Difference between revisions of "Boltzmann distribution"

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$$N = \sum n_i.$$
 
$$N = \sum n_i.$$
  
Then, the N particles will be arranged into microstates given by
+
The N particles are arranged into microstates given by
  
 
$$W = (g_1^{n_1} g_2^{n_2} g_3^{n_3} ...)\frac{N!}{n_1!n_2!n_3! ...}.$$
 
$$W = (g_1^{n_1} g_2^{n_2} g_3^{n_3} ...)\frac{N!}{n_1!n_2!n_3! ...}.$$
  
Solving for lnW, we then get
+
Here, W is the probability of a given distribution. To find the distribution of particles in our system, we will need to find the most probable microstate, or the distribution that maximizes W. To accomplish this, we will equivalently find a maximum in lnW.
 +
 
 +
Solving for lnW, we get
  
 
$$lnW = n_1lng_1 + n_2lng_2 + ... + lnN! - n_1lnn_1 - n_2lnn_2 - ... + n_1 + n_2 + ...$$
 
$$lnW = n_1lng_1 + n_2lng_2 + ... + lnN! - n_1lnn_1 - n_2lnn_2 - ... + n_1 + n_2 + ...$$
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where we implemented Stirling's approximation for large N, $lnN! \approx NlnN - N$, to simplify our expression. Using Lagrange's method of undetermined multipliers, we seek a maximum in
 
where we implemented Stirling's approximation for large N, $lnN! \approx NlnN - N$, to simplify our expression. Using Lagrange's method of undetermined multipliers, we seek a maximum in
  
$$ lnW - \alpha \sum n_i - \beta\sum\epsilon_i\n_i $$
+
$$ lnW - \alpha \sum n_i - \beta\sum\epsilon_in_i, $$
 +
 
 +
where $\alpha$ and $\beta$ are constant values. A maximum in this expression is equivalent to a maximum in lnW or in W because the second and third terms consist of only constant values, where we have defined our total energy and our total number of particles to be constant (isolated ensemble). Thus, we solve for the maximum value of this expression with respect to $n_i$.
 +
 
 +
$$ \frac{d}{d n_i} (lnW - \alpha \sum n_i - \beta\sum\epsilon_in_i) = \frac{d lnW}{dn_i} - \alpha - \beta\epsilon_i = 0 $$
 +
 
 +
Recalling our equation for lnW from earlier,
 +
 
 +
$$ \frac{dlnW}{dn_i} = \frac{d}{dn_i}(n_i + n_i lng_i - n_i lnn_i) $$
 +
$$ \frac{dlnW}{dn_i} = 1 + lng_i - lnn_i - n_i\frac{1}{n_i} $$
 +
$$ \frac{dlnW}{dn_i} = ln\Big(\frac{g_i}{n_i}\Big) $$
 +
 
 +
Plugging this in to our maximum probability expression, we then get
 +
 
 +
$$ ln\Big(\frac{g_i}{n_i}\Big) - \alpha - \beta\epsilon_i = 0 $$
 +
$$ \frac{g_i}{n_i}e^{-\alpha}e^{-\beta\epsilon_i} = 1 $$
 +
$$ n_i = g_i e^{-\alpha}e^{-\beta\epsilon_i} $$
 +
 
 +
Finally, we get
 +
 
 +
$$ \frac{n_i}{n_j} = \frac{g_i}{g_j}e^{-\beta(\epsilon_i - \epsilon_j)}, $$
  
 +
which is a generalized form of the Boltzmann equation. Here, $\beta = \frac{1}{kT}$. The $\alpha$ term in our derivation, though it dropped out of our Boltzmann distribution, is given by $\alpha = \frac{\mu}{kT}$, where $\mu$ is the chemical potential.
  
 
\section{Degeneracies of the hydrogen atom}
 
\section{Degeneracies of the hydrogen atom}

Revision as of 16:39, 23 November 2016

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Boltzmann Distribution

1 Definition

The Boltzmann distribution gives the relative fraction of atoms in two states in thermal equilibrium at a certain temperature, taking into account the degeneracies of these states and the energy difference between states. This distribution applies to large ensembles of atoms such that statistical arguments are valid. The formula for the Boltzmann distribution is given by the following:

where are the number (or number density) of atoms in energy states 1 and 2, respectively, are the respective degeneracies of those energy states (i.e., how many distinct configurations of the atom have that same energy state), is the energy difference between the two states, is Boltzmann’s constant, and is the temperature describing the distribution of states in the system

2 Derivation

Consider an isolated statistical ensemble of N particles with energies given by

and total number of particles

The N particles are arranged into microstates given by

Here, W is the probability of a given distribution. To find the distribution of particles in our system, we will need to find the most probable microstate, or the distribution that maximizes W. To accomplish this, we will equivalently find a maximum in lnW.

Solving for lnW, we get

where we implemented Stirling’s approximation for large N, , to simplify our expression. Using Lagrange’s method of undetermined multipliers, we seek a maximum in

where and are constant values. A maximum in this expression is equivalent to a maximum in lnW or in W because the second and third terms consist of only constant values, where we have defined our total energy and our total number of particles to be constant (isolated ensemble). Thus, we solve for the maximum value of this expression with respect to .

Recalling our equation for lnW from earlier,

Plugging this in to our maximum probability expression, we then get

Finally, we get

which is a generalized form of the Boltzmann equation. Here, . The term in our derivation, though it dropped out of our Boltzmann distribution, is given by , where is the chemical potential.

3 Degeneracies of the hydrogen atom

The energy degeneracies corresponding to each E_n for the hydrogen atom are given by:

Taking the two electron spin states of hydrogen into account, our degeneracies then ultimately become