and terms in f involving squares tions may be neglected. That is, we assume (f – fol/fo « 1. To this approximation es af
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and terms in f involving squares tions may be neglected. That is, we assume (f – fol/fo « 1. To this approximation es af
and terms in f involving squares tions may be neglected. That is, we assume (f – fol/fo « 1. To this approximation es afo afo (41.3) f = fo - To tu m au ax Higher-order effects may be found by an iterative procedure, using in each order the solution to the next lower order when evaluating the parentheses on the right-hand side of (41.2). Now fo is a function of the energy e; the temperature 7; and the chemical potential ; the energy is a function of the velocity. Thus we have they a (41.4) Əfo du Əfo du afo dr + ou dx дz dx 3. [Kittel, Exercise 41.4] Calculate to the second approximation the equation corresponding to (41.3).
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