Root-mean-squared displacement of diffusing molecules

Initially there are N molecules, all at starting position 0. For any particular molecule (which we tag with a label i ), regardless of where the molecule is, it can move only either one step to the left (a distance of –δ) or one step to the right (a distance of +δ). This means that the location of any molecule i after n steps must be related to its immediately previous location (at [ n – 1] steps) by either +δ or –δ:

?i(?)=?i(?–1)±δxi(n)=xi(n-1)±δ
xi(n)=xi(n-1)±δ

What is the average position of all of the molecules after all of them have taken n steps? This is easily calculated by adding the positions of all the molecules and then dividing by the total number of molecules, N . The average position is

⟨??(?)⟩=?1(?)+?2(?)+…+??–2(?)+??–1(?)+??(?)?=∑??=1??(?)?⟨xi(n)⟩=x1(n)+x2(n)+…+xN-2(n)+xN-1(n)+xN(n)N=∑i=1Nxi(n)N
⟨xi(n)⟩=x1(n)+x2(n)+…+xN-2(n)+xN-1(n)+xN(n)N=∑i=1Nxi(n)N

where angled brackets signify the average. Substituting Equation AC.1 for xi ( n ) in the summation gives

⟨??(?)⟩=∑??=1[??(?–1)±?]?={∑??=1??(?–1)?}+{∑??=1(±?)?}⟨xi(n)⟩=∑i=1N[xi(n-1)±δ]N={∑i=1Nxi(n-1)N}+{∑i=1N(±δ)N}
⟨xi(n)⟩=∑i=1N[xi(n-1)±δ]N={∑i=1Nxi(n-1)N}+{∑i=1N(±δ)N}

In Equation AC.3 the first term enclosed in braces is the average position after ( n – 1) steps, < x _i( n − 1)>. The second term in braces is the average of all the steps that all the molecules have just taken. Because the molecules moved randomly, essentially half of the steps must have size +δ and half must have size –δ. Therefore all the steps average to 0—the second term in braces is equal to 0, and Equation AC.3 simplifies to

⟨??(?)⟩=⟨??(?–1)⟩⟨xi(n)⟩=⟨xi(n-1)⟩
⟨xi(n)⟩=⟨xi(n-1)⟩

Equation AC.4 says that the average position of all the molecules remains the same from one step to the next, regardless of how many steps have been taken. This is not surprising because, on average, every +δ step is balanced by a –δ step, the arithmetic average must always be 0. So the average position can never change from x = 0, where all the molecules were initially.

The average position of the molecules is clearly not informative. We need to find a measure that reflects the fact that with time the molecules will spread out in space ( Fig. 2.4 ). One way to get around the fact that there are + and – steps is to square the displacements and then average them. Because squaring any positive or negative number always gives a positive number, averaging the squared displacements guarantees a nonzero result.

Squaring the displacements shown in Equation AC.1 yields

?2?(?)=[??(?–1)±?]2=[??(?–1)±?]×[??(?–1)±?]=?2?(?–1)±2???(?–1)+?2xi2(n)=[xi(n-1)±δ]2=[xi(n-1)±δ]×[xi(n-1)±δ]=xi2(n-1)±2δxi(n-1)+δ2
xi2(n)=[xi(n-1)±δ]2=[xi(n-1)±δ]×[xi(n-1)±δ]=xi2(n-1)±2δxi(n-1)+δ2

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