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## The Partition Function, Physics tutorial

:IntroductionThe Partition functions as a normalization factor, calculate average energy, everything regarding system. The partition function regarding an ideal Monoatomic Gas are as well highlighted and the Sackur-Tetrode formula.

:The Partition Function ZThe Partition Function is stated for all systems for which the Boltzmann distribution applies from the single atoms to macroscopic systems. In another words, the partition function can be stated for all the systems in thermal equilibrium having a heat bath.

That is, Z = Σgie

^{-βε}Here, g = degenacy of the energy level and β = 1/KT (that is, Adiabatic Bulk Modulus)

In quantum mechanics E

_{i }= energy.The probability P(E

_{i}) that a system will be in state ith energy E_{i}is represented by:P(E

_{i}) = ni/N = 1/Z gie^{-βεi}The mean energy E‾ = Σ P

_{i}E_{i}= Σ ε_{i}(1/Z) e^{-βεi}E‾ = 1/Z Σ ε

_{i}e^{-βεi}Pr = Ge

^{s/k}(E_{o}) - β (E_{i})Total Probability = 1

εP

_{r}= Σ Ge^{S/A}(E_{o}) e^{-β (Ei)}= 1εP

_{r }= Ge^{S/k}(E_{o}) Σe^{-β (Ei)}= 1G = [1/(e

^{S/k }Σe^{-βEr})] (e^{S/k }- βEr)εP

_{r}= e^{-βEr}/Σe^{-βEr}=> ΣP

_{r}= 1/2 e^{-βEr}The expression provides us the probability that a system when positioned in a heat bath the system must be in a particular state E

_{i}Remember that the Boltzmann distribution provides the relative probability of a system in thermal equilibrium having a heat bath, to be in a single microstate of the given energy

Pr (microstate of energy E

_{i}) = exp (-E_{i}/KT)That is PF is the 'addition of relative probabilities of all the microstates' and it is applicable in:

Partition Function as the Normalization Factor:

As it is illustrated on the Boltzmann distribution, Z (β) is the sum of relative probabilities of all the microstates or equally, the sum of the relative probabilities of each and every energy value. This is defined only for the system in thermal equilibrium (that signifies it can endanger energy having a temperature bath).

Z(β) =

_{S=1}Σ^{all states exp}(-βE_{S}) =_{j=1}Σ^{all energy level }gj exp (-βE_{j})Here gj = multiplicity of even energy E

_{j}Z(β) =

_{j=1}Σ^{all energy levels }Rel Fr(E_{j})Note that for the system held in a heat both, the system energy is a variable (that is, not a constraint). The system might encompass more energy than the average, or less, all the way down to zero. That is extremely unlikely for most systems, however for mathematical expedience it is still counted as the possibility in Z(β).

As the partition function is the sum of relative probabilities.

This randomness as well reflects the arbitrary zero of energy. Changing the zero energy simply multiplies the PF by a constant without effect on the physics.

The PF serves up as the (inverse of the) normalization factor to change the relative probability of the system being in certain microstate, or having some energy, to an absolute probability:

Pr (State = S) = [Rel Pr (E

_{S})]/[_{r=1}Σ^{all states }Rel Pr (stater)] = exp (-βE_{S})/Z(β)Pr (Energy = E) = [Rel Pr(E)]/[

_{j=1}Σ^{all energy}^{level}Rel Pr(E_{j})] = [Rel Pr(E)]/Z(β)Although this is trivial and simply employs the letter 'Z' for somewhat that is already well familiar.

Partition Function for Average energy:

The other use for Z(β) is to calculate average energy (this is as well trivial and gives no new physical insight). Remember from the fundamental statistics that for any discrete random variable, state E:

(E) = Σ Pr (E

_{i}) E_{i}Thus, for the energy of a system we encompass:

(E) =

_{n=1}Σ^{all states }Pr (E_{s}) E_{s}=_{s=1}Σ^{all states }[Rel Pr (E_{s})/Z(B)]E_{s}= [1/Z(B)]_{j=1}Σ^{all states }e^{-BεS}ε_{S}However purely as a mathematical trick, having no deep physical meaning, we observe that the factor:

e

^{-βE}E_{S}= (δ/δβ) e^{-βE}=> (E) = 1/2(β) Σ -(δ/δβ) e^{-βE}e

^{-βE}E_{S}= (δ/δβ)_{S=1}Σ^{all state}e^{-βE }= -[1δ/Z(β)δβ] Z(β)However this provides us nothing new, as the partition function comprises all the terms required for the basic formula for average value.

From the equation above we could have more simply calculated the average energy from the basic formula.

