## Archive for August, 2009

### Monday Math 87

August 31, 2009

For a sequence an, n=1,2,3,…, the series is known as the Dirichlet series generating function (or sometimes simply the Dirichlet series) for the sequence. For example, if we take the trivial sequence 1,1,1,… (an=1 for all n), then the Dirichlet series is the Riemann zeta function
.
For the alternating sequence 1,-1,1,-1,… (an=(-1)n-1), the Dirichlet series is the Dirichlet eta function
.

Now, consider the case where an=f(n), where f(n) is a multiplicative function.
Then we have Dirichlet series generating function:
.
Using the prime factorization of n, , we see:
.
Now, we procede as we did in this post, which is just the above with multiplicative function . The above can be expressed as a product over the primes:
.
When f(n) is a completely multiplicative function, , so then the series within the summation becomes a geometric series:
.
Note that if our function is the constant function f(n)=1 (it is completely multiplicative), then the above gives us the Euler product for the Riemann zeta function

which we found here.

Note that even if the function is just multiplicative, and not completely multiplicative, it may still be possible to find a non-series expression for the series found in the product over primes. One example can be seen this past post.

### Physics Friday 87

August 28, 2009

Continuing from last week with our simple fermion system, we now consider the system enclosed in walls which are impermiable to our fermions, so that the particle number Ñ is now constant. While this is no longer the grand canonical formalism, the fundamental equation is still valid, as it is an attribute of the system independent of its boundary conditions. However, the chemical potential μ, usually called the Fermi level when speaking of a system of indistinguishable fermions, will not be a constant, but will instead vary with temperature so as to keep Ñ constant. Using our work from last time, we have particle number . While this gives Ñ as a one-to-one function of μ (as Ñ(μ) is an increasing function), it is not one that has an analytic inverse, and solving for μ given Ñ and T must generally be done numerically.

We can also consider the low temperature limit kT→0 (β increasing without bound). Recall that the probability of occupation for a given state in our system is , and since we are dealing with fermions, this gives the occupancy (expectation value for the number of particles in the state): . We see that for μ=εn, the occupancy f(μ,T) of our state is exactly 1/2, independent of the temperature. For εn>μ, increases without bound as β does, so that the probability of occupation goes to zero; for εn<μ, as β increases, so that the probability of occupation goes to one. Thus, as the temperature approaches absolute zero, we find the states with energy below the Fermi level fully occupied and those with energy above the Fermi level empty.
This is what we should expect from a system of fermions. As the temperature approaches absolute zero, the particles should go to the lowest energy state allowed; however, being fermions, the Pauli exclusion principle applies, and thus the ground state can hold only a limited number of particles; when it fills, then the particles begin filling the next level. Then it fills, and so on, until we reach the total number of particles; the threshold between filled and empty states is thus our Fermi level. The Fermi level at a temperature of absolute zero is known as the Fermi energy: .
We also see that as the temperature begins increasing away from absolute zero, the transition at the Fermi level becomes less sharp; the occupancy for states of energy just below the Fermi level decreases and that for states just above the Fermi level increases. Figure 1: Occupancy of a state as a function of state energy at various low temperatures.

In particular, the energy range over which this transfer occurs is on the order of 4kT: expanding in a Taylor series about the point εn=μ, we get .
We should also note that the equation for occupancy as a function of energy possesses inversion symmetry about the point εn=μ, pn,m=1/2.

Now, suppose we have the special case ε1<ε2=ε3, so that ε1 is the ground state, and the excited state is degenerate. Let Ñ=2, so that at absolute zero, the ground state is filled, the excited states are empty, and the Fermi level lies somewhere between
ε1 and ε2. Now, for very low T, we use the approximation for occupancy:
.
In this case, our equation for Ñ becomes:
. Plugging in Ñ=2, we can solve the above for μ, to get

(the ellipsis indicates terms of higher order in kT, which are neglected in our approximation). This illustrates an important general principle: as the temperature increases, the Fermi level is ‘repelled’ by energy regions with higher densities of states.

### Monday Math 86

August 24, 2009

A notable multiplicative function is the Möbius function μ(n) defined as follows:
μ(n)=1 when n is a square-free integer with an even number of distinct prime factors
μ(n)=-1 when n is a square-free integer with an odd number of distinct prime factors
μ(n)=0 when n is not square-free.
Note that μ(1)=1, as 1 has an even number of prime factors; namely, zero.
In terms of the distinct prime factor counting function ω(n) (see here), we can write the Möbius function as:

One can easily confirm that μ(n) is a multiplicative function (but not a completely multiplicative function); if either m or n is not square-free (or both), then mn is not square-free [and μ(mn)=μ(m)μ(n)=0]. If m and n are both square-free and coprime, they have no common prime factors, so mn is square-free and ω(mn)=ω(m)+ω(n) [so ].
In terms of the powers of primes, we see:
.

