Solving a Probability Problem for Thermodynamics: Stirling's Approximation

[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.

 

[Alkatran]

"Thinks in Head"

Let's try this from drunk 1's frame.
Drunk 2 can either move 0 relative, 2 relative, or -2 relative. There's a 1/2 chance of 0 and a 1/4 chance for each 2.
So half the time he doesn't move relative to you, a quarter of the time he moves away (assuming he isn't at the same spot) and the other quarter he moves closer.

Interesting...

 

[tacman]

First of all, it's great that you have already used Stirling's approximation to simplify the expression for the probability in terms of the multiplicity and spin excess. This will make the calculations easier.

To take into account the distance A between the two drunks, we need to consider the different possible paths that the drunks could take in order to end up a distance A apart after M timesteps. One way to do this is to think about the number of forward and backward steps that each drunk takes. For example, if the first drunk takes s1 forward steps and s2 backward steps, and the second drunk takes s3 forward steps and s4 backward steps, then the total number of forward steps is s1 + s3 and the total number of backward steps is s2 + s4.

Since the total number of timesteps is M, we know that s1 + s2 + s3 + s4 = M. Additionally, since the two drunks are a distance A apart, we know that the difference between the total number of forward and backward steps must be equal to A. In other words, we have s1 + s3 - (s2 + s4) = A.

Using these two equations, we can solve for s1 and s2 in terms of s3 and s4, and then substitute these expressions into the formula for the probability that you have already derived. This will give us an expression for the probability in terms of s3 and s4, which we can then sum over all possible values of s3 and s4 to get the total probability.

For part b), we can use a similar approach to calculate the probability that the two drunks meet at the Mth timestep. In this case, we need to consider all possible paths that the two drunks could take in order to meet at the Mth timestep. This will depend on the initial distance L between the two drunks, as well as the number of forward and backward steps taken by each drunk. Again, we can use the equations s1 + s2 + s3 + s4 = M and s1 + s3 - (s2 + s4) = L to solve for s1 and s2 in terms of s3 and s4, and then substitute into the probability expression.

I hope this helps you to approach the problem and solve it. Remember to carefully consider all possible paths and use the equations to simplify the expressions for the probabilities