4.1. Classical Hamiltonian H = m ˙q 2 − m ˙q 2 2 + V (q) = p 2 2m + V (q) . (4.9) Hamilton-Ja cobi equations (4.7) and (4.8) read ˙q = p m (4.10) ˙p = − ∂V ∂q . (4.11) The second equation, which can be rewritten as m¨q = − ∂V ∂q , (4.12) expresses Newton’s second law. 4.1.3 Example L C L C Consider a capacitor having capacitance C connected in parallel to an inductor h aving inductance L.Letq be the charge stored in the c apacitor. The kinetic en ergy in this case T = L ˙q 2 /2 is the energy stored in the inductor, and the potential energy V = q 2 /2C is t he energy stored in the capacitor. The canonical conjugate momentum is given by [see Eq. (4.3)] p = L ˙q,and the Hamiltonian (4.6) is given by H = p 2 2L + q 2 2C . (4.13) Hamilton-Ja cobi equations (4.7) and (4.8) read ˙q = p L (4.14) ˙p = − q C . (4.15) The second equation, which can be rewritten as L¨q + q C =0, (4.16) expresses the requiremen t that the voltage across the capacitor is the same as the one across the inductor. Eyal Buks Thermodynamics and Statistical Physics 129 Chapter 4. Classical Limit of Statistical Mechanics 4.2 Density Function Consider a classical system in thermal equilibrium. The density function ρ (¯q, ¯p) is the probability distribution to find the system in the poin t (¯q, ¯p). The following theorem is given without a proof. Let H (¯q, ¯p) be an Hamilto- nian of a system, and assume that H has the following form H = d X i=1 A i p 2 i + V (¯q) , (4.17) where A i are constants. Then in the classical limit, namely in the limit where Plank’s constant approaches zero h → 0, the density function is given by ρ (¯q, ¯p)=N exp (−βH (¯q, ¯p)) , (4.18) where N = 1 R d¯q R d¯p exp (−βH (¯q, ¯p)) (4.19) is a normalization c onstant, β =1/τ,andτ is the temperature. The notation R d¯q indicates integration over all coordinates, namely R d¯q = R dq 1 R dq 1 · · R dq d , and similarly R d¯p = R dp 1 R dp 1 · · R dp d . Let A (¯q, ¯p) be a variable which depends on the coordinates ¯q and their canonical conjugate momentum variables ¯p. Using the above theorem the average value of A can be calculates as: hA (¯q, ¯p)i = Z d¯q Z d¯pA(¯q, ¯p) ρ (¯q, ¯p) = R d¯q R d¯pA(¯q, ¯p)exp(−βH (¯q, ¯p)) R d¯q R d¯p exp (−βH (¯q, ¯p)) . (4.20) 4.2.1 Equipartition Theorem Assume that the Hamiltonian has the following form H = B i q 2 i + ˜ H , (4.21) where B i is a constant and where ˜ H is independent of q i . Then the following holds  B i q 2 i ® = τ 2 . (4.22) Similarly, assume that the Hamiltonian has the following form Eyal Buks Thermodynamics and Statistical Physics 130 4.2. Density Function H = A i p 2 i + ˜ H , (4.23) where A i is a constant and where ˜ H is independent of p i . Then the following holds  A i p 2 i ® = τ 2 . (4.24) To prove the theorem for the firstcaseweuseEq.(4.20)  B i q 2 i ® = R d¯q R d¯pB i q 2 i exp (−βH (¯q, ¯p)) R d¯q R d¯p exp (−βH (¯q, ¯p)) = R dq i B i q 2 i exp ¡ −βB i q 2 i ¢ R dq i exp (−βB i q 2 i ) = − ∂ ∂β log µ Z dq i exp ¡ −βB i q 2 i ¢ ¶ = − ∂ ∂β log µ r π βB i ¶ = 1 2β . (4.25) The proof for the second case is similar. 