Camille Bélanger-ChampagneLuis Anchordoqui LehmanMcGill College University City University of New York
ThermodynamicsCharged Particle and CorrelationsStatistical Mechanics in Minimum Bias Events at ATLAS
● Physics motivation Statistical Mechanics V • Stefan-Boltzmann law ● Minbias event and track selection November 2014 • Blackbody radiation th ● Azimuthal correlation results February 26 2012, • Bose-Einstein condensation WNPPC 2012 ● Forward-Backward correlation results
Thursday, December 4, 14 1 CLASSICAL ARGUMENTS FOR PHOTON GAS A number of thermodynamic properties of photon gas Overviewcan be determined from purely classical arguments ① Assume photon gas is confined to rectangular box of dimensions Lx Ly Lz 1/3 ⇥ ⇥ further assume that dimensions are all expanded by a factor ☟ that is volume is isotropically expanded by a factor of Cavity modes of electromagnetic radiation have quantized wave vectors 2⇡nx 2⇡ny 2⇡nz even within classical electromagnetic theory ☛ ~k = , , Lx Ly Lz for a given mode "(~k)=~c ~k ⇣ ⌘ | | ↴ energy changes by 1/3 under an adiabatic volume expansion V V ! It follows that ↴ @U @U 1 @U U V = = U and P = = @V S @ S 3 @V 3V ✓ ◆ ✓ ◆ ✓ ◆S
Since U = U ( T,V ) is extensive ☛ we must have P = P ( T ) alone
C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 2 MORE CLASSICAL ARGUMENTS FOR PHOTON GAS
② Since P = P ( T ) aloneOverview ↴ @S @P = using Maxwell relation ☛ @V @T ✓ ◆P ✓ ◆V after invoking the First Law ☛ dU = TdS PdV ☟ @U @U = =3P @V @V ✓ ◆T ✓ ◆P @P = T P @T ✓ ◆V dP 4 It follows that ☛ T =4P P (T )=aT dT ) ☟ a ☛ constant
C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 3 STEFAN-BOLTZMANN LAW ③ Given energy density Overview U/V differential energy flux emitted in direction ✓ relative to surface normal U d⌦ dJ = c cos ✓ d ⌦ ☛ differential solid angle E V 4⇡
Power emitted per unit area
d cU ⇡/2 2⇡ cU 3 J = P = d✓ d sin ✓ cos ✓ = = cP (T ) T 4 E dA 4⇡V 4 V 4 ⌘ Z0 Z0 3 = ca P (T )=aT 4 4 ) Using quantum statistical mechanical considerations we will show that ↴ 2 4 ⇡ kB 8 W Stefan’s constant ☛ = =5.67 10 60c2~3 ⇥ m2K4
C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 4 SURFACE TEMPERATURE OF EARTH 8 We’ll need three lengths: radius of sun ☛ R =6.96 10 m Overview ⇥ radius of earth ☛ R =6.38 106 m 11 radius of earth’s orbit ☛ d =1.50⇥ 10 m ⇥ Assume that earth has achieved a steady state temperature ☛ T balance total power incident upon earth with power radiated by earth 2 2 ⇡R 4 2 R R 4 Power incident upon earth incident = 2 T 4⇡R = ⇡ T P 4⇡d · · d · ✓ ◆ 4 2 Power radiated by earth ☛ radiated = T 4⇡R P R· 1/2 Setting incident = radiated ☛ T = T =0.04817 T P P 2d ✓ ◆ T = 5780 K T = 278.4K ) Mean surface temperature of the earth ☛ T = 287 K Difference is due to fact that earth is not perfect blackbody (object which absorbs all incident radiation upon it and emits according to Stefan’s law) Earth’s atmosphere re-traps fraction of emitted radiation ☛ greenhouse effect
