Chap5 Microcanonical Ensemble v04

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    ME346A Introduction to Statistical Mechanics Wei Cai Stanford University Win 2011

    Handout 5. Microcanonical Ensemble

    January 19, 2011

    Contents

    1 Properties of flow in phase space 2

    1.1 Trajectories in phase space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 One trajectory over long time . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 An ensemble of points flowing in phase space . . . . . . . . . . . . . . . . . . 6

    2 Microcanonical Ensemble 7

    2.1 Uniform density assumption . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.3 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

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    The purpose of this lecture is

    1. To justify the uniform probability assumption in the microcanonical ensemble.

    2. To derive the momentum distribution of one particle in an ideal gas (in a container).

    3. To obtain the entropy expression in microcanonical ensemble, using ideal gas as anexample.

    Reading Assignment: Sethna 3.1, 3.2.

    1 Properties of flow in phase space

    1.1 Trajectories in phase space

    Q: What can we say about the trajectories in phase space based on classicalmechanics?

    A:

    1. Flow line (trajectory) is completely deterministicqi =

    Hpi

    pi = Hqi(1)

    Hence two trajectories never cross in phase space.

    This should never happen, otherwise the flow direction of point P is not determined.2. Liouvilles theorem

    d

    dt

    t+ {, H} = 0 (2)

    So there are no attractors in phase space.

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    This should never happen, otherwise d/dt > 0. Attractor is the place where manytrajectories will converge to. The local density will increase as a set of trajectoriesconverge to an attractor.

    3. Consider a little cube in phase space. (You can imagine many copies of the systemwith very similar initial conditions. The cube is formed all the points representing theinitial conditions in phase space.) Let the initial density of the points in the cube beuniformly distributed. This can be represented by a density field (, t = 0) that isuniform inside the cube and zero outside.

    As every point inside the cube flows to a different location at a later time, the cube istransformed to a different shape at a different location.

    Due to Liouvilles theorem, the density remain = c (the same constant) inside thenew shape and = 0 outside. Hence the volume of the new shape remains constant(V0).

    1.2 One trajectory over long time

    Q: Can a trajectory from on point 1 in phase space always reach any otherpoint 2, given sufficiently long time?

    A: We can imagine the following possibilities:

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    1. We know that the Hamiltonian H (total energy) is conserved along a trajectory.

    So, there is no hope for a trajectory to link 1 and 2 if H(1) = H(2).

    Hence, in the following discussion we will assume H(1) = H(2), i.e. 1 and 2lie on the same constant energy surface: H() = E. This is a (6N 1)-dimensionalhyper-surface in the 6N-dimensional phase space.

    2. The trajectory may form a loop. Then the trajectory will never reach 2 if2 is notin the loop.

    3. The constant energy surface may break into several disjoint regions. If1 and 2 arein different regions, a trajectory originated from 1 will never reach 2.

    Example: pendulum

    4. Suppose the trajectory does not form a loop and the constant energy surface is onecontinuous surface. The constant energy surface may still separate into regions wheretrajectories originated from one region never visit the other region.

    This type of system is called non-ergodic.

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    5. If none of the above (1-4) is happening, the system is called Ergodic. In an ergodicsystem, we still cannot guarantee that a trajectory starting from 1 will exactly gothrough any other part 2 in phase space. (This is because the dimension of the phasespace is so high, hence there are too many points in the phase space. One trajectory,no matter how long, is a one-dimensional object, and can get lost in the phase space,

    i.e. not dense enough to sample all points in the phase space.)But the trajectory can get arbitrarily close to 2.

    At time t1, the trajectory can pass by the neighborhood of2. At a later time t2, thetrajectory passes by 2 at an even smaller distance...

    After a sufficiently long time, a single trajectory will visit the neighborhood of every pointin the constant energy surface.

    This is the property of an ergodic system. Ergodicity is ultimately an assumption,because mathematically it is very difficult to prove that a given system (specified by

    its Hamiltonian) is ergodic.

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    1.3 An ensemble of points flowing in phase space

    Now imagine a small cube (page 1) contained between two constant-energy surfaces H = E,H = E+ E.

