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Physics 112 Mathematical Notes Winter 2000 1. Power Series

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Physics 112 Mathematical Notes Winter 2000 1. Power Series Physics 112 Mathematical Notes Winter 2000 1. Power Series Expansion of the Fermi-Dirac Integral The Fermi-Dirac integral is defined as: ∞ 1 xn−1 dx fn(z) ≡ , Γ(n) z−1ex +1 0 µ/kT where x ≡ /kT and z ≡ e . We wish to derive a power series expansion for fn(z)that is useful in the limit of z → 0. We manipulate fn(z) as follows: ∞ z xn−1 dx fn(z)= Γ(n) ex + z 0 ∞ z xn−1 dx = Γ(n) ex(1 + ze−x) 0 ∞ ∞ z −x n−1 m −x m = e x dx (−1) (ze ) Γ(n) 0 m=0 ∞ ∞ z = (−1)mzm e−(m+1)xxn−1 dx . Γ(n) m=0 0 Using the well known integral: ∞ Γ(n) e−Axxn−1 dx = , An 0 and changing the summation variable to p = m + 1, we end up with: ∞ (−1)p−1zp fn(z)= n . (1) p=1 p which is the desired power series expansion. This result is also useful for deriving additional properties ofthe fn(z). For example, it is a simple matter use eq. (1) to verify: ∂ fn−1(z)=z fn(z) . ∂z Note that the classical limit is obtained for z 1. In this limit, which also corresponds to the high temperature limit, fn(z) z. One additional special case is noteworthy. If 1 z =1,weobtain ∞ (−1)p−1 fn(1) = n . p=0 p To evaluate this sum, we rewrite it as follows: 1 − 1 fn(1) = n n p odd p p even p 1 1 − 1 = n + n 2 n p odd p p even p p even p ∞ ∞ 1 − 1 = n 2 n p=1 p p=1 (2p) ∞ − 1 1 = 1 n−1 n . 2 p=1 p where p even [odd] above means we sum over positive even [odd] integers. The Riemann zeta-function is defined as: ∞ ≡ 1 ζ(n) n . p=1 p Thus, we conclude that ∞ − p−1 ( 1) − 1 fn(1) = n = 1 n−1 ζ(n) . (2) p=0 p 2 2. Asymptotic Expansion of the Fermi-Dirac Integral We again consider the Fermi-Dirac integral, which we rewrite in the following form: ∞ 1 xn−1 dx fn(y) ≡ , Γ(n) ex−y +1 0 where y ≡ µ/kT =lnz. We now wish to derive an asymptotic expansion for fn(y)which is useful in the limit of y →∞. 2 In the limit of y →∞, which corresponds to the low temperature limit, 1 1,x<y, = (3) ex−y +1 0,x>y. This is just the statement that all the energy levels ofthe Fermi gas are singly occupied (i.e., n = 1) below the Fermi energy (F = µ) and are unoccupied (i.e., n =0)abovethe Fermi energy. Thus, in this limit, y n n 1 n−1 y y fn(y) x dx = = . Γ(n) nΓ(n) Γ(n +1) 0 We now want to improve on this approximation. To do so, we subtract this approximate result from the exact integral and study the remainder. First, it is convenient to introduce some notation. The step function is defined as: 1,x>0 , Θ(x)= 0,x<0 . Using this result, eq. (3) implies that: 1 lim =Θ(y − x) . y→∞ ex−y +1 Thus, we examine the following integral: ∞ 1 n−1 hn(y) ≡ − Θ(y − x) x dx . ex−y +1 0 Change variables to w = x − y. Then, ∞ 1 n−1 hn(y)= − Θ(−w) (w + y) dw ew +1 −y ∞ 1 − Θ(−w) (w + y)n−1 dw ew +1 −∞ ∞ n−1 1 w yn−1 − Θ(−w) 1+ dw . ew +1 y −∞ In the second step above, we replaced the lower limit −y with −∞. Since we seek an approximation valid for large y, this should be a good approximation. One can check that (−µ)/kT This behavior is somewhat easier to see by considering the T → 0 limit of n =1/[e +1]. Then, change variables back to x = /kT and y = µ/kT to recover eq. (3). 