# Microcanonical ensemble energy relation to kT

So today I derived the Boltzmann distribution and I stumbled across a particular assertion whose correctness I am wondering about:

In the limit of the microcanonical ensemble (where the energy range is infinitesimally small for microstates), is the following true?

$$dE = kT(dN)$$

Where I am denoting E as total system energy, N as total particles/modes, k as Boltzmann constant and T as temperature.

For reference, here is my derivation:

We take the case of a microcanonical ensemble, a closed system with N particles, W microstates, each with equal probability, P, and fixed energy, E. We can represent W by the combinations of particles, N, divided by the combinations of states (i) that contain particles $n_i$:

$$W = \frac{N!}{n_0!n_1!n_2!\space...}$$

Each state energy is defined as $\epsilon_i$, therefore we can assert that $E = \sum_{i=0}^{\infty}n_i\epsilon_i$ and $N = \sum_{i=0}^{\infty}n_i$. Since each microstate has equal probability, the following is also true: $$WP = 1$$ Basically we want to maximize W here and after some unfortunate math involving taking the derivative of a factorial:

$$dlnW = -\sum_{i}n_i\space dn_i = 0$$

At this point we introduce two more terms involving dN and dE, which are both zero as well:

$$dN = \sum_{i}dn_i = 0$$ $$dE = \sum_{i}\epsilon_i \space dn_i = 0$$

This results in:

$$0 = -\sum_{i}(\alpha + \beta \epsilon_i + ln\space n_i)dn_i$$

Which after a bit more math, ultimately gives:

$$n_i = \frac{N}{\sum_{i}e^{-\beta \epsilon_i}}e^{-\beta \epsilon_i}$$

And the probability of being in state i is $p_i = n_i/N$. This is the desired result and is the Boltzmann Distribution.

We also know that when analyzing with the laws of thermodynamics, $\beta = \frac{1}{kT}$, which also implies the following:

$$\beta \space dE = \frac{1}{kT}dE = \frac{1}{kT}\sum_{i}\epsilon_i \space dn_i = 0$$

But I am reminded of $E = kTN$ which implies $dE = kT(dN)$ that can only be true if: $$\epsilon_i = kT$$ and therefore $$\frac{1}{kT}dE = \frac{1}{kT}\sum_{i}kT \space dn_i = \sum_{i}dn_i = dN$$ Which means that the energy per state is kT, which at first made sense to me. But this makes the Boltzmann Distribution fall apart, going to zero, in fact.

My deduction can't be valid, please correct me where I am wrong.

• Where are you getting the equation E = kTN from? This looks somewhat similar to the "equipartition theorem" which says that in equilibrium, E = 1/2*NkT where N is the total number of degrees of freedom in a system. But you're missing the factor of 1/2. The energy per state in equilibrium should be 1/2*kT. Why do you say that this implies the Boltzmann distribution go to zero? I don't see how that follows. Oct 26 '16 at 0:24
• I think you're right.. I took the E = NkT from the equipartition theorem for modes of a standing wave as N. Not sure why I was thinking that. And it goes to zero because of the denominator of my equation for n_i above, using my assumption about epsilon_i. The infinite sum of e^(-1) is infinity, driving n_i to zero. So in my formulation above, is epsilon_i = 1/2 kT ? Oct 26 '16 at 2:13
• Sorry, I take back what I said about the "energy per state" being 1/2*kT. The (average) energy per particle is 1/2*kT but not the energy per state or even the exact energy of each particle. I've written an answer below explaining more. Oct 26 '16 at 20:48

The equipartition theorem says that the total energy in equilibrium will be $E = \frac{1}{2} NkT$ where N is the number of degrees of freedom (number of particles if each has only 1 d.o.f.).

The i index in your equations runs from 0 to $\infty$ because there are an infinite number of possible states in the system. Their energy levels are represented by the $\epsilon_i$'s. Because there are only N particles, only a finite number of these can be occupied. In other words, there are an infinite number of states $i$ for which $n_i = 0$ and only a finite number where $n_i \neq 0$.

If $\epsilon_i$ were equal to $\frac{1}{2}kT$ for all i, then this would mean all infinitely many states were occupied, leading to obvious problems. For one thing, it would mean $\sum n_i = \infty$ instead of $\sum n_i = N$. So obviously, this is not the case.

Because the total energy of a system in equilibrium is $\frac{1}{2} NkT$, that implies that the average energy per particle is $\frac{1}{2} kT$. However, that does not mean that the energy of each particle is exactly $\frac{1}{2}kT$, just that this is the average of all of them.

So it's not true that either $\epsilon_i = \frac{1}{2} kT$ or even that $\epsilon_i n_i = \frac{1}{2} kT$, it's just that $\frac{1}{N}\sum n_i \epsilon_i = \frac{1}{2} kT$. In words, the average energy per degree of freedom is $\frac{1}{2} kT$.

You write $dE = kT dN$ and then jump to the conclusion that $\epsilon_i = kT$. I'm not sure where this logical leap comes from but I have a guess. With differential equations, one has to be very careful to state what is being held fixed and what is changing, otherwise they could have multiple interpretations. $dE = \frac{1}{2} kT dN$ is correct as long as by that you mean "if I hold the temperature of a system fixed, each additional particle I add to it will add $\frac{1}{2} kT$ amount of energy to the system." But keep in mind that if you are holding the temperature fixed but allowing the energy and number of particles to change, all of the $n_i$'s will change along with that when you add a new particle. In other words, adding 1 more particle causes the rest of the particles to rearrange in whatever way will keep the temperature fixed... some occupation numbers of states may increase or decrease. So it's not that the new additional particle you added ends up having an energy of exactly $\frac{1}{2} kT$.

Instead of holding the temperature of the system fixed, you could choose to hold the energy fixed. Fixing the temperature in terms of the physics means allowing the system to be in contact with a large heat bath at temperature T (the environment). If instead you hold the energy fixed, this means isolating the system from the environment so that no heat is allowed to flow in or out. If you do that, then adding a new particle on top of the N you already have would decrease the temperature. Writing that in terms of an equation looks like $kdT = -2EN^{-2}dN$, which comes from taking the derivative of both sides of $kT = 2E/N$.

Some of this ambiguity can be resolved by using partial derivatives and by indicating what variables are being held fixed.

• thank you for the detailed answer! A key element that you have helped me with here is the fact that for a finite number of particles, N, there has to be a finite number of states with non zero energy. I will be reading your answer over in much detail until I more fully grasp this concept. Oct 26 '16 at 21:01