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Statistical Mechanics

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Presentation on theme: "Statistical Mechanics"— Presentation transcript:

1 Statistical Mechanics
Shihabudheen M M E S Kalladi College Mannarkkad

2 Statistical distributions- Maxwell-Boltzmann statistics –
Distribution of molecular energies in an ideal gas-Average molecular energy- Equipartition theorem Maxwell-Boltzmann speed distribution law- Expressions for rms speed, most probable speed and mean speed

3 The branch of physics called statistical mechanics considers how the overall behavior of a system of many particles is related to the properties of the particles themselves. The statistics obeyed by any system generally belongs to one of the following two categories: Classical Statistics Quantum Statistics. Classical statistics treats particles as distinguishable whereas quantum statistics treats them as indistinguishable.

4 Statistical Distributions
Statistical mechanics deals with the behavior of systems of a large number of particles. We give up trying to keep track of individual particles. If we can’t solve Schrödinger’s equation in closed form for helium (4 particles) what hope do we have of solving it for the gas molecules in this room (10really big number particles). Statistical mechanics handles many particles by calculating the most probable behavior of the system as a whole, rather than by being concerned with the behavior of individual particles.

5 Four balls and two boxes
1 2


7 statistical mechanics determines the most probable way in which a certain total amount of energy E is distributed among the N members of a system of particles in thermal equilibrium at the absolute temperature T. Thus we can establish how many particles are likely to have the energy 1, how many to have the energy 2, and so on. E N  n4  n3  n2  n1

8 The particles are assumed to interact with one another and with the walls of their container to an extent sufficient to establish thermal equilibrium More than one particle state may correspond to a certain energy . If the particles are not subject to the exclusion principle, more than one particle may be in a certain state.

9 Let W be the number of different ways in which the particles can be arranged among the available states to yield a particular distribution of energies The greater the number W, the more probable is the distribution. It is assumed that each state of a certain energy is equally likely to be occupied. (equal a priori probability )

10 The program of statistical mechanics begins by finding a general formula for W for the kind of particles being considered. The most probable distribution, which corresponds to the system being in thermal equilibrium, is the one for which W is a maximum, subject to the condition that the system consists of a fixed number N of particles whose total energy is some fixed amount E. (The No. of particles in a state of energy E) = (The No. of states having energy E) x (probability that a particle occupies the state of energy E).

11 If we know the distribution function, the (probability that a particle occupies a state of energy E), we can make a number of useful calculations. Mathematically, the equation is written g( ) = number of states of energy (statistical weight corresponding to energy ) f( ) = distribution function = average number of particles in each state of energy  = probability of occupancy of each state of energy 

12 In systems such as atoms, only discrete energy levels are occupied, and the distribution g() of energies is not continuous. On the other hand, it may be that the distribution of energies is continuous, or at least can be approximated as being continuous. In that case, we replace g(ε) by g(ε)dε, the number of states between ε and ε+dε. We will find that there are several possible distributions f(ε) which depend on whether particles are distinguishable, and what their spins are.

13 Three different kinds of particles:
Identical particles that are distinguishable. Two particles can be considered distinguishable if their separation is large compared to their de Broglie wavelength. Example: ideal gas molecules. In quantum terms, the wave functions of the particles overlap to a negligible extent. The Maxwell-Boltzmann distribution function holds for such particles.

14 Three different kinds of particles
Identical particles of zero or integral spin that cannot be distinguished one from another because their wave functions overlap. Such particles, called bosons They do not obey the exclusion principle The Bose-Einstein distribution function holds for them. Photons are bosons We shall use Bose-Einstein statistics to account for the spectrum of radiation from a blackbody.

15 Three different kinds of particles
Identical particles with odd half-integral spin that also cannot be distinguished one from another. Such particles, called fermions, They obey the exclusion principle, The Fermi-Dirac distribution function holds for them. Electrons are fermions We shall use Fermi-Dirac statistics to study the behavior of the free electrons in a metal that are responsible for its ability to conduct electric current.

