1

THE IDEA OF STATISTICAL MECHANICS The branch of theoretical physics that deals with the study of the arrange behavior of mechanical system where state of system is uncertain. It is useful in explaining the thermodynamic behavior of large systems. To calculate the thermal properties we use statistical ideas together with a knowledge of laws of mechanics obeyed by the particle that make up a system. Hence the name Statistical Mechanics. The beauty of Statistical Mechanics comes from the very simplicity of the ideas, its uses and the power and generality of the techniques involved. There are two notions of probability that are commonly used statistical and classical probability. CLASSICAL PROBABILIY The more important is classical probability. It will be used to predict what should be seen in the experiment. To calculate classical probability we must list all the possible outcomes of the experimental and assign equal probability to each. But how do we know that these possible outcomes of the experiment correspond to simple or compound ones. If events are simple, it is simple to consign equal probabilities, if they are compound then the probabilities are unequal. We need a way to define a simple event for the outcomes of a single experiment. STATISTICAL PROBABILITY This probability is also very useful. We can obtain statistical probability of the outcome of a measurement from its relative frequency and this is only possible if a well-defined limit to the relative frequency exists and this will happen if the system is in equilibrium. IDEAS ON WHICH STATISTICAL MECHANICS IS BASED 1. State of the system is specified as completely as possible. Each such state will be called a quantum state. 2. In practice, it is impossible to make a complete set of simultaneous measurements, or we can say that we cannot make a complete specification of the state of a system so as to specify quantum state. If we want to use statistical methods, we are forced to use classical probability. 3. We assign the classical probability to each quantum state. It is convenient to imagine an ensemble of systems. The probability that a system is in a particular quantum state is given by the fraction of the ensemble in this state. 4. Suppose there are ‘W’ possible quantum states of the system which satisfy the set of constraints on energy, volume and rest. The essential assumptions of statistical mechanics is that all quantum states are equally probable. The probability that the system is in any of these quantum states is W1. By making this assumption we can calculate average values of any quantity we are interested in, such as average kinetic energy of a particle.

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2

This set of ideas is not complete. There is one extra hypothesis which is needed to construct statistical mechanics. BOLZMANN’S HYPOTHESIS Boltzmann’s hypothesis is that the entropy of a system is related to the probability of its being in a quantum state. The probability is 𝑃 = 𝑊 −1 . If there are W quantum states we can express Boltzmann’s hypothesis in the following way. 𝑆 = 𝜙(𝑊) Where 𝜙(𝑊) is the some unknown function of W. To determine 𝜙(𝑊) we use an approach due to Einstein (1905). Consider two systems A and B which are not interacting so they are independent of each other. Their entropies are, 𝑆𝐴 = 𝜙(𝑊𝐴 ) 𝑆𝐵 = 𝜙(𝑊𝐴 ) Instead of considering two systems separately, we could consider the two systems as a single system −1 with entropy SAB and probability 𝑃𝐴𝐵 = 𝑊𝐴𝐵 𝑆𝐴𝐵 = 𝜙(𝑊𝐴𝐵 ) The total entropy is the system of entropies of two systems. 𝑆𝐴𝐵 = 𝑆𝐴 + 𝑆𝐵 → (1) As two systems are independent, so 𝑆𝐴𝐵 = 𝜙(𝑊𝐴𝐵 ) = 𝜙(𝑊𝐴 𝑊𝐵 ) So equation (1) can be written as 𝜙(𝑊𝐴𝐵 ) = 𝜙(𝑊𝐴 𝑊𝐵 ) + 𝜙(𝑊𝐴 ) + 𝜙(𝑊𝐴 ) The only solution to above equation is 𝜙(𝑊) = 𝐾𝐵 𝑙𝑛𝑊 Hence Boltzmann’s hypothesis leads to a mathematical expression for the entropy involving a new universal constant KB now called Boltzmann’s constant. 𝑆 = 𝐾𝐵 ln⁡(𝑊) Let us illustrate the power of this equation by considering a simple model which could represent an ideal gas or a dilute solution. Consider ‘N’ non interacting molecules moving within a volume ‘V’. Imagine we specify the position of each molecule by surrounding the space into cells each of volume ∆V. The number of ways placing one 𝑉

𝑉

𝑁

particular molecule in the volume is 𝑊 = ∆𝑉. The number of ways of arranging ‘W’ molecules is (∆𝑉) . Since each molecule goes independently of the rest, so according to entropy expression

Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

3

𝑆 = 𝐾𝐵 ln⁡(𝑊) For N molecules 𝑆 = 𝐾𝐵 ln (

𝑉 N ) ⁡ ∆𝑉

According to this equation, the entropy depends on the volume of cell ∆V, something which is arbitrary. Now if we were to alter the volume of the system but to keep ∆V constant, the difference in entropy between initial and final states would be, ∆𝑆 = 𝑆𝑓 − 𝑆𝑖 = 𝑁𝐾𝐵 ln

𝑉𝑓 𝑉𝑖

That is independent of ∆V. Pressure of the system is given by 𝜕𝑆 𝑃 = 𝑇( ) 𝜕𝑉 𝑢 𝑉

As 𝑆 = 𝑁𝐾𝐵 ln 𝑉𝑓 𝑖

Differentiate w.r.t. V 𝜕𝑆 1 𝜕 𝑉 ∆𝑉 1 𝑁𝐾𝐵 = 𝑁𝐾𝐵 . . ( ) = 𝑁𝐾𝐵 . . = 𝑉⁄ 𝜕𝑉 𝜕𝑉 ∆𝑉 𝑉 ∆𝑉 𝑉 ∆𝑉 𝜕𝑆

As 𝑃 = 𝑇 ( )

𝜕𝑉 𝑢

𝑃=𝑇

𝑁𝐾𝐵 𝑉

This is the equation of state for an ideal gas. If 𝑁 = 𝑁𝐴 , then we can write 𝑃𝑉 = 𝑅𝑇⁡or⁡𝑃𝑉 = 𝑛𝑅𝑇

A SIMPLE MODEL OF SPINS ON LATTICE SITES Suppose there are n particles placed on lattice sites with each particle having a spin of

1 2

with associated

magnetic moment 𝜇. Figure shows a possible arrangement of spins Now let us place the spin in magnetic field with magnetic induction B along z-direction. The spin can either point up or down relative to z-axis.

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4 ħ

ħ

The spin-up state has a spin eigenvalue 2 and for spin down it is − 2. Similarly Energy eigenvalue for spin-up particle is ε = −μB Energy eigenvalue for spin-down particle is ε = μB As we have ‘N’ number of particle with each particle has a spin of either up or down. Let n1 particles be spin up and have energy – ε and let n2 particles be spin down and have energy ε. Now the total energy will be U = −n1 ε + n2 ε As N = n1 + n2 (total⁡number⁡of⁡particles)  n2 = N − n1 So total energy in terms of n1 will be 𝑈 = −𝑛1 𝜀 + (𝑁 − 𝑛1 )𝜀  𝑈 = −𝑛1 𝜀 + 𝑁𝜀 − 𝑛1 𝜀  𝑈 = 𝑁𝜀 − 2𝑛1 𝜀 → (𝐴) If we specify U and N and their both n1 and n2 are determined The number of ways of choosing n1 spin-up particles out of N is 𝑊=

𝑁! 𝑛1 ! (𝑁 − 𝑛1 )!

This is degeneracy of the N-particle system with energy U. All these W quantum states satisfy the constraints on system. As entropy is 𝑆 = 𝐾𝐵 ln 𝑊 For spin system, the entropy will be 𝑆 = 𝐾𝐵 ln (

𝑁! ) → (1) 𝑛1 ! (𝑁 − 𝑛1 )!

For small values of ‘N’, W can easily be calculated but for large values of N we use Stirling’s approximation according to which ln 𝑁! = 𝑁 ln 𝑁 − 𝑁 From equation (1) simplify ln (𝑛

𝑁!

) By using Stirling’s approximation and ‘ln’ rules

1 !(𝑁−𝑛1 )!

ln (

𝑁! ) = ln 𝑁! − ln 𝑛! − ln(𝑁 − 𝑛)! 𝑛1 ! (𝑁 − 𝑛1 )!

 ln 𝑊 = 𝑁 ln 𝑁 − 𝑁 − (𝑛1 ln 𝑛1 − 𝑛1 ) − [(𝑁 − 𝑛1 ) ln(𝑁 − 𝑛1 ) − (𝑁 − 𝑛1 )]  ln 𝑊 = 𝑁 ln 𝑁 − 𝑁 − 𝑛1 ln 𝑛1 + 𝑛1 − (𝑁 − 𝑛1 ) ln(𝑁 − 𝑛1 ) + 𝑁 − 𝑛1 Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

5

 ln 𝑊 = 𝑁 ln 𝑁 − 𝑛1 ln 𝑛1 − (𝑁 − 𝑛1 ) ln(𝑁 − 𝑛1 ) By adding and subtracting 𝑛1 ln 𝑁  ln 𝑊 = 𝑁 ln 𝑁 − 𝑛1 ln 𝑛1 − (𝑁 − 𝑛1 ) ln(𝑁 − 𝑛1 ) +⁡𝑛1 ln 𝑁 − 𝑛1 ln 𝑁 Arranging terms    

ln 𝑊 ln 𝑊 ln 𝑊 ln 𝑊

= 𝑁 ln 𝑁 − 𝑛1 ln 𝑁 − 𝑛1 ln 𝑛1 +⁡𝑛1 ln 𝑁 − (𝑁 − 𝑛1 ) ln(𝑁 − 𝑛1 ) = (𝑁 − 𝑛1 ) ln 𝑁 − (𝑛1 ln 𝑛1 −⁡𝑛1 ln 𝑁) − (𝑁 − 𝑛1 ) ln(𝑁 − 𝑛1 ) = (𝑁 − 𝑛1 ) ln 𝑁 − (𝑁 − 𝑛1 ) ln(𝑁 − 𝑛1 ) − (𝑛1 ln 𝑛1 −⁡𝑛1 ln 𝑁) = (𝑁 − 𝑛1 )[ln 𝑁 − ln(𝑁 − 𝑛1 )] − 𝑛1 (ln 𝑛1 − ln 𝑁) 𝑁

 ln 𝑊 = (𝑁 − 𝑛1 ) ln (𝑁−𝑛 ) − 𝑛1 ln 1

 ln 𝑊 = − [𝑛1 ln

𝑛1 𝑁

𝑛1 𝑁 𝑁

− (𝑁 − 𝑛1 ) ln (𝑁−𝑛 )] 1

Multiplying and dividing by ‘N’ 𝑛

𝑛1

𝑛1

𝑁 𝑛1

 ln 𝑊 = − 𝑁 [ 𝑁1 ln  ln 𝑊 = − 𝑁 [ 𝑁 ln

𝑁

𝑁−𝑛1

−(

𝑁

− (1 −

𝑁

) ln (𝑁−𝑛 )] 1

𝑛1 𝑁

𝑁

) ln (𝑁−𝑛 )] 1

𝑎

𝑏

Taking ‘-’ common from second term or using the rule − ln 𝑏 = ln 𝑎 𝑛

𝑛1

𝑛

𝑁 𝑛1

 ln 𝑊 = − 𝑁 [ 𝑁1 ln  ln 𝑊 = − 𝑁 [ 𝑁1 ln

𝑁

+ (1 − + (1 −

𝑛1 𝑁 𝑛1 𝑁

) ln (

𝑁−𝑛1 𝑁

) ln (1 −

)]

𝑛1 𝑁

)]

According to eq (A) 𝑈 = 𝑁𝜀 − 2𝑛1 𝜀 𝑈 2𝑛1 𝜀 =𝜀− 𝑁 𝑁 If we take energy per particle to be independent of N, then

𝑛1 𝑁

is independent of N and ln(𝑊) is

proportional to N. Quantities that are proportional to N are said to be extensive. We have achieved our primary objective to calculate the number of quantum states that satisfy the constraints on the system and so get the entropy. Now let us turn our second main question how we get equation of state.

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6

EQUATION OF STATE To show that we can derive equations of state we shall consider three examples i.e. a spin system, vacancies in a crystal, and crude model of rubber band. 1. SPIN SYSTEM

For a spin system described in previous section we proceed in the following way. From eq (A)

𝑈 = 𝑁𝜀 − 2𝑛1 𝜀   

𝑈

=𝜀−

𝑁 𝑈 𝑁 𝑈

2𝑛1 𝜀 𝑁 2𝑛1

= 𝜀 (1 −

𝑁𝜀

=1−

𝑁

)

2𝑛1 𝑁

𝑈

Let 𝑥 = 𝑁𝜀 2𝑛1

 𝑥 =1−  

2𝑛1 𝑁 𝑛1 𝑁

𝑁

=1−𝑥

=

 1−  1−

1−𝑥

𝑛1 𝑁 𝑛1 𝑁

2

=1− =

1−𝑥

1+𝑥

2

2

As entropy is 𝑆 = 𝐾𝐵 ln 𝑊 For spin system, ln 𝑊 is calculated in previous section which is equal to ln 𝑊 = − 𝑁 [

𝑛1 𝑛1 𝑛1 𝑛1 ln + (1 − ) ln (1 − )] 𝑁 𝑁 𝑁 𝑁

Using this result gives entropy of spin system. 𝑆 = −𝑁𝐾𝐵 [

 Using

𝑛1 𝑁

=

1−𝑥 2

⁡and 1 −

𝑛1

=

𝑛1 𝑛1 𝑛1 𝑛1 ln + (1 − ) ln (1 − )] 𝑁 𝑁 𝑁 𝑁

1+𝑥

in above equation yields us 1−𝑥 1−𝑥 1+𝑥 1+𝑥 𝑆 = −𝑁𝐾𝐵 [( ) ln ( )+( ) ln ( )] 2 2 2 2 𝑁

2

Now as temperature is given by the equation 1 𝜕𝑆 𝜕𝑆 𝑑𝑥 =( ) =( ) 𝑇 𝜕𝑈 𝑣 𝜕𝑥 𝑣 𝑑𝑈 Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

7 𝑈

𝑑𝑥

1

As 𝑥 = 𝑁𝜀, so 𝑑𝑈 = 𝑁𝜀 1 𝜕𝑆 1 =( ) 𝑇 𝜕𝑥 𝑣 𝑁𝜀 As

1−𝑥 1−𝑥 1+𝑥 1+𝑥 ) ln ( )+( ) ln ( )] 2 2 2 2

𝑆 = −𝑁𝐾𝐵 [(



𝜕𝑆 𝜕𝑥

= −𝑁𝐾𝐵

𝜕 𝜕𝑥

[(

1−𝑥 2

) ln (

1−𝑥 2

)+(

1+𝑥 2

) ln (

1+𝑥 2

)]

Using product rule of differentiation 

    

𝜕𝑆

1−𝑥

= −𝑁𝐾𝐵 [( 𝜕𝑥

2

𝜕𝑆

1−𝑥

𝜕𝑥 𝜕𝑆 𝜕𝑥 𝜕𝑆 𝜕𝑥 𝜕𝑆 𝜕𝑥 𝜕𝑆 𝜕𝑥

= −𝑁𝐾𝐵 [( = −𝑁𝐾𝐵 [ = −𝑁𝐾𝐵 [

2 −1 2 −1

1−𝑥

).

1−𝑥 ) 2

+ ln ( ln (

=

𝑁𝐾𝐵 2

1

𝜕𝑆

𝑇

𝜕𝑥 𝑣 𝑁𝜀

ln (

𝑁𝐾𝐵

𝑇

2

1+𝑥

−1 2

2

)

2 1

2 1+𝑥

) − ln (

1−𝑥

1−𝑥 −1 2

)

1

2

2

𝜕

) 𝜕𝑥 (

+ + ln (

) + ln (

1−𝑥

2

+ ln (

1−𝑥 −1

1−𝑥 2

.

1−𝑥

) + ln (

+(

1+𝑥 2

1+𝑥 2

2

).

𝜕

1 (

1+𝑥

) 𝜕𝑥 ln ( 1

2

. + ln (

1+𝑥 ) 2 2

1+𝑥

) + ln ( 1+𝑥 2

2

𝜕

1+𝑥

) 𝜕𝑥 (

2

1

). ] 2

1

). ] 2

)]

2 1+𝑥 2

2

2

1+𝑥

)+(

)]

) ; using ln 𝑎 − ln 𝑏 = ln

𝑎 𝑏

1

Using the value of 1

2 1−𝑥

2

1 (

= −𝑁𝐾𝐵 ( ) [ln (

As ⁡ = ( )

As ⁡ =

2

−1

𝜕

) 𝜕𝑥 ln (

. ln (

𝜕𝑆 𝜕𝑥

, we get

1−𝑥

)

1

1+𝑥 𝑁𝜀

Which can be written as

2𝜀 1−𝑥 ) = ln ( 𝐾𝐵 𝑇 1+𝑥 Taking exponential on both sides

𝑒 2𝜀⁄𝐾𝐵𝑇 =

1−𝑥 1+𝑥

Solving for ‘x’

𝑒 2𝜀⁄𝐾𝐵𝑇 (1 + 𝑥) = 1 − 𝑥  𝑒 2𝜀⁄𝐾𝐵𝑇 + 𝑒 2𝜀⁄𝐾𝐵 𝑇 . 𝑥 = 1 − 𝑥

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)]

8

 𝑥 + 𝑒 2𝜀⁄𝐾𝐵𝑇 . 𝑥 = 1 − 𝑒 2𝜀⁄𝐾𝐵𝑇  𝑥(1 + 𝑒 2𝜀⁄𝐾𝐵𝑇 ) = 1 − 𝑒 2𝜀⁄𝐾𝐵𝑇  𝑥=

1−𝑒 2𝜀⁄𝐾𝐵 𝑇 1+𝑒 2𝜀⁄𝐾𝐵 𝑇

Taking 𝑒 𝜀⁄𝐾𝐵 𝑇 common on both numerator and denominator

 𝑥=  𝑥=

𝑒 𝜀⁄𝐾𝐵 𝑇 (𝑒 −𝜀⁄𝐾𝐵 𝑇 −𝑒 𝜀⁄𝐾𝐵 𝑇) 𝑒 𝜀⁄𝐾𝐵 𝑇 (𝑒 −𝜀⁄𝐾𝐵 𝑇 +𝑒 𝜀⁄𝐾𝐵 𝑇) (𝑒 −𝜀⁄𝐾𝐵 𝑇−𝑒 𝜀⁄𝐾𝐵 𝑇) (𝑒 −𝜀⁄𝐾𝐵 𝑇+𝑒 𝜀⁄𝐾𝐵 𝑇)

As tanh 𝑥 =

𝑒 𝑥 −𝑒 −𝑥 𝑒 𝑥 +𝑒 −𝑥

 𝑥 = −(

. And taking ‘-’ common from above equation.

𝑒 −𝜀⁄𝐾𝐵 𝑇 −𝑒 𝜀⁄𝐾𝐵 𝑇 𝑒 −𝜀⁄𝐾𝐵 𝑇 +𝑒 𝜀⁄𝐾𝐵 𝑇 𝜀

 𝑥 = − tanh ( As 𝑥 =



𝑈 𝑁𝜀 𝑈

𝑁𝜀

𝐾𝐵 𝑇

)

)

, so

= − tanh ( 𝑈

𝜀 𝐾𝐵 𝑇 𝜀

 − = 𝑁 tanh ( 𝜀

)

𝐾𝐵 𝑇 𝜀

 𝑈 = −𝑁𝜀 tanh (

)

𝐾𝐵 𝑇

)

As total energy is 𝑈 = −𝑛1 𝜀 + 𝑛2 𝜀  𝑈 = 𝜀(−𝑛1 + 𝑛2 ) 

𝑈 𝜀

 −

= −𝑛1 + 𝑛2 𝑈 𝜀

= 𝑛1 − 𝑛2

So we can write  −

𝑈 𝜀

= 𝑛1 − 𝑛2 = 𝑁 tanh (

𝜀 𝐾𝐵 𝑇

)

The total magnetic moment moment of the system is the difference between 𝜇𝑛1 , the net magnetic moment of up-spins and the net magnetic moment of the down spins 𝜇𝑛2 . Suppose spins of the particles are aligned by a magnetic induction field B giving rise to an energy upspins of −𝜇𝐵, and energy of down-spins of 𝜇𝐵. The magnetic moment per unit volume, the magnetization, is 𝑀=

𝜇𝑛1 − 𝜇𝑛2 𝜇(𝑛1 − 𝑛2 ) = 𝑉 𝑉

Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

9

As 𝑛1 − 𝑛2 = −

𝑈 𝜀

= 𝑁 tanh (

𝜀 𝐾𝐵 𝑇

), so

Using 𝜀 = 𝜇𝐵 𝑀=

𝜇𝑁 𝜇𝐵 tanh ( ) 𝑉 𝐾𝐵 𝑇

This is the equation of state for the magnetization of the system made up of non-interacting spin 1⁄2 particles in a magnetic field B. 2. VACANCIES IN A CRYSTAL Suppose atoms are taken away from their lattice sites and forced to migrate to the surface of the crystal. Each time this happens, a vacancy at the lattice site is created, something which is called Schottky defect.

Suppose there are ‘n’ such defects in the solid, each with an energy 𝜀, so total energy is 𝑛𝜀. There are N atoms in the crystal and n vacancies arranged at random on N+n lattice sites. The number of configurations of vacancies is

(𝑁+𝑛)! 𝑛!𝑁!

.

As entropy is 𝑆 = 𝐾𝐵 ln 𝑊, so entropy defects will be 𝑆 = 𝐾𝐵 ln

(𝑁 + 𝑛)! 𝑛! 𝑁!

𝑎

Using ln 𝑏 = ln 𝑎 − ln 𝑏

𝑆 = 𝐾𝐵 [ln(𝑁 + 𝑛)! − ln 𝑛! − ln 𝑁!] Using Stirling’s approximation

𝑆 = 𝐾𝐵 [(𝑁 + 𝑛)ln(𝑁 + 𝑛) − (𝑁 + 𝑛) − (𝑛 ln 𝑛 − 𝑛) − (𝑁 ln 𝑁 − 𝑁)]  𝑆 = 𝐾𝐵 [(𝑁 + 𝑛)ln(𝑁 + 𝑛) − 𝑁 − 𝑛 − 𝑛 ln 𝑛 + 𝑛 − 𝑁 ln 𝑁 + 𝑁]  𝑆 = 𝐾𝐵 [(𝑁 + 𝑛)ln(𝑁 + 𝑛) − 𝑛 ln 𝑛 − 𝑁 ln 𝑁] → (𝐴) As initial energy is 𝑈 = 𝑛𝜀, so the temperature is

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10

1 𝜕𝑆 𝜕𝑆 𝜕𝑛 =( ) =( ) 𝑇 𝜕𝑈 𝑣 𝜕𝑛 𝑣 𝜕𝑈 𝑈

As 𝑈 = 𝑛𝜀, so 𝑛 =

𝜀

𝑑𝑛

1

=> 𝑑𝑈 = 𝜀

So the above expression becomes 1 1 𝜕𝑆 = ( ) → (𝐵) 𝑇 𝜀 𝜕𝑛 𝑣 Differentiating equation (A) w.r.t ‘n’

𝜕𝑆 𝜕 [(𝑁 + 𝑛)ln(𝑁 + 𝑛) − 𝑛 ln 𝑛 − 𝑁 ln 𝑁] = 𝐾𝐵 𝜕𝑛 𝜕𝑛 

  

𝜕𝑆

= 𝐾𝐵 [(𝑁 + 𝑛)

𝜕𝑛

𝜕𝑆 𝜕𝑛 𝜕𝑆 𝜕𝑛 𝜕𝑆 𝜕𝑛

𝜕 𝜕𝑛

ln(𝑁 + 𝑛) + ln(𝑁 + 𝑛) 1

𝜕 𝜕𝑛

(𝑁 + 𝑛) − 𝑛

𝜕 𝜕𝑛

ln 𝑛 − ln 𝑛

𝜕 𝜕𝑛

𝑛−

𝜕 𝜕𝑛

𝑁 ln 𝑁]

1

= 𝐾𝐵 [(𝑁 + 𝑛). (𝑁+𝑛) . 1 + ln(𝑁 + 𝑛) . 1 − 𝑛. − ln 𝑛 . 1 − 0] 𝑛

= 𝐾𝐵 [1 + ln(𝑁 + 𝑛) − 1 − ln 𝑛] = 𝐾𝐵 [ln(𝑁 + 𝑛) − ln 𝑛] 𝑎

Using ln 𝑎 − ln 𝑏 = ln 𝑏



𝜕𝑆 𝜕𝑛

= 𝐾𝐵 ln

(𝑁+𝑛) 𝑛

Using this result in equation (B) (𝑁 + 𝑛) 1 1 = 𝐾𝐵 ln 𝑇 𝜀 𝑛 Rearranging the equation as follows

 

1

=

𝑇

𝜀

𝐾𝐵 𝜀

= ln (

𝐾𝐵 𝑇 𝜀⁄𝐾𝐵 𝑇

 𝑒  𝑒  

𝑁 𝑛 𝑛 𝑁

𝜀⁄𝐾𝐵 𝑇

=𝑒 =

ln (

= =

𝑁+𝑛

𝑛 𝑁+𝑛

𝑛 𝑁+𝑛

)

)

𝑛 𝑁

𝑛 𝜀 ⁄ 𝐾𝐵 𝑇

+1 −1

1 𝑒 𝜀⁄𝐾𝐵 𝑇 −1 𝑛

A formula which makes sense only if 𝑁 is much less than 1, for otherwise the solid would be riddled with holes and would collapse. A typical value of 𝜀 is 1⁡𝑒𝑉 which means that at room temperature 𝜀 ⁄𝐾𝐵 𝑇 is about 39 and 𝑛/𝑁 is about 10−17. There are relatively few vacancies in the solid at room temperature. Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

11

3. A MODEL FOR A RUBBER BAND A simple picture of a rubber band is of a single long chain of linking groups of atoms with the links oriented in any direction. When the rubber band is pulled so that the chain of atoms is completely linear there is only one possible arrangement and the entropy is zero. When rubber band is all tangled up there is a huge number of arrangements of the link leading to a large entropy. Heat up a rubber band under tension and it wants to tangle up. Cool it down and it stretches. Let’s simplify matters by assuming that the links can lie in only two directions, either in the direction of increasing Z or in the opposite direction. The links lie in the direction of either +Z or –Z. Start at one end of the chain (the end with smaller value of Z) and count how many lie along +z direction (= 𝑛+ ) and how many lie in the –z direction (= 𝑛− ). There can be many way of ending up with n+ links in the +z direction. Each such arrangement can be represented as a different quantum state. Starting from the end we note a sense of pluses and minuses to label each link. One such state for 15 links is the state 𝜓𝑖 = |+, −⁡, +, +, +, +, −, +, +, +, −, +, −, +, +> There are 11 positive links and 4 negative links. Each link has a length ‘d’. the total extension i.e. the distance from one end of the chain to the other is ‘l’. The work done on the rubber band when it is extended by an amout dl is Fdl where F is tension in the band. The change in internal energy is therefore given by 𝑑𝑈 = 𝑇𝑑𝑆 + 𝐹𝑑𝑙 In microcanonical ensemble, 𝑑𝑈 = 0, so 𝑇𝑑𝑆 = −𝐹𝑑𝑙 𝐹 𝑑𝑆 = − ( ) → (1) 𝑇 𝑑𝑙 𝑢 As 𝑆 = 𝐾𝐵 ln 𝑊, so 𝐹 𝜕 ln 𝑊 = −𝐾𝐵 ( ) 𝑇 𝜕𝑙 𝑢 Let there be n+ lines going to the right and n- going to the left. For a chair of N links, W is given by 𝑊(𝑁, 𝑛+ ) =

𝑁! 𝑛+ ! (𝑁 − 𝑛+ )!

