Topic 6 - Slide 1

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					                 PHY203: Thermal Physics

          Topic 6: Statistics, Entropy & Equilibrium

•   The Einstein Solid
•   Interacting Einstein solids
•   What entropy really means
•   Thermal equilibrium
•   Links between statistical mechanics and thermodynamics
                       Einstein Model of a Solid
•Essentially, this is the familiar “balls and springs”
model: each atom represented by a hard sphere
connected to its neighbours by springs
•Each atom undergoes simple harmonic oscillation
about its equilibrium position
•The Einstein model treats the atoms as quantum
mechanical SHO’s, all with the same frequency 
•Energy levels given by: En = (n + ½)h
•All oscillators independent
•Each atom can oscillate in 3 independent directions
•So, a solid containing N atoms behaves like 3N
independent oscillators
         Pictorial representation of QM oscillator

•In this treatment we’ll represent each
oscillator as shown opposite (parabolic potential
energy with “ladder” of discrete quantised
energy levels)
                                                     n=3
•Measure energies relative to ground state
                                                      n=2
•So each oscillator can have energies in                n=1
discrete multiples of h, call this one unit.
                                                           n=0
•Represent the energy state of the
oscillator by placing an arrow next to the          Oscillator in 2nd excited state,
appropriate energy level                               with 2 units of energy

•Remember (for later), that although we’re
treating the ground state as the zero of
energy, it really has E = h/2
  Microstates of a tiny Einstein solid (3 oscillators!)
       Total energy = 0 units




n=3                 n=3         n=3
 n=2                 n=2         n=2
  n=1                  n=1        n=1
   n=0                  n=0        n=0



                                         (0) = 1
 Microstates of a tiny Einstein solid (3 oscillators!)
Total energy = 1 unit



n=3              n=3      n=3
 n=2              n=2      n=2
   n=1             n=1      n=1
    n=0             n=0      n=0



 n=3
  n=2
                 n=3
                  n=2
                          n=3
                           n=2
                                         (1) = 3
   n=1             n=1      n=1
    n=0             n=0      n=0




 n=3             n=3      n=3
  n=2             n=2      n=2
   n=1             n=1      n=1
    n=0             n=0      n=0
 Microstates of a tiny Einstein solid (3 oscillators!)
Total energy = 2 units


n=3         n=3          n=3            n=3         n=3      n=3
 n=2         n=2          n=2            n=2         n=2      n=2
  n=1         n=1          n=1               n=1      n=1      n=1
   n=0         n=0          n=0               n=0      n=0      n=0




n=3         n=3          n=3            n=3         n=3      n=3
 n=2         n=2          n=2            n=2         n=2      n=2
  n=1         n=1          n=1               n=1      n=1      n=1
   n=0         n=0          n=0               n=0      n=0      n=0




n=3         n=3          n=3            n=3         n=3      n=3
 n=2         n=2          n=2            n=2         n=2      n=2
  n=1         n=1          n=1               n=1      n=1      n=1
   n=0         n=0          n=0               n=0      n=0      n=0




                                  (2) = 6
           Calculation of  for bigger numbers
 New Representation



n=3           n=3      n=3        n=3       n=3       n=3
 n=2           n=2      n=2        n=2       n=2       n=2
  n=1           n=1      n=1        n=1       n=1       n=1
   n=0           n=0      n=0        n=0       n=0       n=0




         Number of oscillators = N, number of lines (|) = N-1
         Number of energy units = number of dots (●) = q
         Total number of symbols = q + (N-1)
     General formula for  for Einstein solid

Total number of microstates for an Einstein solid with N
oscillators and q energy units = number of ways of arranging
lines and dots:


                                 (q  N  1)!
                   (N , q ) 
                                  q !(N  1)!


Example: Calculate the multiplicity (number of microstates)
for an Einstein solid with 30 oscillators and 30 units of
energy……….
    Interacting systems




    Solid A                 Solid B
               Energy
UA, NA, qA, A          UB, NB, qB, B




      Utotal = UA + UB = constant

            total= A B
                               Interacting systems
Consider first 2 tiny interacting Einstein solids, 3 oscillators in
each, sharing a total of 6 units of energy:

                                                          100


        q   q                tota                      80

        A   B    A     B       l                          60

        0    6       1 28           28




                                                 total
                                                          40
        1    5       3 21           63
                                                          20
        2    4       6 15           90
                                                           0
        3    3   10 10             100                          0   1   2   3    4   5   6

                                                                            qA
        4    2   15        6        90
                                          Probability of most likely macrostate
        5    1   21        3        63
                                          (qA=qB=3) = 100/462
        6    0   28        1        28
                                         Probability of least likely macrostate (qA or qB=0)
  7 possible macrostates
                                         = 28/462
                      Interacting systems
•The calculations show that if, for example, the 2 solids are
initially in the qA=1, qB=5 macrostate, at some later time the
system is more likely to be in the qA=2, qB=4 (or qA=3, qB=3)
macrostate than in the qA=0, qB=6 macrostate.