4.2.2 Example Here we calculate the average energy of an harmonic oscillator using both, classical and quantum approac hes. Consider a particle having mass m in a one dimensional para bolic potential given by V (q)=(1/2) kq 2 ,wherek is the spring constant. The kinetic energy is giv en by p 2 /2m,wherep is the canonical momentum variable conjugate to q. The Hamiltonian is given by H = p 2 2m + kq 2 2 . (4.26) In the classical limit the average energy of the system can be easily calculated using the equipartition theorem U = hHi = τ. (4.27) In the quant um treatment the system has energy lev els given by E s = s~ω, where s =0, 1, 2, ,andwhereω = p k/m is the angular resonance fre- quency. The partition function is given by Eyal Buks Thermodynamics and Statistical Physics 131 Chapter 4. Classical Limit of Statistical Mechanics Z = ∞ X s=0 exp (−sβ~ω)= 1 1 − exp (−β~ω) , (4.28) thus the average energy U is given by U = − ∂ log Z ∂β = ~ω e β~ω − 1 . (4.29) Using the expansion U = β −1 + O (β) , (4.30) one finds that in the limit of high temperatures, namely when β~ω ¿ 1, the quantum result [Eq. (4.30)] coincides with the classical limit [Eq. (4.27)]. 4.3 Nyquist Noise Here we employ the equipartition theorem in order to evaluate voltage noise across a resistor. Consider the circuit shown in the figure belo w, which consists of a capacitor having capacitance C, an inductor having inductance L,anda resistor having resistance R, all s erially conn ected. The system is assumed to be in thermal equilibrium at temperature τ.Tomodeltheeffect of thermal fluctuations we add a fictitious voltage source, which produces a random fluctuating voltage V (t). Let q (t) be the charge stored in the capacitor at time t. The classical equation of motion, which is given by q C + L¨q + R ˙q = V (t) , (4.31) represents Kirchhoff’s voltage law. R L C V(t) ~ R L C V(t) ~ Fig. 4.1. Eyal Buks Thermodynamics and Statistical Physics 132 4.3. Nyquist Noise Consider a sampling of the fluctuating function q (t) in the time interval (−T/2,T/2), namely q T (t)= ½ q(t) −T/2 <t<T/2 0else . (4.32) The energy s tored in the capacitor is given by q 2 /2C. Using the equipartition theorem one finds  q 2 ® 2C = τ 2 , (4.33) where  q 2 ® is obtained by averaging q 2 (t), namely  q 2 ® ≡ lim T →∞ 1 T Z +∞ −∞ dtq 2 T (t) . (4.34) Introducing the Fourier transform: q T (t)= 1 √ 2π Z ∞ −∞ dωq T (ω)e −iωt , (4.35) one finds  q 2 ® = lim T →∞ 1 T Z +∞ −∞ dt 1 √ 2π Z ∞ −∞ dωq T (ω)e −iωt 1 √ 2π Z ∞ −∞ dω 0 q T (ω 0 )e −iω 0 t = 1 2π Z ∞ −∞ dωq T (ω) Z ∞ −∞ dω 0 q T (ω 0 ) lim T →∞ 1 T Z +∞ −∞ dte −i ( ω+ω 0 ) t | {z } 2πδ(ω+ω 0 ) = lim T →∞ 1 T Z ∞ −∞ dωq T (ω)q T (−ω) . (4.36) Moreover, using the fact that q(t)isrealonefinds  q 2 ® = lim T →∞ 1 T Z ∞ −∞ dω |q T (ω)| 2 . (4.37) In terms of the po wer spectrum S q (ω)ofq (t), which is defined as S q (ω) = lim T →∞ 1 T |q T (ω)| 2 , (4.38) one finds  q 2 ® = Z ∞ −∞ dωS q (ω) . (4.39) Eyal Buks Thermodynamics and Statistical Physics 133 Chapter 4. Classical Limit of Statistical Mechanics Taking the Fourier transform of Eq. (4.31) yields µ 1 C − iωR − Lω 2 ¶ q(ω)=V ( ω) , (4.40) where V (ω) is the Fourier transform of V (t), namely V (t)= 1 √ 2π Z ∞ −∞ dωV(ω)e −iωt . (4.41) In terms of the resonance frequency ω 0 = r 1 LC , (4 .42) one has £ L ¡ ω 2 0 − ω 2 ¢ − iωR ¤ q(ω)=V (ω) . (4.43) Taking the absolute value squared yields S q (ω)= S V (ω) L 2 (ω 2 0 − ω 2 ) 2 + ω 2 R 2 , (4.44) where S V (ω) is the power spectrum of V (t). Integrating the last result yields Z ∞ −∞ dωS q (ω)= Z ∞ −∞ dω S V (ω) L 2 (ω 2 0 − ω 2 ) 2 + ω 2 R 2 = 1 L 2 Z ∞ −∞ dω S V (ω) (ω 0 + ω) 2 (ω 0 − ω) 2 + ω 2 R 2 L 2 . (4.45) The integrand has a peak near ω 0 ,havingawidth' R/2L. T he Qua lity factor Q is defined as ω 0 Q = R 2L . (4.46) Assuming S V (ω) is a smooth function ne ar ω 0 on the scale ω 0 /Q,andas- suming Q À 1 yield Eyal Buks Thermodynamics and Statistical Physics 134 4.3. Nyquist Noise Z ∞ −∞ dω S q (ω) ' S V (ω 0 ) L 2 Z ∞ −∞ dω (ω 0 + ω) 2 (ω 0 − ω) 2 + ³ 2ωω 0 Q ´ 2 ' S V (ω 0 ) 4ω 4 0 L 2 Z ∞ −∞ dω ³ ω 0 −ω ω 0 ´ 2 + ³ 1 Q ´ 2 = S V (ω 0 ) 4ω 3 0 L 2 Z ∞ −∞ dx x 2 + ³ 1 Q ´ 2 | {z } πQ = S V (ω 0 )πQ 4ω 3 0 L 2 . (4.47) On the other hand, using Eqs. (4.33) and (4.39) one finds Z ∞ −∞ dωS q (ω)=  q 2 ® = Cτ , (4.48) therefore S V (ω 0 )= 4Cω 3 0 L 2 πQ τ, (4.49) or using Eqs. (4.42) and (4.46) S V (ω 0 )= 2Rτ π . (4.50) Th us, Eq. (4.44) can be rewritten as S q (ω)= 2Rτ π 1 L 2 (ω 2 0 − ω 2 ) 2 + ω 2 R 2 . (4.51) Note that the spectral density of V given by Eq. (4.50) is frequency in- dependent. Consider a measurement of the fluctuating voltag e V (t)ina frequency band having width ∆f.Usingtherelation  V 2 ® = Z ∞ −∞ dωS V (ω) , (4.52) one finds that the variance in such a measurement  V 2 ® ∆f is given by  V 2 ® ∆f =4Rτ∆f . (4.53) ThelastresultistheNyquist’snoiseformula. Eyal Buks Thermodynamics and Statistical Physics 135 Chapter 4. Classical Limit of Statistical Mechanics 4.4 Problems Set 4 1. A gas at temperature τ emits a spectral line at wave lengt h λ 0 .The width of the observed spectral line is broadened due to motion of the molecules (this is called Doppler broadening). Show that the relation between spectral line intensity I and wavelength is given by I (λ) ∝ exp " − mc 2 (λ − λ 0 ) 2 2λ 2 0 τ # , (4.54) where c is velocity of fight, and m is mass of a molecule. 2. The circuit seen in the figure below, which contains a resistor R,ca- pacitor C, and an inductor L, is at thermal equilibrium at tempera- ture τ. Calculate the average value  I 2 ® ,whereI is the current in the inductor. RC LRC L 3. Consider a random real signal q(t)varyingintime.Letq T (t)beasam- pling of the signal q(t) in the time interval (−T/2,T/2), namely q T (t)= ½ q(t) −T/2 <t<T/2 0else . (4.55) The Fourier transform is giv en b y q T (t)= 1 √ 2π Z ∞ −∞ dωq T (ω)e −iωt , (4.56) and the power spectrum is given by S q (ω) = lim T →∞ 1 T |q T (ω)| 2 . a) Show that  q 2 ® ≡ lim T →∞ 1 T Z +∞ −∞ dtq 2 T (t)= Z ∞ −∞ dωS q (ω) . (4.57) Eyal Buks Thermodynamics and Statistical Physics 136 4.4. Problems Set 4 b) Wiener-Khinchine Theorem - sho w that the correlation function of the random signal q(t)isgivenby hq (t) q (t + t 0 )i ≡ lim T →∞ 1 T Z +∞ −∞ dtq T (t) q T (t + t 0 )= Z ∞ −∞ dωe iωt 0 S q (ω) . (4.58) 4. Consider a resonator made of a capacitor C, an inductor L,andaresistor R connected in series, as was done in class. Let I (t) be the curren t in the circuit. Using the results obtained in class calculate the spectral density S I (ω)ofI at thermal equilibrium. Show that in the limit of high quality factor, namely when Q = 2 R r L C À 1 , (4.59) your result is consistent with the equipartition theorem applied for the energy stored in the inductor. 