Where 𝑛− = 𝑁 − 𝑛+

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The total extension is 𝑙 = 𝑛+ 𝑑 − 𝑛− 𝑑  𝑙 = (𝑛+ − 𝑛− )𝑑  𝑙 = (𝑛+ + 𝑛+ − 𝑁)𝑑  𝑙 = (2𝑛+ − 𝑁)𝑑 𝑙  = (2𝑛+ − 𝑁)  

𝑑 𝑙

𝑑 𝑙

2𝑛+

= 𝑁(

𝑁 2𝑛+

=(

𝑁𝑑

𝑁

− 1)

− 1)

Let us define the parameter ‘x’ as 𝑥=

𝑙 2𝑛+ =( − 1) 𝑁𝑑 𝑁

 𝑥+1= 

𝑛+

=

𝑁

As 𝑊 = 𝑛

𝑥+1 2 𝑁!

2𝑛+ 𝑁

and similarly

𝑛− 𝑁

𝑥−1

=

2

, Taking ‘ln’

+ !(𝑁−𝑛+ )!

ln 𝑊 = ln (

𝑁! ) 𝑛+ ! (𝑁 − 𝑛+ )!

 ln 𝑊 = ln 𝑁! − ln 𝑛+ ! − ln(𝑁 − 𝑛+ )! Using Stirling’s approximation  ln 𝑊 = 𝑁 ln 𝑁 − 𝑁 − 𝑛+ ln 𝑛+ + 𝑛+ − (𝑁 − 𝑛+ )ln(𝑁 − 𝑛+ ) + 𝑁 − 𝑛+  ln 𝑊 = 𝑁 ln 𝑁 − 𝑛+ ln 𝑛+ − (𝑁 − 𝑛+ )ln(𝑁 − 𝑛+ ) Adding and subtracting 𝑛+ ln 𝑁  ln 𝑊 = 𝑁 ln 𝑁 − 𝑛+ ln 𝑁 − 𝑛+ ln 𝑛+ + 𝑛+ ln 𝑁 − (𝑁 − 𝑛+ )ln(𝑁 − 𝑛+ ) 𝑁  ln 𝑊 = (𝑁 − 𝑛+ ) ln 𝑁 + 𝑛+ ln 𝑛 − (𝑁 − 𝑛+ )ln(𝑁 − 𝑛+ ) +

𝑁

𝑁

 ln 𝑊 = (𝑁 − 𝑛+ ) ln 𝑁−𝑛 ⁡ + 𝑛+ ln 𝑛 +

+

𝑁−𝑛+

 ln 𝑊 = − [(𝑁 − 𝑛+ ) ln (

𝑁

 ln 𝑊 = − [(𝑁 − 𝑛+ ) ln (1 −  ln 𝑊 = −𝑁 [

(𝑁−𝑛+ ) 𝑁

 ln 𝑊 = −𝑁 [(1 − Now using

𝑛+ 𝑁

=

1+𝑥 2

, and

 ln 𝑊 = −𝑁 [(1 −

ln (1 −

𝑛+ 𝑁 𝑛− 𝑁

𝑛+ 𝑁

𝑛+ 𝑁

) ln (1 − =

1+𝑥 2

) + 𝑛+ ln

𝑛+

) + 𝑛+ ln

)+ 𝑛+ 𝑁

𝑛+ 𝑁

)+

ln 𝑛+ 𝑁

]⁡

𝑁 𝑛+ 𝑁

𝑛+ 𝑁

ln

]⁡

]⁡ 𝑛+ 𝑁

]⁡

1−𝑥 2

) ln (1 −

1+𝑥 2

1+𝑥

)+(

2

1+𝑥

) ln (

2

)]⁡

Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

13 1−𝑥

 ln 𝑊 = −𝑁 [(

2

1−𝑥

) ln (

2

1+𝑥

)+(

2

) ln (

1+𝑥 2

)]⁡

Now as the tension F in the rubber band is given as 𝐹 𝜕 = −𝐾𝐵 ( ln 𝑊) 𝑇 𝜕𝑙 𝑢 

𝐹

𝜕

𝑑𝑥

= −𝐾𝐵 (𝜕𝑥 ln 𝑊) . 𝑑𝑙

𝑇

𝑢

𝑙

𝑑𝑥

As 𝑥 = 𝑁𝑑, so

𝑑𝑙

1

= 𝑁𝑑

𝜕 𝜕 1−𝑥 1−𝑥 1+𝑥 1+𝑥 ln 𝑊 = −𝑁 [( ) ln ( )+( ) ln ( )]⁡ 𝜕𝑥 𝜕𝑥 2 2 2 2 Using the product rule of differentiation, we get      As

𝐹 𝑇

𝜕

1−𝑥

𝜕𝑥 𝜕 𝜕𝑥 𝜕

ln 𝑊 = −𝑁 [(



2

) (1−𝑥) .

−1 −1

1−𝑥

ln 𝑊 = −𝑁 [ 2 ln ( 𝜕𝑥 𝜕

𝑁

𝜕𝑥 𝜕 𝜕𝑥

𝑇 𝐹 𝑇

2

2

).

2

1

2

𝑁

+ ln (

2

2

1

1+𝑥

1+𝑥

1−𝑥

).

2

−1

+(

2 1+𝑥

+ 2 + ln (

) + 2 ln (

ln 𝑊 = − 2 [ln (1−𝑥) +ln (

2

1+𝑥 2

2

1

1+𝑥

) (1+𝑥) . 2 + ln (

2

1

) . 2]⁡

1

) . 2]⁡

)]⁡

)]⁡

1+𝑥

ln 𝑊 = − 2 [ln (1−𝑥)]⁡ 𝜕 𝜕𝑥

𝐹

−1

2 1−𝑥 −1

ln 𝑊 = −𝑁 [ 2 + ln (

= −𝐾𝐵 (



2

ln 𝑊) .

𝑑𝑥

𝑢 𝑑𝑙

𝑁

1+𝑥

1

= −𝐾𝐵 (− 2 [ln (1−𝑥)]) . 𝑁𝑑 𝑢

=

𝐾𝐵 2𝑑

1+𝑥

ln (1−𝑥)

𝑙

As 𝑥 = 𝑁𝑑 

𝐹



𝐹

𝑇 𝑇

= =

𝑙 𝑁𝑑 𝑙 1− 𝑁𝑑

𝐾𝐵

1+

𝐾𝐵

𝑁𝑑+𝑙

ln ( 2𝑑

)

ln (𝑁𝑑−𝑙) 2𝑑 𝑙

For small values of 𝑁𝑑, we can make power series expansion of above equation to get in lowest order. 𝐹≈

𝐾𝐵 𝑇𝑙 𝑛𝑑 2

This equation indicates that the tension is proportional to the product of the extension and temperature.

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14

THE CANONICAL ENSEMBLE A SYSTEM IN CONTACT WITH A HEAT BATH Imagine a similar ensemble of similar systems are prepared in the same way with the same number of particles, volume, shape, magnetic field and electric field and so on but energy of each system is not constant. Energy can be passed from one system to its neighbors. Therefore, energy of each system fluctuates. Each system is in thermal contact with the remainder which acts as a heat bath for the system. This ensemble is called the canonical ensemble. The heat bath is any system with large heat capacity so that its temperature doesn’t change when heat energy is added to it. The system and heat bath eventually reach thermal equilibrium at the temperature of heat bath. When the system is in contact with heat bath, it is said to be in canonical state. The combined system, the system being studied together with the heat bath is thermally isolated so that its total energy UT is constant. The entropy of combined system is 𝑆 = 𝐾𝐵 ln 𝑊. Where W is the number of accessible quantum states of the combined system. The system we want to study it let ‘A’ has energy UA and the heat bath has energy 𝑈𝑅 = 𝑈𝑇 − 𝑈𝐴 The number of accessible states of the combined system 𝑊(𝑈𝐴 ) is a function of UA. The volume and number of particles in system A are constant but the energy is allowed to vary 𝑊(𝑈𝐴 ) = 𝑊𝐴 (𝑈𝐴 ) ∗ 𝑊𝑅 (𝑈𝑅 ) 𝑊(𝑈𝐴 ) = 𝑊𝐴 (𝑈𝐴 ) ∗ 𝑊𝑅 (𝑈𝑇 − 𝑈𝐴 ) Where 𝑊𝑅 (𝑈𝑇 − 𝑈𝐴 ) is number of accessible quantum states of the heat bath and 𝑊𝐴 (𝑈𝐴 ) is number of accessible quantum states of a system A. A particular quantum state is chosen 𝜓𝑖 . This quantum state is an eigenstate of the energy operator with eigen value Ei. Here 𝑈𝐴 = 𝐸𝑖 and 𝑊𝐴 (𝑈𝐴 ) = 1 With system A in quantum state 𝜓𝑖 , the total number os quantum states fot the combined system is 𝑊(𝐸𝑖 ) = 1 ∗ 𝑊𝑅 (𝑈𝑇 − 𝐸𝑖 ) Temperature of the heat bath is defined as 1 𝜎 ln(𝑊𝑅 ) = 𝐾𝐵 ( ) 𝑇 𝜎𝑈𝑅 𝑣 Integration of above equation gives 𝑈𝑅

𝑊𝐼𝑅 = 𝑦𝑒 𝐾𝐵 𝑇 Where ‘y’ is integration constant, so as

Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

15

𝑊(𝐸𝑖 ) = 𝑊𝑅 (𝑈𝑇 − 𝐸𝑖 ) = 𝑦𝑒

𝑈𝑇 −𝐸𝑖 𝐾𝐵 𝑇

The basic postulate of statistical mechanics is that all accessible quantum states are equally probable. The total number of accessible quantum states of the combined system is 𝑁𝑡𝑜𝑡𝑎𝑙 = ∑ 𝑊(𝐸𝑗 ) 𝑗

𝑁𝑡𝑜𝑡𝑎𝑙 =

𝑈𝑇 −𝐸𝑖 ∑ 𝑦𝑒 𝐾𝐵 𝑇

=

𝑗

𝐸𝑗 𝑈𝑇 − 𝐾 𝑇 𝐾 𝐵 ∑ 𝑦𝑒 . 𝑒 𝐵𝑇 𝑗

𝑁𝑡𝑜𝑡𝑎𝑙 =

𝑈𝑇 𝐾 𝑦𝑒 𝐵 𝑇

∑𝑒

−

𝐸𝑗 𝐾𝐵 𝑇

𝑗

Where the sum is overall quantum states of A. now the classical probability will be 𝑃𝑖 = As 𝑊(𝐸𝑖 ) = 𝑒

−

𝐸𝑖 𝐾𝐵 𝑇

, and ∑𝑗 ⁡𝑊(𝐸𝑗 ) = ∑𝑗 𝑒

−

𝐸𝑖 𝐾𝐵 𝑇

𝑊(𝐸𝑗 ) ∑𝑗 𝑊(𝐸𝑗 )

, therefore

𝑃𝑖 =

𝑒 −𝐸𝑖 ⁄𝐾𝐵 𝑇 ∑𝑗 𝑒 −𝐸𝑗 ⁄𝐾𝐵 𝑇

As 𝑍 = ∑𝑗 𝑒 −𝐸𝑗 ⁄𝐾𝐵 𝑇 𝑃𝑖 =

𝑒 −𝐸𝑖 ⁄𝐾𝐵 𝑇 → (𝐴) 𝑍

This equation is called Boltzmann’s probability distribution. It tells us the probability that the particular quantum state of the system is occupied. At low temperatures, only quantum states of low energy are occupied as the temperature is increases. Quantum states of higher energy start to have a significant probability of being occupied.

THE PARTITION FUNCTION The denominator in equation (A) is called partition function and is denoted by the symbol Z. The partition function is given by the equation 𝑍 = ∑ 𝑒 −𝐸𝑗 ⁄𝐾𝐵 𝑇 𝑗

Where the sum is taken over all the different quantum states of the system.

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16

Z in an abbreviation of the German word ‘zustandssume’ which means sum over states. As an example consider a very simple system which has three energy levels each of which is non degenerate (non-degenerate means only one quantum state for each level). The energy levels has energies 𝐸1 = 0, 𝐸2 = 1.4 ∗ 10−23 𝐽, 𝐸3 = 2.8 ∗ 10−23 ⁡𝐽. If the heath bath has a temperature 2K, then partition function is 𝑍 = ∑ 𝑒 −𝐸𝑗 ⁄𝐾𝐵 𝑇 𝑗

𝑍=𝑒 𝑍=𝑒

−

0 𝐾𝐵 𝑇

+𝑒

−

−

𝐸1 𝐾𝐵 𝑇

+𝑒

−

𝐸2 𝐾𝐵 𝑇

1.4∗10−23 𝐽 (1.38∗10−23 𝐽/𝐾)∗2

+𝑒 +𝑒

−

𝐸3 𝐾𝐵 𝑇

−

2.8∗10−23 ⁡𝐽 (1.38∗10−23 𝐽/𝐾)∗2

1

𝑍 = 𝑒 0 + 𝑒 −2 + 𝑒 −1 𝑍 = 1 + 0.60 + 0.36 = 1.9744 𝐸

𝐸

𝐵

𝐵

Since 𝐾 2𝑇 = 0.5⁡, 𝐾 3𝑇 = 1. The probability of being in the state of energy E1 is 0.506, of energy E2 is 0.307 and E3 is 0.186 As 𝑃𝑖 =

𝑒 −𝐸𝑖 /𝐾𝐵 𝑇 𝑍

𝑃𝐸 1 =

𝑒 −𝐸1 /𝐾𝐵 𝑇 1 = = 0.506 𝑍 1.9744

𝑃𝐸 2 =

0.60 = 0.307 1.9744

𝑃𝐸 2 =

0.36 = 0.186 1.9744

If there are 𝑔𝑛 quantum states all with the energy 𝐸𝑛 , then Z can be written as −𝐸𝑛

𝑍 = ∑ 𝑔𝑛 𝑒 𝐾𝐵 𝑇 𝑛

Quantity 𝑔𝑛 is called degeneracy of the energy level. For example, consider a system which has two energy levels one of energy 𝐸𝑜 = 0 with a degeneracy of two and the other of energy 𝐸1 = 1.38 ∗ 10−23 𝐽 with a degeneracy of three at 1K. The partition function will be 𝑍 = ∑ 𝑔𝑛

−𝐸𝑛 𝐾 𝑒 𝐵𝑇

𝑛

Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

17

As 𝐸𝑜 = 0 and 𝑔0 = 2 And 𝐸1 = 1.38 ∗ 10−23 𝐽 and 𝑔1 = 3 and 𝑇 = 1𝐾 −𝐸𝑜

−𝐸1

𝑍 = 𝑔𝑜 𝑒 𝐾𝐵 𝑇 + 𝑔1 𝑒 𝐾𝐵 𝑇 = 2𝑒 𝑜 + 3𝑒 1 = 3.104

DEFINITION OF ENTROPY IN THE CANONICAL ENSEMBLE The system under study system A does not have a constant energy when it is in contact with heat bath. The energy of the system fluctuates successive measurements of energy will find it in different energy states. We can calculate entropy of system by using the notion of ensemble of systems. Imagine (𝑀 − 1) replica systems in contact with each other and with the original system these replicas act as a heat bath if we take ‘M’ to be huge. Because of the contact between them, energy can be transferred from one system to its neighbors which in turns transfer energy to their neighbors and so on. Each replica system is identical to the rest, but replicas are taken to be objects which can easily be distinguished from each other by their position. We shall describe the state of each system quantum mechanically by saying that, it is in a particular quantum state 𝜓𝑖 . Let the number of systems in the quantum state 𝜓𝑖 is 𝑛𝑖 out of total number of systems M. There are 𝑛1 systems in quantum state 𝜓1 , 𝑛2 in 𝜓2 and so on. The number of ways of arranging 𝑛1 systems to be in 𝜓1 , 𝑛2 in 𝜓2 is 𝑊=

𝑀! 𝑛1 ! 𝑛2 ! …

Lets take M to be enormous that all 𝑛𝑖 are huge, the entropy for the M systems is given by 𝑆𝑀 = 𝑘𝐵 ln(𝑊)  𝑆𝑀 = 𝑘𝐵 ln (𝑛

𝑀!

)

1 !𝑛2 !…

 𝑆𝑀 = 𝑘𝐵 (ln 𝑀! − ln ∑𝑖 𝑛𝑖 !) Using Stirling’s approximation  𝑆𝑀 = 𝐾𝐵 (𝑀 ln 𝑀 − 𝑀 − ∑𝑖 𝑛𝑖 ln ∑𝑖 𝑛𝑖 + ∑𝑖 𝑛𝑖 )  𝑆𝑀 = 𝐾𝐵 (M⁡ln 𝑀 − 𝑀 − ∑𝑖 𝑛𝑖 ln 𝑛𝑖 + 𝑀)⁡;⁡⁡⁡⁡⁡∑𝑖 𝑛𝑖 = 𝑀  𝑆𝑀 = 𝐾𝐵 (M⁡ln 𝑀 − ∑𝑖 𝑛𝑖 ln 𝑛𝑖 ) By using identity M⁡ln 𝑀 = ∑𝑖 𝑛𝑖 ⁡ln 𝑀, we can write  𝑆𝑀 = 𝐾𝐵 (∑𝑖 𝑛𝑖 ⁡ln 𝑀 − ∑𝑖 𝑛𝑖 ln 𝑛𝑖 ) ∑𝑖 𝑛𝑖

 𝑆𝑀 = 𝑀𝐾𝐵 (  𝑆𝑀 = 𝑀𝐾𝐵

M ∑𝑖 𝑛𝑖 M

⁡ln 𝑀 −

∑𝑖 𝑛𝑖 𝑀

ln 𝑛𝑖 )

(⁡ln 𝑀 − ln 𝑛𝑖 )

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 𝑆𝑀 = −𝑀𝐾𝐵

∑𝑖 𝑛𝑖 M

(ln 𝑛𝑖 − ⁡ln 𝑀) 𝑛

𝑛

 𝑆𝑀 = −𝐾𝐵 𝑀 (∑𝑖 M𝑖 ln ( 𝑀𝑖 )) As M tends to infinity so the ratio 𝑛𝑖 ⁄𝑀 tends to the probability of finding the system in the quantum states 𝜓𝑖 . So 𝑆𝑀 = −𝐾𝐵 𝑀 ∑ 𝑃𝑖 ln 𝑃𝑖 𝑖

Hence the entropy per system can be written as 𝑆=

𝑆𝑀 −𝐾𝐵 𝑀 ∑𝑖 𝑃𝑖 ln 𝑃𝑖 = = −𝐾𝐵 ∑ 𝑃𝑖 ln 𝑃𝑖 𝑀 𝑀 𝑖

This is the formula for average entropy of a system in the canonical ensemble, which is when it is in contact with heat bath. This expression for entropy simplifies when all the probabilities are equal. If there are W quantum states, the probabilities are 𝑃𝑖 = 𝑊 −1 => 𝑃𝑖 =

1 𝑊

So 𝑊

𝑆 = −𝐾𝐵 ∑

𝑊

𝑃𝑖 ln 𝑃𝑖 = −𝐾𝐵 ∑

𝑖=1

1 1 ln ( ) 𝑊 𝑖=1 𝑊

𝑆 = 𝐾𝐵 ln(𝑊) Which is Boltzmann’s formula for the entropy.

THE BRIDGE TO THERMODYNAMICS THROUGH Z The partition function Z is not just the normalization constant needed to get the sum of the probabilities equal to one. But it is much more important than that. Its importance arises because it enables us to make a direct connection between quantum states of system and its thermodynamic properties such as free energy, the pressure, and the entropy. We can calculate all the thermodynamic properties of a system from the knowledge of partition function. HELMHOLTZ FREE ENERGY Helmholtz Free Energy is a thermodynamic potential that measures the useful work obtainable from closed thermodynamic system at constant temperature. In canonical ensemble the probability of being in quantum state 𝜓𝑖 is 𝑃𝑖 = 𝑒 (−𝐸𝑖 ⁄𝐾𝐵 𝑇) ⁄𝑍 When the probabilities are not equal, the entropy is given as

Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

19

𝑆 = −𝐾𝐵 ∑ 𝑃𝑖 ln 𝑃𝑖 𝑖

As 𝑃𝑖 = 𝑒 (−𝐸𝑖 ⁄𝐾𝐵 𝑇) ⁄𝑍 Taking ‘ln’ on both sides ln 𝑃𝑖 = ln 𝑒 (−𝐸𝑖 ⁄𝐾𝐵 𝑇) − ln 𝑍 ln 𝑃𝑖 =

−𝐸𝑖 − ln 𝑍 𝐾𝐵 𝑇

Putting it in the equation for entropy −𝐸𝑖 𝑆 = −𝐾𝐵 ∑ 𝑃𝑖 ( − ln 𝑍) 𝐾𝐵 𝑇 𝑖

 𝑆 = 𝐾𝐵 ∑𝑖 𝑃𝑖 (  𝑆 = 𝐾𝐵 ∑𝑖  𝑆 = ∑𝑖

𝑃𝑖 𝐸𝑖 𝐾𝐵 𝑇

𝑃𝑖 𝐸𝑖 𝑇

𝐸𝑖 𝐾𝐵 𝑇

+ ln 𝑍)

+ 𝐾𝐵 ∑𝑖 𝑃𝑖 ln 𝑍

+ 𝐾𝐵 ∑𝑖 𝑃𝑖 ln 𝑍

Now mean or average energy is given as ̅ = ∑ 𝑃𝑖 𝐸𝑖 𝑈 𝑖

Using this value, we get 𝑆=

̅ 𝑈 + 𝐾𝐵 ∑ 𝑃𝑖 ln 𝑍 𝑇 𝑖

Multiplying by ‘T’ ̅ + 𝐾𝐵 𝑇 ∑ 𝑃𝑖 ln 𝑍 𝑇𝑆 = 𝑈 𝑖

  

̅ = 𝐾𝐵 𝑇 ∑𝑖 𝑃𝑖 ln 𝑍 𝑇𝑆 − 𝑈 ̅ − 𝑇𝑆 = −𝐾𝐵 𝑇 ∑𝑖 𝑃𝑖 ln 𝑍 𝑈 ̅ − 𝑇𝑆 = −𝐾𝐵 𝑇 ln 𝑍 ⁡⁡ ∴ ∑𝑖 𝑃𝑖 = 1 𝑈

̅ − 𝑇𝑆 is the average value of the Helmholtz free energy, F, a function of state. So we can The quantity 𝑈 relate free energy to the partition function as 𝐹 = −𝐾𝐵 𝑇 ln 𝑍 Where KB is Boltzmann’s constant and Z is the partition function.

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𝑍=

−𝐸𝑗 𝐾 ∑ 𝑒 𝐵𝑇 𝑗

Once we know all the quantum states of a system and their energy eigenvalues, we can calculate Z and hence obtain the free energy of the system.

THE CONDITION FOR THERMAL EQUILIBRIUM Suppose the system of N atoms has a series of energy eigenvalues with the ground state having an energy 𝐸𝑜 = 0. The system is in contact with a heat bath at temperature T so that the probability of finding it in quantum state 𝜓𝑖 is given by Boltzmann’s probability distribution. 𝑃𝑖 = 𝑒 (−𝐸𝑖 ⁄𝐾𝐵 𝑇) ⁄𝑍  𝑃0 = 𝑒 (−𝐸0 ⁄𝐾𝐵 𝑇) ⁄𝑍. As 𝐸0 = 0  𝑃0 = 𝑒 0 ⁄𝑍 = 1/𝑍 The ratio of probability of being in quantum state 𝜓𝑖 to that of being in the ground state is −𝐸𝑖

−𝐸𝑖 𝑃𝑖 𝑒 𝐾𝐵 𝑇 𝑍 = ∗ = 𝑒 𝐾𝐵 𝑇 𝑃0 𝑍 1

This equation must be correct for all quantum states for otherwise the distribution would not be the Boltzmann’s distribution. Taking natural log on both sides of the above equation ln

𝑃𝑖 −𝐸𝑖 = 𝑃0 𝐾𝐵 𝑇 𝑃

 𝑇 ln 𝑃 𝑖 = 0

 𝑇=

−𝐸𝑖

𝐾𝐵 −𝐸𝑖

𝑃 𝐾𝐵 ln( 𝑖 )

→ (𝐵)

𝑃0

FOR EXAMPLE: Suppose there are four degenerate energy levels with energies 0, 1.9 ∗ 10−20 𝐽, 3.6 ∗ 10−20 𝐽⁡𝑎𝑛𝑑⁡5.2 ∗ 10^ − 20𝐽. The system is observed repeatedly and it is found that the probabilities of being in these levels are 𝑃0 = 0.598, 𝑃1 = 0.264, 𝑃2 = 0.150, and⁡𝑃3 = 0.088. is this system in thermal equilibrium and if so, what is the temperature? Solution: 𝐸0 = 0, ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝐸1 = 1.9 ∗ 10−20 𝐽⁡ 𝐸2 = 3.6 ∗ 10−20 𝐽, ⁡𝐸3 = ⁡5.2 ∗ 10−20𝐽 Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

21

And 𝑃0 = 0.498, 𝑃1 = 0.264⁡ 𝑃2 = 0.150⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑃3 = 0.088 As energy levels are non-degenerate so we can use equation (B) 𝑇=

−𝐸𝑖 𝑃 𝐾𝐵 ln (𝑃𝑖 ) 0

For 𝐸1 → 𝑃1 𝑇=

−𝐸1 −1.9 ∗ 10−20 𝐽 −1.9 ∗ 10−20 𝐽 1.9 ∗ 10−20 𝐽 = = = 𝐾 = 2169𝐾 𝑃 0.264 1.38 ∗ 10−23 𝐽/𝐾(−0.634) 0.875 ∗ 10−23 𝐾𝐵 ln (𝑃1 ) 𝐾𝐵 ln ( ) 0.498 0

For 𝐸2 → 𝑃2 −𝐸2 −3.6 ∗ 10−20 𝐽 −3.6 ∗ 10−20 𝐽 3.6 ∗ 10−20 𝐽 𝑇= = = = 𝐾 = 2173𝐾 𝑃 0.150 1.38 ∗ 10−23 𝐽/𝐾( ) ∗ 10−23 𝐾𝐵 ln (𝑃2 ) 𝐾𝐵 ln ( ) 0.498 0 For 𝐸3 → 𝑃3 −𝐸3 −5.2 ∗ 10−20 𝐽 −5.2 ∗ 10−20 𝐽 3.6 ∗ 10−20 𝐽 𝑇= = = = 𝐾 = 2174𝐾 −23 𝑃 0.088 ) ∗ 10−23 𝐾𝐵 ln (𝑃3 ) 𝐾𝐵 ln (0.498) 1.38 ∗ 10 𝐽/𝐾( 0 Since the temperatures are approximately same within the accuracy of the probabilities that are given, the system can be assumed to be in thermal equilibrium at about 2172 K.

THERMODYNAMIC QUANTITIES FROM ln (Z) Consider a small reversible change in F in which the number of particles is kept constant. Such a change can be written as 𝑑𝐹 = −𝑃𝑑𝑉 − 𝑆𝑑𝑇 So 𝑃 = −(

𝜕𝐹 ) 𝜕𝑉 𝑇

Similarly 𝑆=(

𝜕𝐹 ) 𝜕𝑇 𝑣

As 𝐹 = −𝐾𝑏 𝑇 ln(𝑍), Putting in the equation of P

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22

𝑃 = 𝐾𝐵 (

𝜕 𝑇 ln(𝑍)) → (1) 𝜕𝑉 𝑇

𝜕 𝑆 = 𝐾𝐵 ( 𝑇 ln(𝑍)) → (2) 𝜕𝑇 𝑉 These are the expressions for pressure and entropy. The definition of entropy allows us to derive a new formula for the heat capacity. Since đ𝑄𝑟𝑒𝑣 = 𝑇𝑑𝑆, we have for reversible change in temperature at constant volume. 𝜕𝑆 𝐶𝑉 = 𝑇 ( ) → (𝑎) 𝜕𝑇 𝑉 𝐶𝑃 = 𝑇 (

𝜕𝑆 ) 𝜕𝑇 𝑃

From equation (1) we get isothermal compressibility, K a measure of how the volume of a system changes when the pressure is altered at constant temperature. 𝜕𝑃 𝐾 −1 = −𝑉 ( ) 𝜕𝑉 𝑇 𝜕𝐹

As 𝑃 = − (𝜕𝑉) , so 𝑇

𝐾

−1

𝜕 2𝐹 = −𝑉 ( 2 ) 𝜕𝑉 𝑇

It means the inverse of compressibility is proportional to the second derivative of the free energy w.r.t. volume. Similarly we can obtain heat capacity at constant volume from equations (a) and (3) 𝜕𝑆 𝐶𝑉 = 𝑇 ( ) 𝜕𝑇 𝑉 𝜕𝐹

As 𝑆 = − (𝜕𝑇) , so 𝑉

𝜕𝐹 𝐶𝑉 = 𝑇 ( 2 ) 𝜕𝑇 𝑉 The heat capacity at constant volume is proportional to the second derivative of the free energy w.r.t. temperature. Quantities which are proportional to second derivative of free energy w.r.t. some variable are called response functions. The pressure, entropy, heat capacity and isothermal compressibility all can be calculated once we know ln 𝑍 ̅ − 𝑇𝑆 Now as 𝐹 = 𝑈

Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

23

̅ = 𝐹 + 𝑇𝑆  𝑈 The average internal energy can be calculated by using 𝐹 = −𝐾𝐵 𝑇 ln(𝑍) And 𝑆 = 𝐾𝐵 (

𝜕 𝑇 ln(𝑍)) 𝜕𝑇 𝑉

𝜕 ̅ = −𝐾𝐵 𝑇 ln(𝑍) + 𝐾𝐵 𝑇 ( 𝑇 ln(𝑍)) 𝑈 𝜕𝑇 𝑉 ̅ = 𝐾𝐵 𝑇 ( 𝜕 𝑇 ln(𝑍)) − 𝐾𝐵 𝑇 ln(𝑍)  𝑈 𝜕𝑇 𝑉



̅ = 𝐾𝐵 𝑇 [𝑇 𝜕 ln(𝑍) + ln(𝑍) 𝜕 𝑇] − 𝐾𝐵 𝑇 ln(𝑍)  𝑈 𝜕𝑇 𝜕𝑇 ̅ = 𝐾𝐵 𝑇 2 ( 𝜕 ln(𝑍)) + 𝐾𝐵 𝑇 ln(𝑍) − 𝐾𝐵 𝑇 ln(𝑍)  𝑈 𝜕𝑇 𝑉

𝜕

̅ = 𝐾𝐵 𝑇 ( ln(𝑍))  𝑈 𝜕𝑇 2

𝑉

TWO LEVEL SYSTEM Consider simplest possible quantum state in which there are just two non degenerate quantum states with energies 𝜀⁡𝑎𝑛𝑑 − 𝜀. An example is a single particle of spin 1⁄2 having energy eigenvalues 𝜇𝐵⁡𝑎𝑛𝑑 − 𝜇𝐵 where 𝜇 Is magnetic moment of the particle. 1

Imagine a system to be a particle with 2 spin. The particle is replicated M times to build up on ensemble of systems. We assume that each particle with its spin is distinguishable from the rest. As the partition function is 𝑍 = ∑𝑒

−

𝜀𝑗 𝐾𝐵 𝑇

𝑗

For two level system, it will become 𝜀

−𝜀

𝑍 = 𝑒 𝐾𝐵 𝑇 + 𝑒 𝐾𝐵 𝑇 As 𝐹 = −𝐾𝐵 Tln(𝑍) 𝜀

−𝜀

 𝐹 = −𝐾𝐵 𝑇 ln (𝑒 𝐾𝐵𝑇 + 𝑒 𝐾𝐵 𝑇 )  𝐹 = −𝐾𝐵 Tln [𝑒

𝜀 𝐾𝐵 𝑇

(1 +

All Our Notes Are Belong To You

𝑒 −𝜀⁄𝐾𝐵 𝑇 𝑒 𝜀⁄𝐾𝐵 𝑇

)]

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24 𝜀

 𝐹 = −𝐾𝐵 Tln [𝑒 𝐾𝐵𝑇 (1 + 𝑒 −2𝜀⁄𝐾𝐵 𝑇 )]  𝐹 = −𝐾𝐵 𝑇 [ln 𝑒  𝐹 = −𝐾𝐵 𝑇 [

𝜀

𝐾𝐵 𝑇

𝜀 𝐾𝐵 𝑇

+ ln(1 + 𝑒 −2𝜀⁄𝐾𝐵 𝑇 )]

+ ln(1 + 𝑒 −2𝜀⁄𝐾𝐵 𝑇 )]

 𝐹 = −𝜀 − 𝐾𝐵 𝑇 ln(1 + 𝑒 −2𝜀⁄𝐾𝐵 𝑇 ) This is free energy for two level system Now entropy can be calculated as 𝜕𝐹 𝑆 = −( ) 𝜕𝑇 𝑉 𝜕

 𝑆 = − [𝜕𝑇 (−𝜀 − 𝐾𝐵 𝑇 ln(1 + 𝑒−2𝜀⁄𝐾𝐵𝑇 ))] 𝜕

 𝑆 = − [𝜕𝑇 (−𝐾𝐵 𝑇 ln(1 + 𝑒−2𝜀⁄𝐾𝐵𝑇 ))] 𝜕

 𝑆 = 𝜕𝑇 𝐾𝐵 𝑇 ln(1 + 𝑒−2𝜀⁄𝐾𝐵𝑇 ) Using product rule 𝜕𝑇

𝜕

 𝑆 = 𝐾𝐵 ln(1 + 𝑒−2𝜀⁄𝐾𝐵𝑇 ) 𝜕𝑇 + 𝑇 𝜕𝑇 𝐾𝐵 ln(1 + 𝑒−2𝜀⁄𝐾𝐵𝑇 )

1 𝜕 (1 + 𝑒−2𝜀⁄𝐾𝐵𝑇 ) 1+𝑒−2𝜀⁄𝐾𝐵𝑇 𝜕𝑇 𝐾𝐵 𝑇 −2𝜀⁄𝐾𝐵 𝑇 . 2𝜀 𝐾𝐵 ln(1 + 𝑒−2𝜀⁄𝐾𝐵𝑇 ) + (1+𝑒−2𝜀 ⁄𝐾𝐵𝑇 𝑒 ) 𝐾𝐵 𝑇2 2𝜀⁄𝑇 𝐾𝐵 ln(1 + 𝑒−2𝜀⁄𝐾𝐵𝑇 ) + (1+𝑒−2𝜀⁄𝐾𝐵𝑇 ) 𝑒−2𝜀⁄𝐾𝐵𝑇 2𝜀⁄𝑇 𝐾𝐵 ln(1 + 𝑒−2𝜀⁄𝐾𝐵𝑇 ) + (𝑒2𝜀⁄𝐾𝐵𝑇 +1)

 𝑆 = 𝐾𝐵 ln(1 + 𝑒−2𝜀⁄𝐾𝐵𝑇 ) + 𝐾𝐵 𝑇  𝑆=  𝑆=  𝑆=

Similarly heat capacity at constant volume is 𝜕𝑆 𝐶𝑉 = 𝑇 ( ) 𝜕𝑇 𝑉 2𝜀⁄𝑇

𝜕

 𝐶𝑉 = 𝑇 𝜕𝑇 [𝐾𝐵 ln(1 + 𝑒−2𝜀⁄𝐾𝐵𝑇 ) +

(𝑒2𝜀⁄𝐾𝐵𝑇 +1)

]

Using product and division rule of derivatives 2𝜀

𝐾𝐵 −2𝜀 ⁄𝐾𝐵 𝑇 (1+𝑒 )

 𝐶𝑉 = 𝑇 [

. 𝑒−2𝜀⁄𝐾𝐵 𝑇 .

2𝜀 𝐾𝐵 𝑇2

⁡+

2𝜀

2𝜀

2𝜀

2𝜀

(𝑒𝐾𝐵 𝑇 +1)(− 2)+( )(𝑒𝐾𝐵 𝑇 ) 𝑇 𝑇 𝐾𝐵 𝑇2 (𝑒2𝜀⁄𝐾𝐵 𝑇 +1)

2

]

Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

25

 𝐶𝑉 =

−2𝜀 2𝜀 𝐾 𝑒 𝐵𝑇. 2 𝑇 𝑇[ (1+𝑒−2𝜀⁄𝐾𝐵 𝑇 ) −2𝜀

2𝜀

⁡−

2𝜀 ( 2) 𝑇

(𝑒2𝜀⁄𝐾𝐵 𝑇 +1)

2𝜀

2𝜀

+

2𝜀 4𝜀2 𝐾 (𝑒 𝐵 𝑇 ) 𝐾𝐵 𝑇3 2] (𝑒2𝜀⁄𝐾𝐵 𝑇 +1)

−2𝜀

2𝜀

(𝑒𝐾𝐵 𝑇 )( 2 )+( 2)−( 2)−( 2 )(𝑒𝐾𝐵 𝑇 ) 𝑇 𝑇 𝑇 𝑇

 𝐶𝑉 = 𝑇 [

 𝐶𝑉 =

 𝐶𝑉 =  𝐶𝑉 =  𝐶𝑉 =  𝐶𝑉 =  𝐶𝑉 =  𝐶𝑉 =

(1+𝑒−2𝜀⁄𝐾𝐵 𝑇 )(1+𝑒2𝜀⁄𝐾𝐵 𝑇 )

⁡+

2𝜀 4𝜀2 𝐾𝐵 𝑇 ) ( 𝑒 𝐾𝐵 𝑇3 2] 2𝜀⁄𝐾𝐵 𝑇

(1+⁡𝑒

)

4𝜀2 (𝑒2𝜀⁄𝐾𝐵 𝑇 ) 𝐾𝐵 𝑇3 𝑇[ 2] 2𝜀⁄𝐾𝐵 𝑇

(1+⁡𝑒

)

4𝜀2 (𝑒2𝜀⁄𝐾𝐵 𝑇 ) 𝐾𝐵 𝑇2 (1+⁡𝑒2𝜀⁄𝐾𝐵 𝑇 )

2

4𝜀2 (𝑒2𝜀⁄𝐾𝐵 𝑇 ) 𝐾𝐵 𝑇2 (1+⁡𝑒4𝜀⁄𝐾𝐵 𝑇 +⁡2𝑒2𝜀⁄𝐾𝐵 𝑇 ) 4𝜀2 𝐾𝐵 𝑇2 (𝑒−2𝜀⁄𝐾𝐵 𝑇 )(1+⁡𝑒4𝜀⁄𝐾𝐵 𝑇 +⁡2𝑒2𝜀⁄𝐾𝐵 𝑇 ) 4𝜀2 2𝜀⁄𝐾𝐵 𝑇

𝐾𝐵 𝑇2 (𝑒−2𝜀⁄𝐾𝐵 𝑇 +⁡𝑒

+2)

4𝜀2 𝜀

𝐾𝐵 𝑇2 .4 𝑐𝑜𝑠ℎ2(𝐾 𝑇) 𝐵 𝜀2 𝜀

𝐾𝐵 𝑇2 𝑐𝑜𝑠ℎ2(𝐾 𝑇) 𝐵

So this is the heat capacity for two level system and is commonly referred as Schottky heat capacity. Now the average internal energy 𝑇𝑆 + 𝐹 for two level system is ̅ = 𝑇𝑆 + 𝐹 𝑈 Where 𝐹 = −𝜀 − 𝐾𝐵 𝑇 ln(1 + 𝑒 −2𝜀⁄𝐾𝐵 𝑇 ) and 𝑆 = 𝐾𝐵 ln(1 + 𝑒−2𝜀⁄𝐾𝐵𝑇 ) + ̅ = 𝑇 [𝐾𝐵 ln(1 + 𝑒−2𝜀⁄𝐾𝐵𝑇 ) + 𝑈

̅= 𝑈

2𝜀 (𝑒2𝜀⁄𝐾𝐵 𝑇 +1)

2𝜀⁄𝑇 (𝑒2𝜀⁄𝐾𝐵𝑇 +1)

2𝜀⁄𝑇 ] − 𝜀 − 𝐾𝐵 𝑇 ln(1 + 𝑒−2𝜀⁄𝐾𝐵𝑇 ) (𝑒2𝜀⁄𝐾𝐵𝑇 + 1)

−𝜀

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26

̅= 𝑈 ̅= 𝑈 ̅= 𝑈 ̅= 𝑈 ̅= 𝑈

2𝜀−𝜀(𝑒2𝜀⁄𝐾𝐵 𝑇 +1) (𝑒2𝜀⁄𝐾𝐵 𝑇 +1)

2𝜀−𝜀−𝜀(𝑒2𝜀⁄𝐾𝐵 𝑇 ) (1+⁡𝑒2𝜀⁄𝐾𝐵 𝑇 )

𝜀−𝜀𝑒2𝜀⁄𝐾𝐵 𝑇 (1+⁡𝑒2𝜀⁄𝐾𝐵 𝑇 )

𝜀(1−𝑒2𝜀⁄𝐾𝐵 𝑇 ) (1+⁡𝑒2𝜀⁄𝐾𝐵 𝑇 )

−𝜀(𝑒2𝜀⁄𝐾𝐵 𝑇 −1) (𝑒2𝜀⁄𝐾𝐵 𝑇 +1)

̅ = −𝜀 tanh (  𝑈

𝜀 𝐾𝐵 𝑇

)

For N-two level systems, we get ̅ = −𝑁𝜀 tanh ( 𝑈

𝜀 ) 𝐾𝐵 𝑇

SINGLE PARTICLE IN ONE DIMENSIONAL BOX Imagine a system which is simply a box with a simple particle in it. The box is replicated with (𝑀 − 1) times to build up the Gibbs ensemble. Assuming that each particle is localized in box and each box is distinguishable from rest. To determine thermodynamics of each particle let us start by considering the quantum states of a particle of mass ‘m’ moving along the x-direction. Schrodinger’s wave equation in one dimension is −

ħ2 𝜕 2 𝜙(𝑥) + 𝑉(𝑥)𝜙(𝑥) = 𝜀𝜙(𝑥) 2𝑚 𝜕𝑥 2

Where 𝑉(𝑥)⁡is PE and 𝜙(𝑥) is a single particle and 𝜀 is the energy. The particle is confined in one dimensional box of length L. P.E is zero in range 0 < 𝑥 < 𝐿 and infinite elsewhere. Solution of Schrodinger’s equation which satisfies boundary conditions 𝜙(𝑥 = 𝐿) = 0 and 𝜙(𝑥 = 0) = 0 is a standing wave of the form 𝑛𝜋𝑥 ) 𝐿

