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Statistical Mechanics of Blue and Orange Bacteria

  1. Mar 22, 2012 #1
    1. The problem statement, all variables and given/known data
    500 blue and 500 orange bacteria are placed in a growth medium. Each bacterium divides every hour. A predator eats exactly 1000 bacteria per hour irrespective of color.

    a) What is the ultimate probability distribution for the colors of bacteria in the growth medium?

    b) How long will it take to reach this equilibrium state?

    c) If the predator has a 1% preference to blue bacteria how does this change the final distribution?



    2. Relevant equations

    Nb(t)=2tNb0- number consumed...
    (Not sure how to express that probabilistically)
    same for No

    I also thought about using the different ways the 1000 consumed bacteria could be chosen from each of the orange and blue but they are indistinguishable (right?) so that does not matter.

    3. The attempt at a solution

    a) It seems the most probable solution is that the distribution would remain equal but this is an unstable equilibrium, a small perturbation in either direction will be exponentiated through the growth process. Consequently I think that the ultimate probability will be all of one type, either blue or orange, equal probability for each option. I am not sure how to show this rigorously though.

    Except if it takes an hour before the bacteria divides into 2, it seems the predator would consume all 1000 (500 orange and 500 blue) in the first hour before any are able to divide. I figured this was an error in the phrasing of the question but is this significant?

    b) A long time... Not sure how to rigorously determine this.

    c) Now the final distribution will be all orange because the number of blue will asymptotically approach 0 as time progresses.
     
  2. jcsd
  3. Jul 12, 2012 #2
    This is exactly what i thought!

    I looked up this problem in the book by Yung Kuo Lim on the thermodynamics and statistics book. 2nd problem in the stats section.
    http://depts.washington.edu/chemcrs/bulkdisk/chem552A_win10/homework_Homework2_Solution_Part_1.pdf [Broken]

    The solution there uses of number of ways of choosing a specific color. Please check that solution and comment.
     
    Last edited by a moderator: May 6, 2017
  4. Jul 12, 2012 #3
    I did find and consider this solution after posting my question. I disagree with the way this solution treats the evolution of the problem. Me and a college each wrote a small program to simulate this process and both of us independently found that the probability diverges to either all orange or all blue if let run long enough, each type with equal probability. He went on to calculate a probability evolution to determine the needed iterations to reach equilibrium. The details of this were a bit beyond what I could easily explain (I would need to go review what he did).

    The key point to realize is there is a finite probability that the predator eats all the remaining of one type and as the ratio of one to the other shifts from 1:1, that probability increases. If let iterate enough, the solution falls into the state of all one type and becomes stable.

    What Lim is describing would better be phased as, two bacteria, 5000 of each type, are allowed to grow for a long time before introducing a predator which eats each of the two types indiscriminately much quicker than they reproduce. When the predator has returned the population to the total initial 10000 bacteria, what is the distribution of each of the two types?

    The problem with this new question is the answer is too obvious from the beginning, not as good for a qualification exam.

    Below is my [unelegant] matlab/octave code if you are interested in playing with it.


    iter=100;
    result=zeros(iter,3);
    for k=1:iter
    n=1000;
    r=500;
    g=n-r;
    i=1;
    while i<4000 %r>0 && g>0
    r=2*r;
    g=2*g;
    for j=1:n
    if rand<r/(r+g);
    r=r-1;
    else
    g=g-1;
    end
    end
    % fprintf('after %d steps there are %d red and %d green\n',i,r,g);
    i=i+1;
    if r<0
    g=n;
    r=0;
    elseif g<0
    r=n;
    g=0;
    end
    end
    result(k,:)=[r,g,i];
    end
    result

    figure;
    hist(result(:,1),100)
    title('red bacteria after 100 hours');
    h = findobj(gca,'Type','patch');
    set(h,'FaceColor','r')%,'EdgeColor','w')
    % green is just the opposite
     
    Last edited: Jul 12, 2012
  5. Jul 12, 2012 #4
    Hey thanks a lot.

    The crucial point is that the system reaches a standstill due to the points 0 and 10000. The arguments that were used to reach the books expression dont hold

    I have been trying to convince people of this and you have gone ahead and written a code.
    you are awesome sir.
     
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