    As all points in the cube flows inphase space. The cube transformsinto a different shape but its volumeremains V0.

    The trajectories of many non-linearsystems with many degrees of free-dom is chaotic, i.e. two trajecto-

    ries with very similar initial con-ditions will diverge exponentiallywith time.

    Q. How can the volume V0 remainconstant while all points in the orig-inal cube will have to be very farapart from each other as time in-creases?

    A: The shape ofV0

    will become verycomplex, e.g. it may consists ofmany thin fibers distributed almostevery where between the two con-stant energy surface.

    At a very late time t, (, t) stillhas the form of

    (, t) =

    C V00 / V0 (3)

    except that the shape of V0 is dis-tributed almost every where be-tween constant energy surface.

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    A density function (, t) corresponds to an ensemble of points in phase space. Suppose wehave a function A() defined in phase space. (In the pendulum example, we have consideredA = 2.)

    The average of function A() over all points in the ensemble is called the ensemble average.

    If changes with time, then the ensemble average is time dependent

    A(t)

    d6NA() (, t) (4)

    From experience, we know that many system will reach an equilibrium state if left alone fora long time. Hence we expect the following limit to exist:

    limt

    A(t) = Aeq (5)

    Aeq is the equilibrium ensemble average.Q: Does this mean that

    limt

    (, t) = eq()? (6)

    No. In previous example, no matter how large t is,

    (, t) =

    C V00 / V0 (7)

    The only thing that changes with t is the shape of V0. The shape continues to transformwith time, becoming thinner and thinner but visiting the neighborhood of more and morepoints in phase space.

    So, limt

    (, t) DOES NOT EXIST!

    Whats going on?

    2 Microcanonical Ensemble

    2.1 Uniform density assumption

    In Statistical Mechanics, an ensemble (microcanonical ensemble, canonical ensemble, grandcanonical ensemble, ...) usually refers to an equilibrium density distribution eq() that doesnot change with time.

    The macroscopically measurable quantities is assumed to be an ensemble average over eq().

    Aeq

    d6NA() eq() (8)

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    In the microcanonical ensemble, we assume eq to be uniform inside the entire regionbetween the two constant energy surfaces, i.e.

    eq() = mc() = C E H() E+ E

    0 otherwise

    (9)

    There is nothing micro in the microcanonical ensemble. Its just a name with an obscurehistorical origin.

    Q: How do we justify the validity of the microcanonical ensemble assumption, given thatlimt

    (, t) = mc() (recall previous section)?

    A:

    1. As t increases, (, t) becomes a highly oscillatory function changing volume rapidlybetween C and 0, depending on whether is inside volume V0 or not.

    But if function A() is smooth function, as is usually the case, then it is reasonable toexpect

    limt

    d

    6N

    A() (, t) =

    d

    6N

    A() mc() (10)In other words, lim

    t(, t) and eq() give the same ensemble averages.

    2. A reasonable assumption for eq() must be time stationary, i.e.

    eqt

    = {eq, H} = 0 (11)

    Notice thatmc() = [(H() E) (H() E E)] C (12)

    where (x) is the step function.

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    Because mc is a function of H {mc, H} = 0.

    Hencemc

    t = 0 (13)The microcanonical ensemble distribution mc is stationary!.

    3. The microcanonical ensemble assumption is consistent with the subjective probabilityassignment. If all we know about the system is that its total energy H (which shouldbe conserved) is somewhere between E and E + E, then we would like to assignequal probability to all microscopic microstate that is consistent with the constraintE H() E+ E.

    2.2 Ideal Gas

    (ensemble of containers each having Nideal gas molecules)

    Ideal gas is an important model in statis-tical mechanics and thermodynamics. Itrefers to N molecules in a container. Theinteraction between the particles is suffi-ciently weak so that it will be ignored inmany calculations. But conceptually, theinteraction cannot be exactly zero, other-wise the system would no longer be ergodic a particle would never be able to trans-fer energy to another particle and to reachequilibrium when there were no interac-tions at all.

