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 λ ☟ that is volume is isotropically expanded by a factor of λ 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 λ 1/3 under an adiabatic volume expansion V λV − ! It follows that ↴ @U @U 1 @U U V = λ = 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✓ dφ sin ✓ cos ✓ = = cP (T ) σT 4 E dA 4⇡V 4 V 4 ⌘ Z0 Z0 3 σ = ca P (T )=aT 4 4 ) Using quantum statistical mechanical considerations we will show that ↴ 2 4 ⇡ kB 8 W Stefan’s constant ☛ σ = =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 ⊕ σT 4⇡R = ⇡T P 4⇡d · · d · ⊕− ✓ ⊕− ◆ 4 2 Power radiated by earth ☛ radiated = σT 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) λ L ··· ☛ λ ☛ 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/"19. 90$739)9 4⇡V 2 substituting (86) in (87) ☛ g(⌫)d⌫ = 3 ⌫ d⌫ >L c L >!L .D-8L 1L 41<<#+L 86D.#<L #.L >!L 8.L >1L 9 #.L*9 #<L48141;>$1.+L>1L #<L >!L.D-8L1L51<#>#EL<><L 1L C.Luis B.-Champagne Anchordoqui &.? 9<L %/L>!L %/C9F+L )9 >2L 9 *9 >!>L #29 G&>!%.L L<!++L1L>!%)/<<L2 *9 1L Thursday, December 4, 14 .L1@.@L3LL<4"8LG#>"L8#D<L*9 7 D<>#AD>'.L 6D>#1.L 18L *9 GL1>#.L .LLE(8@#.L<1+#L>!8L8L>!8L>J4<L1LGE<L1.L+1.$>D#.+LG#>!L E+1&>JL .L>G2L>:.<E8=LG#>!LE+1#>JL ++L 8L4814>L#.L>!L<-L (:>#1.L !.L ++L >!:L GE<L 8L >*0L &.>1L 1D.>L 6DA#1.L 1-<L L L 89 9 #.L!L1<#,+>18L1L>!L<<-+JLE#8@<LG#>!L#><L1G.L86D.JL.L HL 8L 1.<#;#.L .L <<-+JL 1L +#.8L 1<#++>1;<L >!8L -D<>L L .L D448L +#-(>L>1L>!L7D.JL <4>:D-LL -I#-D-L 86D/JL #<L >8K -(.L81-L>!LBL>!>L>!8L8L1.+JL4!1/1.<L 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 ⇠