Editing Elastic collision (section)

===Relativistic derivation using hyperbolic functions===

Using the so-called ''parameter of velocity'' <math>s</math> (usually called the [[rapidity]]),

<math display="block">\frac{v}{c}=\tanh(s),</math>
we get
<math display="block">\sqrt{1-\frac{v^2}{c^2}}=\operatorname{sech}(s).</math>

Relativistic energy and momentum are expressed as follows:
<math display="block">\begin{align}
E &= \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}} = m c^2 \cosh(s) \\
p &= \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}=m c \sinh(s)
\end{align}</math>

Equations sum of energy and momentum colliding masses <math>m_1</math> and <math>m_2,</math> (velocities <math>v_1, v_2, u_1, u_2</math> correspond to the velocity parameters <math>s_1, s_2, s_3, s_4</math>), after dividing by adequate power <math>c</math> are as follows:
<math display="block">\begin{align}
m_1 \cosh(s_1)+m_2 \cosh(s_2) &= m_1 \cosh(s_3)+m_2 \cosh(s_4) \\
m_1 \sinh(s_1)+m_2 \sinh(s_2) &= m_1 \sinh(s_3)+m_2 \sinh(s_4)
\end{align}</math>

and dependent equation, the sum of above equations:
<math display="block">m_1 e^{s_1}+m_2 e^{s_2}=m_1 e^{s_3}+m_2 e^{s_4}</math>

subtract squares both sides equations "momentum" from "energy" and use the identity <math display="inline">\cosh^2(s)-\sinh^2(s)=1,</math> after simplifying we get:
<math display="block">2 m_1 m_2 (\cosh(s_1) \cosh(s_2)-\sinh(s_2) \sinh(s_1)) = 2 m_1 m_2 (\cosh(s_3) \cosh(s_4)-\sinh(s_4) \sinh(s_3))</math>

for non-zero mass, using the hyperbolic trigonometric identity <math display="inline">\cosh(a-b)=\cosh(a)\cosh(b)-\sinh(b)\sinh(a),</math> we get:
<math display="block">\cosh(s_1-s_2) = \cosh(s_3-s_4)</math>

as functions <math>\cosh(s)</math> is even we get two solutions:
<math display="block">\begin{align}
s_1-s_2 &= s_3-s_4 \\
s_1-s_2 &= -s_3+s_4
\end{align}</math>
from the last equation, leading to a non-trivial solution, we solve <math>s_2</math> and substitute into the dependent equation, we obtain <math>e^{s_1}</math> and then <math>e^{s_2},</math> we have:
<math display="block">\begin{align}
e^{s_1} &= e^{s_4}{\frac{m_1 e^{s_3}+m_2 e^{s_4}} {m_1 e^{s_4}+m_2 e^{s_3}}} \\
e^{s_2} &= e^{s_3}{\frac{m_1 e^{s_3}+m_2 e^{s_4}} {m_1 e^{s_4}+m_2 e^{s_3}}}
\end{align}</math>

It is a solution to the problem, but expressed by the parameters of velocity. Return substitution to get the solution for velocities is:
<math display="block">\begin{align}
v_1/c &= \tanh(s_1) = {\frac{e^{s_1}-e^{-s_1}} {e^{s_1}+e^{-s_1}}} \\
v_2/c &= \tanh(s_2) = {\frac{e^{s_2}-e^{-s_2}} {e^{s_2}+e^{-s_2}}}
\end{align}</math>

Substitute the previous solutions and replace:
<math>e^{s_3}=\sqrt{\frac{c+u_1} {c-u_1}}</math> and <math>e^{s_4}=\sqrt{\frac{c+u_2}{c-u_2}}, </math> after long transformation, with substituting:
<math display="inline"> Z=\sqrt{\left(1-u_1^2/c^2\right) \left(1-u_2^2/c^2\right)} </math>
we get:
<math display="block">\begin{align}
v_1 &= \frac{2 m_1 m_2 c^2 u_2 Z+2 m_2^2 c^2 u_2-(m_1^2+m_2^2) u_1 u_2^2+(m_1^2-m_2^2) c^2 u_1} {2 m_1 m_2 c^2 Z-2 m_2^2 u_1 u_2-(m_1^2-m_2^2) u_2^2+(m_1^2+m_2^2) c^2} \\
v_2 &= \frac{2 m_1 m_2 c^2 u_1 Z+2 m_1^2 c^2 u_1-(m_1^2+m_2^2) u_1^2 u_2+(m_2^2-m_1^2) c^2 u_2} {2 m_1 m_2 c^2 Z-2 m_1^2 u_1 u_2-(m_2^2-m_1^2) u_1^2+(m_1^2+m_2^2) c^2}\,.
\end{align}</math>