Minimum Bias Events at ATLAS

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1 Camille Bélanger-Champagne McGill University Lehman College City University of New York Thermodynamics Charged Particle and Statistical Correlations Mechanics in Minimum Bias Events at ATLAS Thermodynamics II 4 September 2014 February 26 th 2012, WNPPC 2012 Physics motivation Heat capacity Minbias event and track selection Heat machines Azimuthal Second Law correlation of Thermodynamics results Forward-Backward Carnot Theorem correlation results Clausius inequality Entropy 1

2 HEAT CAPACITY In most cases adding heat to system leads to increase of its T Heat capacity C = Q dt (28) Heat capacity is proportional to system size and so is extensive variable Introduce specific quantities heat and heat capacity per kilomole q Q n c = C n = q dt (29) Heat capacity depends on the type of the process 2

3 HEAT CAPACITY (isobaric and isochoric processes) If heat is added to system while volume is kept constant dv =0 we obtain the isochoric heat capacity If we keep a constant pressure we obtain the isobaric heat capacity In isochoric case no work is done C V = dp =0 C P = Q dt Q dt! V! P (30) so heat fully converts into internal energy and temperature increases In the isobaric case system usually expands upon heating (31) and a negative work is done on it This leads to smaller increase of U and thus smaller increase of T 3

4 HEAT CAPACITY (isothermal and adiabatic processes) In isothermal process system receives or lose heat but dt =0 C T = ±1 In adiabatic process Q =0 but the temperature changes C S =0 Subscript S refers to entropy state function conserved in reversible adiabatic processes 4

5 ISOCHORIC HEAT CAPACITY Let us rewrite first law of thermodynamics (25) with (17) in form Considering energy as a function of du = Q = du + V T dt + Combining this with the previous equation V dt + @V T! dv (33) T + P # dv (34) At constant volume this equation yields C Q =(@U/@T ) V dt! V so (35) 5

6 ISOBARIC HEAT CAPACITY To find isobaric heat capacity C P we must take into account that at constant P the in (32) changes because of thermal expansion, V P dt Inserting this into (34) we obtain Q = " C @T! P # dt (37) P so C P = C V T + P (38) 6

7 MAYER S RELATION Because of of negligibly weak interaction between particles energy of ideal gas depends only on T (@U/@V ) T =0 From the equation of state (5) we obtain (@V/@T ) V = nr/p Substituting these in (38) we obtain Mayer s relation for ideal gas C P = C V + nr (39) or c P = c V + R for heat capacities per kilomole In terms of number of particles N Mayer s relation becomes C P = C V + Nk B (40) or c P = c V + k B for heat capacities per particle 7

8 ADIABATIC PROCESS OF IDEAL GAS For ideal gas internal energy is a function of temperature only du = V dt = C V dt In adiabatic process Substituting these two results in (25) and using (17) we obtain C V dt = PdV Either P or V can be eliminated with help of (5) (44) (45) C V dt = nrt dv V (46) This can be integrated if temperature dependence C V (T ) is known 8

9 For perfect gas C V PERFECT GAS = const ln T = integration of (46) yields nr C V ln V + const (47) or equivalently TV nr/c V = const (48) it is convenient to introduce C P C V (49) with help of Mayer s relation (39) C V = nr 1 C P = nr 1 (50) Adiabatic equation (48) becomes TV 1 = const (51) Using (5) we can rewrite (51) as or else as PV = const (52) TP 1/ 1 = const (53) 9

10 Using (52) WORK DONE IN ADIABATIC PROCESS W 12 = Z V2 V 1 PdV= const Z V2 V 1 dv V = 1 1 (V 1 2 V 1 1 ) (54) const = P 1 V 1 = P 2 V 2 Substituting into (54) W 12 = 1 1 (P 2 V 2 P 1 V 1 ) (55) Using (5) and (50) one can simplify this formula to W 12 = nr 1 (T 2 T 1 )=C V (T 1 T 2 ) (56) 10

11 ALTERNATIVELY... According to first law of thermodynamics in adiabatic process work is equal to change of internal energy W 12 = U 1 U 2 For ideal gas U = U(T ) and du/dt = C V Z Internal energy of ideal gas is given by U(T )= C V (T ) dt (57) For perfect gas C V = const U(T )=C V T + U 0 U 0 = const (58) U 1 U 2 = C V (T 1 T 2 ) (59) so that (56) follows 11

12 HEAT MACHINES Heat machines were a major application of thermodynamics in XIX century Basis of their understanding is first law of thermodynamics (25) Heat machine includes two reservoirs with different temperatures and a system that exchanges heat with two reservoirs and does work in a cyclic process There are three types of heat machines: engines, refrigerators and heat pumps Engines of XIX century used steam as a source of heat (obtained by heating water by fire) Contemporary motors use fuel that burns and generates heat directly Refrigerators and heat pumps also use agents other than water 12

13 HEAT ENGINE During one cycle system receives heat from hot reservoir gives heat Q 1 to cold reservoir and makes work Q 2 W 13

14 Integrating (25) over cycle EFFICIENCY OF HEAT ENGINE Efficiency of engine ratio of output energy to input energy = W Q 2 I U = du =0 ( Q 1 is lost energy) In cyclic processes internal energy of system does not change: I 0= ( Q W) =Q W = Q 2 Q 1 W (60) (61) (62) so that the work done by the system is W = Q 2 Q 1 Inserting this into (60) Note that <1 To make efficiency =1 Q 1 Q 2 as high as possible we should minimize Q 1 (63) 14

