The differential cross section is measured in certain frame should depend on the frame geometry. and the angle from one frame is different from other. and it is related by:

$\frac{d\sigma}{d\Omega} = \frac{d \sigma}{d\Omega '} \frac{d\Omega '}{d\Omega}$

where

$\frac{d\Omega '}{d \Omega} = \frac{sin(\theta ')}{sin(\theta )} \frac{d\theta '}{d\theta} \frac{d \phi'}{d \phi} = \frac{ d cos(\theta ' )}{d cos(\theta ) }$

Lets define the scattering situation in center of momentum frame with a dash and the before scattering is capital letter, after scattering is small letter.

$P_1' = (E_1' , P', 0 ) \rightarrow p_1' = ( e_1', p_1' cos(\theta '), p_1' sin (\theta ') )$

$P_2' = ( E_2', - P' , 0 )$

$P_1 = (E_1, P,0 ) \rightarrow p_1 = ( e_1, p_1 cos( \theta ) , p_1 sin(\theta) )$

$P_2 = ( m_2 , 0, 0 )$

From the particle 2, we can define the Lorentz transform from center of momentum frame to lab frame. Using hyperbolic expression of Lorentz transform, it is just a negative rotation in the hyperbolic space.

$m_2 = E_2' cosh( \chi ) + P' sinh( \chi )$

$0 = E_2' sinh( \chi ) - P' cosh( \chi )$

these give:

$tanh( \chi ) = \beta = P'/E_2'$

Thus, the relation of the particle 1 after the scattering is :

$p_1 cos( \theta ) = e_1' sinh (\chi ) + p_1' cosh( \chi) cos( \theta ')$

$p_1 sin( \theta ) = p_1' sin( \theta ')$

Thus, the angle in lab and CM frame is related by:

$tan( \theta ) = \frac{sin( \theta ' ) }{cosh(\chi ) (\tau + cos ( \theta '))}$

where

$\tau = \frac{e_1 '}{p_1'} \frac{ P'}{ E_2'}$

for elastic scattering, $P' = p_1'$ and $e_1' = E_1'$. and in low speed, $E_1' : E_2' = m_1 : m_2$ and $cosh( \chi ) = 1$, then $\tau$ is simply the mass ratio.  it reduced to classical form:

$tan (\theta ) = sin( \theta ' ) / ( \tau + cos( \theta ' ) )$

the cosine of the lab frame is :

$cos( \theta ) = \frac{cosh(\chi)(\tau + cos(\theta '))}{\sqrt{ cosh^2(\chi )(\tau+cos( \theta '))^2 + sin^2 (\theta ')}}$

$\frac{d cos(\theta) }{d cos(\theta ')} =\frac{ cosh(\chi)(1+ \tau cos(\theta ') }{ \sqrt{cosh^2(\chi )(\tau + cos( \theta '))^2 + sin^2 (\theta ')}^3}$

In classical limit:

$\frac{d cos(\theta) }{d cos(\theta ')} =\frac{ (1+ \tau cos(\theta ') }{ \sqrt{1+ \tau^2 + 2 \tau cos( \theta ')}^3}$

There fore, the different cross sections are related :

$\frac{d \sigma}{d\Omega}_{lab} = \frac{d\sigma}{d \Omega }_{CM}$$\frac{ \sqrt{cosh^2(\chi )(\tau + cos( \theta '))^2 + sin^2 (\theta ')}^3}{cosh(\chi)(1+ \tau cos(\theta ')) }$

notices that the $\tau$ and $cosh(\chi )$ is related, they are not independent. and there is a singular point.

This graph is m1 = 1, elastic scattering from P = 0 ( red) , 1 ( orange ), 2 ( green), 3 (blue), 4 ( purple) , 5 ( black). In P= 0 case, it is not mean there is no collision, it is just a classical curve from small momentum.