    Consider an ensemble of gas contain-ers containing ideal gas particles (mono-atomic molecules) that can be describedby the microcanonical ensemble.

    Q: What is the velocity distributionof on gas particle?

    The Hamiltonian of N-ideal gas molecules:

    H({qi}, {pi}) =3Ni=1

    p2i2m

    +Ni=1

    (xi) (14)

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    where (x) is the potential function to represent the effect of the gas container

    (x) =

    0 if x V (volume of the container) if x / V (15)

    This basically means that xi has to stay within volume V and when this is the case, we canneglect the potential energy completely.

    H({qi}, {pi}) =3Ni=1

    p2i2m

    (16)

    The constant energy surface H({qi}, {pi}) = E is a sphere in 3N-dimensional space, i.e.,3Ni=1

    p2i = 2mE = R2 (17)

    with radius R =

    2mE.

    Lets first figure out the constant C in the microcanonical ensemble,

    mc() =

    C E H() E+ E0 otherwise

    (18)

    Normalization condition:

    1 =

    d6N mc() =

    EH()E+E

    d6NC =

    (E+ E) (E)

    C (19)

    where(E) is the phase space volume of region H(

    ) E and(E + E) is the phasespace volume of region H() E+ E. This leads to

    C =1

    (E+ E) (E) (20)

    How big is (E)?

    (E) =

    H()E

    d6N = VN 3N

    i=1 p2i2mE

    dp1 dpN (21)

    Here we need to invoke an important mathematical formula. The volume of a sphere of

    radius R in d-dimensional space is,1

    Vsp(R, d) = d/2 Rd

    (d/2)!(22)

    1It may seem strange to have the factorial of a half-integer, i.e. (d/2)!. The mathematically rigorousexpression here is (d/2 + 1), where (x) is the Gamma function. It is defined as (x)

    0tx1 et dt.

    When x is a positive integer, (x) = (x 1)!. When x is not an integer, we still have (x + 1) = x(x).12

    =

    . Hence

    32

    = 1

    2

    ,

    52

    = 3

    4

    , etc. We can easily verify that Vsp(R, 3) = 43 R3 andVsp(R, 2) = R2.

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    The term behind VN is the volume of a sphere of radius R =

    2mE in d = 3N dimensionalspace. Hence,

    (E) = VN 3N/2R3N

    (3N/2)!(23)

    limE0

    (E+ E) (E)E

    =(E)

    E

    =3N

    2

    1

    E

    (2mE)3N/2

    (3N/2)!VN

    =(2m)3N/2 E3N/21 VN

    (3N/2 1)! (24)

    In the limit of E 0, we can write

    C = 1E

    E(E+ E) (E)

    =1

    E (3N/2 1)!

    (2m)3N/2 E3N/21 VN(25)

    Q: What is the probability distribution of p1 the momentum of molecule i = 1 in thex-direction?

    A: The probability distribution function forp1 is obtained by integrating the joint distributionfunction mc(q1,

    , q3N, p1,

    , p3N) over all the variables except p1.

    f(p1) =

    dp2 dp3N dq1dq2 dq3Nmc(q1, , q3N, p1, , p3N)

    =

    2mE

    3Ni=1 p

    2i 2m(E+E)

    dp2 dp3NVNC

    =

    2mEp21

    3Ni=2 p

    2i 2m(E+E)p21

    dp2 dp3NVNC

    =

    Vsp

    2m(E+ E) p21, 3N 1

    Vsp

    2mEp21, 3N 1

    VNC(26)

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    In the limit of E 0,Vsp

    2m(E+ E) p21, 3N 1

    Vsp

    2mEp21, 3N 1

    E

    =

    E Vsp2mEp21, 3N 1=

    E

    (3N1)/2(2mEp21)(3N1)/2

    3N12

    !

    =3N 1

    2

    2m

    2mEp21(3N1)/2(2mEp21)(3N1)/2

    3N12

    !

    = 2m(3N1)/2(2mEp21)3(N1)/2

    3(N1)2

    !

