Difference between geometrical cross section / total cross section and differential cross section (scattering) - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2022-01-25T23:22:19Z https://physics.stackexchange.com/feeds/question/631361 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/631361 1 Difference between geometrical cross section / total cross section and differential cross section (scattering) imbAF https://physics.stackexchange.com/users/275029 2021-04-22T21:21:02Z 2021-04-22T23:46:24Z <p>I am trying to understand the difference between the 3 above mentioned terms but i have it hard to picture each of them. I know that the geometrical cross section would be the physical area for which we would have a collision (billiard spheres) and i know that the scattering cross section if we have a charged nucleus and an electron colliding with it, will be bigger then the physical cross section of the nucleus. But what i cannot understand are the following 3 things:</p> <p>How does the total cross section in case of a scattering event looks like? Like how to visualize that.</p> <p>How is it related to the differential cross section (geometrically). I assume if i integrated in spherical coordinates the diff. cross section I should get the total one.</p> <p>And most importantly ,we were given the following formula without actually showing the logic of how it was derivated:</p> <p>Total rate <span class="math-container">$W_r$</span> of scattering events <span class="math-container">$dN_s$</span> per unit of time:</p> <p><span class="math-container">$W_r = dN_s/dt = J \cdot N_t \cdot \sigma_{tot}$</span>, where</p> <p>J is the flux density, <span class="math-container">$N_t$</span> is the number of the nuclei of the target object and <span class="math-container">$\sigma_{tot}$</span> is the total cross section.</p> <p>If there are links/ docs with detailed explanation to the above 3 questions (specially the last one) and about the Rutherford scattering, that would also help!</p> https://physics.stackexchange.com/questions/631361/-/631363#631363 1 Answer by rfl for Difference between geometrical cross section / total cross section and differential cross section (scattering) rfl https://physics.stackexchange.com/users/243047 2021-04-22T22:00:58Z 2021-04-22T22:00:58Z <blockquote> <p>How does the total cross section in case of a scattering event looks like? Like how to visualize that.</p> </blockquote> <p>For a billiard ball, it is the size of the ball, i.e. the geometric cross-section. If you picture the billiard ball being magnetic, then you can see that also other magnetic balls could be deflected even if they don't hit the ball mechanically. So the total cross section of a magnetic billiard ball for the scattering off another magnetic billiard ball is larger than its geometric cross-section.</p> <blockquote> <p>How is it related to the differential cross section (geometrically). I assume if i integrated in spherical coordinates the diff. cross section I should get the total one.</p> </blockquote> <p>Exactly, that's the relation.</p> <blockquote> <p>Rate equation...</p> </blockquote> <p>Consider what impacts the number of scatter events that you expect, simply by the units of these things. The stronger your beam of incident particles, the more scatter events. That's a linear dependence. Thus the <span class="math-container">$J$</span> factor. Units are projectiles per area per time. For the number of scatter centers, same thing, if you got twice as many you'll get twice as many scatters, thus the <span class="math-container">$N_t$</span> factor. Units are just the number that you put in the way of your beam. Then the cross section, as above, in units of area. Multiply that together, gives units of events per time, i.e. your scatter rate.</p>