Skip to main content

Model Equations

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 rir_i is the vector (rix,riy,riz)(r^x_i, r^y_i, r^z_i) position of the ithi^{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, tijct^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 rij=(rij∙rij)1/2={(rijx)2+(rijy)2+(rijz)2}1/2r_{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, tijc≡(tij−t0)t^c_{ij} \equiv (t_{ij} - t_0).

(3)uij2(tijc)2+2bijtijc+rij2−σij2=0(3) u^2_{ij}(t^c_{ij})^2 + 2b_{ij}t^c_{ij} + r^2_{ij} - \sigma^2_{ij} = 0

(4)tijc={−bij−(bij2−uij2(rij2−σij2))1/2}/uij2≡{−bij−Dij1/2}/uij2(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)bij=rij∙uij(5) b_{ij} = r_{ij} \bullet u_{ij}

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

(7)Dij=(bij2−uij2(rij2−σij2))≡(7) D_{ij} = (b^2_{ij} - u^2_{ij}(r^2_{ij} - \sigma^2_{ij})) \equiv discriminant. Note that uiju_{ij} in m/s is a large number while rijr_{ij} is very small.

<table width="75%">
<tr>
<td width="25%" align="center">
![](./DMD_Model_Eq_1.jpg)
<td width="25%" align="center">
![](./DMD_Model_Eq_2.jpg)
<td width="25%" align="center">
![](./DMD_Model_Eq_3.jpg)
</tr>
<tr>
<td width="25%" align="center">
$b_{ij} > 0$(forget about it)
<td width="25%" align="center">
$D_{ij} < 0$(forget about it)
<td width="25%" align="center">
$D_{ij} > 0$(schedule it)
</tr>
</table>

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

(8)△uimj=−2(bijc/σ2)rijc(mi+mj)=−△ujmi(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}

<table width="100%">
<tr>
<td width="80%" align="left">
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 $y = u^f_i; x = u^f_i; u^f_i = u^i_i cos \theta; u^f_j = u^i_i sine\theta$

<td width="20%" align="center">
![](./DMD_Model_Eq_4.jpg)
</tr>
</table>

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