5. A classical system is described using a set of coordinates {q 1 ,q 2 , , q N } and the corresponding canonically conjugate variables {p 1 ,p 2 , , p N }. The Hamiltonian of the system is given by H = N X n=1 A n p s n + B n q t n , (4.60) where A n and B n are positive constants and s and t are even positive integers. Show that t he average energy of the system in equilibrium at temperature τ is given by hUi = Nτ µ 1 s + 1 t ¶ , (4.61) 6. A sma ll hole of area A is made in the walls of a vessel of volume V con- taining a classical ideal gas of N particles of mass M each in equilibrium at temperature τ. Calculate the number of particles dN, which escape through the opening during the infinitesimal time interval dt. 7. Consider an ideal gas of Fermions ha ving mass M andhavingnointernal degrees of freedom at temperature τ. The velocity of a particle is denoted as v = q v 2 x + v 2 y + v 2 z . Calculate the quan t ity hvi ¿ 1 v À (the symbol hi denoted av eraging) in the: a) classical limit (high temperatures). Eyal Buks Thermodynamics and Statistical Physics 137 Chapter 4. Classical Limit of Statistical Mechanics b) zero temperature. 8. Consider an ideal gas of N molecules, each of mass M, contained in a cen tr ifuge of radius R and length L rotating with angular velocity ω about its axis. Neglect the effect of grav ity. The system is in equilibrium at temperature τ =1/β. Calculate the particle density n (r) as a function of the radial distance from the axis r (where0≤ r ≤ R) . 9. Consider an ideal classical gas of p articles having mass M and having no internal degrees of freedom at temperature τ.Letv = q v 2 x + v 2 y + v 2 z . be the velocity of a particle. Calculate a) hvi b) p hv 2 i 10. A mixture of t wo classical ideal gases, consisting of N 1 and N 2 particles of mass M 1 and M 2 , respectively, is enclosed in a cylindrica l vessel of heigh t h and area of bottom and top side S. The vessel is placed in a gravitational field having acceleration g. The system is in thermal equi- librium at temperature τ. Find the pressure exerted on the upper wall of the cylinder. 4.5 Solutions Set 4 1. Let λ bethewavelengthmeasuredbyanobserver,andletλ 0 be the wave length of the emitted light in the reference frame where the molecule is at rest. Let v x be the velocity of the molecule in the direction of the ligh t ray from the molecule to the observer. Due to Doppler effect λ = λ 0 (1 + v x /c) . (4.62) The probability distribution f (v x ) is proportional to f (v x ) ∝ exp µ − mv 2 x 2τ ¶ , (4.63) thus using v x = c (λ − λ 0 ) λ 0 , (4.64) the probability distribution I (λ) is proportional to I (λ) ∝ exp " − mc 2 (λ − λ 0 ) 2 2λ 2 0 τ # . (4.65) 2. The energy s tored in the inductor is U L = LI 2 /2. Using the eq uipartition theorem hU L i = τ/2, thus  I 2 ® = τ L . (4.66) Eyal Buks Thermodynamics and Statistical Physics 138 [...]... , = exp M 2 2 (4 .93 ) thus N M 2 i exp h n (r) = 2 2L exp M R2 1 2 à M 2 r2 2 ả (4 .94 ) 9 In the classical limit the probability distribution of the velocity vector v satises ả à M v2 , (4 .95 ) f (v) exp 2 where v = |v| Eyal Buks Thermodynamics and Statistical Physics 143 Chapter 4 Classical Limit of Statistical Mechanics a) By changing the integration variable x= M v2 2 (4 .96 ) one nds 2 dvv... (4. 89) b) Using the identity ZF d n = n+1 F n+1 0 Eyal Buks Thermodynamics and Statistical Physics 142 4.5 Solutions Set 4 one nds 1 hvi = v à F R d 0 à F R 0 d 0 1 2 F F 3 2 2 2 F 3 = 2 = ả à F R 1/2 ả d ả2 9 8 (4 .90 ) 8 The eect of rotation is the same as an additional external eld with potential energy given by 1 U (r) = M 2 r2 , 2 (4 .91 ) thus n (r) = A exp [U (r)] = A exp à M 2 2 r 2 ả , (4 .92 )... ) (4. 