Markov Chains: Ehrenfest's Diffusion Problem:

There are SIX molecules shared by TWO chambers:

1. Determine P (Probability) matrix and write it.

2. If there are 2 molecules in Chamber 1 initially, how many molecules will be in Chamber 1 after 5 seconds?

NOTE: The images above are an example of an Ehrenfest's Diffusion Problem involving FOUR molecules. Let me know if you would like more images.

system. To make the problem computationally manageable, we assume n = 2 so there are 4 molecules in the system. The matrix for the system consists of 5 states: 0, 1, 2, 3, 4 molecules in chamber 1. Thus it has the form P= End of Step 0 2 3 4 0 0 1 0 0 0 1 0 0 1 4 0 3 4 Start 2 0 0 1 0 1 0 of IN Step 3 0 0 3 0 1 4 4 4 0 0 O 0 0 As developed in the Weather Regime Example above, we define the initial states with a row vector similar to the initial state for the weather regimes as shown in Eas. (4.3.1.) - (4.3.4). The S-component row vector is written as RV, where all elements are zero except the ith element which is 1. The 5 initial state elements in vector form are given by RV = (1,0,0,0,0) - no molecules in chamber 1 4.6.1 RV, = (0,1,0,0,0)-- 1 molecule in chamber 1 4.6.2 4.6.3 RV = (0,0,1,0,0) --- 2 molecules in chamber 1 RV, = (0,0,0.1,0) - 3 molecules in chamber 1 4.6.4 RV = (0,0,0,0,1) - 4 molecules in chamber 1 4.65 Thus, the product RVP End Step Probabilities 466 where End Step Probabilities is a five-component row vector where the ordered elements 1, 2,3,4,5 give the probabilities of o. 1, 2, 3, 4 molecules in chamber 1, respectively We need to explain how the metri elements were obtained. But before that explanation, let us test the validity of the initial state vectors and the elements in the matrix We first consider the situation where there are initially no molecules in chamber 1. The initial row vector RV = (1.0.0.0.0) represents this situation. When a molecule is chosen at random from the two chambers, it is 100% certain that the molecule comes from chamber since chamber 1 has no molecules. At the end of a second, the step size, we have 1 molecule in chamber 1 and 3 molecules in chamber. The product RV - - (1,0,0,0,0) P - (0.1,0,0,0 4,6,7 Indicates there is molecule in chamber 1. Thus, the first row of the matrixis correct. Now let us assume there is 1 matrixin chamber 1. Then the end of step results RV (0,1,0,0,0) -4.0.0.0 which indicates there is a 25% chance that chamber 1 contains O molecules and a 75% chance that chamber contains Imolecule. Continuing this process, the end stage results are found to be correct But in this case, different from the weather regime case, the next step results are not ven. They must be derived from probability arguments. We make these ruments for to molecules in chamber 1 at Inaltime, and secondly for the case where there is a side colecule in chamber at initial time To determine the end stage elements of row 3, thone probabilities associated with a start with no molecules in chamber 1, we know for sure that the molecule chosen at the initial second must come from chamber and it goes to chamber 1. Thus, chamber 1 has a single molecule at the end of the first second-the first step, and chamber has 3 molecules at the end of the first second. The number 1 must therefore reside in element of the first row indicating there is 1 element in chamber 1 The other 4 mers in the first row indicate that there is ox probability or no possibile chance for chamber to have 0, 2, 3 or 4 molecules in it after the end of step 1 Now we argue the case where there is initially 1 molecule in chamber 1. The result we obtain here is not influenced by what we found for the case above When a molecule is chosen at random, there is a 25% chance it comes from chamber and a 75% chance it comes from chamber 2. if it comes from chamber 2, there is a chance that there will be 2 molecules in chamber 1 at the end of the step. But if the molecule is chosen from chamber 1, there is a 25% chance that chamber 1 has no molecules in it. Thus, the end of-step probabilities for this case is 25% chance that chamber has no molecules and a 75% chance that chamber 1 has 2 molecules. The row 2 therefore becomes 0,0,0) as shown in the matte. It is left to the student to argue the cases for 2, 3, and 4 molecules in chamber 1 atinital time We now ask the question. If there are 3 moleodies in chamber I at initial time, what are the probabilities of molecule mumbers in chamber after 10 seconds. The newer is by RV (0,0,0, 1.0).pl 469 = (0,0.4995,0,0.5005,0) This result indicates that after 10 seconds there is a 50% chance of 1 molecule in chamber 1 and a 50% chance of 3 molecules in chamber 1. The probabilities become steady after about 15 seconds. Its interesting to consider the matrix associated with chamber 2 and the probabilities of molecules in chamber 2 after 10 seconds. That is left as an exercise. 4.7 Random Walk Random walk is an important problem in physics that appears in many guises. Our omos from Nohel Laureate Subramanyan Chandrasekhar system. To make the problem computationally manageable, we assume n = 2 so there are 4 molecules in the system. The matrix for the system consists of 5 states: 0, 1, 2, 3, 4 molecules in chamber 1. Thus it has the form P= End of Step 0 2 3 4 0 0 1 0 0 0 1 0 0 1 4 0 3 4 Start 2 0 0 1 0 1 0 of IN Step 3 0 0 3 0 1 4 4 4 0 0 O 0 0 As developed in the Weather Regime Example above, we define the initial states with a row vector similar to the initial state for the weather regimes as shown in Eas. (4.3.1.) - (4.3.4). The S-component row vector is written as RV, where all elements are zero except the ith element which is 1. The 5 initial state elements in vector form are given by RV = (1,0,0,0,0) - no molecules in chamber 1 4.6.1 RV, = (0,1,0,0,0)-- 1 molecule in chamber 1 4.6.2 4.6.3 RV = (0,0,1,0,0) --- 2 molecules in chamber 1 RV, = (0,0,0.1,0) - 3 molecules in chamber 1 4.6.4 RV = (0,0,0,0,1) - 4 molecules in chamber 1 4.65 Thus, the product RVP End Step Probabilities 466 where End Step Probabilities is a five-component row vector where the ordered elements 1, 2,3,4,5 give the probabilities of o. 1, 2, 3, 4 molecules in chamber 1, respectively We need to explain how the metri elements were obtained. But before that explanation, let us test the validity of the initial state vectors and the elements in the matrix We first consider the situation where there are initially no molecules in chamber 1. The initial row vector RV = (1.0.0.0.0) represents this situation. When a molecule is chosen at random from the two chambers, it is 100% certain that the molecule comes from chamber since chamber 1 has no molecules. At the end of a second, the step size, we have 1 molecule in chamber 1 and 3 molecules in chamber. The product RV - - (1,0,0,0,0) P - (0.1,0,0,0 4,6,7 Indicates there is molecule in chamber 1. Thus, the first row of the matrixis correct. Now let us assume there is 1 matrixin chamber 1. Then the end of step results RV (0,1,0,0,0) -4.0.0.0 which indicates there is a 25% chance that chamber 1 contains O molecules and a 75% chance that chamber contains Imolecule. Continuing this process, the end stage results are found to be correct But in this case, different from the weather regime case, the next step results are not ven. They must be derived from probability arguments. We make these ruments for to molecules in chamber 1 at Inaltime, and secondly for the case where there is a side colecule in chamber at initial time To determine the end stage elements of row 3, thone probabilities associated with a start with no molecules in chamber 1, we know for sure that the molecule chosen at the initial second must come from chamber and it goes to chamber 1. Thus, chamber 1 has a single molecule at the end of the first second-the first step, and chamber has 3 molecules at the end of the first second. The number 1 must therefore reside in element of the first row indicating there is 1 element in chamber 1 The other 4 mers in the first row indicate that there is ox probability or no possibile chance for chamber to have 0, 2, 3 or 4 molecules in it after the end of step 1 Now we argue the case where there is initially 1 molecule in chamber 1. The result we obtain here is not influenced by what we found for the case above When a molecule is chosen at random, there is a 25% chance it comes from chamber and a 75% chance it comes from chamber 2. if it comes from chamber 2, there is a chance that there will be 2 molecules in chamber 1 at the end of the step. But if the molecule is chosen from chamber 1, there is a 25% chance that chamber 1 has no molecules in it. Thus, the end of-step probabilities for this case is 25% chance that chamber has no molecules and a 75% chance that chamber 1 has 2 molecules. The row 2 therefore becomes 0,0,0) as shown in the matte. It is left to the student to argue the cases for 2, 3, and 4 molecules in chamber 1 atinital time We now ask the question. If there are 3 moleodies in chamber I at initial time, what are the probabilities of molecule mumbers in chamber after 10 seconds. The newer is by RV (0,0,0, 1.0).pl 469 = (0,0.4995,0,0.5005,0) This result indicates that after 10 seconds there is a 50% chance of 1 molecule in chamber 1 and a 50% chance of 3 molecules in chamber 1. The probabilities become steady after about 15 seconds. Its interesting to consider the matrix associated with chamber 2 and the probabilities of molecules in chamber 2 after 10 seconds. That is left as an exercise. 4.7 Random Walk Random walk is an important problem in physics that appears in many guises. Our omos from Nohel Laureate Subramanyan Chandrasekhar