C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 5 QUANTUM WAVES INSIDE A BOX
Wave function for quantumOverview waves in one-dimensional box ☛ =A sin kx 2⇡ n⇡ k = = n =1, 2, 3, (85) L ··· ☛ ☛ de Broglie wavelength L box dimension 2L substituting c = ⌫ in (85) ☛ n = ⌫ c 2V 1/3 For a cube V = L3 ☛ n = ⌫ (86) c
2 2 2 2 + n = nx + ny + nz ☛ nx,ny,nz Z 2 2 2 2 1/2 Possible values occupy first octant of sphere of radius n =(nx + ny + nz)
C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 6 DENSITY OF STATES OF QUANTUM WAVES INSIDE A BOX Overview g(⌫ ) d ⌫ ☛ number of possible frequencies in range (⌫,⌫ + d⌫)
n ⌫ g(⌫) d⌫ ☛ number of possible set of integers in (n, n + dn) / ) 9 9 within shell of thickness dn of octant of sphere of radius n
(9
1 ⇡ g(⌫)d⌫ = 4⇡n2dn = n2dn (87) 8 2 9 93!&'9- 9 5"$%+ 339)9- 9,9-5,69- 99 4/" 1 9. 90$739)9 4⇡V 2 substituting (86) in (87) ☛ g(⌫)d⌫ = 3 ⌫ d⌫ >L c L >!L .D-8L 1 L 41<<#+L 86D.# D<>#AD>'. L 6D>#1.L 18L *9 GL1>#.L .LLE(8@#. L<1+#L>!8L8L>!8L>J4 E+1&>JL .L>G2L>:. L L 89 9 #.L!L1<#,+>18L1 L>!L<<-+JLE#8@ D448L +#-(>L>1L>!L 7D.JL <4>:D-LL -I#-D-L 86D/JL # 9 PHOTON STATISTICS Photons are bosonsOverview of spin 1 and hence obey Bose-Einstein statistics N 1 number of photons per quantum state ☛ j fj = = " /kT gj e j 1 N(") 1 for continuos distribution ☛ f(")= = g(") e"/kT 1 N(⌫) 1 photon energy is ☛ f(⌫)= = g(⌫) eh⌫/kT 1 C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 8 PLANCK RADIATION FORMULA photon gas ☛ two polarizationOverview states 8⇡V g(⌫)d⌫ = ⌫2d⌫ c3 Energy u ( ⌫ ) d ⌫ in range ( ⌫ , ⌫ + d⌫) number of photons in this range times energy h ⌫ of each ☟ u(⌫) d⌫ = N(⌫)d⌫ h⌫ ⇥ Substituting N ( ⌫ )= g ( ⌫ ) f ( ⌫ ) ↴ 8⇡hV ⌫3 d⌫ u(⌫)d⌫ = Planck’s radiation formula ☛ c3 eh⌫/kT 1 C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 9 ULTRAVIOLET CATASTROPHE Using Overview h⌫ eh⌫/kT 1= + (h2) kT O take classical limit h 0 ! 8⇡kT 2 uclass(⌫)= limu(⌫)=V ⌫ h 0 c3 ! In classical electromagnetic theory total energy integrated over all frequencies diverges ultraviolet catastrophe ☛ divergence comes from large ⌫ part of integral which in optical spectrum is ultraviolet portion Bose-Einstein factor imposes effective ultraviolet cutoff on frequency integral total energy is finite kT/h ☟ 5 4 8⇡ (kT) 15 4 3 u(T )=3P (T )= =7.56464 10 (T/K) erg/cm 15h3c3 ⇥ 1J 107 erg = 6.24 109 GeV ⌘ ⇥ C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 10 SPECTRAL DENSITY Define spectral densityOverview u(⌫,T) 15 h (h⌫/kT )3 ⇢ (⌫,T)= = 4 E u(T ) ⇡ kT eh⌫/kT 1 so that ⇢"(⌫,T) d⌫ is fraction of electromagnetic energy between frequencies ⌫ and ⌫ + d⌫ (under equilibrium conditions) 1 d⌫ ⇢" (⌫,T)=1 Z0 Maximum occurs when s h ⌫ /k T satisfies ⌘ B d s3 s s =0 s =3 s =2.82144 ds e 1 ) 1 e ) ⇣ ⌘ C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 11 5.5. PHOTON STATISTICS 195 BLACKBODY RADIATIONOverview AT THREE TEMPERATURES Figure 5.2: Spectral density ρε(ν,T) for blackbody radiation at three temperatures. 