Camille Bélanger-Champagne Lehman McGill College University City University of New York Thermodynamics Charged Particle and Correlations Statistical Mechanics in Minimum Bias Events at ATLAS Statistical Mechanics V November 214 February 26 th 212, WNPPC 212 Physics motivation Stefan-Boltzmann law Minbias event and track selection Blackbody radiation Azimuthal correlation results Bose-Einstein condensation Forward-Backward correlation results 1
CLASSICAL ARGUMENTS FOR PHOTON GAS A number of thermodynamic properties of photon gas can be determined from purely classical arguments 1 Assume photon gas is confined to rectangular box of dimensions 1/3 further assume that dimensions are all expanded by a factor L x L y L z that is volume is isotropically expanded by a factor of Cavity modes of electromagnetic radiation have quantized wave vectors even within classical electromagnetic theory for a given mode "( ~ k)=~c ~ k energy changes by It follows that V @U @V S = @U @ 1/3 S ~ k = 2 nx L x under an adiabatic volume expansion = 1 3 U and P = @U @V, 2 n y L y V! Since U = U(T,V ) is extensive we must have P = P (T ) alone S, 2 n z L z V = U 3V 2
MORE CLASSICAL ARGUMENTS FOR PHOTON GAS 2 Since P = P (T ) alone using Maxwell relation @S @V P = @P @T V It follows that after invoking the First Law @U @V T dp dt T @U = =3P @V P @P = T P @T V =4P ) P (T )=at 4 a du = TdS constant PdV 3
Given energy density Power emitted per unit area STEFAN-BOLTZMANN LAW U/V differential energy flux emitted in direction dj E = c U V d cos 4 relative to surface normal d differential solid angle J E = dp da = cu 4 V Z /2 d Z 2 d sin cos = cu 4 V = 3 4 cp (T ) T 4 = 3 4 ca ) P (T )=at 4 Stefan s constant = 2 kb 4 6c 2 ~ 3 =5.67 1 8 W m 2 K 4 Using quantum statistical mechanical considerations we will show that 4
SURFACE TEMPERATURE OF EARTH We ll need three lengths: radius of sun R =6.96 1 8 m radius of earth R =6.38 1 6 m radius of earth s orbit d =1.5 1 11 m Assume that earth has achieved a steady state temperature balance total power incident upon earth with power radiated by earth 2 Power incident upon earth P incident = R2 R R 4 d 2 T 4 4 R 2 = T 4 d Power radiated by earth Setting P incident = P radiated T = P radiated = T 4 4 R 2 R 2d 1/2 T T = 578 K ) T = 278.4 K = 287 K Mean surface temperature of the earth T Difference is due to fact that earth is not perfect blackbody T =.4817 T (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 5
QUANTUM WAVES INSIDE A BOX Wave function for quantum waves in one-dimensional box =A sin kx k = 2 = n L n =1, 2, 3, (85) de Broglie wavelength L box dimension substituting c = n = 2L in (85) c For a cube V = L 3 n = 2V 1/3 c (86) n 2 = n 2 x + n 2 y + n 2 z n x,n y,n z 2 Z + Possible values occupy first octant of sphere of radius n =(n 2 x + n 2 y + n 2 z) 1/2 6
DENSITY OF STATES OF QUANTUM WAVES INSIDE A BOX g( ) d number of possible frequencies in range (, + d ) n / ) g( ) d number of possible set of integers in (n, n + dn) within shell of thickness dn of octant of sphere of radius n g( )d = 1 8 4 n2 dn = 2 n2 dn (87) substituting (86) in (87) g( )d = 4 V c 3 2 d 7
PHOTON STATISTICS Photons are bosons of spin 1 and hence obey Bose-Einstein statistics number of photons per quantum state f j = N j g j = 1 e " j/kt 1 for continuos distribution f(") = N(") g(") = 1 e "/kt 1 photon energy is f( ) = N( ) g( ) = 1 e h /kt 1 8
Energy PLANCK RADIATION FORMULA photon gas two polarization states u( )d in range g( )d = 8 V c 3 (, + d ) 2 d number of photons in this range times energy u( ) d = N( )d h h of each Substituting N( ) =g( ) f( ) Planck s radiation formula u( )d = 8 hv c 3 apple 3 d e h /kt 1 9
Using take classical limit ULTRAVIOLET CATASTROPHE In classical electromagnetic theory total energy integrated over all frequencies diverges ultraviolet catastrophe divergence comes from large part of integral Bose-Einstein factor imposes effective ultraviolet cutoff on frequency integral total energy is finite e h /kt h! 1= h kt + O(h2 ) u class ( ) = lim h! u( ) =V 8 kt c 3 2 u(t )=3P (T )= 8 5 (kt) 4 which in optical spectrum is ultraviolet portion kt/h 15h 3 c 3 =7.56464 1 15 (T/K) 4 erg/cm 3 1 1J 1 7 erg = 6.24 1 9 GeV
