irfpy.util.collision

Handles collision.

Note

YF has old module handling ratherford scattering. I haven’t ported it, but can be done on request.

irfpy.util.collision.binary_collision(mass1, pos1, vel1, mass2, pos2, vel2, collision_length)

Calculate the binary collision

Parameters:

• mass1 – Mass of the particle 1

• pos1 – Postion vector of the particle 1

• vel1 – Velocity vector of the particle 1

• mass2 – Mass of the particle 2

• pos2 – Postion vector of the particle 2

• vel2 – Velocity vector of the particle 2

irfpy.util.collision.binary_collision_stationary_vector1(mass1, pos1, vel1, mass2, collision_length)

Slightly generalized version of binary_collision_stationary()

Calculate the final velocity after the binary collision.

This function accepts the position and velocity vectors of particle 1. Particle 2 should be origin, and stationary. If the negative relative velocity (particles 1 and 2 separates as time goes), a collision will not happen.

Other conditions are hold as binary_collision_stationary(). (e.g. Particle 2 should be at origin, and stationary.)

irfpy.util.collision.binary_collision_stationary(mass1, speed1, mass2, impact_parameter, collision_length)

Cancluate a binary collision in one-particle stationary system.

Parameters:

• mass1 – Mass of moving particle, m1

• speed1 – Speed of the moving particle, v1. Positive.

• mass2 – Mass of stationary particle, m2

• impact_parameter – The impact parameter, d. Positive.

• collision_length – The radius of the cross section, or r1+r2 where r1 is the radius of the moving particke and r2 is the radius of the stationary particle.

Returns:

(V1, W1, V2, W2)

A binary collision is simulated here. Situation is that a flying (incoming) particle 1 has mass1 with speed 1 (positive y component) in the (negative x, positive y) quadrant. The stationary particle 2 at the origin will be collided. The returned is the final velocity (x, y components). If the impact paramter is more than the collision_length, the particle will not collide.

Theory

A binary collision is characterized by the two particles with masses, m1 and m2.

Let’s be in a system where the second particle with m2 is in rest at the origin. This system change be chosen without losing the generality.

First particle (with m1) flies from infinite -x, toward the origin, with a speed of v1 along x axis. Just by rotation, we can select a frame that the first particle is in the (-x, +y) quadrant, center in z=0 plane, with v1>0 with only vx component.

Then, after the collision, the motion of two particles should be confined in x-y plane since no force acting in the z direction. Assume the resulting velocity vector as (V1, W1) for particle m1, and (V2, W2) for particle m2.

Momentum conservation will give you

$$m_1{v}_1=m_1{V}_1+m_2{V}_2$$

and

$$0=m_1{W}_1+m_2{W}_2$$

Energy conservation will give you

$$m_1{v}_1^2=m_1{V}_1^2+m_1{W}_1^2+m_2{V}_2^2+m_2{W}_2^2$$

The force only act on the collision, so that the particle m2 should have the following geometric relation of velocity.

$$W_2=-\frac{d}{\sqrt{(r_1+r_2{)}^2-d^2}}{V}_2$$

Solving above four equations to derive the V1, W1, V2 and W2. For simplicity, we use the following notations.

$$\begin{array}{rl}\mu & =\frac{m_2}{m_1}\\ \delta & =\frac{d}{\sqrt{(r_1+r_2{)}^2-d^2}}\\ \alpha & =(\mu +1)(1+{\delta }^2)\end{array}$$

The solutions are

$$\begin{array}{rl}{V}_1& =v_1(1-\frac{2\mu }{\alpha })\\ W_1& =\frac{2v_1\mu \delta }{\alpha }\\ V_2& =\frac{2v_1}{\alpha }\\ W_2& =-\frac{2v_1\delta }{\alpha }\end{array}$$

Todo

To check the above formulation is correct. In particular the fourth equation should be checked. How to? Probably we may need another formulation using center of gravity system…