15 CARNOT CYCLE Carnot cycle consists of two isotherms T 1 and T 2 and two adiabats (working body being ideal gas) Q =0 Q =0 Cycle goes clockwise work done by system W = I PdV > 0 Q 2 Heat is received on the isothermal path AB at T = T 2 Q 1 Heat is given away on isothermal path CD at T = T 1 DA There is no heat exchange on the adiabatic paths BC and 15

16 For ideal gas Using (20): Using (51): T 2 V B 1 = T 1 V CARNOT ENGINE U = U(T ) du = Q W =0 along isotherms C 1 T 2 V A U = const Q 2 = Q AB = W AB = nrt 2 ln V B V A Q 1 = Q CD = W CD = nrt 2 ln V C V D > 0 1 = T 1 V D 1 (64) (65) Dividing these equations by each other T 2 (63) gives Carnot formula for temperature scale Q 1 Q 2 = T 1 =1 V B /V A = V C /V D 16 T 1 T 2 (66) (67)

17 Efficiency EFFICIENCY OF CARNOT ENGINE becomes close to 1 as T 1! 0 practically it is impossible to realize In standard engines T 1 is temperature at normal conditions T 1 = 300 K T 2 of hot reservoir must essentially exceed T 1 e.g. for T 2 = 600 K =0.5 In practice processes in heat engines deviate from Carnot cycle this leads to further decrease of efficiency 17

18 REFRIGERATOR Refrigerators are inverted heat engines Work is done on system that extracts heat from cold reservoir (eventually lowering its temperature) and gives heat Q 2 to hot reservoir the environment Q 1 (environment at T 2 ) 18

19 Efficiency of refrigerator EFFICIENCY OF REFRIGERATOR c = output input = Q 1 W = Q 1 Q 2 Q 1 (68) For Carnot refrigerator with help of (66) c = T 1 T 2 T 1 (69) Efficiency of a refrigerator can be very high if T 1 and T 2 are close This is the situation when the refrigerator starts to work As T 1 decreases well below T 2 (environmental 300 K ) efficiency becomes small 19

20 HEAT PUMP Similar to refrigerator but interpretation of reservoirs 1 and 2 changes 1 is the environment from where heat is being pumped whereas 2 is the reservoir that is being heated (e.g. a house) (environment at T 1 ) 20

21 Efficiency EFFICIENCY OF HEAT PUMP c = output input = Q 2 W = Q 2 Q 2 Q 1 (70) For Carnot heat pump efficiency becomes inverse of that of Carnot engine d = T 2 T 2 T 1 (71) Efficiency of heat pump is always greater than 1 T 1 T 2 d If and are close to each other becomes large This characterizes the initial stage of the work of a heat pump After increases well above efficiency becomes close to 1 Heating a house in winter involves T 2 T 1 T K d 10 while In reality there are losses that lower heat pump efficiency T K 21

22 THE SECOND LAW OF THERMODYNAMICS Postulate of Lord Kelvin: A transformation whose only final result is to extract heat from a source at fixed temperature and transform that heat into work is impossible Postulate of Clausius: A transformation whose only result is to transfer heat from a body at a given temperature to a body at a higher temperature is impossible These postulates which have been repeatedly validated by empirical observations constitute the Second Law of Thermodynamics 22

23 CARNOT S THEOREM Efficiency of any reversible heat engine operating between heat reservoirs with temperatures T 1 and T 2 is equal to the efficiency of Carnot engine =1 T 1 /T 2 while the efficiency of any irreversible heat engine is lower than this To prove Carnot s theorem assume that wicked awesome wonder engine has an efficiency greater than Carnot engine Key feature of Carnot engine is its reversibility we can go around its cycle in opposite direction creating a Carnot refrigerator Envision wonder engine to driving a Carnot refrigerator 23

24 WONDER ENGINE DRIVING CARNOT S REFRIGERATOR 24

25 , We assume figure NOTHING BEATS A CARNOT ENGINE W Q 2 = wonder > Carnot = W0 Q 0 2 W = Q 2 Q 1 = Q 0 2 Q 0 1 = W 0 (72) According to second law of thermodynamics, Difference of efficiencies Q 2 Q (73) 0 = W Q 2 W 0 Q 0 2 = W Q 2 W Q 2 = W (Q0 2 Q 2 ) Q 2 Q 0 2 apple 0 (74) that proves Carnot s theorem In reversible case no heat flows from hot to cold reservoir and 0 = 25

26 INFINITESIMALLY CARNOT CYCLE Rewrite (74) using (63) for (arbitrary body) and (67) for (ideal gas) 0 =1+ Q 1 Q 2 1+ T 1 T 2 = Q 1 Q 2 + T 1 T 2 apple 0 (75) Since Q 2 > 0 Q 1 T 1 + Q 2 T 2 apple 0 (76) For an infinitesimally narrow Carnot cycle Q 1 T 1 + Q 2 T 2 apple 0 (77) 26

27 CLAUSIUS INEQUALITY Arbitrary curve in PV plane can be decomposed (to arbitrary accuracy) as combination of Carnot cycles isotherms adiabats T = 5 3 I Q = TdS Q T apple 0 Clausius inequality (78) with equality holding if all cycles are reversibles 27

28 ENTROPY Consequence of second law is existence of entropy: a state thermodynamic equilibrium whose differential is given by Q = TdS (79) S being a state function does not change in any reversible cyclic process: I Q T =0 Since Q is extensive so is S Units of entropy are [S] =J/K 28

Minimum Bias Events at ATLAS

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