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I am trying to understand this paper but I am not able to do it. Please help. See the paper here. First of all how could they write collision time $t_0$ as given there?. I am not able to get it even after reading it many times. At least can anybody suggest something that explains 3rd and 4th paragraph in the paper?

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  • $\begingroup$ Your link is broken, the paper is behind a paywall, and the paper’s result is not what you say it is. $\endgroup$ – G. Smith May 17 at 17:48
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    $\begingroup$ I've taken my best guess at fixing up the wordos, typos, and mis-punctuations herein, but you should check my work. $\endgroup$ – dmckee May 17 at 18:16
  • $\begingroup$ Thanks.. I have fixed those eŕrors. My need is only to know how they wrote a collission time.. from velority and average separation.. after that it is okay. But can I expect any help from here? Or where else I should go? $\endgroup$ – fahd May 19 at 1:18
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First of all how could they write collision time $t_0$ as given there?

So the average separation is $a_0$, that means that there is one particle per volume $a_0^D$, where $D$ is the dimension of the space (the authors ignore numerical factors everywhere). The radius of the sphere is $r_0$, therefore its cross-section (effective "area" of collision) is $r_0^{D-1}$, therefore, the sphere collides with all spheres in the volume $u_0 r_0^{D-1}$ per unit time (where $u_0$ is the rms sphere velocity). As there is one sphere per volume $a_0^D$, the collision time is $\frac{a_0^D}{u_0 r_0^{D-1}}$.

The rest of the first few paragraphs is pretty straightforward. I would not like to rewrite the paragraphs here.

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  • $\begingroup$ I can't get why did we multiply the area $r_0^{D-1}$ with that distance which it travels in unit time $\endgroup$ – fahd May 21 at 6:58
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    $\begingroup$ @fahd : A sphere travels distance $u_0$ in unit time. For the resulting trajectory - a straight-line segment of length $u_0$ - let us consider all points that are located not farther than $r_0$ from this trajectory. The volume that such points form is $u_0 r_0^{D-1}$ (check it for $D=2$ to understand that). All spheres in this volume will collide with the initial sphere in unit time (again, the authors ignore numerical factors, and taking into account the movement of the spheres does not change result). $\endgroup$ – akhmeteli May 21 at 12:00
  • $\begingroup$ Okay, now something is clear for me. I think as we can write $ time = total distance /velociy(distance travelled in unit time)$ here we are writing it as $total volume/volume traversed in unit time$. But I think if we are writing so the upper term would be $a_0*r_0^{D-1}$ isn't it? $\endgroup$ – fahd May 24 at 15:10
  • $\begingroup$ since particle is not going to cover whole the volume $a_0^{D}$ but only the volume in which diretion it is moving, i.e $a_0r_0^{D-1}$. If this is true can't we just write first collission time as $\frac{a_0}{u_0}$ ? $\endgroup$ – fahd May 25 at 0:32
  • $\begingroup$ @fahd: Sorry, according to the rules (physics.stackexchange.com/help/how-to-answer), I should "avoid trying to answer questions which...require too much guidance for [me] to answer in full". $\endgroup$ – akhmeteli May 25 at 3:41

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