5.5.4 Distribution of blackbody radiation C.Luis B.-Champagne Anchordoqui 2 Thursday,Recall December that the 4, frequency14 of an electromagnetic wave of wavevector k is ν = c/λ = ck/2π.Thereforethenumber12 of photons (ν,T) per unit frequency in thermodynamic equilibrium is (recall there are two polarization states) NT 2 V d3k V k2 dk (ν,T) dν = 3 !ck/k T = 2 !ck/k T . (5.82) N 8π · e B 1 π · e B 1 − − We therefore have 8πV ν2 (ν,T)= 3 hν/k T . (5.83) N c · e B 1 − Since a photon of frequency ν carries energy hν,theenergyperunitfrequency (ν) is E 8πhV ν3 (ν,T)= 3 hν/k T . (5.84) E c · e B 1 − Note what happens if Planck’s constant h vanishes, as it does in the classical limit. The denominator can then be written hν ehν/kBT 1= + (h2) (5.85) − kBT O and 8πkBT 2 CL(ν,T)= lim (ν)=V ν . (5.86) E h 0 E · c3 → In classical electromagnetic theory, then, the total energyintegratedoverallfrequenciesdiverges.Thisisknown as the ultraviolet catastrophe,sincethedivergencecomesfromthelargeν part of the integral, which in the optical spectrum is the ultraviolet portion. With quantization, theBose-Einsteinfactorimposesaneffectiveultraviolet cutoff kBT/hon the frequency integral, and the total energy, as we found above, is finite: ∞ π2 (k T )4 E(T )= dν (ν)=3pV = V B . (5.87) E · 15 (!c)3 !0 BOSE-EINSTEIN GAS In macroscopic limit N and V Overview!1 !1 so that concentration of particles n = N/V is constant energy levels of system become so finely quantized that become quasi continuous Bose-Einstein continuum distribution is N(") 1 f(")= = (88) g(") e(" µ)/kT 1 Initial concern is determining how chemical potential varies with temperature Adopt convention of choosing ground state energy to be zero At T =0 all N bosons will be in the ground state Setting " =0 in (88) we see that if f ( " ) is to make sense µ<0 ) Furthermore ☛ µ =0 at temperature of absolute zero and only slightly less than zero at non-zero temperatures assuming N to be large number C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 13 HIGH TEMPERATURE LIMIT At high temperatures ☛Overview in classical limit of dilute gas MB distribution applies (" µ)/kT f(")=e 2⇡mkT 3/2 Z with µ = kT ln Z = 2 V N h ✓ ◆ ☟ ✓ ◆ µ 2⇡mkT 3/2 V = ln kT h2 N "✓ ◆ # 4 For 1 kilomole of a boson gas comprising He atoms at standard T and P 27 23 3/2 µ 2⇡(6.65 10 )(1.38 10 )(273) 22.4 = ln ⇥ ⇥ kT (6.63 10 34)2 6.02 1026 ( ⇥ ✓ ⇥ ◆) = 12.43 µ = 0.29 eV C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 14 VALIDITY OF DILUTE GAS APPROXIMATION Average energy of idealOverview monoatomic gas atom " =(3/2)kT =0.035 eV (" µ)/kT =1.5 + 12.4 = 13.9 substituting this value in (88) 7 f(")=9 10 ⇥ confirming validity of approximation In classical limit 3/2 ↴ µ/kT 2⇡mkT V e = h2 N ✓ ◆ positive number that increases with temperature and decreases with particle density N/V It turns out that below some temperature T B ideal Bose-Einstein gas undergoes Bose-Einstein condensation A macroscopic number of particles N N falls into ground state 0 ⇠ These particles are called Bose-Einstein condensate C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 15 LOW-TEMPERATURE LIMIT 1 Setting energy of ground state to zero ☛ Overview N0 = µ e 1 Resolving this for µ 1 kBT µ = kBT ln 1+ = N0 ⇠ N0 ⇣ ⌘ Since N 0 is very large ☛ µ =0 below Bose condensation temperature For µ =0 ☛ easy to calculate number of particles in excited states Nex 1 ⇢(") N = d" ex e " 1 Z0 Total number of particles ☛ N = N0 + Nex In three dimensions ☛ using density of states ⇢ ( " ) given by (39) V 2m 3/2 1 p" V 2m 3/2 1 1 px N = d" = dx ex 2 2 " 2 2 3/2 x (89) (2⇡) ~ 0 e 1 (2⇡) ~ 0 e 1 ⇣ ⌘ Z ⇣ ⌘ Z C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 16 MATHEMATICAL INTERLUDE s 1 1 x Overviewdx = (s) ⇣(s) ex 1 Z0 ( s ) ☛ gamma-function satisfying (n + 1) = n (n)=n! (1/2) = p⇡, (3/2) = ⇡/2 ⇣ ( s ) ☛ Riemann zeta function p 1 s ⇣(s)= k kX=1 ⇣(1) = ,⇣(3/2) = 2.612 ⇣(5/2) = 1.341 1 ⇣(5/2) =0.5134 ⇣(3/2) C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 17 BOSE-EINSTEIN CONDENSATE 3/2 V 2mkBT (89) yields ☛ OverviewNex = (3/2) ⇣(3/2) (2⇡)2 ~2 ⇣ (increasing⌘ with temperature) At T = T B one has N ex = N that is ↴ V 2mk T 3/2 N = B B (3/2) ⇣(3/2) (90) (2⇡)2 ~2 ⇣ ⌘ and thus condensate disappears ☛N 0 =0 From equation above follows ↴ ~2 (2⇡)2 n 2/3 N kBTB = n = (91) 2m (3/2) ⇣(3/2) V ⇣ ⌘ that is ☛ T n2/3 B / In typical situations ☛ T B < 0 .1 C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 18 TEMPERATURE EVOLUTION 31 Overview N / N N0 / N ex T /TB FIG. 10: Condensate fraction N (T )/N and the fraction of excited particles N (T )/N for the ideal Bose-Einstein gas. Condensate fraction0 N 0 ( T ) /N and fraction of excitedex particles Nex(T )/N and thus the condensate disappears, N0 =0. From the equation above follows 2/3 2 2 C.Luis B.-Champagne Anchordoqui ! (2π) n N 2 kBTB = ,n= , (231) Thursday, December 4, 14 2m !Γ(3/2)ζ(3/2)" V 19 2/3 that is TB ∝ n . In typical situations TB < 0.1 K that is a very low temperature that can be obtained by special methods such as laser cooling. Now, obviously, Eq. (229) can be rewritten as N T 3/2 ex = ,TT (232) N T ≤ B # B $ The temperature dependence of the Bose-condensate is given by N N T 3/2 0 =1 ex =1 ,TT . (233) N − N − T ≤ B # B $ One can see that at T = 0 all bosons are in the Bose condensate, while at T = TB the Bose condensate disappears. Indeed, in the whole temperature range 0 T TB one has N0 N and our calculations using µ =0inthis temperature range are self-consistent. ≤ ≤ ∼ Above T there is no Bose condensate, thus µ = 0 and is defined by the equation B $ ∞ ρ(ε) N = dε . (234) ˆ eβ(ε µ) 1 0 − − This is a nonlinear equation that has no general analytical solution. At high temperatures the Bose-Einstein (and also Fermi-Dirac) distribution