Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 1

Physics 141.Lecture 22.

Collisions 2016: The Movie.

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 2

Images from Lawrence Berkeley National Laboratory

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 3

Physics 141.Lecture 22.

• Course Information.

• Entropy.

• Quick review of topics discussed during our last lecture.

• Heat capacity.

• The Maxwell-Boltzmann velocity distribution of gas molecules.

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 4

Physics 141.Course information.

• Homework # 10 is due on Friday 12/2 at 12 pm.• Homework set # 11 is due on Tuesday December 13 at

noon. This set is optional.• Note: your homework grade is based on the best 10

homework grades out of a total of 11.• Exam # 3 will take place on Tuesday December 6 at 8 am in

Hoyt. This exam will cover the material discussed inChapters 10, 11, and 12 and equilibrium.

• There will be no recitations and no office hours this week .• Let us now have a look at the status of experiment # 5.

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 5

Analysis of experiment # 5.Timeline (more details during next lectures).

• ✔11/14: collisions in the May room• ✔ 11/20: analysis files available.• ✔ 11/21: each student determines his/her best

estimate of the velocities before and after thecollisions (analysis during regular lab periods).

• 11/23: complete discussion and comparison ofresults with colliding partners and submit finalresults (velocities and errors) to professorWolfs.

• 11/25: professor Wolfs compiles results,determines momenta and kinetic energies, anddistributes the results.

• 11/28: office hours by lab TA/TIs to help withanalysis and conclusions.

• 12/2: students submit lab report # 5.

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 6

Distributing 4 units of energy among three degrees of freedom.

• Consider an atom with threedegrees of freedom; each degreeof freedom has a vibrationalcharacter with the samecharacteristic frequency.

• For this system we find:• 3 ways: 4:0:0 configuration.• 6 ways: 3:1:0 configuration.• 3 ways: 2:2:0 configuration.• 3 ways: 1:1:2 configuration.

• What is the probability to see thedifferent configurations?

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 7

Distributing energy.The fundamental assumption.

• In order to determine the probability to observe a certainconfiguration, we rely on the fundamental assumption ofstatistical mechanics to make this determination:

A fundamental assumption in statistical mechanics is that in our stateof microscopic ignorance, each microstate (microscopic distributionof energy) corresponding to a given macrostate (total energy) isequally probable.

• For example N = 3:• 15 microstates; probability of each one is 1/15.• 3 ways: 4:0:0 configuration (20% probability).• 6 ways: 3:1:0 configuration (40% probability).• 3 ways: 2:2:0 configuration (20% probability).• 3 ways: 1:1:2 configuration (20% probability).

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 8

Distributing energy.Two N = 3 atoms.

• Consider now a system with twoatoms, each with three degrees offreedom.

• The number of states for n = 1, 2,3, and 4 quanta in a given atomare easily determined:

• n = 1: 100, 010, 001• n = 2: 200, 110, 101, 020, 011,

002• n = 3: 300, 210, 201, 120, 111,

102, 030, 021, 012, 003• n = 4: 400, 310, 301, 220, 211,

202, 130, 121, 112, 103, 040, 031,022, 013, 004

• The most likely microstate is thusthe 2:2 state.

Atom 1 Atom 2 # states

n = 4 n = 0 15 x 1

n = 3 n = 1 10 x 3

n = 2 n = 2 6 x 6

n = 1 n = 3 3 x 10

n = 0 n = 4 1 x 15

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 9

Distributing energy.Arranging quanta.

• Extending our study to more complex systems (with moredegrees of freedom) is not too difficult.

• If we want to distribute q quanta amount N one-dimensionaloscillators we find that the number of possible ways is equalto

• Note: q! = q x (q - 1) x (q - 2) x (q - 3) x …. x 2 x 1.

# =

q + N −1( )!q! N −1( )!

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 10

Distributing energy.Arranging quanta.

• We can verify the previousequations by considering the casewhere we have up to 4 quanta tobe arranged among 3 one-dimensional oscillators.

• Note: 0! = 1.• Why?• n! = n x (n-1)! If 0! was 0, 1!

would be 0, etc. etc.

• The previous equation predictsthe correct number of states forthis system.

q N #

0 3 2!/(0! 2!) = 1

1 3 3!/(1! 2!) = 3

2 3 4!/(2! 2!) = 6

3 3 5!/(3! 2!) = 10

4 3 6!/(4! 2!) = 15

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 11

Distributing energy.Number of microstates.

• The number of possible microstatesquickly becomes very larger when weincrease the number of oscillators.

• Consider two blocks, brought incontact. One block has 300oscillators (100 atoms) and the secondblock has 200 oscillators (67 atoms).

• When we distribute 100 quanta ofvibrational energy among the blocks,we find that the maximum number ofstats occurs when 60 quanta are givento the first block and 40 to the secondblock.

