Solving a Probabilities Problem in Thermodynamics: Stuck? Let's Figure It Out!

[jessawells]

i'm stuck trying to figure out this probabilities problem for my thermodynamics class. the question is:

consider an idealized drunk, restricted to walk in one dimension (eg. back and forward only). the drunk takes a step every second, and each pace is the same length. let us observe the drunk in discrete timesteps, as they walk randomly - with equal probability - back or forward.

a) suppose we have 2 non-interacting drunks who start out in the same location. What is the probability that the drunks are a distance A apart after M timesteps? (use stirling's approximation if you need to)

b)suppose instead that the 2 drunks started a distance L apart. Find the probability that the drunks meet at precisely the Mth timestep.


i know that the probability of a binary model system is given by:
P = multiplicity of system / total # of accessible states
= g (M, s) / 2^M

where g is the multiplicity and s is the spin excess (# of forward steps - # of backward steps")

= M! / [(1/2M+s)! (1/2M-s)! 2^M]

using stirling's approx. for large M, this becomes,

P (M,s) = sqrt[2/M(pi)] exp[-2s^2/M]


i'm not sure where to go from here and I'm really confused. the formula i wrote takes care of the M timesteps, but how do i factor in the distance A? how should i go about doing this question? I would appreciate any help. Thanks.

 

[anathema]

Thank you for your question. The probability of the drunks being a distance A apart after M timesteps can be found by considering the total number of accessible states for the system. In this case, the system can be thought of as a random walk, where each step has equal probability of being forward or backward.

To find the total number of accessible states, we can think of the system as a series of steps, where each step can be either forward or backward. This can be represented as a binary number, where 0 represents a backward step and 1 represents a forward step. For example, if we have 4 timesteps, the binary number 0101 would represent two backward steps followed by two forward steps.

The number of possible binary numbers of length M is 2^M. However, not all of these numbers will result in the drunks being a distance A apart. To find the total number of accessible states, we need to consider the condition that the drunks are a distance A apart. This can be represented as the condition that the sum of the steps taken by the two drunks is equal to A.

For example, if A = 2 and M = 4, the possible binary numbers that would result in the drunks being 2 steps apart are 0011, 0101, and 0110. This is because the sum of the steps in these numbers is equal to 2.

Therefore, the total number of accessible states for the system is the number of binary numbers that satisfy the condition of the drunks being a distance A apart. This can be found using the binomial coefficient, which is given by:

C(M, (M+A)/2) = M! / [(M+A)/2)! ((M-A)/2)!]

Using this, the probability of the drunks being a distance A apart after M timesteps is given by:

P(A,M) = C(M, (M+A)/2) / 2^M

For part b), the probability of the drunks meeting at precisely the Mth timestep can be found by considering the number of ways in which the two drunks can take steps to reach the same point after M timesteps. This can be represented as the condition that the sum of the steps taken by the two drunks is equal to L.

Using the same method as above, the total number of accessible states for this condition is given by the binomial

 

[wais]

Hi there,

I can understand your confusion with this problem. It can be quite challenging to incorporate both the time steps and distance A in the probability calculation. However, I will try to break it down for you and hopefully, it will make more sense.

Firstly, let's define the variables in the problem:

M = number of timesteps
A = distance between the two drunks
L = initial distance between the two drunks
s = spin excess (# of forward steps - # of backward steps)

Now, let's focus on part a) of the problem. We want to find the probability that the drunks are a distance A apart after M timesteps. To do this, we need to consider all the possible paths that the drunks can take to reach this distance A.

Let's say that the first drunk takes x number of forward steps and y number of backward steps. This means that the second drunk will take (M-x) forward steps and (M-y) backward steps to reach the distance A. Since each step is of equal length, we can say that:

x + (M-x) = A
and
y + (M-y) = A

Solving these equations, we get x = (M+A)/2 and y = (M-A)/2. Now, we can use these values to calculate the multiplicity of the system using the formula you mentioned above:

g(M,s) = (M!)/[(1/2(M+A))!(1/2(M-A))!2^M]

Substituting this in the formula for probability, we get:

P(A,M) = g(M,s)/2^M = (M!)/[(1/2(M+A))!(1/2(M-A))!2^M * 2^M]

Simplifying this, we get:

P(A,M) = (M!)/[(1/2(M+A))!(1/2(M-A))!]

Now, for part b) of the problem, we want to find the probability that the drunks meet at precisely the Mth timestep. This means that they must be at the same location after M timesteps, which is A = 0. Using the same approach as above, we can calculate the probability as:

P(0,M) = (M!)/[(1/2(M+0))!(1/2(M-0))!]

Simplifying this, we get