anarres: (Default)
[personal profile] anarres
My particle physics exam is later today! I generally react to exam stress in funny ways, like procrastinating a lot and not being able to sleep. So..

There is a type of fundamental particles called quarks. Each quark can come in one of three "colours" (nothing to do with what the word "colour" means in everyday life). If you smash an electron and a positron together in a particle collider, you get a shower of particles, which can be divided into two broad types: hadrons (things made of quarks) and leptons (things that aren't made of quarks). Using the laws of particle physics you can predict what the ratio of hadrons produced to leptons produced should be. The actual experimentally measured ratio is three times higher than what you would expect if quarks did not come in three colours, so this ratio is taken to be evidence of the existence of colour.

This argument is based on the principle that if species Z can decay either to A or to B, it will decay faster than if it could just decay to A, all other things being equal. This principle comes up in physics a lot and it is always presented as being obvious, but it didn't seem that obvious to me. So I wrote a python script to simulate particle decay (notice the word "procrastination" in the title of this post :-P ). It's funny how writing a simulation can clarify your understanding of a system, because as soon as I started writing the pseudocode in my notebook I realized that for every time step, I would do two (simulated with a random number generator) die rolls, one to see if Z decays to A and one to see if Z decays to B, so yes, obviously a particle with more decay routes has a greater chance to decay (all other factors being equal) and it was silly of me not to realize that before.

But anyways I wrote a script (procrastination!) that does this:

For every time-step:
Z -> A with probability p
Z -> B with probability p
A -> Z with probability p
B -> Z with probability p

I let it run to see what the equilibrium concentrations of A, B, and Z would be. (By equilibrium concentrations, I mean that no matter what initial amounts of A, B and Z you start with, if you let the simulation run long enough you eventually get to a certain ratio that stays constant thereafter).

I thought the equilibrium concentrations would be: 50% Z, 25% A, 25% B, because I thought that SYSTEM 1 was equivalent to a simpler system:

For every time-step:
Z -> C with probability p
C -> Z with probability p where C = {either A or B}.

I think it's clear that SYSTEM 2 would lead to equal amounts of Z and C. But SYSTEM 1 didn't do what I expected - instead it went to a third each Z, A, and B! I can't figure out why it does this - somehow SYSTEM 1 knows that A and B are not the same, but how does it know this? It seems that A and B do the same thing and behave as if they were the same thing, but somehow the mysterious Laws of Mathematics can tell them apart.

So either I wrote the python script wrong, or my understanding of the system is wrong, dunno. I'll have another look at it later... (It's silly to write scripts when you're a) sleep-deprived and b) not going to have time to debug them, it just leads to confusion.)

Date: 2010-05-26 06:47 pm (UTC)
pauamma: Cartooney crab holding drink (Default)
From: [personal profile] pauamma
Disclaimer: the following is drawn from long-ago memories of discrete statistics, and may have no resemblance to particle physics, real or simulated, but what you observed comes straight out of your simulationequations:
For every time-step:
Z -> A with probability p
Z -> B with probability p
A -> Z with probability p
B -> Z with probability p

Let rA(t), rB(t), rZ(t) be the ratios of A, B, and Z at time t. Then:
rA(t+dt) = rA(t)*(1-p)+rZ(t)*p
rB(t+dt) = rB(t)*(1-p)+rZ(t)*p
rZ(t+dt) = rZ(t)*(1-2*p)+rA(t)*p+rB(t)*p

At equilibrium, rA(t+dt)=rA(t) and ditto for B and Z. Thus:
rA(t) = rA(t)*(1-p)+rZ(t)*p
rB(t) = rB(t)*(1-p)+rZ(t)*p
rZ(t) = rZ(t)*(1-2*p)+rA(t)*p+rB(t)*p

Moving all variables to the left side:
rA(t)*(1+p-1)-rZ(t)*p = 0
rB(t)*(1+p-1)-rZ(t)*p = 0
rZ(t)*(1+2*p-1)-rA(t)*p-rB(t)*p = 0

Simplifying and dividing by p (if p = 0, any initial distribution is trivially stationary):
rA(t)-rZ(t) = 0
rB(t)-rZ(t) = 0
2*rZ(t)-rA(t)-rB(t) = 0

Thus: rA(t)=rB(t)=rZ(t)


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