69) ư đ 4 Using I () = iq () and q 2 = C one nds for the case Q 1 Z Z ư 2đ ư đ 2 I = (4.70) d SI () ' 0 d Sq () = 2 q 2 = , 0 L in agreement with the equipartition theorem for the energy stored in the inductor LI 2 /2 5 Calculate for example Eyal Buks Thermodynamics and Statistical Physics Z + | 0 1 = lim d eit qT ()qT () T T Z 0 = d eit Sf () 0 0 d 0 qT ( 0 )ei (t+t ) 1 39 0 dt... eMl gh ) (4.102) Thermodynamics and Statistical Physics 144 Nl nl (0) = S h R exp (Ml gz) dz = 0 Eyal Buks 4.5 Solutions Set 4 Using the equation of state p = n , where n = N/V is the density, one nds that the pressure on the upper wall of the cylinder is given by p = (n1 (h) + n2 (h)) à ả M1 N1 M2 N2 g = + exp (M1 gh) 1 exp (M2 gh) 1 S Eyal Buks Thermodynamics and Statistical Physics (4.103) 145... given by fF () = 1 , 1 + exp [( à) ] (4.84) where = 1/ and à is the chemical potential The velocity v of such an orbital is related to the energy by = Mv 2 2 (4.85) The 3D density of state per unit volume is given by 1 g () = 2 2 à 2M ~2 ả3/2 1/2 (4.86) Thus Eyal Buks Thermodynamics and Statistical Physics 141 Chapter 4 Classical Limit of Statistical Mechanics R R 1 d g () fF () v d g () fF... thus Eyal Buks 0 dvv2 vf (v) = N A hvi 4V Thermodynamics and Statistical Physics (4. 79) 140 4.5 Solutions Set 4 In the classical limit à ả Mv 2 f (v) exp , 2 (4.80) thus, by changing the integration variable x= Mv2 2 (4.81) one nds 2 dvv3 exp Mv 2 0 hvi = R Â Ă 2 exp Mv 2 dvv 2 0 à ả1/2 R 2 0 dxx exp (x) R = 1/2 exp (x) M 0 dxx à ả1/2 8 , = M R (4.82) and NA dN = dt 4V à 8 M ả1/2 (4.83) 7 The... dxx exp (x) R = 1/2 exp (x) M 0 dxx à ả1/2 8 = M R (4 .97 ) (4 .98 ) b) Similarly 2 dv v4 exp Mv ư 2đ 2 0 Â Ă v = R 2 exp Mv 2 dv v 2 0 R 2 0 dx x3/2 exp (x) R = M 0 dx x1/2 exp (x) 2 3 , = M2 R (4 .99 ) thus p hv 2 i = à 3 M ả1/2 = à 3 8 ả1/2 hvi (4.100) 10 For each gas the density is given by nl (z) = nl (0) exp (Ml gz) , where l {1, 2}, 0 z h and the normalization constant is found from the requirement... 1 39 0 dt ei(+ )t {z } 2(+0 ) Chapter 4 Classical Limit of Statistical Mechanics ư đ t Bn qn = = = R t t Bn qn exp (Bn qn ) dqn R t exp (Bn qxn ) dqn R t t 0 R Bn qn exp (Bn qn ) dqn t 0 exp (Bn qxn ) dqn Z Â Ă d t log exp Bn qn dqn d 0 (4.71) where = 1/ Changing integration variable t x = Bn qn , dx = (4.72) t1 tBn qn dqn , (4.73) leads to ã á Z ư đ 1 d 1 t Bn qn = log (Bn ) t t1 x t 1 ex... ả (4.75) 6 Let f (v) be the probability distribution of velocity v of particles in the gas The vector u is expressed in spherical coordinates, where the z axis is chosen in the direction of the normal outward direction v = v (sin cos , sin sin , cos ) (4.76) By symmetry, f (v) is independent of and The number dN is calculated by integrating over all possible values of the velocity of the leaving . that the voltage across the capacitor is the same as the one across the inductor. Eyal Buks Thermodynamics and Statistical Physics 1 29 Chapter 4. Classical Limit of Statistical Mechanics 4.2 Density. heorem for the e ne rg y stored in the inductor LI 2 /2. 5. Calculate for example Eyal Buks Thermodynamics and Statistical Physics 1 39 Chapter 4. Classical Limit of Statistical Mechanics B n q t n đ = R B n q t n exp. (4 .94 ) 9. In the classical limit the probability distribution of the velocity vector v satisfies f (v) ∝ exp µ − Mv 2 2τ ¶ , (4 .95 ) where v = |v|. Eyal Buks Thermodynamics and Statistical Physics