16 Statistics Classical statistics Quantum statistics
Maxwell-Boltzmann Statistics Quantum statistics Bose-Einstein statistics Fermi-Dirac statistics

17 The Maxwell-Boltzmann distribution:
We know that classical particles which are identical but far enough apart to be distinguishable obey Maxwell-Boltzmann statistics. Or Two particles can be considered distinguishable if their separation is large compared to their de Broglie wavelength. The Maxwell-Boltzmann distribution function states that the average number of particles fMB(ε) in a state of energy ε in a system of particles at the absolute temperature T is

18 The is the distribution function for ‘n’ no. of particles in ‘g’ states.
The number of particles having energy ε at temperature T is A is like a normalization constant; we integrate n(ε) over all energies to get N, the total number of particles. ε is the particle energy, k is Boltzmann's constant

19 Molecular Energies in an ideal Gas
We assume a continuous distribution of energies so that, the no. of particles having energies between ϵ and ϵ+dϵ is given by where g(ϵ) dϵ is the no of states in the energy range ϵ and ϵ+dϵ. g(ϵ) is called density of states. It turns out to be easier to find the number of momentum states corresponding to a momentum p, and transform back to energy states.

20 Corresponding to every value of momentum there is a value of energy.
Momentum is a 3-dimensional vector quantity. Every (px,py,pz) corresponds to some energy. We count how many momentum states are there in a region of space and then transform to the density of energy states. We need to find how many momentum states are in this spherical shell.

21 The Maxwell-Boltzmann distribution is for classical particles, so we write
The number of momentum states in a spherical shell from p to p+dp is proportional to 4πp2dp (the volume of the shell). Thus, we can write the number of states having momentum between p and p+dp as where B is a proportionality constant, which we will worry about later.

22 Because each p corresponds to a single ε,
Now, so that and The constant C contains B and all the other proportionality constants lumped together.

23 To find the constant C, we evaluate
where N is the total number of particles in the system. Now The result is

24 so that This is the number of molecules having energy between ε and ε+dε in a sample containing N molecules at temperature T. “It forms the basis of the kinetic theory of gases, which accurately explains many fundamental gas properties, including pressure and diffusion.” “The Maxwell-Boltzmann distribution also finds important applications in electron transport and other phenomena.”

25 Here’s a plot of the distribution:

26 Continuing with the article, the total energy of the system is
Evaluation of the integral gives This is the total energy for the N molecules, so the average energy per molecule is exactly the result you get from elementary kinetic theory of gases.

27 Calculation:

28 Results of Kinetic Theory
KE of individual particles is related to the temperature of the gas: ½ mv2 = 3/2 kT Where v is the average velocity.

29 Boltzmann Distribution
Demonstrated that there is a wide range of speeds that varies with temperature.

30 Equipartition of Energy
Each degree of translational freedom takes ½ kT. KEx +KEy+KEz = ½ kT + ½ kT + ½ kT KEtotal = 3/2 kT This is true for single point masses that possess no structure. Each new DOF requires ½ kT of energy. Each new DOF contributes ½ kT to the total internal energy of the gas. This is the Equipartition Theorem.

31 Equipartition of Energy
For molecules, i.e. multi-atom particles, there are added degrees of freedom.

32 Internal Energy of Di-Atom
Three translational DOF + 2 rotational DOF = 5 DOF. Each DOF contributes ½ kt, so the internal energy of a diatomic gas is, U = 5/2 NkT, For a gas of N molecules.

33 Here’s a plot (number having a given speed vs. speed):
Because ε = mv2/2, we can also calculate the number of molecules having speeds between v and v + dv. The result is “We” (Beiser) call this n(v). Here’s a plot (number having a given speed vs. speed):

34 The speed of a molecule having the average energy comes from solving
for v. The result is vrms is the speed of a molecule having the average energy . It is an rms speed because we took the square root of the square of an average quantity.

35 The average speed can be calculated from
The result is Comparing this with vrms, we find that

36 You can also set dn(v) / dv = 0 to find the most probable speed
You can also set dn(v) / dv = 0 to find the most probable speed. The result is The subscript “p” means “most probable.” Summarizing the different velocity results:

37 Plot of velocity distribution again:

38 Example 9.4 Find the rms speed of oxygen molecules at
0 m/s? 10 m/s? 100 m/s? 1,000 m/s? 10,000 m/s? You need to know that an oxygen molecule is O2. The atomic mass of O is 16 u (1 u = 1 atomic mass unit = 1.66x10-27 kg).


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