-[1δ/Z(β)δβ] Z(β) = (δ/δβ) ln Z(β)

This consists of the benefit; however that ln Z (β) ≡ free energy, therefore there is certain physical significance to this and it leads directly to:

(E) = - (δ/δβ) ln Z(β) = (δ/δβ) A (β)

Here A(β) is the free energy of the system

Free Energy and the Partition Function:

Whenever considering the possible macrostate of a system, what actually matters is the most probable macrostate of a system, and to be meaningful, it should be the only macrostate having any important chance of occurring.

To calculate the free energy from the PF, we make use of the general conclusion that, for macroscopic systems, the partition function is dominated through the one most probable macrostate.

Mathematically, calculate the compute sum

Z ≡ Sum of all the state relative probabilities.

As the most probable state and the complete sum are equivalent, we do whatever simpler. The equilibrium microstate of a system is simply the most probable macrostate in thermal equilibrium; this in turn is that microstate having the largest number of likely microstates. That is, largest product of (Boltzmann factor) x (# microstates).

ε

_{most likely }= max (g j ε^{-βεj})j = all energies

Thus in principle, we can determine the equilibrium macrostate via first recognizing all the macrostates, their probabilities, and the macrostate for each and pick that microstate having the largest number of likely microstates. This is only meaningful whenever the most probable microstate consists of the probability close to 1, so that all the other macrostates are so unlikely they can be ignored. For systems having a large number of subsystems, N (example: large number of particles), this condition is true only one macrostate consists of any reasonable chance of occurring.

We can now exhibit that for systems having a large number of component subsystems (that is, system of large N) there is a single approximation to the Helmholtz free energy of the system from the partition function.

A(β) ≈ -KT ln Z (β) (Large system)

Here is why: for large N, there will be just a few terms in the partition function that dominate not only all the other terms, however all the other terms combined.

Such few terms all have extremely nearly the same energy, therefore they can be lumped altogether as a single multiplicity for that energy.

Z(β) =

_{j=1}Σ^{all energy level }gj exp(-E_{j}/KT) ≈_{m}Σ^{state near maximum }gm exp(-Emax/KT)Here, E

_{max}is the dominant energy in the sum. This signifies that the probability of the system being at E_{max}is essentially 1:Pr(E

_{m}) = Σg_{m }exp (E_{max}/KT)/Z(β) = Z(β)/Z(β) = 1This is as we anticipate for systems of large 'N' in thermal equilibrium; there is just 1 realistic macrostate. Now if Z(β) consists of only one significant energy E

_{max}than we can rewrite Z(β) is Z(T) to follow the reasoning of the free energy. We can substitute the sum of the few g_{m}close to E_{max}having a single g_{eff}, as they all contain essentially the similar energy, E_{max}.S(ε

_{max}) ≡ L ln g_{eff}is the entropy of the macrostate having energy E_{max}.= exp [- A(T)/KT)]

Here A(T) is the free energy of the macrostate at temperature 'T'.

Once again, we are just quantifying how multiplicity states energy, as before, and measuring multiplicity, in units of energy.

Resolving for A(T) yields the outcome above.

A(T) ≈ -KT ln Z (T) (Large system)

Note that as 'Z' is stated only up to the multiplicative constant, A(T) is stated only up to an additive constant that is for all time true for energy (that is, free or otherwise). Note that for small systems, like atoms, 'temperature' is not defined and thus, neither is free energy. The above approximation just works for large systems, where the thermal equilibrium macrostate is sharply stated.

The partition function is computed for a given set of constraints, just similar to the different types of free energies.

For a system constrained as for the Helmholtz free energy (fixed T. no work done) we calculate Z(T) in such constraints, and A = - KT ln Z provides Helmholtz free energy. Whenever the system had different constraints like those of Gibbs free energy (fixed T and P) then Z(T) is different and G = - KT ln Z provides Gibbs free energy.

:Partition Function of an Ideal Monoatomic GasAssume that an ideal monoatomic gas comprising of 'N' particles, each of mass 'm' and occupying a volume 'V'.

This signifies that the energy of the system is completely translational. That is, the potential energy is zero as intermolecular forces are absent.