simplifies to the Boltzmann distribution and µ can be easily found. Equation (217)with the help of Eq. (231) can be rewritten in the form 3/2 βµ 1 T e− = 1, (235) ζ(3/2) T % # B $ so that the high-temperature limit implies T TB. Let us consider now the energy of the ideal Bose% gas. Since the energy of the condensate is zero, in the thermodynamic limit one has ∞ ερ(ε) U = dε (236) ˆ eβ(ε µ) 1 0 − − 32 CHEMICAL POTENTIAL Overview T /TB µ /(kBTB ) FIG. 11: Chemical potential µ(T ) of the ideal Bose-Einstein gas. Dashed line: High-temperature asymptote corresponding to the Boltzmann statistics.Dashed line: High-temperature asymptote corresponding to Boltzmann statistics C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 20 C /(NkB ) T /TB FIG. 12: Heat capacity C(T )oftheidealBose-Einsteingas. in the whole temperature range. At high temperatures, T T , the Boltzmann distribution applies, so that U is ! B given by Eq. (71) and the heat capacity is a constant, C =(3/2) NkB. For T ∞ ερ(ε) U = dε (237) ˆ eβε 1 0 − that in three dimensions becomes 3/2 3/2 3/2 V 2m ∞ ε V 2m 5/2 U = 2 2 dε βε = 2 2 Γ(5/2)ζ(5/2) (kBT ) . (238) (2π) ! ˆ0 e 1 (2π) ! ! " − ! " INTERNAL ENERGY Since energy of condensateOverview is zero 1 "⇢(") U = d" e (" µ) 1 Z0 For T T B ☛ Boltzmann distribution applies ➣ U is given by (71) and heat capacity is constant C =(3/2) NkB 1 "⇢(") For T T 3/2 (5/2) ⇣(5/2) T 3/2 3 ⇣(5/2) U = NkBT = NkBT T TB TB (3/2) ⇣(3/2) TB 2 ⇣(3/2) ⇣ ⌘ ⇣ ⌘ C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 21 32 T /TB µ /(kBTB ) HEAT CAPACITY OF IDEAL BOSE-EINSTEIN GAS Overview FIG. 11: Chemical potential µ(@TU) of the ideal Bose-EinsteinT 3/ gas.2 15 Dashed⇣(5/ line:2) High-temperature asymptote corresponding to the Boltzmann statistics.CV = = NkB T TB @T V TB 4 ⇣(3/2) ⇣ ⌘ ⇣ ⌘ C /(NkB ) T /TB C.Luis B.-Champagne Anchordoqui FIG. 12: Heat capacity C(T )oftheidealBose-Einsteingas. 2 Thursday, December 4, 14 22 in the whole temperature range. At high temperatures, T T , the Boltzmann distribution applies, so that U is ! B given by Eq. (71) and the heat capacity is a constant, C =(3/2) NkB. For T ∞ ερ(ε) U = dε (237) ˆ eβε 1 0 − that in three dimensions becomes 3/2 3/2 3/2 V 2m ∞ ε V 2m 5/2 U = 2 2 dε βε = 2 2 Γ(5/2)ζ(5/2) (kBT ) . (238) (2π) ! ˆ0 e 1 (2π) ! ! " − ! " EQUATION OF STATE entropy Overview T 3/2 CV (T 0) 2 T 5 ⇣(5/2) S = dT 0 = CV = NkB T TB 0 T 0 3 TB 2 ⇣(3/2) Z ⇣ ⌘ free energy T 3/2 ⇣(5/2) F = U TS = NkBT T TB TB ⇣(3/2) pressure ⇣ ⌘ @F @F @T 3 F 2 T F P = = B = B = @V T @TB T @V T 2 TB 3 V V ⇣ ⌘ ⇣ ⌘ ⇣ ⌘ ⇣ ⌘⇣ ⌘ T 3/2 ⇣(5/2) equation of state PV = NkBT T TB TB ⇣(3/2) ⇣ ⌘ compared to PV = Nk B T at high temperatures 3/2 P of ideal Bose gas with condensate contains additional factor (T/TB) < 1 This is because particles in condensate are not thermally agitated and thus they do not contribute into pressure C.Luis B.-Champagne Anchordoqui 2 Thursday, December 4, 14 23