Random nature of multiple scattering

1. Aug 7, 2014

vrinda mukund

I have some confusion about multiple scattering.

We always say that the problem of single scattering is always deterministic in nature.But while modeling the problem of multiple scattering, we take that the problem is stochastic in nature. I don't understand why. Why multiple scattering/collision is considered as stochastic?

We can exactly say what will occur when a single electron or a photon collides with an atom/molecule/aerosol. We can use the cross section values for various possible processes like ionization, dissociation etc to find the probability of occurrence of different processes. But while studying the case where millions of electrons or photons are coming and colliding with molecules or atoms, why we say that processes that occur will be random ?

2. Aug 7, 2014

3. Aug 7, 2014

DrDu

I don't think this is true. Many effects like the propagation of light in a medium as described by the refractive index are due to non-stochastic multiple scattering.

4. Aug 7, 2014

Simon Bridge

I thought there was a stochastic approach to refractive index, but I cannot find it :(
We don't normally use the refractive index thing at the individual photon level though do we?

... but that's a good point anyway. Not all multiple collisions get the stochastic treatment.
Light transport gets the stochastic treatment for very complicated geometries - like in the case of transport through tissue.

In this way it is kinda like the "randomness" of a thrown die - while strictly a deterministic system, it is just too difficult to calculate and we'd need to know the parameters involved to an inconvenient level of accuracy anyway.

5. Aug 7, 2014

vrinda mukund

Randomness confusion

@Simon Bridge
Sir, I am talking about the situation where millions of monoenergetic electrons coming and colliding with neutral molecules. We have cross section values availbale for different possible processes. Using these cross section values, we can find out the probability of occurance of each of the possible outcome. But still, when we model such a process, we usually use some stochastic model. The reason that is said is, multiple collision is a random process. But i don't understand. The same set of cross section and hence same set of probabilities are valid, no matter whether it is single electron or million electrons. So, if we can predict exactly which process an electron will do, using the probability values calculated using crosss sections, we can determine, which process each and every electron in a swarm of millions of electrons will do. So how randomness is coming into picture? Kindly help ...

6. Aug 7, 2014

Simon Bridge

I know what you are talking about and my earlier statements stand.

... there's the problem: we can't.
We can get a good idea for scattering off a single target atom - but then the result has to be carried through to the next target ... which atom will be the next scattering center? What is the incoming electron energy?
You end up with a branching tree of possibilities - and for each possible 2nd scatterer, there are a host of possible 3rd scatterers ... the number of steps in the calculation, and the level of uncertainty in what happens, increases exponentially.

This is how you get randomness from determinism - which, I think, is your actual question.
Add to that - the actual interactions are QM events - so each individual scattering in inherently statistical in nature.

Many people have a picture of electrons and atoms being like balls and pegs in a pachinko game - this is not the case.

7. Aug 8, 2014

DrDu

I think the point is that in this example, we only have a statistic of the position of the molecules which scatter. If the molecules sit on well defined positions, like in a crystal, there will be coherent multiple scattering.

8. Aug 8, 2014

vrinda mukund

@DrDu
Sir,I am studying the situation where molecules are freely and uniformly distributed; not on any crystal lattice. Could you please elaborate 'coherent multiple scattering ' ? In the condition where molecules are uniformly and freely distributed,is this coherent multiple scattering valid ?

@Simon Bridge
Sir, what you mean by this "the result has to be carried through to the next target " ? Also in the case of molecules freely and uniformly distributed, what is the relevance of "which molecule is the next scattering centre" . As i said incoming electron energy is also fixed.Also could you elaborate on "for each possible 2nd scatterer, there are a host of possible 3rd scatterers" ?

Regards
Vrinda

9. Aug 8, 2014

DrDu

Multiple scattering from the molecules in a gas is neither completely coherent nor incoherent. The incoherence is a consequence of the molecules forming a statistical ensemble (Maxwell-Boltzmann) and not a pure state. At T=0, scattering would be completely coherent.

10. Aug 8, 2014

Simon Bridge

Think of all the things that can happen to the electron at the first time it scatters.
Whatever the outcome of the first scatter - the final state of the electron is what meets the next atom. There is a choice of possibilities - so you get a branching probability tree.

Send an identical electron through immediately after it, and it will follow a different trajectory.

Where you are talking about a beam of electrons all having a slightly different incoming path and all interacting with each other, any chance of anticipating individual electrons is lost.

But didn't you say you were considering a large number of neutral atoms as the target - not molecules? It will be best if you stick to one example at a time.

Free-uniform distribution is itself a statistical description. So you are thinking of something like scattering through a gas target rather than something like a crystalline solid?

In order to be able to have a deterministic path for the electron, you need to also be able to track the locations of the atoms in the gas. Have you heard of the "three body problem"?

... The energy of the electron is not fixed after it has encountered the first atom, and nor is it's momentum.
As you have stated before, there are a range of different interactions possible - each with a different outcome.