The energy of a particle in the ith cell is represented by:

ε

_{i}= P^{2}/2mAnd the single particle partition function is:

Z = Σ

_{i}e^{-βεi}= Σ_{i }exp [- (-βP_{i}^{2})/2m]If energy is a continuous variable, then we can rewrite it as:

Z = 1/h

^{3}∫∫∫∫∫∫ exp [- β/2m (P_{x}^{2}+ P_{y}^{2}+ P_{z}^{2})] dx dy dz dpx dpy dpzIntegration over the space variable provides 'V' so that:

Z = V/h

^{3}∫_{-∞}∫^{∞}∫ exp {-β/2m ((P_{x}^{2}+ P_{y}^{2}+ P_{z}^{2})} dpx dpy dpzIt must be noted that the three integrals are similar and it is sufficient to compute any one of them.

By computing and rearranging we get,

Z

_{N}= (Σ_{i}e^{-βEi})^{N}= Z^{N}Here Z is given by E, Therefore,

Z

_{N}= Z^{N }= (V^{N}/h^{3}N) (2mπ/β)^{3N/2}Having got an expression for the partition function for a N-particle ideal gas, we can now compute for others.

The Sacker-Tetrode:The correct method to treat a system of indistinguishable particle is to use the quantum statistics.

Though, we can employ an ad-hoc procedure recommended by Gibbs. This is termed as correct Boltzmann counting. We start by looking at the equation for the thermodynamic probability W. Treating the particles as indistinguishable amounts to dividing W by N!. Therefore,

W = 1/π

_{i=1}ni!In the 'P' function for an ideal gas made up of 'N' indistinguishable particles, this is reflected in the form.

Z

_{N}^{C}= (Z_{N}) correct = Z_{N}/N! = V^{N}/N! (2πmK_{B}T/h^{2})^{3N/2}Take cogent of both sides, we obtain:

ln Z

_{N}^{C}= N [ln V + 3/2 ln(K_{β}T) + 3/2 ln (2πm/h^{2})] - ln N!For large N

_{1}we can make use of stirlings approximation ln N! = N ln N - N to getln Z

_{N}^{C}= N [ln(V/N) + 3/2 ln (K_{β}T) + 3/2 ln (2πm/h^{2}) + 1]As U and P based on the derivative of Z, the presence of the factor N! leaves their expression unchanged. Though, for entropy, the expression is changed to:

S = N K

_{B }[ln (VT^{3/2}/N) + δ_{o}]With δ

_{o}= 1 + δ = ln {(2πmK_{β}/h^{2})^{3/2}e^{5/2}}Hence,

S = NK

_{B}ln {(V/Nλ^{3}_{dβ}) π^{3/2 }e^{5/2}}Where λ

_{dB}is de Broglie Wavelength related by the gaseous particles at temperature 'T'.This equation is termed as sacker - Tetrode Equation. This has been established experimentally that this is the right expression for the entropy of an ideal monoatomic gas at high temperature.

:Diatomic GasesAssume that a diatomic molecule such as HCL. It might be treated as a two particle (atom) system hold through inter-atomic forces all along the line joining the particles. Assume that the masses of atoms be m

_{1}and m_{2}(supposed to be point like) and separated by a distance 'r'.This is termed as the dumb-bell model of the molecule and is depicted in the figure shown below:

Let us select x-axis all along the line joining the masses. The moments of inertia regarding the two axes at right angles to the line joining m

_{1}and m_{2 }and passing via the centre of gravity C_{2}is represented by:I

_{y}= I_{z}= I = (m_{1}m_{2}/m_{1}+m_{2}) r^{2}= μr^{2}where μ = m_{1}m_{2}/m_{1}+m_{2}Is the reduced mass of the molecule. The moment of inertia about the line connecting the molecule is taken to be equivalent to zero. The kinetic energy of the molecule is:

E = 1/2 I

_{y}W^{2}_{y}+ 1/2 I_{z}W^{2}_{z}= 1/2 I(W^{2}_{y }+ W^{2}_{z})Whenever the bonding is not perfectly rigid, such atoms can vibrate about their corresponding equilibrium positions. The simplest supposition is that each and every atom executes simple harmonic motion. The motion of such atoms can be reduced to the harmonic vibration of a single point mass 'μ' about an equilibrium position.

Therefore, for a diatomic molecule, we can encompass two vibrational degrees of freedom, apart from the translational and rotational degrees of freedom.

The total number of degrees of freedom:

f = f

_{tran}+ f_{rot}+ f_{vib}f = 3 + 2 + 2 = 7

As each and every degree of freedom in classical physics is related by the energy (K

_{β}T/2). We found out that:E‾ = 7/2 K

_{B}TSo that the heat capacity for the gas made of 'N' particles is C

_{V}= 7/2 R and the ratio of head capacitiesγ = 9/7 = 1.29

It represents that the heat capacity of a gas is constant; independent of temperature and similar for all gases. And γ, for a diatomic gas is less than the value for the monatomic gas.

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