𝜙𝑛 (𝑥) = 𝐴 sin (

First boundary condition is automatically satisfied by such a wavefunction but for second condition n must be positive non zero integer. Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

27

If n is negative, then the wavefunction is linearly dependent, but we want independent solution. Corresponding to independent quantum states −3𝜋𝑥 3𝜋𝑥 𝜙−3 (𝑥) = 𝐴 sin ( ) = −𝐴 sin ( ) = −𝜙3 (𝑥) 𝐿 𝐿 Same wavefunction as of 𝜙3 apart from – sign, it is not independent solution. The energy eigenvalue for wavefunction 𝜙𝑛 (𝑥) is ħ2 𝜋 2 𝑛2 𝜀𝑛 = ( ) = 𝛼𝑛2 2𝑚𝐿2 ħ2 𝜋 2

Where 𝛼 = 2𝑚𝐿2 The quantity 𝛼 sets the scale of energy For an electron in a length of centimeter, the value of 𝛼 is 𝛼 = 6.0 ∗ 10−34 𝐽 Or in terms of temperature 𝛼⁄𝐾 = 4.4 ∗ 10−11 𝐾, which is very small. 𝐵⁡

If we do experiments at 300K, the quantity 𝛾 = 𝛼/𝐾𝐵 is about 1.5 ∗ 10−13. Now our primary objective is to work out partition function and then thermodynamic properties. The partition function for translational motion in one dimension is ∞

𝑍𝑡𝑟𝑎𝑛𝑠 = ∑ 𝑒

𝜀 − 𝑛 𝐾𝐵 𝑇

𝑛=1 ∞

𝑍𝑡𝑟𝑎𝑛𝑠 = ∑ 𝑒 −𝛾𝑛

2

𝑛=1

Where the factor 𝛾 is minute ∞ 2

𝑍𝑡𝑟𝑎𝑛𝑠 = ∫ 𝑒 −𝛾𝑛 𝑑𝑛 0

Let 𝑥 = 𝛾𝑛2



𝑥 𝛾

= 𝑛2

 𝑛= 

𝑑𝑛 𝑑𝑥

√𝑥 √𝛾 1

=

2√𝛾

1

𝑥 −2

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28 ∞

𝑍𝑡𝑟𝑎𝑛𝑠 = ∫ 𝑒

∞ −𝛾𝑛2

𝑑𝑛 = ∫ 𝑒 −𝑥

0

0

1 2√𝛾

𝑥

−

1 2 𝑑𝑥

=

1 2√𝛾

∞

1

∫ 𝑥 2−1 . 𝑒 −𝑥 𝑑𝑥 0

∞

As ∫0 𝑥 𝑛−1 . 𝑒 −𝑥 𝑑𝑥 = Γ(𝑛) 𝑍𝑡𝑟𝑎𝑛𝑠 =

1

1 Γ( ) 2√𝛾 2

1

As Γ (2) = √𝜋, so 𝑍𝑡𝑟𝑎𝑛𝑠

1

1

𝜋 2 𝐿2 𝑚𝐾𝐵 𝑇 2 √𝜋 = =( ) =( ) √𝜋 = 4𝛾 2𝜋ħ2 2√𝛾 2 √𝛾 1

Now the total final result can be written as 𝑍𝑡𝑟𝑎𝑛𝑠 = 2𝜋ħ2

Where 𝜆𝐷 = (𝑚𝐾 𝑇)

𝐿 𝜆𝐷

1 2

𝐵

The quantity 𝜆𝐷 is called de Broglie wavelength. According to de Broglie, the wavelength of a particle is related to the momentum p as 𝜆 = ℎ/𝑝. There is a distribution of the momentum or particle due to thermal excitation and therefore a distribution of wavelengths. The thermal de Broglie wavelength is the thermal average wavelength of a free particle of momentum ‘p’ in a three dimensional system. Now, the free energy of a single gas particle in a length C is 𝐹 = −𝐾𝐵 𝑇 ln(𝑍) 1

𝐿2 𝑚𝐾𝐵 𝑇 2 𝐾𝐵 𝑇 𝐿2 𝑚𝐾𝐵 𝑇 𝐹 = −𝐾𝐵 𝑇 ln ( ) = − ln ( ) 2𝜋ħ2 2 2𝜋ħ2 𝜕𝐹

As entropy is 𝑆 = − (𝜕𝑇)

𝑉

𝜕𝐹 𝐾𝐵 𝑇 2𝜋ħ2 𝐿2 𝑚𝐾𝐵 𝐿2 𝑚𝐾𝐵 𝑇 𝐾𝐵 = . 2 . + ⁡ ln ( ) 𝜕𝑇 2 𝐿 𝑚𝐾𝐵 𝑇 2𝜋ħ2 2𝜋ħ2 2 

𝜕𝐹

𝐾𝐵



𝜕𝐹

2 𝐾𝐵

= 𝜕𝑇 𝜕𝑇

=

 𝑆=

2

𝐾𝐵 2

𝐿2 𝑚𝐾𝐵 𝑇 𝐾𝐵

+ ⁡ ln (

)

2𝜋ħ2 2 𝐿2 𝑚𝐾𝐵 𝑇

(1 + ln (

(1 + ln (

2𝜋ħ2 𝐿2 𝑚𝐾𝐵 𝑇 2𝜋ħ2

))

))

Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

29 𝜕𝑆

As heat capacity is 𝐶𝑉 = 𝑇 (𝜕𝑇)

𝑉

𝜕𝑆 𝜕 𝐾𝐵 𝐿2 𝑚𝐾𝐵 𝑇 = [ln ( ) + 1] 𝜕𝑇 𝜕𝑇 2 2𝜋ħ2 

𝜕𝑆

= 𝜕𝑇

𝐾𝐵



𝜕𝑆

2 𝐾𝐵

𝜕𝑇

=

2𝜋ħ2

[𝐿2 𝑚𝐾

𝐵

. 𝑇

𝐿2 𝑚𝐾𝐵 2𝜋ħ2

]

2𝑇

Now 𝐶𝑉 = 𝑇 (

𝜕𝑆 𝐾𝐵 𝐾𝐵 ) =𝑇 = 𝜕𝑇 𝑉 2𝑇 2

Translational motion of a single particle in one dimension gives a contribution to the heat capacity of 𝐾𝐵 ⁄2

SINGLE PARTICLE IN A THREE DIMENSIONAL BOX The same method can be used to describe quantum states of a particle in a cube of volume 𝐿3 . Wavefunction of that cube is 𝑛1 𝜋𝑥 𝑛2 𝜋𝑦 𝑛3 𝜋𝑧 ) sin ( ) sin ( ) 𝐿 𝐿 𝐿

𝜙𝑙 (𝑥, 𝑦, 𝑧) = 𝐴 sin (

A wavefunction describes a standing wave in all 3-dimensions of a wave which vanishes when 𝑥, 𝑦, or⁡𝑧 = 0⁡where⁡𝑥, 𝑦, or⁡𝑧 = 1 We can denote quantum state by 𝜙𝑖 = |𝑛1 , 𝑛2 , 𝑛3 > The energy of free particle of mass ‘m’ involves these three quantum numbers 𝑛1 , 𝑛2 , and⁡𝑛3 The quantum state is given by the solution of ħ2 2 ∇ 𝜙(𝑥, 𝑦, 𝑧) + 𝑉(𝑥, 𝑦, 𝑧) + 𝜙𝑖 (𝑥, 𝑦, 𝑧) = ∑ 𝜙𝑖 (𝑥, 𝑦, 𝑧) 2𝑚 𝑖

The P.E is zero inside box and infinite elsewhere. The three quantum numbers 𝑛1 , 𝑛2 , and⁡𝑛3 are all positive integers. Energy eigenvalues are 𝜀=

ħ2 𝜋 2 2 (𝑛 + 𝑛22 + 𝑛32 ) 2𝑚𝐿2 1

So the partition function for translational motion is

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30 ∞

∞

∞ 2

2

2

𝑍𝑡𝑟𝑎𝑛𝑠 = ∑ ∑ ∑ 𝑒 −𝛾(𝑛1 +𝑛2 +𝑛3 ) 𝑛1 =1 𝑛2 =1 𝑛3 =1

We can write this as ∞

∞

𝑍𝑡𝑟𝑎𝑛𝑠 = [ ∑ 𝑒

−𝛾𝑛12

][∑ 𝑒

𝑛1 =1

∞ −𝛾𝑛22

𝑛2 =1

𝑍𝑡𝑟𝑎𝑛𝑠 = ∑ 𝑒

−𝛾𝑛12

𝑛1 =1

𝑛3 =1 1

∞

∞

2

] [ ∑ 𝑒 −𝛾𝑛3 ]

=∫𝑒

−𝛾𝑛12

0

𝑚𝐾𝐵 𝑇 2 𝑑𝑛1 = 𝐿 ( ) 2𝜋ħ2

Which is calculated in previous section. So 1

𝑍𝑡𝑟𝑎𝑛𝑠

1

1

3

𝑚𝐾𝐵 𝑇 2 𝑚𝐾𝐵 𝑇 2 𝑚𝐾𝐵 𝑇 2 𝑚𝐾𝐵 𝑇 2 3 = 𝐿( ) . 𝐿 ( ) . 𝐿 ( ) = 𝐿 ( ) 2𝜋ħ2 2𝜋ħ2 2𝜋ħ2 2𝜋ħ2

Now from 𝑍𝑡𝑟𝑎𝑛𝑠 we can calculate all the thermodynamics of a single particle in a box of volume = L3 . As free energy is 𝐹 = −𝐾𝐵 𝑇 ln(𝑍𝑡𝑟𝑎𝑛𝑠 ) 3

 𝐹=

𝑚𝐾 𝑇 2 −𝐾𝐵 𝑇 ln [𝑉 ( 2𝜋ħ𝐵2 ) ] 3

𝑚𝐾 𝑇

 𝐹 = −𝐾𝐵 𝑇 [ln 𝑉 + 2 ln ( 2𝜋ħ𝐵2 )] The pressure is 𝑃 = −(

𝜕𝐹 ) 𝜕𝑉 𝑇

𝜕𝐹 𝜕 3 𝑚𝐾𝐵 𝑇 1 −𝐾𝐵 𝑇 = [−𝐾𝐵 𝑇⁡ (ln 𝑉 + ln ( ))] = −𝐾𝐵 𝑇. . 1 = 2 𝜕𝑉 𝜕𝑉 2 2𝜋ħ 𝑉 𝑉 𝑃=

𝐾𝐵 𝑇 ⁡𝑜𝑟⁡𝑃𝑉 = 𝐾𝐵 𝑇 𝑉

Formula for the contribution to the pressure of a single particle in a gas For N particles, equation will be 𝑃𝑉 = 𝑁𝐾𝐵 𝑇⁡(provided⁡that⁡particle⁡doesn′ tinteract) Where N is Avogadro’s number, then 𝑃𝑉 = 𝑁𝐴 𝐾𝐵 𝑇 For an ideal gas 𝑅 = 𝑁𝐴 𝐾𝐵 Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

31

 𝑃𝑉 = 𝑅𝑇 As entropy is 𝑆 = −(

𝜕𝐹 ) 𝜕𝑇 𝑉

3 𝑚𝐾𝐵 𝑇 𝐹 = −𝐾𝐵 𝑇 [ln 𝑉 + ln ( )] 2 2𝜋ħ2 𝜕𝐹 3 𝑚𝐾𝐵 𝑇 3 = −𝐾𝐵 [ln(𝑉) + ln ( ) + ] 𝜕𝑇 2 2𝜋ħ2 2 And so 3 𝑚𝐾𝐵 𝑇 3 𝑆 = 𝐾𝐵 [ln(𝑉) + ln ( ) + ] 2 2𝜋ħ2 2 And as heat capacity is 𝐶𝑉 = 𝑇 (

𝜕𝑆 ) 𝜕𝑇 𝑉

So 𝜕𝑆 3 2𝜋ħ2 𝑚𝐾𝐵 3𝐾𝐵 = 𝐾𝐵 [ . . ]= 2 𝜕𝑇 2 𝑚𝐾𝐵 𝑇 2𝜋ħ 2𝑇 Which means 𝐶𝑉 = 𝑇

3𝐾𝐵 3 = 𝐾𝐵 2𝑇 2 3

In 3-dimensions, a single particle gives a contribution of 2 𝐾𝐵 to the heat capacity. 3

For N-non interacting particles, heat capacity will be 2 𝑁𝐾𝐵

EXPRESSIONS FOR HEAT AND WORK Heat and work are very simple concepts in statistical mechanics; the addition of small amount of heat corresponds to changing the probabilities slightly keeping energy levels fixed. The addition of small amount of work corresponds to changing energy levels slightly keeping the probabilities fixed. Any process which cause a shift in energy level can give rise to work. Examples of such processes include changing the volume, the magnetic induction field, or electric field, for all these processes can change the energy levels of the system. The average internal energy is defined to be

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32

̅ = ∑ 𝑃𝑖 𝐸𝑖 𝑈 𝑖

Now let us suppose the system starts in its canonical state because it is in equilibrium with a heat bath at temperature T. The probabilities of the system being in each quantum state are given by 𝑒 −𝐸𝑖 ⁄𝐾𝐵 𝑇 𝑃𝑖 = 𝑍 Where 𝐸𝑖 is the energy eigenvalue for quantum state 𝜓1 . An infinitesimal process carried out quasistatistically leads to changes in the average energy of the system of ̅ = ∑ 𝑃𝑖 𝐸𝑖 𝑈 𝑖

̅ = ∑𝑖 𝑑𝑃𝑖 𝐸𝑖 + ∑𝑖 𝑃𝑖 𝑑𝐸𝑖  𝑑𝑈 ̅ = đ𝑄𝑟𝑒𝑣 − 𝑑𝑊𝑟𝑒𝑣 . As small change in internal energy of reversible processes can be written as 𝑑𝑈 Imagine the external parameters are constant; the volume is kept constant and the magnetic and electric fields are kept constant; all the constraints we can make on the system are imposed. We cannot alter the external parameters which means energy levels are fixed so 𝑑𝐸1 = 0, so the only way we can increase internal energy of the system is to add heat in an infinitesimal process. The addition of heat occurs when energy levels are fixed and probabilities are altered somehow. It follows that the work done on the system is given by 𝑑𝑊 = ∑ 𝑃𝑖 𝑑𝐸𝑖 𝑖

Reversible work is given by the weighted average of changes in the energy levels for constant probabilities. This statement gives us a very simple picture of the work done on a system. Both the energy levels and probabilities are altered when system undergoes a change. If the change in energy levels and probabilities are not small, then we cannot identify heat and work separately. The identification of heat and work can only be made for infinitesimal changes in the energy levels and probability. We can use this idea to calculate the work done when the volume of a system change. Consider the change in the energy levels of gas atoms in a cube of volume 𝑉 = 𝐿3 . The single particle energy levels are 𝜀𝑖 =

ħ2 𝜋 2 2 ħ2 𝜋 2 2 2 2) (𝑛 (𝑛1 + 𝑛22 + 𝑛32 ) + 𝑛 + 𝑛 = 2 3 2 2𝑚𝐿2 1 2𝑚𝑉 3

We cannot change energy levels by changing the volume. Let us shrink the cube systematically in each dimension. This changes the energy levels, since 𝜕𝜀𝑖 ħ2 𝜋 2 2 𝜕 −2 (𝑛1 + 𝑛22 + 𝑛32 ) = 𝑉 3 𝜕𝑉 2𝑚 𝜕𝑉 Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

33

  

𝜕𝜀𝑖 𝜕𝑉 𝜕𝜀𝑖 𝜕𝑉 𝜕𝜀𝑖 𝜕𝑉

2 ħ2 𝜋 2

=− . =− =−

(𝑛12 + 𝑛22 + 𝑛32 ).