Width is proportional to 1/√q or 1/√N

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 12

Distributing energy.Achieving equilibrium.

• Consider bringing the two blocks incontact when the first block has 90quanta of vibrational energy and thesecond block only 10.

• The process of exchange of energy isa random (statistical) process, thedirection of transfer will be in thedirection in which the number ofpossible states increases.

• The exchange of energy will continueeven when the most probabledistribution is reached, but at thatpoint there will only be smallfluctuation around the most probabledistribution.

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 13

Distributing energy.Number of states: different representations.

• Since the number of states of asystem is enormous (even forsmall system with a few hundredoscillators) it is often moreconvenient to look at the naturallogarithm of the number of states.

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 14

Entropy.• Since the evolution of the system

depends on the number ofmicrostates, the concept of entropyis introduced.

• The entropy S of one of the blockswe have been discussing is definedas

S = k lnΩ

where k is the Boltzmann constant(1.4 x 10-24 J/K).

• The entropy of the total system isthe sum of the entropy of the blocksthat make up the system.

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 15

Entropy.

• The most probable configuration ofthe system of the two blocks is theconfiguration for which the entropyhas a maximum.

• When we bring two blocks ofdifferent temperature in contact,energy exchange will take placeuntil the system has achievedthermal equilibrium.

• Thermal equilibrium is defined asthe energy distribution whichmaximizes the entropy of thesystem (in this case the two blocks).

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 16

Entropy and the Second Law of Thermodynamics.

• To achieve thermal equilibrium, thesystem will maximize its entropy.

• The most likely evolution of thesystem is the focus of the secondlaw of thermodynamics:

If a closed system is not inequilibrium, the most probableconsequence is that the entropy ofthe system will increase.

• Note: even when the two blocks arein thermal equilibrium, there maystill be exchange of energy betweenthe blocks, but the time averagedenergy exchange will be zero.

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 17

Entropy and the Second Law of Thermodynamics.

• If we know the entropy as functionof for example the number ofvibrational quanta of block 1, wecan express the condition forequilibrium as

or

dSdq1

=dS1

dq1

+dS2

dq1

= 0

dS1

dq1

= −dS2

dq1

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 18

3 Minute 03 Second Intermission.

• Since paying attention for 1 hourand 15 minutes is hard when thetopic is physics, let’s take a 3minute 03 second intermission.

• You can:• Stretch out.• Talk to your neighbors.• Ask me a quick question.• Enjoy the fantastic music.• Solve a WeBWorK problem.

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 19

Entropy and the temperature.

• Since q1 + q2 = constant (assumingthere is no exchange of energy withthe environment surrounding the twoblocks), the condition for equilibriumcan be rewritten as

• Since thermal equilibrium is achievedwhen the temperature of the blocksare the same, the slope of the entropycurves must be related to thetemperature of the blocks.

dS1

dq1

=dS2

dq2

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 20

Entropy and the temperature.

• The temperature of a system isdefined as

• The temperature defined in thismanner is expressed in units of Kelvin(K) since the internal energy ismeasured in units of Joules and theentropy is measured in units of J/K.

• Using the definition of temperaturewe conclude that block 1 initially hasa higher temperature than block 2.

1T=

dSdEint

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 21

Entropy and temperature.

• The maximum entropy is reachedwhen

or

• Assuming that each oscillator hasthe same energy quantization, thiscondition is equivalent to

dSdq1

=dS1

dq1

+dS2

dq1

= 0

dS1

dq1

= −dS2

dq1

=dS2

−dq1( ) =dS2

dq2

1T1

= 1T2

⇔ T1 = T2

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 22

The Boltzmann distribution.• Question:

What does the fundamental theoryof statistical mechanics tell usabout the probability to see asingle atom with a specificenergy?

• Answer:A lot.

• Consider a single oscillator incontact with 300 other oscillators.

• The entropy of this system isequal to the entropy of the 300oscillators.

• If the energy of the singleoscillator increases, the numberof states decreases.

Energy of300 oscillators Energy of

1 oscillator

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 23

The Boltzmann distribution.• The the entropy curve is close to

linear at small ΔE and its slope isequal to dS/dE1.

• We can approximate thedependence of the entropy on ΔEin the following manner:

• The number of states of thesystem, as function of ΔE, cannow be determined:

Ω = elnΩ = eS / k = eA/ ke−ΔE / kT

S = A− dS

dE ΔE = A− ΔET

Energy of300 oscillators Energy of

1 oscillator

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 24

The Boltzmann distribution.

• Since the probability isproportional to the number ofstates, we conclude that:

The probability of finding amicroscopic system to be in astate with energy ΔE above theground state of the system isproportional to e-ΔE/kT.

• This probability distribution iscalled the Boltzmanndistribution.

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 22, Page 25

Enough physics for a day!