Forest UCM NLM GalileanTans

The C.M. frame is often chosen to theoretically calculate cross-sections even though experiments are conducted in the Lab frame. In such cases you will need to transform cross-sections between two frames.

The total cross-section should be frame independent

[math]\displaystyle{ \sigma_{C.M.} = \sigma_{Lab} }[/math]

or

[math]\displaystyle{ \sigma(\theta) d \Omega = \sigma(\psi) d \Omega^{\prime} }[/math]

where

[math]\displaystyle{ \theta }[/math] is in the CM frame and [math]\displaystyle{ \psi }[/math] is in the Lab frame.

A non-relativistic transformation

[math]\displaystyle{ \sigma(\theta) d \Omega = \sigma(\psi) d \Omega^{\prime} }[/math]

[math]\displaystyle{ \sigma(\theta) 2 \pi \sin(\theta) d \theta = \sigma(\psi) 2 \pi \sin (\psi) d \psi }[/math]

[math]\displaystyle{ \Rightarrow \sigma(\psi) = \frac{\sin(\theta)}{\sin(\psi)} \frac{d \theta}{d \psi} \sigma(\theta) }[/math]

The transformation is governed by the dependence of [math]\displaystyle{ \theta }[/math] on [math]\displaystyle{ \psi }[/math] [math]\displaystyle{ \left( \frac{d \theta}{d \psi} \right ) }[/math]

Lets return back to our picture of the scattering Process

if we superimpose the vectors [math]\displaystyle{ \vec{v}_1 }[/math] and [math]\displaystyle{ \vec{v}_1^{\prime} }[/math] we have

Trig identities (non-relativistic Gallilean transformation) tell us

[math]\displaystyle{ v_1 \sin(\psi) = v_1^{\prime} \sin(\theta) }[/math]

[math]\displaystyle{ v_1 cos(\psi) = v_{cm} + v_1^{\prime} \cos(\theta) }[/math]

solving for [math]\displaystyle{ \psi }[/math]

[math]\displaystyle{ \tan(\psi) = \frac{\sin(\psi)}{\cos(\psi)} = \frac{v_1^{\prime} \sin(\theta)/v_1}{\frac{v_{CM}}{v_1} + \frac{v_1^{\prime} \cos(\theta)}{v_1} } = \frac{\sin(\theta)}{\cos(\theta) + \frac{v_{CM}}{v_1^{\prime}}} }[/math]

For an elastic collision only the directions change in the CM Frame: [math]\displaystyle{ u_1^{\prime}= v_1^{\prime} }[/math] & [math]\displaystyle{ u_1^{\prime}= v_2^{\prime} }[/math]

From the definition of the C.M.

[math]\displaystyle{ \vec{v}_{CM} = \frac{m_1 \vec{u}_1 + m_2 \vec{u}_2}{m_1+m_2} = \frac{m_1}{m_1+m_2} \vec{u}_1 }[/math]

conservation of momentum in CM Frame [math]\displaystyle{ \Rightarrow }[/math]

[math]\displaystyle{ m_1 u_1^{\prime} = - m_2 u_2{\prime} }[/math]

[math]\displaystyle{ \Rightarrow v_1^{\prime} = u_1^{\prime} = \frac{-m_2}{m_1} u_2^{\prime} }[/math]

Gallilean Coordinate transformation

[math]\displaystyle{ \vec{u}_1 = \vec{u}_1^{\prime} + \vec{v}_{CM} = \vec{u}_1^{\prime} + \frac{m_1}{m_1+m_2} \vec{u}_1 }[/math]

[math]\displaystyle{ \Rightarrow u_1{\prime} = \left [ 1 - \frac{m_1}{m_1 + m_2} \right ] u_1 = \frac{m_2}{m_1+m_2}u_1 }[/math]

[math]\displaystyle{ \Rightarrow v_1^{\prime} = u_1^{\prime} =\frac{m_2}{m_1+m_2} u_1 }[/math]

another expression for [math]\displaystyle{ \psi }[/math]

using the above gallilean transformation we can do the following

[math]\displaystyle{ \frac{v_{CM}}{v_1^{\prime}}= \frac{\frac{m_1}{m_1+m_2} u_1}{\frac{m_2}{m_1+m_2} u_1} = \frac{m_1}{m_2} }[/math]

or

[math]\displaystyle{ \tan(\psi) = \frac{\sin(\theta)}{\cos(\theta) + \frac{m_1}{m_2}} }[/math]

after a little trig substitution

[math]\displaystyle{ \Rightarrow \frac{m_1}{m_2} = \frac{sin(\theta - \psi)}{\sin(\psi)} = }[/math] constant

now use the chain rule to find [math]\displaystyle{ \frac{d \theta}{d \psi} }[/math]

[math]\displaystyle{ f \equiv \frac{sin(\theta - \psi)}{\sin(\psi)} = }[/math] constant

[math]\displaystyle{ df = 0 = \frac{ \partial f}{\partial \psi} d \psi + \frac{ \partial f}{\partial \theta} d \theta }[/math]

[math]\displaystyle{ \Rightarrow \frac{d \theta}{d \psi} = \frac{-\frac{ \partial f}{\partial \psi} }{\frac{ \partial f}{\partial \theta} } }[/math]

[math]\displaystyle{ -\frac{ \partial f}{\partial \psi} = \frac{\cos(\theta - \psi)}{\sin(\psi)} + \frac{\sin(\theta - \psi)}{\sin(\psi)} }[/math]

[math]\displaystyle{ \frac{ \partial f}{\partial \theta }= 1 + \frac{\sin(\theta - \psi) \cos(\psi)}{\cos(\theta - \psi) \sin(\psi)} }[/math]

after substitution:

[math]\displaystyle{ \sigma(\psi) = \frac{\sin(\theta)}{\sin(\psi)} \frac{d \theta}{d \psi} \sigma(\theta) }[/math]

[math]\displaystyle{ =\frac{\sin(\theta)}{\sin(\psi)} \left [ 1 + \frac{\sin(\theta - \psi) \cos(\psi)}{\cos(\theta - \psi) \sin(\psi)} \right ] \sigma(\theta) }[/math]

For the above equation to be more useful one would prefer to recast it in terms of only [math]\displaystyle{ \psi }[/math] and masses.

[math]\displaystyle{ \sigma(\psi) = \frac{\left [ \frac{m_1}{m_2}\cos(\psi) + \sqrt{1-\left ( \frac{m_1 \sin(\psi) }{m_2} \right )^2 }\right ]}{\sqrt{1 - \left ( \frac{m_1 \sin(\psi)}{m_2}\right )^2 }}\sigma(\theta) }[/math]

Forest_UCM_NLM#Galilean_Transformations