# Derivation of mazwell boltzman statistics.

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.

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Please use the underscore key "_" instead of the word "subscribe". It's standard notation and much easier to read. "e_1" instead of "e(subscribe 1)" and "e^1" instead of "e(superscribe 1)"

stevendaryl
Staff Emeritus
Please use the underscore key "_" instead of the word "subscribe". It's standard notation and much easier to read. "e_1" instead of "e(subscribe 1)" and "e^1" instead of "e(superscribe 1)"

Even better, write [ i t e x ] e_1 [ / i t e x ] or [ i t e x ] e^1 [ / i t e x ] (without the spaces) to make $e_1$ and $e^1$

dlgoff
Gold Member
Even the https://www.physicsforums.com/Nexus/editor/superscript.png [Broken] and https://www.physicsforums.com/Nexus/editor/subscript.png [Broken] from the "Go Advanced" method is better. e.g. e1 or e1

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stevendaryl
Staff Emeritus
Here's what I think you're saying, but it's hard to read:

Let $\Omega = \dfrac{N!}{n_1! n_2! \ldots n_r!}$

Using Sterling's approximation, we can write this as:

$ln(\Omega) = ln(N!) - \sum_i n_i ln(n_i) - n_i$

Now, we want to maximize this subject to the constraints:
• $\sum_i n_i = N$, the total number of particles.
• $\sum_i n_i E_i = E$, the total energy.

So we use Lagrange multipliers, and maximize
$ln(\Omega) - a (\sum_i n_i) - b (\sum_i n_i E_i)$
$= ln(N!) - \sum_i (n_i ln(n_i) - n_i - a n_i - b n_i E_i)$

Now, to maximize, we vary each of the $n_i$ and set the variation to zero. The variation is:

$\sum_i \delta n_i (ln(n_i) - a - b E_i)$

For this to be zero regardless of the values of $\delta n_i$, each term in the summation must be zero. So for every $i$, we must have:

$ln(n_i) - a - b E_i = 0$

Even if ln(ni)-a-be¡ doesn't equal to zero, then there is chance that summation will be zero doesn't it?why don't we consider that case?

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Bill_K
Even if ln(ni)-a-be¡ doesn't equal to zero, then there is chance that summation will be zero doesn't it?why don't we consider that case?

Because, as stevendaryl said, the summation must be zero for arbitrary δni, and this can only be true if each term is zero separately.

$\sum_i \delta n_i (ln(n_i) - a - b E_i)$

For this to be zero regardless of the values of δni, each term in the summation must be zero.

Say for example we have ab+cd +ef =0
We have three cases
1st is (a,c,e)=0
2nd one is (b,c,e)=0
3rd is say for example let ab=3 , cd=5 and ef=-8

What we have done is regardless if value of del(n¡), the second part will zero, yeah i get that but what about other possible cases and why did we took this particular line and preceded?

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stevendaryl
Staff Emeritus
Say for example we have ab+cd +ef =0
We have three cases
1st is (a,c,e)=0
2nd one is (b,c,e)=0
3rd is say for example let ab=3 , cd=5 and ef=-8

What we have done is regardless if value of del(n¡), the second part will zero, yeah i get that but what about other possible cases and why did we took this particular line and preceded?

The equation we want to solve is:

$\sum_i \delta n_i (ln(n_i) - a - b E_i) = 0$

We want this to be true for EVERY possible value of $\delta n_i$. So let's pick a particular value of $\delta n_i$:

$\delta n_{17} = 1$
$\delta n_0 = \delta n_1 = \ldots = \delta n_{16} = \delta n_{18} = \ldots = 0$

So for that value of $\delta n_i$, the sum simplifies to:

$ln(n_{17}) - a - b E_{17} = 0$

The only way to make the sum equal to zero for every possible value of $\delta n_i$ is if every term is zero.

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You made a mistake the total number of particle is constant therefore if dn_17th is 1 there must be another state dn_i=(-1) that is when you remove on particle from a state you have to add it up to a another system so that
Summation of dn_i=0

stevendaryl
Staff Emeritus
You made a mistake the total number of particle is constant therefore if dn_17th is 1 there must be another state dn_i=(-1) that is when you remove on particle from a state you have to add it up to a another system so that
Summation of dn_i=0

You're forgetting what the whole point of lagrange multipliers is: Introducing lagrange multipliers allows you to treat the parameters as independent, for the purposes of minimization.

stevendaryl
Staff Emeritus
You're forgetting what the whole point of lagrange multipliers is: Introducing lagrange multipliers allows you to treat the parameters as independent, for the purposes of minimization.

There's one way to see how this works. To start off with, you have $r$ parameters, $n_1, n_2, \ldots n_r$ and you have two constraints: the total number of particles, and the total energy. So there are $r-2$ independent parameters.

Now, we introduce two new parameters, the two Lagrange multipliers. We don't introduce any new constraints. So now the situation is that we have $r+2$ parameters, and we have $2$ constraints. So there are $r$ independent parameters. We can let those $r$ independent parameters be $n_1, n_2, \ldots n_r$. Then the two lagrange multipliers are determined by those.

So introducing lagrange multipliers allows us to treat the $n_i$ as independent.

stevendaryl
Staff Emeritus
There's one way to see how this works. To start off with, you have $r$ parameters, $n_1, n_2, \ldots n_r$ and you have two constraints: the total number of particles, and the total energy. So there are $r-2$ independent parameters.

Now, we introduce two new parameters, the two Lagrange multipliers. We don't introduce any new constraints. So now the situation is that we have $r+2$ parameters, and we have $2$ constraints. So there are $r$ independent parameters. We can let those $r$ independent parameters be $n_1, n_2, \ldots n_r$. Then the two lagrange multipliers are determined by those.

So introducing lagrange multipliers allows us to treat the $n_i$ as independent.

We can work out the very simplest case: just two energy levels $e_1$ and $e_2$. In this case, we don't need lagrange multipliers, because the constraints uniquely determine $n_1$ and $n_2$:

$n_1 + n_2 = N$
$n_1 e_1 + n_2 e_2 = E$

So

$n_1 = \dfrac{N e_2 - E}{e_2 - e_1}$
$n_2 = \dfrac{E - N e_1}{e_2 - e_1}$

The Lagrange multiplier approach gives us:

$ln(n_1) - a - b e_1 = 0$
$ln(n_2) - a - b e_2 = 0$

So

$n_1 = e^{a+b e_1}$
$n_2 = e^{a + b e_2}$

We still have the same constraints:

$n_1 + n_2 = N$
$n_1 e_1 + n_2 e_2 = E$

but these become constraints on $a$ and $b$:

$e^{a + b e_1} + e^{a + b e_2} = N$
$e_1 e^{a + b e_1} + e_2 e^{a + b e_2} = E$