0
$\begingroup$

The question is in the title: How is the cross section related to the probability of a process?

If I have the total cross section $\sigma$ of a process how is this related to the probability of this process? What about the differential cross section?


context copied from linked question :

I have to build a MonteCarlo from scratch and I find myself stuck.

I have to simulate a proton losing energy into an absorber and producing delta-rays.

If it were a photon I could use the formula $p=\mu \exp(-\mu s)$ were $\mu$ is the macroscopic cross section $s$ is the space between two collisions a nd $p$ is the probability of a collision. In this case I could generate a random number $p$ and invert the formula to get $s$.

In the case of the proton, I can use the Bethe-Block to simulate the loss of energy between two collisions.

But my problem is: which is the probability that a proton collide with an electron of the absorber? It should be proportional to the cross section of the process, but which is the cross section and how is it actually connected with the probability of a collision?

$\endgroup$
  • $\begingroup$ It depends on the process. But let us make a ratio: $\frac{d\sigma}{\sigma\cdot d\Omega}\cdot d\Omega$. It is the probability to scatter in the solid angle $d\Omega$. $\endgroup$ – Vladimir Kalitvianski Jul 28 '18 at 13:51
  • $\begingroup$ @VladimirKalitvianski So the probability that the scattering transfers and energy $dE$ is $\dfrac{1}{\sigma}\, \dfrac{d\sigma}{dE}$, is it correct? $\endgroup$ – mattiav27 Jul 28 '18 at 14:09
  • $\begingroup$ @VladimirKalitvianski and given a cross section which is the probability that there is a collision at the position x? $\endgroup$ – mattiav27 Jul 28 '18 at 14:16
  • $\begingroup$ Did you check Wikipedia? $\endgroup$ – Qmechanic Jul 28 '18 at 15:00
  • $\begingroup$ @Qmechanic I did, but it confuses me. I need to find the probability that a collision happens at the position x: I need it for a MonteCarlo, see my other question: physics.stackexchange.com/q/419889/46926 $\endgroup$ – mattiav27 Jul 28 '18 at 15:12
1
$\begingroup$

Total cross section $\sigma$

dartboard The cross-section of an event (e.g. collision) is the fraction of a total area that results in the event. This can be a true area (i.e. throwing darts at a dart board) or some analogous area ( i.e. meeting a person at a party). The probability of the event is then: $$ p = \frac{\sigma}{A}$$ Where $p$ is the probability of the event occurring, $A$ is the total area it can occur over, and $\sigma$ is the cross-section of the event. It is important to note that the "dart" is randomly thrown into the area.

In the example of darts, you could calculate the cross-section of the bull's-eye (the red center of the dartboard) by either:

  1. Measuring the diameter of the bull's-eye and then estimating it's area based on the assumption it is circular.
  2. Throw many darts, one at a time, at the dart board in a random fashion and count the fraction that land in the bull's-eye. This fraction will be equal to the fraction of area that makes up the bull's-eye.

Differential Cross-section

Differential cross-section is just the cross-section as a function of some 'impact parameters'. The impact parameters are whatever you need to uniquely describe the event: the energy or angle of a collision, or the 'overlap' of interests two people at a party have. Whatever the impact parameters, you integrate over their entire values to get the total cross section from the differential cross section: $$ \sigma = \int \frac{ d\sigma }{ d\Omega } d\Omega $$

Where $\sigma$ and $d\sigma / d\Omega$ are the total and differential cross section, respectively. Sometimes the differential cross-section is just $d\sigma$.

$\endgroup$
  • $\begingroup$ OP has elaborated so my answer no longer covers the scope of the question. $\endgroup$ – cms Jul 28 '18 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.