Random Walk Problem: Finding Relative Dispersion

[Goldenlemur]

Ok I have a varaition of the random problem as follows. We have a container with volume V and N particles. We consider a subvolume v and n particles. The probability of particles being inside v is (v/V)

Ok I found the mean of n (mean number of molecules in v)

< n > = N*v*(1/V)

Then they ask to find the relative dispersion in mean number of molecules in v

relative dispersion = ([1-(v/V)]/ (< n > + [1-(v/V)]))

1) Next they ask conisder relative dispersion when v << V

Well the relative dispersion then becomes,

relative dispersion = 1/< n > ; one over the mean of n

2) Then consider relative dispersion when v appoarching V

relative dispersion = 0

I am not sure what is the physical meaning of 1 and 2 so not sure if I'm doing the problem right. I think it is... I have the following reason for 2 since the subvolume is appoarcing the oringal volume of the containter then the probability of particle in v becomes one therefore the dispersion from the mean vanishes... Can some give me some guidence?

 

[Greg Bernhardt]

It sounds like you are on the right track. When v << V, the relative dispersion is 1/<n>, which means that the average number of particles inside the subvolume is much less than the total number of particles in the container. When v is approaching V, the relative dispersion is 0, which means that there is no variation from the mean, since the probability of particles being inside the subvolume is 1.

 

[GSXR750]

Your understanding of the problem is correct. The relative dispersion is a measure of how spread out the data is from the mean. In this case, the mean number of molecules in v is affected by the size of v and the total number of particles N. As v approaches V, the relative dispersion becomes smaller because the probability of particles being in v becomes higher and the data becomes less spread out. On the other hand, when v is much smaller than V, the relative dispersion becomes large because the probability of particles being in v becomes smaller and the data becomes more spread out. This makes sense intuitively since a smaller v means there is less space for the particles to be in, so they are more likely to be concentrated in certain areas.