- #36
etotheipi
There's also a nice way to find the magnitude ##V## of the relative velocity between two particles of masses ##m_1## and ##m_2## and 4-momenta ##p_1^a## and ##p_2^a##. If we work in the rest frame of ##m_1##, then ##p_1^a = (m_1, \vec{0})## and ##p_2^a = (\frac{m_2}{\sqrt{1-V^2}}, \vec{p}_2')## the product of the two 4-momenta can be written\begin{align*}
(p_1)_a (p_2)^a &= \frac{m_1 m_2}{\sqrt{1-V^2}} \implies V = \sqrt{1- \frac{(m_1m_2)^2}{[(p_1)_a (p_2)^a]^2}}
\end{align*}Let's now switch to any arbitrary frame of reference, in which ##p_1^a = (E_1, \vec{p}_1)## and ##p_2^a = (E_2, \vec{p}_2)##. We can write ##(p_1)_a (p_2)^a## as\begin{align*}
(p_1)_a (p_2)^a &= E_1 E_2 - \vec{p}_1 \cdot \vec{p}_2 \\ \\
&= \frac{m_1 m_2}{\sqrt{1-v_1^2}\sqrt{1-v_2^2}} - \frac{m_1 m_2 \vec{v}_1 \cdot \vec{v}_2}{\sqrt{1-v_1^2}\sqrt{1-v_2^2}} \\ \\
&= \frac{m_1m_2 (1-\vec{v}_1 \cdot \vec{v}_2)}{\sqrt{(1-v_1^2)(1-v_2^2)}}
\end{align*}Hence, we can determine ##V## in terms of ##\vec{v}_1## and ##\vec{v}_2##,\begin{align*}
V = \sqrt{1- \frac{(1-v_1^2)(1-v_2^2)}{(1-\vec{v}_1 \cdot \vec{v}_2)^2}} = \frac{\sqrt{v_1^2 -2 \vec{v}_1 \cdot \vec{v}_2 + v_2^2 - \left( v_1^2 v_2^2 - [\vec{v}_1 \cdot \vec{v}_2]^2 \right) }}{1-\vec{v}_1 \cdot \vec{v}_2}
\end{align*}However, ##v_1^2 -2 \vec{v}_1 \cdot \vec{v}_2 + v_2^2 = (\vec{v}_1 - \vec{v}_2)^2##, and from vector analysis we also know that\begin{align*}
(\vec{v}_1 \times \vec{v}_2)^2 = (\vec{v}_1 \times \vec{v}_2) \cdot (\vec{v}_1 \times \vec{v}_2) &= (\vec{v}_1 \cdot \vec{v}_1)(\vec{v}_2 \cdot \vec{v}_2) - (\vec{v}_1 \cdot \vec{v}_2)(\vec{v}_2 \cdot \vec{v}_1) \\
&= v_1^2 v_2^2 - [\vec{v}_1 \cdot \vec{v}_2]^2
\end{align*}which results in\begin{align*}
V = \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}
\end{align*}which is, of course, symmetric in the two particles!
(p_1)_a (p_2)^a &= \frac{m_1 m_2}{\sqrt{1-V^2}} \implies V = \sqrt{1- \frac{(m_1m_2)^2}{[(p_1)_a (p_2)^a]^2}}
\end{align*}Let's now switch to any arbitrary frame of reference, in which ##p_1^a = (E_1, \vec{p}_1)## and ##p_2^a = (E_2, \vec{p}_2)##. We can write ##(p_1)_a (p_2)^a## as\begin{align*}
(p_1)_a (p_2)^a &= E_1 E_2 - \vec{p}_1 \cdot \vec{p}_2 \\ \\
&= \frac{m_1 m_2}{\sqrt{1-v_1^2}\sqrt{1-v_2^2}} - \frac{m_1 m_2 \vec{v}_1 \cdot \vec{v}_2}{\sqrt{1-v_1^2}\sqrt{1-v_2^2}} \\ \\
&= \frac{m_1m_2 (1-\vec{v}_1 \cdot \vec{v}_2)}{\sqrt{(1-v_1^2)(1-v_2^2)}}
\end{align*}Hence, we can determine ##V## in terms of ##\vec{v}_1## and ##\vec{v}_2##,\begin{align*}
V = \sqrt{1- \frac{(1-v_1^2)(1-v_2^2)}{(1-\vec{v}_1 \cdot \vec{v}_2)^2}} = \frac{\sqrt{v_1^2 -2 \vec{v}_1 \cdot \vec{v}_2 + v_2^2 - \left( v_1^2 v_2^2 - [\vec{v}_1 \cdot \vec{v}_2]^2 \right) }}{1-\vec{v}_1 \cdot \vec{v}_2}
\end{align*}However, ##v_1^2 -2 \vec{v}_1 \cdot \vec{v}_2 + v_2^2 = (\vec{v}_1 - \vec{v}_2)^2##, and from vector analysis we also know that\begin{align*}
(\vec{v}_1 \times \vec{v}_2)^2 = (\vec{v}_1 \times \vec{v}_2) \cdot (\vec{v}_1 \times \vec{v}_2) &= (\vec{v}_1 \cdot \vec{v}_1)(\vec{v}_2 \cdot \vec{v}_2) - (\vec{v}_1 \cdot \vec{v}_2)(\vec{v}_2 \cdot \vec{v}_1) \\
&= v_1^2 v_2^2 - [\vec{v}_1 \cdot \vec{v}_2]^2
\end{align*}which results in\begin{align*}
V = \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}
\end{align*}which is, of course, symmetric in the two particles!
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