In this page we will discuss some basic concepts of the elastic collision of two particles. These calculations can be useful in a Monte Carlo context in order to evaluate the energy transfer and the particles trajectories.
Index:
Introduction
Let’s consider a system of two particles: the projectile and the target. we will consider a steady target and a projectile impinging on it. First some notation:
Capital letters for the target as: M (mass), V (velocity)
Lower case letters for the projectile: m (mass), v (velocity)
We will make use of two reference systems: Laboratory frame (subscript L), Center of mass frame (subscript c)
Finally to distinguish from before and after the collision ww will use the symbol ” ‘ “, for example projectile velocity after the collision in Lab frame is: \vec v'_L .
center of mass velocity
For a set of particle we can define the center of mass position as:
\vec r_{cm} = \frac{\sum_i m_i \vec r_i}{\sum_i m_i}Consequently the center of mass velocity (\vec v_{cm} ) is:
\vec v_{cm} = \frac{d\vec r_{cm}}{dt}Through momentum conservation is possible to obtain \vec v_{cm} . Indeed, calling p the momentum we have:
\vec p_{tot} = \sum_i \vec p_i = \sum_i m_i \frac{d\vec r_i}{dt} \frac{\sum_i m_i}{\sum_i m_i} = \sum_i m_i \vec v_{cm}Then in the simple case of two particles we have:
(m + M)\vec v_{cm} = m \vec v_L + M\vec V_LSince we assumed a stationary target V_L = 0 then, defining B = M/m results:
\vec v_{cm} = \frac{1}{1+B}\vec v_LIn elastic collisions total kinetic energy is conserved, this mean that center of mass velocity remain constant.
velocities in Com frame
In center of mass reference system, target and projectile velocities are easly related through the momentum balance ( \vec p_{cm} = 0 ):
m\vec v_c + M \vec V_c = 0
From these comes:
\vec V_c = -\frac{m}{M} \vec v_c
Imposing internal kinetic energy conservation we see that velocities in COM frame remains constants.
energy exchange (target point of view)
Kinetic energy of the target depends from the square modulus of the velocity, so we start from it:
V^2_L = \frac{2T'_L}{M} = \vec V'_L \bullet \vec V'_L = (\vec v_{cm} + \vec V'_c)\bullet (\vec v_{cm} + \vec V'_c) = v^2_{cm} + V'^2_c + 2v_{cm} V'_c cos(\pi - \theta_c)
Where the argument in the cosine is the angle between \vec v_{cm} and \vec V'_c . The figure below can be useful to understand the velocities directions.
Since target is at rest V_c = v_{cm} = V'_c, moreover, center of mass velocity can be written in therms of the projectile speed in Lab system before the collision, by doing so we can make appear the projectile energy before the collision (E). Finally we get the target recoil energy:
T'_L = \left [\frac{4EmM}{(m+M)^2}\right] \frac{1-cos\theta_c}{2}
The term in brakets is the maximum amount of energy which can be transferred to the target, and it has a maximum for B = 1 meaning m = M.
scattering angles
From the figure B we can easly derive the scattering angle of the target, since V'_c and v_{cm} have same length, \gamma must be: \gamma = \frac{\pi - \theta_c}{2}
Again from figure B, using some trigonometric considerations with the formulas obtained before, we can derive the following relation for the scattering angle of the projectile in Lab frame:
cos\theta_L = \frac{1+Bcos\theta_c}{\sqrt{B^2 +2Bcos\theta_c +1}}