- #1

- 153

- 1

## Main Question or Discussion Point

Let number of energy levels be e(subscribe 0), e(subscribe 1)...........e(subscribe r)

N be the total number of particles and n(subscribe i) be the number of particle in with energy level where i =0,1,2,.......r

Let Ω be the number of ways attaining a given microstate

Ω=N!/[ n(subscribe 0)!*n(subscribe 1)!*..........n(subscribe r)! ]

For the most probable state Ω should be maximum i.e.,

dΩ/d(n subscribe i) =0

lnΩ=ln[N!]-ln [summation of n(subscribe i)! ] where i=0,1,.......r.

Applying stirlings approximation

lnΩ={ N*ln[N]-N }- [summation of n(subscribe i)*ln [n(subscribe i)]-n(subscribe i)

Taking derivative with respective n(subscribe i)

Let n(subscribe i)=n(i)

(1/Ω)*dΩ/dn(i)=-[summation of ln(n(i)) ]

At maximum dΩ/dn(i) =0 => dΩ =- summation of lnn(i) dn(i) =0

=> summation of lnn(i) dn(i)=0 let this be first equation

Total energy is constant i.e.,

Summation of n(i)*e(i) = constant, => summation of e(i)dn(i)=0 ( 2nd equation) because the summation of change in the total energy level of each state is to zero

Total number of particle is constant

Summation of dn(i)=0 be the third equation

By lagranges method of undetermined multiplication

Multiply 2nd equation by b and 3rd equations by a and add the three equations

We get => summation of [ ln(n (i)) +a + b*e(i) ] d( ni ) =0

And the next step is [ln(n(i)) +a + b*e(i) ] =0....., how is it?

We can't divide both sides dn(i) since the equation is in summation.

N be the total number of particles and n(subscribe i) be the number of particle in with energy level where i =0,1,2,.......r

Let Ω be the number of ways attaining a given microstate

Ω=N!/[ n(subscribe 0)!*n(subscribe 1)!*..........n(subscribe r)! ]

For the most probable state Ω should be maximum i.e.,

dΩ/d(n subscribe i) =0

lnΩ=ln[N!]-ln [summation of n(subscribe i)! ] where i=0,1,.......r.

Applying stirlings approximation

lnΩ={ N*ln[N]-N }- [summation of n(subscribe i)*ln [n(subscribe i)]-n(subscribe i)

Taking derivative with respective n(subscribe i)

Let n(subscribe i)=n(i)

(1/Ω)*dΩ/dn(i)=-[summation of ln(n(i)) ]

At maximum dΩ/dn(i) =0 => dΩ =- summation of lnn(i) dn(i) =0

=> summation of lnn(i) dn(i)=0 let this be first equation

Total energy is constant i.e.,

Summation of n(i)*e(i) = constant, => summation of e(i)dn(i)=0 ( 2nd equation) because the summation of change in the total energy level of each state is to zero

Total number of particle is constant

Summation of dn(i)=0 be the third equation

By lagranges method of undetermined multiplication

Multiply 2nd equation by b and 3rd equations by a and add the three equations

We get => summation of [ ln(n (i)) +a + b*e(i) ] d( ni ) =0

And the next step is [ln(n(i)) +a + b*e(i) ] =0....., how is it?

We can't divide both sides dn(i) since the equation is in summation.

Last edited: