How to derive relative velocity in relativistic two-particle scattering?

Question

While calculating the scattering cross section for one particle scattering we use the formula $$J_i=\rho v$$ where $\rho$ is the density of the incoming beam and $v$ is the velocity of the incoming beam.

But for two-particle scattering we use instead of $v$, a relative velocity for the incoming flux: $$v_{rel}=\frac{\sqrt{(p_1.p_2)^2-m_1^2.m_2^2}}{E_1E_2}$$ where $p_i$ and $E_i$ denote, for $i=1,2$, the $3-$momenta and energies of the particle whose masses are $m_1$ and $m_2$. In a collinear frame, i.e., if $p_1$ and $p_2$ are along the same line, this reduces to$$v_{rel}=\Bigg|\frac{p_1}{E_1}-\frac{p_2}{E_2}\Bigg|$$ where this coincides for relative velocities in case for non-relativistic speeds. I have tried finding a derivation but have been unable to do so.

Answer 1

This is an old post, and you have probably figured this out, but I will provide an answer to make the post complete.

Consider four-momentum of two objects, object 1 in frame 1, and object 2 in frame 2, with the relative velocity $v_{\rm rel}$ between them. Recall from special relativity in natural untis (c=1) that $$ p_{1}^\mu = (E_1, \vec{p}_1) = (\gamma_1 m_1, \gamma_1m_1\vec{v}_{1}), \quad \gamma_1 = \frac{1}{\sqrt{1 - v_{1}^2}} $$ where $v_{1}$ is the relative velocity between frame 1 and some other inertial reference frame (e.g. when boosting from one frame to another). Consider the inner product between the four-vectors $p_1$ and $p_2$ in the rest-frame of $p_2$, i.e. $p_2=(E_2, \vec{0})$, when boosting between the two frames:
$$ p_1\cdot p_2 = \frac{m_1m_2}{\sqrt{1-v_{\rm rel}^2}}. $$ Here, $v_{\rm rel}$ is the relative velocity between frame 1 and frame 2. Solving for $v_{\rm rel}$ gives $$ v_{\rm rel} = \frac{\sqrt{(p_1\cdot p_2)^2 - m_1^2m_2^2}}{p_1\cdot p_2} $$ which is manifestly Lorentz-invariant, as the product $p_1\cdot p_2$ and masses $m_1,m_2$ are Lorentz-invariant. Writing the poduct $p_1\cdot p_2$ in an arbitrary frame, we have $$ p_1\cdot p_2 = E_1E_2 - \vec{p}_1\cdot \vec{p}_2 = \gamma_1\gamma_2m_1m_2 - \gamma_1\gamma_2m_1m_2\vec{v}_1\cdot \vec{v}_2 = \gamma_1\gamma_2m_1m_2(1 - \vec{v}_1\cdot \vec{v}_2) $$ where we substituted $E_1=\gamma_1m_1$, $\vec{p}_1=\gamma_1m_1\vec{v}_1$ and similary for $E_2$, $\vec{p}_2$. We would like to substitute this into the expression for $v_{\rm rel}$ above, but first we recall some vector-identities: $$ \begin{align} (\vec{a}\times \vec{b})^2 &= (\vec{a}\times\vec{b})_i(\vec{a}\times\vec{b})_i \\ &= \epsilon_{ijk}\epsilon_{ilm}a_jb_ka_lb_m \\ &= (\delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl})a_jb_ka_lb_m \\ &= a_ja_jb_kb_k - a_jb_ja_kb_k \\ &= a^2b^2 - (\vec{a}\cdot\vec{b})^2 \end{align} $$ so we can write $$ \begin{align} (p_1\cdot p_2)^2 &= (m_1m_2\gamma_1\gamma_2)^2(1 - 2(\vec{v}_1\cdot \vec{v}_2) + (\vec{v}_1\cdot\vec{v}_2)^2) \\ &= (m_1m_2\gamma_1\gamma_2)^2(1 - 2(\vec{v}_1\cdot \vec{v}_2) + v_1^2v_2^2 - (\vec{v}_1\times\vec{v}_2)^2) \\ &= (m_1m_2\gamma_1\gamma_2)^2((\vec{v}_1 - \vec{v}_2)^2 + 1 - v_1^2 - v_2^2 + v_1^2v_2^2 - (\vec{v}_1\times\vec{v}_2)^2) \\ &=(m_1m_2\gamma_1\gamma_2)^2((\vec{v}_1 - \vec{v}_2)^2 + (1-v_1^2)(1-v_2^2) - (\vec{v}_1\times\vec{v}_2)^2) \\ &= (m_1m_2\gamma_1\gamma_2)^2\left((\vec{v}_1 - \vec{v}_2)^2 + \frac{1}{\gamma_1^2\gamma_2^2}- (\vec{v}_1\times\vec{v}_2)^2\right) \end{align} $$ Then we get $$ (p_1\cdot p_2)^2 - m_1^2m_2^2 = (m_1m_2\gamma_1\gamma_2)^2\left((\vec{v}_1 - \vec{v}_2)^2 - (\vec{v}_1\times\vec{v}_2)^2\right) $$ Inserting into $v_{\rm rel}$ gives $$ v_{\rm rel} = \frac{\sqrt{(p_1\cdot p_2)^2 - m_1^2m_2^2}}{p_1\cdot p_2} = \frac{\sqrt{(\vec{v}_1-\vec{v}_2)^2 - (\vec{v}_1\times\vec{v}_2)^2}}{1-\vec{v}_1\cdot\vec{v}_2} $$

Now we derived two common expressions for the relative velocity in natural units. For colinear motion, i.e. $\vec{v}_1\| \vec{v}_2$, $v_{\rm rel}$ trivially reduces to $$ v_{\rm rel} = \frac{|v_1-v_2|}{1-v_1v_2} = \frac{E_1E_2|v_1-v_2|}{p_1\cdot p_2} = \frac{|p_1E_2-p_2E_1|}{p_1\cdot p_2} $$ For the scattering cross-section, we recognize the numerator of $v_{\rm rel}$ from the flux in the cross-section, but we still have the denominator. It is not the relative velocity that is used for the cross-section, it is the Møller velocity, which is defined as the numerator of the relative velocity.

Source: Landau, L.D.; Lifshitz, E.M. (2002) [1939]. The Classical Theory of Fields. Course of Theoretical Physics. Vol. 2 (4th ed.).

Answer 2

As a supplement to the answer by @QuantonPhysics , here's a derivation that uses rapidity, which already assumes that we are working in a spacetime plane coplanar with the two 4-momenta, $\tilde p_1$ and $\tilde p_2$.

\begin{align*} \tilde p_1 \cdot \tilde p_2 &= m_1 m_2 \cosh\theta_{rel} =m_1 m_2\frac{1}{\sqrt{1-\tanh^2\theta_{rel}}} =m_1 m_2\frac{1}{\sqrt{1-v_{rel}^2}} \end{align*} So, \begin{align*} (\tilde p_1 \cdot \tilde p_2)^2(1-\tanh^2\theta_{rel}) &= m_1^2 m_2^2\\ (\tilde p_1 \cdot \tilde p_2)^2-(\tilde p_1 \cdot \tilde p_2)^2\tanh^2\theta_{rel} &= m_1^2 m_2^2\\ -(\tilde p_1 \cdot \tilde p_2)^2\tanh^2\theta_{rel} &= -(\tilde p_1 \cdot \tilde p_2)^2+m_1^2 m_2^2\\ \tanh^2\theta_{rel} &= \frac{(\tilde p_1 \cdot \tilde p_2)^2-m_1^2 m_2^2}{(\tilde p_1 \cdot \tilde p_2)^2}\\ v_{rel}=\tanh\theta_{rel} &= \frac{\sqrt{(\tilde p_1 \cdot \tilde p_2)^2-m_1^2 m_2^2}}{(\tilde p_1 \cdot \tilde p_2)}\\ &= \frac{\sqrt{( m_1 m_2\cosh\theta_{rel})^2-m_1^2 m_2^2}}{( m_1 m_2\cosh\theta_{rel})}\\ &= \frac{\sqrt{( m_1 m_2\sinh\theta_{rel})^2}}{( m_1 m_2\cosh\theta_{rel})}\\ &= \frac{( m_1 m_2\sinh\theta_{rel})}{( m_1 m_2\cosh\theta_{rel})}\\ \end{align*} as possibly expected.

If we begin with the last line, \begin{align*} v_{rel}=\tanh\theta_{rel} &= \frac{( m_1 m_2\sinh\theta_{rel})}{( m_1 m_2\cosh\theta_{rel})}\\ &= \frac{( m_1 m_2\sinh(\theta_2-\theta_1))}{( m_1 m_2\cosh\theta_{rel})}\\ &= \frac{m_1 m_2(\sinh\theta_2\cosh\theta_1-\sinh\theta_1\cosh\theta_2)}{( m_1 m_2\cosh\theta_{rel})}\\ &= \frac{p_2 E_1 - p_1 E_2}{( m_1 m_2\cosh\theta_{rel})}\\ v_{rel} &= \frac{p_2 E_1 - p_1 E_2}{ \tilde p_1\cdot \tilde p_2}\\ \end{align*}