3 the error we make at this step is exponentially small. In the limit oflarge y we expand (1 + w/y)n−1 in a power series using the binomial theorem: n−1 ∞ p w w 1+ = C(n − 1,p) y p=0 y ∞ p 1 n(n − 1)(n − 2) ···(n − p) w = . n p=0 p! y Note that this formula is valid even when n is not a positive integer. (For positive integer n, the sum stops after p = n − 1.) If n is not a positive integer, the above expansion converges only when |w/y| < 1. This raises a concern, since we are about to insert this expansion into an integral where w varies from −∞ to +∞. No matter how large y is, this step appears to be illegal in the range of w-integration where |w| > |y|. Nevertheless, we shall close our eyes to this problem and proceed. One can show that this step cannot result in a convergent expansion ofthe integral (when n is not a positive integer). However, the resulting expression is a valid asymptotic expansion. Although the proofofthis statement lies beyond the scope ofthis note, it should be noted that the phenomenon just described is quite common in the derivation ofasymptotic expansions. Thus, we insert the expansion of(1 + w/y)n−1 into the integral, and we interchange the sum and integral (this step also needs mathematical justification). The result is: ∞ n−1 ∞ y n(n − 1)(n − 2) ···(n − p) 1 p hn(y) − Θ(−w) w dw . (4) n yp p! ew +1 p=0 −∞ At this point, focus on the integrand. Define F (w) ≡ (ew +1)−1 −Θ(−w). For positive values of w,Θ(−w)=0sothatF (w)=1/(ew + 1) while 1 −e−w −1 F (−w)= − 1= = = −F (w) . e−w +1 e−w +1 ew +1 That is, F (w) is an odd function of w. Thus, for even p, the integral in eq. (4) vanishes. For odd p,sincewpF (w) is an even function, we can set the lower limit of integration to 0 and multiply by 2 without changing the value ofthe integral. Thus, setting p =2j − 1, with j =1, 2, 3,..., we end up with ∞ ∞ 2yn−1 n(n − 1)(n − 2) ···(n − 2j +1) w2j−1 hn(y) dw . n y2j−1(2j − 1)! ew +1 j=1 0 4 The integral above is related to the Riemann zeta-function; using eq. (2), ∞ xn−1 dx =Γ(n)fn(z =1) ex +1 0 1 =Γ(n)ζ(n) 1 − . 2n−1 For positive integer n,Γ(n)=(n − 1)!,soweendupwith n ∞ − − ··· − 2y n(n 1)(n 2) (n 2j +1) − 1 hn(y) 2j 1 2j−1 ζ(2j) . n j=1 y 2 The last step is to note the relation between fn(y)andhn(y): yn 1 fn(y)= + hn(y) . Γ(n +1) Γ(n) Inserting our result for hn(y) and using the fact that Γ(n +1)=nΓ(n), we arrive at the desired asymptotic expansion for fn(y)fory 1: n ∞ − − ··· − y n(n 1)(n 2) (n 2j +1) − 1 fn(y) 1+2 2j 1 2j−1 ζ(2j) . Γ(n +1) j=1 y 2 Note that the values of ζ(2j) are known for positive integer j: (−1)j+1(2π)2j ζ(2j)= B2j , 2(2j)! where the B2j are Bernoulli numbers. For convenience, we display the first two terms ofthe asymptotic expansion of fn(y): yn π2n(n − 1) 1 fn(y) 1+ + O . Γ(n +1) 6y2 y4 Bernoulli numbers are defined via the Taylor series of the function x/(ex − 1). Specifically, x ∞ B xn n . ex − = n 1 n=0 ! By explicitly computing this Taylor series, one finds that B0 =1,B1 = −1/2, B2 =1/6, B3 =0, B4 = −1/30,... In general, for n =1, 2, 3,..., B2n+1 =0. 5 3. Power Series Expansion of the Bose-Einstein Integral The Bose-Einstein integral is defined as: ∞ 1 xn−1 dx gn(z) ≡ , Γ(n) z−1ex − 1 0 where x ≡ /kT and z ≡ eµ/kT . For a Bose gas with a conserved particle number, one must have µ<0 which means that 0 ≤ z<1. The photon gas possesses no conserved particle number; hence, µ =0orz =1. We wish to derive a power series expansion for gn(z) that is useful in the limit of z → 0. The steps are very similar to those ofsection 1. We manipulate gn(z) as follows: ∞ z xn−1 dx gn(z)= Γ(n) ex − z 0 ∞ z xn−1 dx = Γ(n) ex(1 − ze−x) 0 ∞ ∞ z = e−xxn−1 dx (ze−x)m Γ(n) 0 m=0 ∞ ∞ z = zm e−(m+1)xxn−1 dx . Γ(n) m=0 0 Evaluating the remaining integral and putting p = m + 1 yields the desired power series expansion for the Bose-Einstein integral: ∞ zp gn(z)= n . (5) p=1 p This result is also useful for deriving additional properties of the gn(z).Forexample,itis a simple matter use eq. (5) to verify: ∂ gn−1(z)=z gn(z) . (6) ∂z As noted above, the allowed region of z corresponds to 0 ≤ z ≤ 1(wherez =1 corresponds to the case ofa photon gas). The power series for gn(z) converges for all 6 allowed values of z. As in the Fermi-Dirac case, in the classical limit (i.e.,forz 1), which also corresponds to the high temperature limit, gn(z) z.
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