4.1. Brownian motion[[Student version, December 8, 2002]] 103
0 0.2 0.40.60.8 1.0.10.20.30.40 0.20.4 0.60.8 1.0.10.20.30.40 0.2 0.4 0.6 0.8 1.0.10.20.30.4P(x)x xxabc4 coins: 4 coins: 36 coins:
Figure 4.3: (Experimental data.) Behavior of the binomial distribution. (a)Four coins were tossed, and the
fractionxthat came up heads was recorded. The histogram shows the result for a sample of 57 such trials. Since
this is a discrete distribution, the bars have been normalized so that the sum of their heights equals one. (b)Another
sample of 57 tosses of 4 coins. (c)This time 36 coins were tossed, again 57 times. The resulting distribution is much
narrower than (a,b); we can say with greater certainty that “about half” our coin tosses will come up heads if the
total number of tosses is large. The bars are not as tall as in (a,b) because the same number of tosses (57) is now
being divided between a larger number of bins (37 rather than 5). [Data kindly supplied by R. Nelson.]
likely to end up somewhere in the middle. But where?
One way to answer the question would be to proceed as in the Example on page 101, finding
the probability of arriving at various locations. Another way would be to list explicitly all the pos-
sible outcomes for a 10 000-toss sequence, then make a histogram of the corresponding frequencies
(Figure 4.3). But there is a better way.
Suppose each step is of lengthL.Thusthe displacement of stepjiskjL,wherekjis equally
likely to be±1. Call the position afterjstepsxj;the initial position isx 0 =0(see Figure 4.4a).
Thenx 1 =k 1 L,and similarly the position afterjsteps isxj=xj− 1 +kjL.
Wecan’t say anything aboutxjbecause each walk is random. Wecan,however, make definite
statements about theaverageofxj overmany different trials: For example, Figure 4.4b shows
that〈x 3 〉=0.The diagram makes it clear why we got this result: In the average over all possible
outcomes, those with net displacement to the left will cancel the contributions of their equally likely
analogs with net displacement to the right.
Thus the average displacement of a random walk is zero. But this doesn’t mean we won’t go
anywhere! The Example on page 101 showed that the probability of ending right where we started
issmall,not large, for largeN.Toget a meaningful result, recall the discussion in Section 3.2.1:
Foranideal gas,〈vx〉=0but〈vx^2 〉 =0.Following that hint, let’s compute〈xN^2 〉in our problem.
Figure 4.4 shows such a computation, yielding〈x 32 〉=3L^2.
Your Turn 4b
Repeat this calculation for a walk of four steps, just to make sure you understand how it works.Admittedly, the math gets tedious. Instead of exhaustively listing all possible outcomes, though,
wecan note that
〈(xN)^2 〉=〈(xN− 1 +kNL)^2 〉=〈(xN− 1 )^2 〉+2L〈xN− 1 kN〉+L^2 〈(kN)^2 〉. (4.3)In the last expression, the final term just equalsL^2 ,because (±1)^2 =1.For the middle term, note