# Derivation for the most probable macrostate for distinguishable particles using lagrange's method of undetermined multipliers

We have an expression for $\Omega$ (occupation of each macrostate) in terms of $n_i$ (occupation numbers) . We want to find the $n_i$ which maximises $\Omega$.

We now that

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

Using stirlings approximation we can write this as

$$=(Nln[N]-N)-\sum_i(n_iln[n_i]-n_i)$$

We are subject to two constraints 1. Total number of particles remain constant

$$N=constant: \sum_idn_i=0$$

1. Total internal energy remains constant

$$U=constant: \sum_i\epsilon_i dn_i=0$$

We now use lagrange's method of undetermined multipliers.

$$d(ln[\Omega]=-d(\sum_i(n_iln[n_i]-n_i))=0$$

$$=-\sum_i dn_i(ln[n_i]+\frac{n_i}{n_i}-1)=0$$

This implies that

$$\sum_iln[n_i]dn_i=0$$

So, we have three things which must be zero simultaneously. We can multiple zero by any factor and still have zero so,

$$\sum_iln[n_i]dn_i=0, \alpha\sum_i\epsilon_i dn_i=0, \beta\sum_idn_i=0$$

We can combine these expressions to write our overall maximisation problem as:

$$\sum_i(ln[n_i]+\alpha+\beta\epsilon_i)dn_i=0 \forall \alpha,\beta$$

Now the part I raise issue with is: apparently the only way for this to be solved is that every term in the sum has to be zero. So

$$ln[n_i]+\alpha+\beta\epsilon_i=0 \forall i,$$ but specific $\alpha$ and $\beta$.

However, surely, we are doing a sum here, there is no requirement that this be zero at all points. For instance, the sum may be zero, but part of the sum might be positive, but this is canceled by a negative part.

Anyway, just to finish the derivation, this rearranges to

$$ln[n_i]=-\alpha-\beta\epsilon_i$$

Solving for $n_i$

$$n_i=e^{-\alpha-\beta\epsilon_i}=Ae^{-\beta\epsilon_i}$$

Giving

$$N=\sum_in_i=A\sum_i^{\beta\epsilon_i}$$

This is because $dn_i$ can be arbitrary. You get an infinite number of equations by choosing different $dn_i$. For these equations to be statisfied simultaneously, you need the coefficient to be zero. \begin{equation} \ln n_i + \alpha + \beta \epsilon_i = 0 \end{equation} Note because you have included Lagrange multipliers, $dn_i$ can be treated as independent.
• Why can $dn_i$ be arbitrary? For our differential calculus to work to find the maximum we must have the smallest size possible - as occupation number must go in integers, we must go in integer steps. If it isn't this size, wouldn't we have some sort of error in the maximum? We might skip over where n_i =0? – Tomi Apr 13 '16 at 18:38
• Usually the occupation number is huge, so $\delta n / n$ is infinitesimally small. Therefore you can think of it as approaching to continuum limit. – Kevin Ye Apr 13 '16 at 19:19