    (27)

    Returning to f(p1), and only keep the terms that depend on p1,

    f(p1)

    2mEp21 3(N1)

    2

    1 p21

    2mE

    3(N1)2

    (28)

    Notice the identity

    limn

    1 +

    x

    n

    n= ex (29)

    and that N N 1 in the limit of large N. Hence, as N ,

    f(p1)

    1 23N

    3N

    2E

    p212m

    3N/2 exp

    p

    21

    2m

    3N

    2E

    (30)

    Using the normalization condition

    dp1 f(p1) = 1 (31)

    we have,

    f(p1) =1

    2m(2E/3N)exp

    p

    21

    2m

    3N

    2E

    (32)

    Later on we will show that for an ideal gas (in the limit of large N),

    E =3

    2

    N kB T (33)

    where T is temperature and kB is Boltzmanns constant. Hence

    f(p1) =1

    2mkBTexp

    p

    21/2m

    kBT

    (34)

    Notice that p21/2m is the kinetic energy associated with p1. Hence f(p1) is equivalentto Boltzmanns distribution that will be derived later (in canonical ensemble).

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    2.3 Entropy

    Entropy is a key concept in both thermodynamics and statistical mechanics, as well as ininformation theory (a measure of uncertainty or lack of information). In information theory,

    if an experiment has N possible outcomes with equal probability, then the entropy isS = kB log N (35)

    In microcanonical ensemble,

    S(N ,V ,E) = kB log

    number of microscopic states betweenthe constant energy surfaces:

    E H() E+ E

    (36)

    For an ideal gas,

    S(N ,V ,E) = kB log(E+ E)

    (E)

    N! h3N(37)

    The numerator inside the log is the volume of the phase space between the two constantenergy surfaces. h is Plancks constant, which is the fundamental constant from quantummechanics.

    Yes, even though we only discuss classical equilibrium statistical mechanics, a bare minimumof quantum mechanical concepts is required to fix some problems in classical mechanics.

    We can view this as another evidence that classical mechanics is really just an approximationand quantum mechanics is a more accurate description of our physical world. Fortunately,these two terms can be intuitively understandable without working with quantum mechanicsequations. The following are the justifications of the two terms in the denominator.

    1. N! term: Quantum mechanics says that the gas molecules are all identical or indistin-guishable. Even though we would like to label molecules as 1, 2,...,N there is reallyno way for us to tell which one is which! Therefore, two molecular structures withcoordinates: x1 = (1, 2, 3), x2 = (4, 5, 6) and x1 = (4, 5, 6), x2 = (1, 2, 3) are indistin-guishable from each other.

    Swapping the location between two molecules does not give a new microscopic state.

    2. h3N term: h = 6.626

    1034 Js is Plancks constant.

    The numerator, (E), is the phase space volume and has the unit of (momentum distance)3N.

    The term inside log has to be dimensionless, otherwise, the magnitude of entropy woulddepend on our choices for the units of length, time, mass, and etc, which would beclearly absurd.

    h has the unit of momentum distance. Therefore h3N has exactly the right unit tomake the entire term inside the log dimensionless.

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    The uncertainty principle in quantum mechanics states that we cannot measure boththe position and the momentum of any particle to infinite accuracy. Instead, theirerror bar must satisfy the relation:

    qi pi h for any i = 1, , 3N (38)

    Therefore, (E)/(N! h3N) gives us the number of distinguishable states contained in-

    side a phase space volume of (E).

    We can show that the entropy expression for the ideal gas in microcanonical ensemble is

    S(N ,V ,E) = NkB

    log

    V

    N

    4mE

    3Nh2

    3/2+

    5

    2

    (Sackur-Tetrode formula) (39)

    We will derive the Sackur-Tetrode formula later. (Stirlings formula is used to derive it.)

    Define number density = NV , and de Broglie wavelength

    =h

    4mE/3N

    =

    h2mkBT

    (40)

    then

    S(N ,V ,E) = NkB

    5

    2 log(3)

    (41)

    In molecular simulations, the microcanonical ensemble is usually referred to as the

    N V E ensemble.

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