### Physics Friday 86

August 21, 2009

Let us consider a model quantum system with three permitted spacial ‘orbital’ states; a particle in state n=1,2, or 3 has energies ε1, ε2, and ε3, respectively. Now suppose are particles are spin-½ fermions. Then the spin projection has two possible states, ms=½ and ms=-½; these are designated “up” and “down”. Thus, there are six possible states (n,ms), with n=1,2,3, and ms=±½. Since the particles are fermions, no more than one particle may be found in each of these states at a given time.
Now, let us consider this system in contact with a thermal reservoir with a fixed temperature T, and a reservoir of our fermions with Gibbs potential μ; this is to say, in the grand canonical ensemble. Compute the grand canonical partition sum. To compute it, we note that the grand canonical partition sum factors across non-interacting states, like the canonical partition sum does. Thus, $\mathcal{Z}=z_{1,-\frac{1}{2}}z_{1,\frac{1}{2}}z_{2,-\frac{1}{2}}z_{2,\frac{1}{2}}z_{3,-\frac{1}{2}}z_{3,\frac{1}{2}}$. Now, each orbital state partition sum has only two terms: one for the empty state, with E=0 and N=0; and one for the occupied state, with E=εn and N=1. Thus, (assuming no magnetic field, so energy is independent of the spin direction): $z_{n,m_s}=e^{-\beta(0-\mu\cdot0)}+e^{-\beta(\epsilon_n-\mu\cdot1)}=1+e^{-\beta(\epsilon_n-\mu)}$.
Now, we could also pair the states with the same spatial orbit n (but differing spin alignments): $z_{n}=z_{n,-\frac{1}{2}}z_{n,\frac{1}{2}}=\left(1+e^{-\beta(\epsilon_n-\mu)}\right)^2=1+2e^{-\beta(\epsilon_n-\mu)}+e^{-2\beta(\epsilon_n-\mu)}$.
We can interpret this expanded product in terms of the four possible states of n: the empty state, two singly occupied states (of opposite spin), and one doubly-occupied state.
The probability that the state (n,ms) is empty is thus $\frac{1}{z_{n,m_s}}$, and the probability that it is occupied is $p_{n,m}=\frac{e^{-\beta(\epsilon_n-\mu)}}{z_{n,m_s}}=\frac{1}{e^{\beta(\epsilon_n-\mu)}+1}$.

Now, our grand canonical partition sum is $\mathcal{Z}=\left[1+e^{-\beta(\epsilon_1-\mu)}\right]^2\left[1+e^{-\beta(\epsilon_2-\mu)}\right]^2\left[1+e^{-\beta(\epsilon_3-\mu)}\right]^2$. Thus, the grand potential is $\Psi=-\frac{\ln\mathcal{Z}}{\beta}=-\frac{2}{\beta}\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]$.
Now, the mean number of particles ⟨N⟩ can be found from Ψ: $\begin{array}{rcl}\langle{N}\rangle&=&-\frac{\partial\Psi}{\partial\mu}\\&=&\frac{2}{\beta}\frac{\partial}{\partial\mu}\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]\\&=&\frac{2}{\beta}\left[\frac{\partial}{\partial\mu}\ln\left(1+e^{-\beta(\epsilon_1-\mu)}\right)+\frac{\partial}{\partial\mu}\ln\left(1+e^{-\beta(\epsilon_2-\mu)}\right)+\frac{\partial}{\partial\mu}\ln\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]\\&=&\frac{2}{\beta}\left[\frac{\beta{e^{-\beta(\epsilon_1-\mu)}}}{1+e^{-\beta(\epsilon_1-\mu)}}+\frac{\beta{e^{-\beta(\epsilon_2-\mu)}}}{1+e^{-\beta(\epsilon_2-\mu)}}+\frac{\beta{e^{-\beta(\epsilon_3-\mu)}}}{1+e^{-\beta(\epsilon_3-\mu)}}\right]\\&=&\frac{2}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{2}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{2}{e^{\beta(\epsilon_3-\mu)}+1}\end{array}$
Note that this is the same result we get were we to sum the probability of occupation over all six states, as we should expect.
Similarly, we could find the energy via $U=-\frac{\partial}{\partial\beta}(\ln\mathcal{Z})+\mu\langle{N}\rangle$, or we can use the occupation probabilities to get: $U=\sum_{n,m}E_{n,m}p_{n,m}=\frac{2\epsilon_1}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{2\epsilon_2}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{2\epsilon_3}{e^{\beta(\epsilon_3-\mu)}+1}$.
We can also find entropy via $S=k\left(\ln\mathcal{Z}+\beta{U}-\beta\mu\langle{N}\rangle\right)$ (see here), getting: $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$.

### A Question

August 18, 2009

Does anyone know what the current international law and policy says concerning the legal status of, and conflict with, extraterrestrial or other non-human intelligences? Would the Geneva Conventions apply to alien invaders?

Addendum: Seeing the comments, I feel I must clarify. No, I don’t think that UFO sightings are alien visitors, nor do I believe that any contact has been made. I am referring to where international law and treaty currently stands with regards to a purely hypothetical future situation. More precisely, something like Robert J. Freitas Jr.’s analysis, only with regards to international rather than United States law.