Define spectral density SPECTRAL DENSITY so that E (, T )= u(, T ) u(t ) " (, T ) d (under equilibrium conditions) = 15 4 h kt (h /kt ) 3 e h /kt 1 is fraction of electromagnetic energy between frequencies and + d Z 1 d " (, T )=1 Maximum occurs when d ds s 3 e s 1 s h /k B T = ) s 1 e satisfies s =3 ) s =2.82144 11
BLACKBODY RADIATION AT THREE TEMPERATURES 12
In macroscopic limit BOSE-EINSTEIN GAS N!1 V!1 and 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 f(") = N(") g(") = 1 e (" µ)/kt 1 Initial concern is determining how chemical potential varies with temperature Adopt convention of choosing ground state energy to be zero (88) T = N " = f(") ) µ< At all bosons will be in the ground state Setting in (88) we see that if is to make sense Furthermore µ = at temperature of absolute zero and only slightly less than zero at non-zero temperatures assuming 13 N to be large number
HIGH TEMPERATURE LIMIT At high temperatures in classical limit of dilute gas MB distribution applies µ = kt ln Z N µ kt = f(") =e (" ln with " 2 mkt h 2 µ)/kt Z = 3/2 V N 2 mkt # h 2 3/2 V For 1 kilomole of a boson gas comprising 4 He atoms at standard T and P µ kt = ln = 12.43 ( apple2 (6.65 1 27 )(1.38 1 23 )(273) (6.63 1 34 ) 2 µ =.29 ev 3/2 22.4 6.2 1 26 ) 14
VALIDITY OF DILUTE GAS APPROXIMATION Average energy of ideal monoatomic gas atom substituting this value in (88) In classical limit (" µ)/kt =1.5 + 12.4 = 13.9 e µ/kt = f(") =9 1 7 2 mkt h 2 3/2 V N " =(3/2)kT =.35 ev confirming validity of approximation 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 These particles are called Bose-Einstein condensate falls into ground state 15
LOW-TEMPERATURE LIMIT Setting energy of ground state to zero Resolving this for µ µ = k B T ln N = 1+ 1 N = k B T N 1 e µ 1 Since is very large µ =below Bose condensation temperature For N µ = easy to calculate number of particles in excited states Z 1 N ex = d" (") e " 1 N ex Total number of particles N = N + N ex (") In three dimensions using density of states given by (39) N ex = V 2m 3/2 (2 ) 2 ~ 2 Z 1 d" p " e " 1 = V (2 ) 2 2m ~ 2 3/2 1 3/2 Z 1 dx p x e x 1 (89) 16
MATHEMATICAL INTERLUDE Z 1 dx xs 1 e x 1 = (s) (s) (s) gamma-function satisfying (n + 1) = n (n) =n! (s) Riemann zeta function (s) = 1X k=1 (1/2) = p, (3/2) = p /2 k s (1) = 1, (3/2) = 2.612 (5/2) = 1.341 (5/2) (3/2) =.5134 17
(89) yields BOSE-EINSTEIN CONDENSATE V 2mkB T 3/2 (2 ) 2 ~ 2 N ex = At T = T B one has N ex = N that is N = V 2mkB T 3/2 B (2 ) 2 ~ 2 (3/2) (3/2) (increasing with temperature) (3/2) (3/2) and thus condensate disappears N = (9) From equation above follows k B T B = ~2 2m (2 ) 2 n (3/2) (3/2) 2/3 n = N V (91) that is T B / n 2/3 In typical situations T B <.1 18
TEMPERATURE EVOLUTION N / N N ex / N T /T B Condensate fraction N (T )/N and fraction of excited particles N ex (T )/N 19
CHEMICAL POTENTIAL T /T B µ /( k B TB ) Dashed line: High-temperature asymptote corresponding to Boltzmann statistics 2
Since energy of condensate is zero Z 1 INTERNAL ENERGY For T T B Boltzmann distribution applies is given by (71) For T < T B U = U = d" and heat capacity is constant Z 1 µ = " (") e (" µ) 1 U C =(3/2) Nk B d" " (") e " 1 in three dimensions U = V 2m 3/2 (2 ) 2 ~ 2 Z 1 d" " 3/2 e " 1 = V (2 ) 2 2m ~ 2 3/2 (5/2) (5/2) (k B T ) 5/2 using (9) this can be rewritten as U = Nk B T T T B 3/2 (5/2) (5/2) (3/2) (3/2) = Nk BT T T B 3/2 3 2 (5/2) (3/2) T apple T B 21
HEAT CAPACITY OF IDEAL BOSE-EINSTEIN GAS C V = @U @T V = Nk B T T B 3/2 15 4 (5/2) (3/2) T apple T B C /( NkB ) T /T B 22
EQUATION OF STATE entropy S = Z T dt C V (T ) T = 2 3 C V = Nk B T T B 3/2 5 2 (5/2) (3/2) T apple T B free energy F = U TS = Nk B T pressure @F P = @V T equation of state compared to = @F @T B T PV = Nk B T @TB @V PV = Nk B T T T B 3/2 (5/2) (3/2) T apple T B 23 T = 3 2 F T B T T B 3/2 (5/2) (3/2) at high temperatures P of ideal Bose gas with condensate contains additional factor 2 3 T B V T apple T B = F V (T/T B ) 3/2 < 1 This is because particles in condensate are not thermally agitated and thus they do not contribute into pressure