DMD:Background:The DMD Model Equations

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The relationship between time and position gives:

(1) r_{ij} = r_i - r_j = (r^0_i + u_i(t-t_0)) - (r^0_j + u_j(t-t_0))
= (r^x_{ij}, r^y_{ij}, r^z_{ij})

where r_i is the vector (r^x_i, r^y_i, r^z_i) position of the i^{th} particle and  u_i is the vector velocity.
Note: html to make bold typeface for vectors in math mode is unknown to this author. Sorry.
At the time of contact, t^c_{ij}, we have ,

(2) \sigma^2_{ij} = (r^c_{ij})^2 = \{(r^0_i + u_it^c_{ij}) - (r^0_j + u_jt^c_{ij})\}^2
\equiv \{(r_{ij} + u_{ij}t^c_{ij})\}^2 = \{r_{ij} \bullet r_{ij} + 2t^c_{ij}r_{ij} \bullet u_{ij} + u_{ij}(t^c_{ij})^2\}

Where r_{ij} = (r_{ij} \bullet r_{ij})^{1/2} = \{ (r^x_{ij})^2 + (r^y_{ij})^2 + (r^z_{ij})^2 \}^{1/2}

The relation leads to a quadratic equation for the ij collision time, t^c_{ij} \equiv (t_{ij} - t_0).

(3) u^2_{ij}(t^c_{ij})^2 + 2b_{ij}t^c_{ij} + r^2_{ij} - \sigma^2_{ij} = 0
(4) t^c_{ij} = \{-b_{ij} - (b^2_{ij} - u^2_{ij}(r^2_{ij} - \sigma^2_{ij}))^{1/2} \} / u^2_{ij} \equiv \{-b_{ij} - D^{1/2}_{ij}\} / u^2_{ij}

Where :
(5) b_{ij} = r_{ij} \bullet u_{ij}

(6) u_{ij}^2 = (u_i - u_j)^2

(7) D_{ij} = (b^2_{ij} - u^2_{ij}(r^2_{ij} - \sigma^2_{ij})) \equiv discriminant. Note that u_{ij} in m/s is a large number while r_{ij} is very small.

DMD Model Eq 1.jpg

DMD Model Eq 2.jpg

DMD Model Eq 3.jpg

b_{ij} > 0(forget about it)

D_{ij} < 0(forget about it)

D_{ij} > 0(schedule it)



At the time of the collision, the velocities of the particles change according to,

(8) \frac{\vartriangle u_i}{m_j} = \frac{-2(b^c_{ij} / \sigma^2)r^c_{ij}}{(m_i + m_j)} = \frac{-\vartriangle u_j}{m_i}

We can derive this formula by assuming that particle j is stationary (reference frame) and particle i is moving on the x-axis with equal mass. The j-direction after collision is given by the line of action r^c_{ij}, since that is all j feels about momentum change. Conservation of momentum means that u^f_i + u^f_j = u^i_i with the geometric interpretation of a sum of vectors in the form of a triangle. Conservation of energy gives (u^f_i)^2 + (u^f_j)^2 = (u^i_i)^2. The Pythagorean theorem applied to the conservation of energy means that this triangle must be a right triangle. Therefore, we can rotate the coordinate system such that y = u^f_i; x = u^f_i; u^f_i = u^i_i cos \theta; u^f_j = u^i_i sine\theta

DMD Model Eq 4.jpg

Note that r^c_{ij} and b^c_{ij} must be updated to the point of collision before computing the velocity changes.