For a monatomic gas as the target, a "scatterer" is an atom in the gas.
When a single electron enters the region of the target, there is a probability it will interact with an atom before it exits.
Once it has interacted with one for the first time, there are a range of possible final states for the electron (and for the atom interacted with) ... then there is a possibility that it will have another interaction before leaving the gas.
If you want to avoid a stochastc description of this process, you need to track the scatterers.

An example would be the case where the atoms are arranged in a crystal lattice - and the energies are low enough not to significantly disturb the system. In this case we would not normally use a stochastic description for the resulting scattering.

I think you need to be more specific in your examples.
Pick a specific case where you see stochastic methods being used and you don't understand why this is not deterministic.

11. Aug 9, 2014

vrinda mukund

Okay, then i will be more specific with my problem.

A beam of (say, 1 million) monenergetic (say 1 keV) electrons are coming and colliding molecules, all of same species. The electrons can do either an elastic collison or an inelastic collision with molecules. If its is an inelastic collision, again the possibilities are, the molecule can be ionized or dissociated or excited.
These are all the possibilites.

Now i would like to bring cross sections(xs) into picture. Molecule will be having cross sections for each of these reactions at different energies. Using the cross sections we can calculate probabiliites for different possible reactions. For example, at 1 keV

Probability of elastic collision =(xs for elastic collision at 1 keV)/ (Total scatering xs at 1 keV)

Probability of ionization =(xs for ionization at 1 keV)/ (Total inelastic xs at 1 keV) and so on....

Now let us arbitararily assign some numerical values for these probabilites so that it will be easy to dissuss:
probability of elastic collision at 1 keV : p(elastic) = 0.4
p(inelastic) = 0.6
p(ionization) = 0.5
p(dissociation) = 0.3
p(excitation)=0.2

From these probability values it is very clear that if an electron of 1 keV comes and hits the molecule, the molecule will be ionized. the electron will lose some energy (say 100 eV)as a result of the process. Remainning energy in electron = 900 eV. This electron can again do further collision. We can calculate the probability value for each reaction in the same way at 900 eV and can again predict which process will happen. So the problem is clearly deterministic.

Now if another electron of 1 keV comes and hits the molecule, again the same same set of cross sections and same set of probabilities are valid, again we can predict. In this way we can predict what each electron among the one million electrons will do. So how come randomness is coming into picture ?
Kindly help !!!!!!!!!

12. Aug 9, 2014

DrDu

Let me see if I understand you correctly: You are starting from a set of probabilities for different processes and ask how probability comes into play?

13. Aug 9, 2014

Simon Bridge

Yeah - is this the distinction between randomness and determinism you are exploring or the distinction between regular probabilities and stochastic processes?

... I didn;t think that was possible. Can you provide a reference to where an beam of 1eV electrons is produced? Usually you get a distribution of energies?

... so you are not talking about one electron undergoing multiple collisions but of one molecule in the target undergoing multiple collisions?
In that case, the state of the molecule is the stochastic variable... like with Brownian motion.

The result of each collision depends on the result of previous collisions (which determines the state of the molecule - vibrational, rotational, translational, and electronic/ionic) as well as the reaction crossections, which are, themselves, statistical. Now - you show me how, after many such interactions, you can get a deterministic description of the molecule's state over time? How about just as a function of the number of collisions already undergone?

You can try it more simply for something more idealized.
One large object is constantly bombarded by lots of small objects - treat as classically idealized elastic collisions if you like - even restrict it to 1 or 2 dimensions - should keep things simple. Your task is to find the deterministic function of the position of the big object as a function of time - or even as a function of the number of collisions.

Last edited: Aug 9, 2014
14. Aug 9, 2014

Staff: Mentor

The sum of these "probabilities" equals 2.0. It should equal 1.0 (100%).

15. Aug 9, 2014

vrinda mukund

@Simon Bridge
"is this the distinction between randomness and determinism you are exploring or the distinction between regular probabilities and stochastic processes? "

What i want to know is, how the scenario changes from determininstic to stochastic nature when we move from single scattering to multiple scattering.

"Can you provide a reference to where an beam of 1eV electrons is produced? Usually you get a distribution of energies?"

Experimentally it is possible . I can generate beam of 1 kev electrons and make them collide with molecules.

"so you are not talking about one electron undergoing multiple collisions but of one molecule in the target undergoing multiple collisions? "

No, I am talking about the situation where there are a large number of electrons and large number of molecules. Each one of these electrons can keep on colliding with these molecules untill the energy of each and every electron is completely degraded. So i am not considereng the case where one molecule undergoing multiple collision, but each and every electron in the incident beam doing multiple collsions.

16. Aug 10, 2014

vrinda mukund

@ jtbell

That is not the way to add up the probability. Probability of a collision = probabilty of elastic collision + probabilty of inelastic collision = 0.6 + 0.4 =1

Again for inelastic collion, adding the probabilties of possible processes, total probability of inelastic collison = 0.5 +0.3 +0.2 = 1