### Monday Math 85

August 17, 2009

In number theory, an arithmetic function f(n) is called a multiplicative function if it possesses the property that f(ab)=f(a)f(b) for any positive integers a and b which are coprime (gcd(a,b)=1). Note that since gcd(n,1)=1 for any positive integer n, we see that for any multiplicative function, f(1)=1.
If the property f(ab)=f(a)f(b) holds for all positive integers a and b, coprime or not, then the function is said to be completely multiplicative.
One example of a multiplicative function is Euler’s totient function φ(n) (some others may be found here).
Now, consider the prime factorization of n: $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$. Plugging this into a multiplicative function, we see: $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$.
Thus, a multiplicative function is determined completely by its values for the powers of the prime numbers. This simplifies the computation of multiplicative functions. For example, consider φ(n). For a power of a prime pk, k>0, the only positive integers less than or equal to this number which aren’t coprime with it are multiples of p: p, 2p, 3p, … , (pk-1)p=pk; this is pk-1 numbers, so the totient is $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$. This allows us to compute the totient of any number from it’s prime factorization, e.g.:
φ(720)=φ(24325)
=φ(24)φ(32)φ(5)
=24-1(2-1)·32-1(3-1)·51-1(5-1)
=8·6·4=192

Let f(n) and g(n) be multiplicative functions, and let h(n)= f(n)g(n) be their product. Then for any coprime positive integers a and b, $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$,
so the product of two multiplicative functions is multiplicative.

### Physics Friday 85

August 14, 2009

One of the simplest problems solved by applying the time-independent Schrödinger equation is the one-dimensional “particle in a box”, where the potential V(x) is zero for 0<x<L, and infinite everywhere else. Then we have ψ(x)=0 for x≤0 and xL, and for 0<x<L, Schrödinger equation becomes $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$
with boundary conditions ψ(0)=ψ(L)=0. The general solution to the differential equation, which can be rewritten as $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$, is $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$, where $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$, or solving for energy, $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$. The boundary condition ψ(0)=0 tells us that B=0, and the condition ψ(L)=0 thus tells us that $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$, and thus $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$, n a positive integer. Thus, in terms of that n, $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$, and $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$, and the energy is quantized. Normalizing via $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$, we get normalized wavefunction $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$.
Now, let us consider the momentum. The one-dimensional quantum momentum operator is $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$, and we see $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$; our energy eigenstates are obviously not momentum eigenstates. However, note that the eigenfunctions of $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$ are of the form $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$ (with eigenvalues $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$). Noting that $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$, we see that our energy eigenstate $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$ is the linear combination of a state of momentum $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$ and a state of momentum $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$, with equal probabilities for these two possible measurements. Compare this to the classical particle in a one dimensional box, where it has kinetic energy $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$, half the time travelling with momentum of magnitude p in one direction, the other half moving in the opposite direction with momentum of magnitude p.

Now, let us consider a three-dimensional cubic box:
V(x, y, z)=0 when 0<x<L, 0<y<L, and 0<z<L, and is infinite at all other points.
Then we have for the interior of our cube: $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$,
with boundary conditions $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$.
using the definition of nabla for Cartesian coordinates: $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$,
and using separation of variables via $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$,
we get $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$ $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$ $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$
with $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$. These are just copies of the one-dimensional problem, so we obtain normalized 3d wavefunction $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$, where nx, ny, and nz are positive integers. The energy is $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$. A given state is defined entirely by the triplet of positive integers nx, ny, nz. Note that except for the ground state nx=ny=nz=1 ( $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$), all of the energy levels are degenerate; all states for which $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$ have the same value will have the same energy.

[Note also that the proper linear combination of the plane wave with momentum $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$ and the seven other plane waves formed by reflections in x, y, z, and combinations thereof, will produce our energy eigenfunction.
Again compare to the possible momentum values of a classical particle in a box making perfectly elastic collisions with the walls.]

### Goo

August 11, 2009

I just got in from watching the G.I. Joe movie. Is it wrong of me that my first thought afterwards was wanting to make some red goo of my own?

### Monday Math 84

August 10, 2009

Last week, I showed that for a fraction that converts to a pure repeating decimal, one with a denominator coprime with 10, the number of digits in the repeat is equal to the smallest whole number n such that 10n-1 is divisible by the fraction denominator d. In terms of modular arithmetic, this means that 10n≡1 (mod d). This is called the multiplicative order of 10 modulo d, and denoted as ordd(10) or Od(10). Let us make a chart of the multiplicative orders of 10 modulo the first few whole numbers coprime with 10:

d ordd(10)
3 1
7 6
9 1
11 2
13 6
17 16
19 18
21 6

You might note that in several cases, all prime, ordd(10) is d-1; however, this is obviously not a general rule. For the other primes, such as 3, though, we have ordd(10) dividing d-1; this does hold for d a prime. