Collision sphere

Equivalent considerations for nonstatic, sheared systems demonstrate the kinematical possibility of such shearing motions. This requires, inter alia, that the distance between any two sphere centers remains larger than 2a. The static viewpoint can be generalized to such circumstances as follows Rather than considering the lattice deformation, it suffices to examine the deformed collision sphere. The latter body 3 is defined as the set of points... 

As in the kinetic theory of gases, a selected molecule is represented as a collision sphere of radius equal to the sum of the radii of the colliding molecules. The neighbors with which it collides are mass points. The uncorrected frequency of collision is given by the usual equation of the kinetic theory of gases ... 

The radial distribution function is defined as before and is evaluated at the point ok on the collision sphere ... 

The radial distribution function implicitly takes account of the enhancement of the number density in the neighborhood of the collision sphere due to the excluded volume of the other molecules making up the system. 

Figure A3.1.3. The collision cylinder for collisions between particles with velocities v and v. The origin is placed at the centre of the particle with velocity v and the z-axis is in the direction of v - v. The spheres indicate tire range, a, of the intennolecular forces.

Figure A3.1.9. Hard sphere collision geometry in the plane of the collision. Here a is the diameter of the spheres.

The first molecular dynamics simulation of a condensed phase system was performed by Alder and Wainwright in 1957 using a hard-sphere model [Alder and Wainwright 1957]. In this model, the spheres move at constant velocity in straight lines between collisions. All collisions are perfectly elastic and occur when the separation between the centres of... 

We are concerned with bimolecular reactions between reactants A and B. It is evident that the two reactants must approach each other rather closely on a molecular scale before significant interaction between them can take place. The simplest situation is that of two spherical reactants having radii Ta and tb, reaction being possible only if these two particles collide, which we take to mean that the distance between their centers is equal to the sum of their radii. This is the basis of the hard-sphere collision theory of kinetics. We therefore wish to find the frequency of such bimolecular collisions. For this purpose we consider the relatively simple case of dilute gases. 

If the transition state theory is applied to the reaction of two hard spheres, the result is identical with that of simple collision theory. - pp Because transition state theory is an equilibrium theory, it can be inferred that collision theory is also an equilibrium theory. 

Sphere I weighs 1 lb and is traveling at 2 ft/s in the positive x direction when it strikes sphere 2, weighing 5 lb and traveling in the negative x direction at 1 ft/s. What will be the final velocity of the system if the collision is (a) plastic, or (b) Elastoplaslic with e = 0.5 ... 

In general, the distribution function changes in time because of the underlying motion of the hard-spheres. Consider first the nonphysical case where there are no collisions. Phase-space conservation, or Louiville s Theorem [bal75], assures us that... 

Now let us add the possibility of collisions. Before we proceed, we make the following two assumptions (1) only binary collisions occur, i.e. we rule out situations in which three or more hard-spheres simultaneously come together (which is a physically reasonable assumption provided that the gas is sufficiently dilute), and (2) Boltzman s Stosszahlansatz, or his molecular chaos assumption that the motion of the hard-spheres is effectively pairwise uncorrelated i.e. that the pair-distribution function is the product of individual distribution functions ... 

Now, equation 9.27 must of course be modified in the presence of collisions. Since, by definition, the rate of change of f x, v, t) (Px(Pv is equal to the increase in hard-spheres in d xd v as a result of collisions, we have, in general, that... 

Consider a sphere-sphere collision. Let the incoming and outgoing distributions be given by /i, /2 and /i, /2, respectively. Also, assume that the pre- and postcollision velocity-space regions occupied by hard-spheres 1 and 2 are given by cPv, cPv and cPv2, cPv2, respectively. 

Boltzman s H-Theorem Let us consider a binary elastic collision of two hard-spheres in more detail. Using the same notation as above, so that v, V2 represent the velocities of the incoming spheres and v, V2 represent the velocities of the outgoing spheres, we have from momentum and energy conservation that... 

Fig. 9.6 Collision of two hard-spheres shown in a reference frame in which sphere 1 is at rest n is a unit vector along the line of the two centers of the colliding spheres.

In the relative coordinate system in which the first sphere is at rest, a collision can occur only if the center of the second sphere lies within the collision cylinder" as shown in figure 9.6. Now, the volume of the cylinder is equal to b d4> db u 5t. Prom figure 9.6 it should be clear that... 

Suppose that k — k(x, v) is some quantity associated with a hard-sphere such that in any collision between spheres that takes place at position x, we have that... 

Figure 7-3 shows an experimental study of the collision of hard spheres. The experimenter imparts energy of motion to the white ball (see Figure 7-3A, 7-3B). He does so by doing work by striking the ball with the end of a cylindrical stick (a cue). The amount of energy of motion (kinetic energy) received by the ball is fixed by the amount of work done. If the ball is struck softly (little work being done), it moves slowly. If the ball is struck hard (much work being done), it moves rapidly. The kinetic energy of the white ball appears because work was done—the amount of work, IV, determines and equals the amount of kinetic energy, (KE),. In symbols,...

The angle of deflection for the collision of rigid elastic spheres may be obtained from Fig. 1-3. The minimum distance of approach is... 

For hard sphere collisions, v(v) would be proportional to v9 and the mean free path independent of v A(v) is an equivalent mean free path for a- general force law. Cf. S. Chapman and T. G. Cowling, The Mathematical Theory of Non- Uniform Oases, pp. 91 and 348, Cambridge University Press, 1958. 

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