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Root Mean Squared (RMS) Displacement
Diffusion is a process that is relatively inefficient over large differences. This fact is nicely illustrated by comparing the linear (slope = 1) vs. square-root dependence of diffusion distance over time. In the linear case, as time increases, distance increases by equal increments. However, when distance increases as the square root of time then the distance traveled is less as time increases.
How do we know this is true?
Consider the diffusion of a solute in one dimension where there are S number of solute molecules that can move either one step to the left (a distance of – d) or one step to the right (+d) from position zero.
Therefore the location of any molecule (n) after (a) steps must be related to it’s previous position at (a-1) by either +d or – d.
or x^ n ( a )^ xn ( a ^1 ) d
So then, what is the average position of all the molecules after all of them have taken (a) steps?
Average position = S
x (^) 1 ( a ) x 2 ( a ) x 3 ( a ) xn 1 ( a ) xn ( a )
The solution is obtained by simply summing all of the positions of the molecules and dividing by the total (S).
Using sigma notation: S
d
S
x a
S
x a
S
n
S
n
n
S
n
n^
Since the molecules move randomly, half of the steps must have been +d and the other half -d
-d +d
-d +d
-d +d
-d +d
Time
0
Distance
d t
d √t
Time
0
Distance
dd tt
ddd √√√ttt
thus:^1 ^0
S
d
S
n
So that the average position must be: x^ n (^ a )^ xn ( a ^1 )
Based on the previous analysis, the average displacement does not provide a quantitative description of diffusion. The reason for this is because the average number of +d and – d steps are equal within the population and so the average will be zero.
To avoid this problem, the square of the displacements is determined and then an average is calculated. Since the square of any number is always positive, the averaged square displacement will always be non-zero.
Thus: x^2^ n^ (^ a )[ xn^2 ( a ^1 ) d ]^2 [ xn^2 ( a ^1 )^2 dnx ( a ^1 ) d^2 ]
To determine the average, sum each term for all S molecules then divide by S to give the average for each term:
S
d
S
d xa
S
x a
S
x a
S
n
S
n
n
S
n
n
S
n
n^
1
2 1 1
2 1
(^2) ( ) ( 1 ) 2 ( 1 )
The term on the left side of the equals sign is the averaged squared displacement after (a) steps. The first term on the right of the equals sign is the averaged squared displacement after (a-1) steps.
Using the same logic as above, the ±2dnx^2 (a-1) term must result in a value of zero since the number of + and – steps are equal.
Therefore:
x^2 ( a ) x^2 ( a 1 ) d^2 x^2 ( a ) x^2 ( a ) x^2 d^2 x^2 ( a ) ad^2
n n n n n n
Although a step-wise analysis has been useful to this point, in reality it is impossible to monitor all of the steps that molecules will take. Thus, an expression in the context of elapsed time (t) is more useful than counting steps.
If each step is taken in an interval of t, then t = a t so that a = t/ t
Hence: ( ) ()
2
2 22 t
t
d
x n a ad xn
Note the units of d^2 /t: remember that d = length of each step (cm) and t = sec. So d^2 /t has units of cm^2 /sec. These are the same units as the diffusion coefficient (D)!!
