Example 2.11: Standard deviation Consider the probability density u of Eq. (2.50a) and the average |u|

Example 2.11: Standard deviation σ

Consider the probability density Πu of Eq. (2.50a) and the average |u| of Eq. (2.50b). Evaluate correspondingly

(u − |u|)2 =



∞

−∞

(u − |u|)2Πudu = σ2 (2.69)

(the quantity σ being called standard deviation). Show that generally:

(u − |u|)2i = (2i)!

2ii!

σ2i ,

(u − |u|)2i+1= 0.

⎫⎬ ⎭

i

=

1, 2, .

.

.

.

(

2.70)

(Answer: σ2 = kT/m).