v_{rel}=\sqrt{(p_1p_2)^2-m_1^2m_2^2}$

Question

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?

Answer 1

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).