                            100


                            80


                            60
                   total




                            40


                            20


                             0
                                  0   1   2   3    4   5   6

                                              qA


 •Implies a “preferred” direction for the energy transfer process,
 reminiscent of 2nd law of thermodynamics
                          A bigger system
Now let solid A contain 300 oscillators, solid B 200 oscillators, and
let the total energy be 100 units:




  •As expected, probability distribution “sharpens up” as numbers get bigger,
  becomes increasingly unlikely to find deviations from equilibrium, or to get
  energy flowing in the “wrong” direction
  •Gaussian distribution for physically realistic numbers (>~1022): see SPS
  derivation & Schroeder2.3
                           2nd law revisited
•When our two bodies are placed in contact, energy is exchanged between
them until they find themselves in the most probable macrostate (ie the one
with the greatest multiplicity or number of microstates )
•Once they have arrived at this macrostate, spontaneous flow of energy
between the bodies ceases
•Essentially, then, behind the apparent mystique of the 2nd law we have a
simple fact:

“Things do what they’re most likely to do according to the standard rules of
probability and statistics”!

•We can restate the 2nd law along the lines of:

“Any large system in equilibrium will be found in the macrostate with the
largest ”

Or…….” tends to increase”       (Schroeder p. 74)
                              Entropy and 
•The entropy form of the 2nd law says (more or less) that “the entropy of the
universe tends to increase”
•We have deduced from our statistical arguments that “ tends to increase”
•Suggests that entropy is somehow related to 
•However, if we have 2 systems, A and B, then Stotal=SA + SB, but total=AB
•So, S cant be simply proportional to 


Everything’s OK if we have:
                                  S  ln 

                              S  k B ln 
         Boltzmann’s Constant
                   S  k B ln 



                                             Boltzmann’s
                                             tombstone (Central
                                             Cemetery, Vienna)




This equation provides the essential link between
thermodynamics and statistics………………..
                Conditions for equilibrium
Consider 2 systems, isolated from the rest of the universe,
separated by a partition


                                          UT = U1 + U2 = constant
             System 1      System 2       VT = V1 + V2= constant
             U1, V1, N1    U2, V2, N2     NT = N1 + N2 = constant
                                          ST = S1 + S2



 General principle: total entropy ST is maximised at equilibrium
                     Thermal equilibrium
•If the partition is rigid, impermeable, but thermally conducting,
then the systems can only interact by exchange of heat.
•Therefore only U1 and U2 can vary as equilibrium is
approached (but remember U1+U2 = constant)
•Choose U1 as single independent variable


                         Q                  Maximise ST by setting:
            System 1         System 2
                                                 ST
            U1, V1, N1       U2, V2, N2               0
                                                 U 1
               Thermal equilibrium
        ST   S 1 S 2 S 1 S 2 U 2
                                   0
        U 1 U 1 U 1 U 1 U 2 U 1

                                       U 2
          U 2  UT  U 1 , therefore         1
                                       U 1
                       S 1 S 2
                           
                       U 1 U 2

We know that at equilibriu m, T1  T2, , suggesting that
                S
T is related to
                U

To see how, consider dependence of S on U for our
interacting Einstein solids……………………………………
                          A bigger system
Now let solid A contain 300 oscillators, solid B 200 oscillators, and
let the total energy be 100 units:




  •As expected, probability distribution “sharpens up” as numbers get bigger,
  becomes increasingly unlikely to find deviations from equilibrium, or to get
  energy flowing in the wrong direction
  •Gaussian distribution for physically realistic numbers (>~1022): see SPS
  derivation
Entropy vs energy for solids A & B

                                     •UA = hqA ; UB = hqB
                                     •Note that axes for qA and qB
                                     go in opposite directions
                                     •So, in indicated region,
                                     energy of A increases
                                     spontaneously as equilibrium
                                     is approached
                                     •Energy of B decreases as
                                     equilibrium is approached
                                     •So, in region indicated,
                                     TA<TB

   S A S B
       
   U A U B                               1 S 
                                               
                                          T  U V ,N
Very Important Equations


     S  k B ln 


     1  S 
                 (fundamental definition
     T  U V ,N   of absolute temperature)

				
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