Multiphase Thermodynamics and Transport Phenomena - Appunti, Esercizi e Domande

B

A Taylor expansion for a function f(x) where x is in a neighborhood of x is given by:

0

()

() = ( ) + ( − ) +...

0 0

∂Ω 1

Since → and returning to the

= β = Ω ( ) = Ω ( ) − β

∂

exponential form → .

Ω ( ) = Ω ( )(− β )

This relationship is replaced within the probability function:

Ω ( − ) Ω ( )(−β )

= =

∑ Ω ( − ) ∑ Ω ( )(−β )

=1 =1

Since Ω (E ) does not depend on the index of the summation it can be extracted and

B TOT

simplified with the numerator, thus obtaining the canonical distribution: 29

(−β ) .

=

∑ (−β )

=1

Microstates do not all have the same probability of occurring as probability is an

exponential function of energy.

Microstates with lower energy are the most likely to occur.

FUNDAMENTAL RELATION of

PARTITION FUNCTION

The derivatives of the partition function Q with respect to are shown:

⎡ ⎤

⎢ ⎥

∂ ∑ (−β )

⎢ ⎥

∂ ⎣ ⎦ → the derivative of a summation is equal to the summation of the

=

∂β ∂β

∂(−β )

derivatives → → I calculate the derivative with the rule of composite

∑ ∂β

functions → ∑ (− )(− β ) =− ∑ (− β )

⎡ ⎤ ∂ (−β )

2 ∂ ∂ ∂

∂

⎢ ⎥

= ( ) = − ∑ (− β ) =− ∑ =

⎢ ⎥

2 ∂β ∂β ∂β ∂β

∂β ⎣ ⎦

2

= ∑ (− β )

So to summarize:

∂

the first derivative

●​ =− ∑ (− β )

∂β

2 2

∂

the second derivative

●​ = ∑ (− β )

2

∂β

MEAN ENERGY - THERMAL EQUATION of STATE

The average value of the variable y is calculated from the product of one of its possible

values yK and the probability of the occurrence of a microstate with y = yK → the

(−β )

probability of occurrence of a microstate is .

=

∑ (−β )

=1

∑ (−β )

If the property is energy, the average energy is given by =1

=

∑ (−β )

=1

where E depends on the volume and the number of particles, while β depends on the

R

temperature. Since in a canonical ensemble the volume and the number of particles are fixed,

by fixing the temperature the average energy is also fixed. Fixing the temperature still defines

the canonical ensemble, since the total energy E is not constrained, only the average energy

is. 30

∑ (−β )

As just shown, the average energy is given by .

=1

=

∑ (−β )

=1

It can be noted that both the numerator and the denominator can be replaced with relations

∑ (−β ) ∂

−

involving the partition function: .

=1 ∂β

= =

∑ (−β )

=1

By isolating the key relation, we obtain ∂

− ∂

∂β

= − ∂β

which is a thermal equation of state since it relates the energy to the temperature, the

volume, and the number of particles.

ENERGY DISPERSION

The energy dispersion represents the variance of the energy, that is,how much the energy of

a system can fluctuate around its average value. 2

2

The energy dispersion is defined as .

( ∆ ) = ( − )

2 2 2

2 2

Expanding the square of the binomial: .

( − ) = ( − 2 + ) = − 2() +

The average value of an average property is the value itself

2 2 2 2 2

2 2

2 2

− 2() + = − 2 + = − 2 + = −

where: ∂

−

is obtainable from the partition function: .

∂β

●​ =

2

∑ (−β ) ( )

∂

2 2 ∂

2 2

1 ∂ ∂

●​ since

∂β

= = = = ∑ (− β )

2 2

∂β

∂β ∂β

∑ (−β )

2

∂

1

In order to find a simpler expression for it is necessary to start from the expression:

2

∂β

1

∂ ∂ → calculating the derivative →

( )

∂β ∂β 2

2 2 ( )

1 1 ∂ 1 ∂ 1 1

∂ ∂

∂ ∂ ∂ →

( )= + (− ) = −

2 2

2

∂β ∂β ∂β ∂β ∂β

∂β ∂β

2

2 ( )

1 1 1

∂

∂ ∂ ∂ and express the inverse formula for the term we need to

( )= −

2

∂β ∂β ∂β

∂β 2

2 ( )

∂ 1 1

∂

1 ∂ ∂

simplify: .

= ( )+

2

∂β ∂β ∂β

∂β ∂

2 −

( )

2 2 ∂ 1 1 1

∂ ∂ ∂

The expression for is where so by

∂β

= ( )+ = =−

∂β ∂β ∂β ∂β

2

2 ∂

substituting it .

=− +

∂β

The energy dispersion is therefore given by:

2 2 2

2 2 ∂

( ∆ ) = − = (− + ) −

∂β 31

2 ∂

(

∆

) =− ∂β ∂

The energy dispersion, being a squared term, is always positive, therefore must be a

∂β

negative quantity.

This can be stated: 1

●​ Qualitatively: considering the definition of absolute temperature when the

β =

temperature increases, β decreases, while the average energy increases (from the

3 ∂

equation of state ). → The two variations are opposite → hence, ​ is

=

2 ∂β

negative. 1

●​ Mathematically: knowing that ​, its infinitesimal variation is

β =

1 1 1 1 −1 and substituting this into the formula for the

β = ( ) = =

2

2 2 2

∂ ∂

dispersion, we get which is a positive quantity

( ∆ ) =− (− ) =

∂ ∂

MECHANICAL EQUATION of STATE

The infinitesimal quantity of macroscopic work is

and that is the mean on a canonical distribution, of the macroscopic work on every

microstate:

We multiply and divide by - obtaining ∂ (−β )

If we calculate the derivative of it’s equal to

∂

∂ (−β ) ∂ which is the same expression on the numerator of the

= (− β)(− β )

∂ ∂

macroscopic work, that becomes: 32

In the numerator, it is possible to interchange the summation with the derivative, and the

definition of the derivative of Q appears, while in the denominator the definition of Q is

already present. ()

1 ()

In general so we obtain that the infinitesimal quantity of macroscopic

=

()

work is and from the definition we also know that

so we can obtain a mechanical EoS (relation between temperature,

pressure, volume and number of particles):

1 ∂ 1 ∂

→

− =− − =−

.

β ∂ β ∂

In the canonical ensemble, through the definition of Q we can obtain:

∂

− ∂

●​ :

thermal EoS dividing Q with respect to ∂β

= =− ∂β

1 ∂

●​ mechanical EoS deriving Q with respect to volume: − =−

β ∂

ENTROPY as FUNCTION of Q

A quantity can be expressed through its partial derivatives. In the case of a reversible process,

the logarithm of the partition function Q is given by:

∂ ∂ (1)

= + β

∂ ∂β

∂ ∂

We can substitute the EoS just derived and in (1):

= &bet