3 2𝑚 2 ħ2 𝜋 2

(

3𝑉 2𝑚𝑉 2⁄3 2𝜀𝑖

1 𝑉 5⁄3

) (𝑛12 + 𝑛22 + 𝑛32 )

3𝑉

By changing the volume of the system, each of the single particle energy levels change. Since the total energy 𝐸𝑖 of quantum state is made up of a sum of these single particle energy levels, we find that 𝜕𝐸𝑖 2𝐸𝑖 =− 𝜕𝑉 3𝑉 We can do this by changing the volume of the gas. This work done is given by 𝜕𝐸𝑖 −𝑃𝑑𝑉 = ∑ 𝑃𝑖 𝑑𝐸𝑖 = ∑ 𝑃𝑖 ( ) 𝑑𝑉 𝜕𝑉 𝑖

As

𝜕𝐸𝑖 𝜕𝑉

𝑖

2𝐸

= − 3𝑉𝑖 , so −𝑃𝑑𝑉 = ∑ 𝑃𝑖 (− 𝑖

 𝑃𝑉 =

̅ 2𝐸𝑖 2 2𝑈 ) 𝑑𝑉 = − ∑ 𝑃𝑖 𝐸𝑖 𝑑𝑉 = − 𝑑𝑉 3𝑉 3𝑉 3𝑉 𝑖

̅ 2𝑈 3

Work can be done in other ways. Any quantity which couples to the energy levels can be used to do work. Suppose we have a gas of atoms in a magnetic induction field. The electrons in each atoms couple to the magnetic induction field through their orbital motion and their spin. The coupling of magnetic induction field to the energy levels of gas atoms is called Zeeman Effect. Such a system can do a magnetic work. When the magnetic induction field changes, all the energy levels move and work is done. We could also apply the electric fields to the gas of atoms. The shift of the energy of gas atoms with electric field is called the Stark effect. Any interaction which will couple to the energy levels can give rise to work. We could apply both electric and magnetic field to the system and do both electrical and magnetic work. The changes in energy levels induced by fields can be used to obtain expressions for different states of work. From these expressions, we can derive equation of state for the system. Suppose we have a system in which energy levels depend on the volume of sample V, and on a uniform magnetic induction field which acts only along the z-direction 𝐵 = 𝐵𝑧 𝑘̂ Infinitesimal changes in energy level 𝐸𝑖 (𝑉, 𝐵2 ) arise from two processes. (1) We can change volume keeping magnetic induction constant All Our Notes Are Belong To You

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34

(2) We can change magnetic induction and keep volume constant. For small changes in both quantities 𝜕𝐸𝑖 𝜕𝐸𝑖 𝑑𝐸𝑖 = ( ) 𝑑𝑉 + ( ) 𝑑𝐵𝑧 𝜕𝑉 𝜕𝐵𝑧 As 𝑑𝑊 = ∑ 𝑃𝑖 𝑑𝐸𝑖 𝑖

The work now has two terms 𝜕𝐸𝑖 𝜕𝐸𝑖 𝑑𝑊 = ∑ 𝑃𝑖 𝑑 [( ) 𝑑𝑉 + ( ) 𝑑𝐵𝑧 ] 𝜕𝑉 𝜕𝐵𝑧 𝑖

Since the first form involves the change in volume, it must be just −𝑃𝑑𝑉. The second term is the magnetic work. 𝜕𝐸𝑖 đ𝑊 = ∫ 𝐻. 𝑑𝐵𝑑 3 𝑟 = ∑ 𝑃𝑖 ( ) 𝑑𝐵𝑧 𝜕𝐵𝑧 𝑖

the magnetic field can be written as 𝐻=

𝐵 −𝑀 𝑀𝑜

Where M is the magnetic moment per unit volume or magnetization. Suppose for simplicity, that the magnetic field H as well as B is uniform over space and lies along 2direction, then ∑ 𝑃𝑖 ( 𝑖

𝜕𝐸𝑖 ) 𝑑𝐵𝑧 = ∫ 𝐻. 𝑑𝐵𝑑 3 𝑟 𝜕𝐵𝑧

Putting the value of H in this equation ∫(

𝐵 1 − 𝑀) . 𝑑𝐵𝑑 3 𝑟 = ∫ 𝐵𝑧 𝑑𝐵𝑧 𝑑 3 𝑟 − ⁡ ∫ 𝑀2 𝑑𝐵2 𝑑 3 𝑟 𝑀𝑜 𝑀𝑜

This equation can be used to calculate the magnetization of a simple system. In the absence of magnetic field 1 𝜕𝐸𝑖 𝑀𝑧 = − ∑ 𝑃𝑖 ( ) 𝑉 𝜕𝐵𝑧 𝑣 𝑖

For a single spin ½ system with energy leves 𝜇𝐵𝑧 and −𝜇𝐵𝑧 .

Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

35

𝑀𝑧 =

𝜇 𝜇𝐵𝑧 tanh ( ) 𝑉 𝐾𝐵 𝑇

ROTATIONAL ENERGY LEVELS FOR DIATOMIC MOLECULES Consider a molecule such as carbon monoxide , made up of two different atoms and one carbon and one oxygen separated by a distance d. Such a molecule can exist in quantum states of different orbital angular momentum with each state having energy. 𝜀=

ħ𝑙(𝑙 + 1) 2𝐼

Here 𝐼 = 𝑀𝑑 2 is the moment of inertia of molecule about an axis through its center of mass and 𝑙 = 0, 1, 2 … is the quantum number associated with orbital angular momentum. Each energy level of the rotating molecule has a degeneracy 𝑔𝑙 = (2𝑙 + 1). This means there is one 𝑙 = 0 state, three 𝑙 = 1 states, five 𝑙 = 2 states etc. As partition function is 𝑍 = ∑𝑒

−

𝐸𝑗 𝐾𝐵 𝑇

𝑗

In case of degeneracy, the partition function is 𝑍 = ∑ 𝑔𝑛

𝐸 − 𝑛 𝐾 𝑒 𝐵𝑇

𝑛

Here 𝑔𝑛 = (2𝑙 + 1), 𝐸𝑙 =

ħ𝑙(𝑙+1) 2𝐼

So partition function will become ∞

𝑍 = ∑(2𝑙 + 1)𝑒

−

ħ𝑙(𝑙+1) 2𝐼𝐾𝐵 𝑇

𝑙=0 ħ2

For large values of 𝐼𝐾 𝑇 , only the first few terms in the summation are important. We can truncate the 𝐵

series after the first two terms. ħ2

𝑍 = 1 + 3𝐼𝐾𝐵 𝑇 + ⋯ Taking ln on both sides ħ2

ln 𝑍 = ln (1 + 3𝐼𝐾𝐵 𝑇 ) = − 3𝑒

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−

ħ2 𝐼𝐾𝐵 𝑇

+⋯

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36

As 𝐹 = 𝐾𝐵 𝑇 ln(𝑍)  𝐹 = −3𝐾𝐵 𝑇𝑒

−

ħ2 𝐼𝐾𝐵 𝑇

As ∞

𝑍 = ∑(2𝑙 + 1)𝑒

−

ħ2 2𝐼𝐾𝐵 𝑇

𝑙=0

For small values of

ħ2 𝐼𝐾𝐵 𝑇

(for higher temperatures) the sum can be replaced with the little error by an

integral. ∞

𝑍 = ∫ (2𝑙 + 1)𝑒

−

ħ2 2𝐼𝐾𝐵 𝑇 𝑑𝑙

0

Let 𝑦 = 𝑙(𝑙 + 1) 

𝑑𝑦 𝑑𝑙

= 2𝑙 + 1 ∞

𝑍 = ∫ 𝑦𝑒 0

−

ħ2 𝑦 2𝐼𝐾𝐵 𝑇

∞

2

ħ 𝑦 𝑑𝑦 − 2𝐼𝐾 𝐵 𝑇 𝑑𝑦 =∫ 𝑒 𝑦 0

ħ2 𝑦

Let 𝑥 = 2𝐼𝐾 𝑇, 𝐵



𝑑𝑥 𝑑𝑦

ħ2

= 2𝐼𝐾

𝐵𝑇

ħ2

 𝑑𝑥 = 2𝐼𝐾

𝐵𝑇

𝑑𝑦 ∞

2

ħ 𝑦 2𝐼𝐾𝐵 𝑇 ħ2 − 2𝐼𝐾𝐵 𝑇 . 𝑍= ⁡∫ 𝑒 𝑑𝑦 ħ2 2𝐼𝐾𝐵 𝑇 0

 𝑍=

2𝐼𝐾𝐵 𝑇

 𝑍=

2𝐼𝐾𝐵 𝑇

ħ2

∞

∫0 𝑒 −𝑥 𝑑𝑥

ħ2

As 𝐹 = −𝐾𝐵 𝑇 ln(𝑍), so 𝐹 = −𝐾𝐵 𝑇 ln (

2𝐼𝐾𝐵 𝑇 ) ħ2

𝜕𝐹

As entropy is 𝑆 = − (𝜕𝑇) , so 𝑉

𝑆 = 𝐾𝐵

𝜕 2𝐼𝐾𝐵 𝑇 𝑇 ln ( 2 ) 𝜕𝑇 ħ

Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

37

1

 𝑆 = 𝐾𝐵 [𝑇. 2𝐼𝐾𝐵 𝑇 . ħ2

2𝐼𝐾𝐵 ħ2

2𝐼𝐾𝐵 𝑇

 𝑆 = 𝐾𝐵 [1 + ln (

ħ2

2𝐼𝐾𝐵 𝑇

+ ln (

ħ2

)]

)]

As heat capacity is 𝐶𝑉 = 𝑇 (

𝜕𝑆 ) 𝜕𝑇 𝑉

𝜕𝑆 1 2𝐼𝐾𝐵 𝐾𝐵 = 𝐾𝐵 ( . 2 )= 2𝐼𝐾𝐵 𝑇 ħ 𝜕𝑇 𝑇 ħ2 So 𝐶𝑉 = 𝑇.

𝐾𝐵 = 𝐾𝐵 𝑇

It follows that the contribution to the heat capacity at constant volume for high temperature is KB.

VIBRATIONAL ENERGY LEVELS FOR DIATOMIC MOLECULES To a good approximation, the vibrational states of a diatomic molecule are those of a harmonic oscillator. There is an attractive potential 𝑉(𝑟) holding the atoms together, a potential which has a minimum at a separation of ‘d’. For a small displacement from minimum, the potential is of the form (𝑟 − 𝑑)2 𝑑 2 𝑉(𝑟) 𝑉(𝑟) ≈ 𝑉(𝑑) + ( ) ⁡ 2 𝑑𝑟 2 𝑟=𝑑 That is the P.E is that of a simple harmonic oscillator. The energy levels of a one dimensional harmonic oscillator are given by 1 𝜀 = ħ𝜔 (𝑛 + ) 2 Where 𝜔⁡is angular frequency of vibration and n is is a positive integer or zero. As ∞

𝑍 = ∑𝑒

𝜀 − 𝑛 𝐾𝐵 𝑇

𝑛=0

So 1 ħ𝜔(𝑛+ ) 2 − ∑ 𝑒 𝐾𝐵 𝑇 ∞

𝑍=

𝑛=0

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38

 𝑍=𝑒  𝑍=

ħ𝜔 2𝐾𝐵 𝑇

−

+𝑒

3ħ𝜔 2𝐾𝐵 𝑇

−

+𝑒

5ħ𝜔 2𝐾𝐵 𝑇

−

+ ⋯⁡

− ħ𝜔 𝑒 2𝐾𝐵 𝑇 − ħ𝜔 1−𝑒 𝐾𝐵 𝑇

ħ𝜔 For high temperatures i.e. 𝑇 ≫ ⁡ 𝐾 𝑇, the partition function is approximately 𝐵

𝑍=

𝐾𝐵 𝑇 ħ𝜔

As 𝐹 = −𝐾𝐵 𝑇 ln(𝑍) ħ𝜔

− 𝑒 2𝐾𝐵𝑇

 𝐹 = −𝐾𝐵 𝑇 ln (

ħ𝜔 − 1−𝑒 𝐾𝐵𝑇

 𝐹 = −𝐾𝐵 𝑇 [ln 𝑒  𝐹 = −𝐾𝐵 𝑇 [− 𝐹=

ħ𝜔 2

ħ𝜔 2𝐾𝐵 𝑇

−

ħ𝜔 2𝐾𝐵 𝑇

)

− ln (1 − 𝑒

− ln (1 − 𝑒 ħ𝜔 𝐾𝐵 𝑇

−

+ 𝐾𝐵 𝑇 ln (1 − 𝑒

ħ𝜔 𝐾𝐵 𝑇

−

ħ𝜔 𝐾𝐵 𝑇

−

)]

)]

)⁡

So the entropy will be 𝜕𝐹 𝑆=( ) 𝜕𝑇 𝑉

Here

𝜕𝐹 𝜕𝑇

= 𝐾𝐵 [𝑇.

1 ħ𝜔 −𝐾 𝑇 (1−𝑒 𝐵 )

.𝑒

ħ𝜔 𝐾𝐵 𝑇

−

.

−ħ𝜔 𝐾𝐵 𝑇 2

+ ln (1 − 𝑒

𝑆 = −𝐾𝐵 ln (1 −

ħ𝜔 − 𝐾 𝑒 𝐵𝑇 )

ħ𝜔 𝐾𝐵 𝑇

−

)]

ħ𝜔⁄𝑇

+

ħ𝜔

(𝑒 𝐾𝐵𝑇 −1)

𝑒 𝜕𝑆

Now 𝐶𝑉 = 𝑇 (𝜕𝑇)

𝑉

𝜕𝑆 = −𝐾𝐵 𝜕𝑇

1 ħ𝜔 − (1 − 𝑒 𝐾𝐵 𝑇 )

. (−𝑒

−

ħ𝜔 𝐾𝐵 𝑇 ) (

ħ𝜔 ) +⁡ 𝐾𝐵 𝑇 2

ħ𝜔 ħ𝜔 −ħ𝜔 ħ𝜔 ħ𝜔 ) (𝑒 𝐾𝐵 𝑇 − 1) ( 2 ) − ( 𝑇 ) (𝑒 𝐾𝐵 𝑇 ) (− 𝑇 𝐾𝐵 𝑇 2 ħ𝜔 (𝑒 𝐾𝐵 𝑇

2

− 1)

Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

39

MINIMIZING THE FREE ENERGY One of the simplest and most powerful technique for calculating the equilibrium properties of a system comes from the observation that the free energy is at minimum when the system is at equilibrium. All we have to do is obtaining an expression for the free energy as a function of a parameter (or set of parameters) and then locate the minimum in free energy. This procedure generates an equation of state. The value of this approach lies not only in the range of problems to which it can successfully be applied, but also in generation of physical intuition in guessing the solution to very difficult problems. We concentrate here on easy ones. MINIMIZING THE HELMHOLTZ FREE ENERGY First we prove the idea that a system of constant volume which is displaced from thermal equilibrium evolves so that the Helmholtz Free Energy tends to a minimum. In other words, Helmholtz Free Energy is a minimum for a system held at constant volume and temperature. Consider system ‘A’ to be in thermal contact with a large heat bath or reservoir, which is at temperature ‘T’ given by, 1 ∂ ln(WR ) =( ) KBT ∂UR V

Where UR is internal energy of reservoir. Integrating the above equation gives us WR = γeUR⁄KB T

Where γ is integration constant and is independent of UR . When system ‘A’ is placed in contact with heat bath, the joint number of accessible states is W = WA WR

So WR = γeUR⁄KB T W = WA . γeUR ⁄KBT

Now WA can be written as

WA = e(KBTln(WA))⁄KBT As K B T will be cancelled with K B T in denominator and eln(WA ) = 𝑊𝐴 All Our Notes Are Belong To You

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40

Using the above expression, W can be written as W=e

(KB Tln(WA ))⁄KB T

UR K . γe BT

= γe(UR+KBTln(WA))⁄KBT

As total energy of the joint system is UT = UA + UR => UR = UT − UA Substituting this value W = γe(UT−UA+KBTln(WA))⁄KB T The combined system is in equilibrium when 𝐾𝐵 ln(𝑊) is maximum. The idea is to maximize ln⁡(𝑊) by varying UA. To do this, we must minimize the quantity, UA − K B Tln(WA ) UA − TSA = → (1) KBT KBT Where SA = K B ln⁡(WA ) The temperature of system A is the same as that of heat bath because they are in thermal equilibrium. Hence we can replace T by TA in eq (1). It follows that we minimize the Helmholtz free energy, FA = UA − TA ⁡SA. The system is in equilibrium when total energy is maximum. It follows that the system in in equilibrium when free energy of system A is minimum. We can derive an equation of state for the system from the condition that the free energy is minimum when the system is in equilibrium with a heat bath. By way of illustration, consider n vacancies in a solid with each vacancy raising the energy by an amount ‘𝜀’. The solid is composed of ‘N’ atoms on ‘N+n’ lattice sites. The number of ways of arranging the vacancies in the lattice sites is

(N+n)! N!n!

.

So entropy will be

S = K B ln (

(N + n)! ) → (2) N! n!

By definition of Helmholtz free energy, we know that

F = U − TS As there are ‘n’ vacancies and each vacancy raise energy by an amount ε, so total energy is U = nε. So we can write,

F = nε − TS Now putting the value of S in the above equation i.e.

 F = nε − K B T ln (

(N+n)! N!n!

)

 F = nε − K B T[ln(N + n)! − lnN! n!] Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

41

 F = nε − K B T[ln(N + n)! − ln N! − ln n!] Now using Stirling’s approximation i.e. ln N! = N ln N − N⁡

 F = nε − K B T[(N + n) ln(N + n) − (N + n) − (N ln N − N⁡) − (n ln n − n)]  F = nε − K B T[(N + n) ln(N + n) − N − n − N ln N + N − n ln n + n]  F = nε − K B T[(N + n) ln(N + n) − N ln N − n ln n] Adding and subtracting n ln N

F = nε − K B T[(N + n) ln(N + n) − N ln N − n ln N − n ln n + n ln N]  F = nε − K B T[(N + n) ln(N + n) − (N + n) ln N − (n ln n − n ln N)]  F = nε − K B T[(N + n) ln(N + n) − (N + n) ln N − n(ln n − ln N)]  F = nε − K B T[(N + n)(ln(N + n) − ln N) − n(ln n − ln N)]

Using ln a − ln b = ln

a b

 F = nε − K B T [(N + n) ln (

N+n

n

N

N

) − n ln ( )⁡]

Multiplying and dividing by N

 F = nε − K B T. N [(

N+n

N+n

n

n

N

N

N

N

) ln (

) − ln ( )⁡] → (A)

Now we minimize ‘F’ by varying n to find the equilibrium number of vacancies. Differentiating eq (A) with respect to ‘n’

∂F ∂ ∂ N+n N+n n n ) ln ( ) − ln ( )⁡] = nε − K B T. N [( ∂n ∂n ∂n N N N N ∂F

Setting ∂n = 0, we get

0 = ε − K B TN [

∂ N+n N+n ∂ n n ( ) ln ( )− ln ( )⁡] ∂n N N ∂n N N

Using product rule of differentiation

 0 = ε − K B TN [(

N+n N

1

1

N+n

1

n

1

1

n

1

N

N

N

N

⁄N N

N

N

. N+n . + ln ( N

1

 0 = ε − K B TN [ + ln (

N+n

1

1

N

N

n

1

N

N

) − + ⁡ ln ( ) . ⁡]

N 1

N+n

N 1

N

N

N

 0 = ε − K B TN [ ln (

) ) − ( . n . + ⁡ ln ( ) . )⁡]

n

) + ⁡ ln ( )⁡] N

1

Taking N, common which will be cancelled out with N and then equation becomes  0 = ε − K B T [ln (

N+n N

n

) + ⁡ ln (N)⁡]

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42 a

Using ln a − ln b = ln b, we get N+n N ⁄ )⁡] n

 0 = ε − K B T [⁡ln (

N+n N

= ε − K B T [⁡ln (

N

 ε = K B T [⁡ln ( 

ε KB T

= [⁡ln (

N

N+n

. n )⁡] = ε − K B T [⁡ln (

n

)⁡]

N+n n

)⁡]

N+n

)⁡]

n

Taking exponent on both sides

eε ⁄ KB T =

N+n N = + 1⁡ n n

 e ε⁄ K B T − 1 =  n=

N

N n

eε⁄KB T −1

This is the same result we obtained in chapter 4 by using techniques of microcanonical ensemble.

MINIMIZING THE GIBBS FREE ENERGY The same technique can be used to find condition for a system which is kept out at a constant pressure, not at constant volume. We show that Gibbs free energy is minimum for a system held at constant pressure and temperature. The definition of pressure is ∂ ln(W) P = KBT ( ) ∂V T Which by integration and rearrangement gives us PV = ln W KBT Which can be further solved and can be written in terms of W as W = ePV⁄KB T For heat bath of energy UR, volume VR, pressure PR, and at temperature T, the number of accessible states of heat bath is

WR = γeUR⁄KBT . W As W = ePV⁄KB T For P = PR , and V = VR Caution: Read at your own risk. This is just an incomplete version, so too many errors are expected. Your suggestions are always appreciated. It is also available in soft form. A PDF can be downloaded from phylib.wordpress.com

43

W = e P R VR ⁄ K B T So WR can now be written as

WR = γeUR⁄KBT . ePRVR⁄KBT WR = γe(UR+PRVR)⁄KBT → (A) Suppose that the system attached to the heat bath has volume VA and energy UA . The total energy UT = UA + UR is fixed as is the total volume VT = VA + VR Substituting the values of UR and VR in equation (A), we get

WR = γe[UT −UA+PR(VT −VA )]⁄KB T The total number of states of the combines system (A plus heat bath) is given by W = WA (UA . VA ) ∗ WR → (B) WA can be expressed as

WA = eKB T ln(WA )⁄KB T Using this value of WA and value of WR from equation, we get

W = eKBT ln(WA )⁄KBT . γe[UT−UA +PR(VT−VA)]⁄KBT  W = ⁡γe[UT−UA+PR(VT−VA)+KBT ln(WA )]⁄KBT

Gibbs Free Energy 𝐺 = 𝐹 + 𝑃𝑉 As 𝐹 = 𝑈 − 𝑇𝑆 𝐺 = 𝑈 − 𝑇𝑆 + 𝑃𝑉

For system A The system is in equilibrium when the entropy is maximum. The idea is to maximize ln W for constant pressure PR , and to do this, the quantity 𝐺 = 𝑈𝐴 − 𝑃𝐴 𝑉𝐴 − 𝑇𝐴 𝑆𝐴 (UA + PR VA − TSA ) must be minimized. The system is in thermal and mechanical equilibrium so that TA = T and PA = PR . In order to maximize W, we must minimize Gibbs free energy G = UA + PA VA − TA SA of system A. To illustrate the method, let us reconsider vacancies in a solid but this time allow the pressure of vacancies to increase the volume of the solid. For simplicity, let us suppose that both atoms and vacancies occupy the same volume. As the number of vacancies increases, the volume of solid expands. Suppose there are n vacancies in the lattice with N atoms or N+n lattice sites. The energy of n vacancies is nε, and the volume occupied is (N + n)V where V is the volume occupied by each atom or vacancy. The Gibbs free energy is: G = F + PV As F = U − TS, so G = U − TS + PV

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44 (N+n)!

As U = nε, and S = K B ln (

N!n!

), and V = (N + n)V, so G can be written as

(N + n)! G = nε − K B T ln ( ) + PV(N + n) N! n! a

Using ln b = ln a − ln b and ln ab = ln a + ln b G = nε − K B T [ln(N + n)! ⁡ − ln N! − ln n!]⁡ + PV(N + n) Using Stirling’s approximation G = nε − K B T[(N + n) ln(N + n) − N − n − N ln N + N − n ln n + n] + PV(N + n)  G = nε − K B T[(N + n) ln(N + n) − N ln N − n ln n] + PV(N + n) Adding and subtracting n ln N G = nε − K B T[(N + n) ln(N + n) − N ln N − n ln N∗ − n ln n + n ln N∗ ] + PV(N + n) a

Using ln a − ln b = ln b G = nε − K B T [(N + n) ln (

N+n n ) − n ln ( )] + PV(N + n) N N

Multiply and divide by N (N + n) N+n n n G = nε − K B TN [ ln ( ) − ln ( )] + PV(N + n) N N N N ∂G

To minimize G by varying ‘n’ and differentiate above equation w.r.t ‘n’ an set ∂N = 0.

(N + n) ∂G ∂ ∂ N+n n n ∂ ) − ln ( )] + = nε − K B TN [ ln ( PV(N + n) ∂N ∂N ∂N N N N N ∂N  0 = ε − K B TN [  0 = ε − K B TN [

∂ (N+n)

ln (

∂N N (N+n) N

 0 = ε − K B TN [(

N+n N

1

N+n

1

n

1

1

n

1

N

N

N

N

⁄N N

N

N

 0 = ε − K B T ln (

N+n

1

N

N

N

) ) − . n . − ⁡ ln ( ) . ⁡] + PV

1

n

1

N

N

) − − ⁡ ln ( ) . ⁡] + PV N

1

n

N

N

N n

N

) − − ⁡ ln ( )⁡] + PV

) − ⁡ ln ( )⁡] + PV

N N+n N

n

ln ( )] + PV

1

N+n

N N+n

∂N N

N+n

N

 0 = ε − K B T [ln (

n

ln ( )] + PV

1

)−

N

. N+n . + ln (

1

 0 = ε − K B T [ln (

∂ n

∂N N N ∂ n n

N+n

 0 = ε − K B TN [ + ln (  0 = ε − K B TN [ln (

)−

N

ln (

N

N 1

N+n

N

. )⁡] + PV n

) ⁡ + PV

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45

 K B T ln (  ln (   

n N

n

n

N+n

N+n

n N

N+n

n

)=

=e

) = ε + PV

ε+⁡PV

KB T ε+⁡PV⁄KB T

+ 1 = eε+⁡PV⁄KBT = eε+⁡PV⁄KBT − 1

 n=

N eε+⁡PV⁄KB T −1

From the definition of Gas F + PV and from the relation dF = −SdT − PdV, it follows that

dG = dF + VdP + PdV  dG = −SdT − PdV + VdP + PdV  dG = VdP − SdT So average volume of solid is

V=(

∂G 1 ) = Nv [1 + ε+⁡PV⁄K T ] B −1 ∂P T,n e

The presence of thermally excited vacancies makes the solid expand.

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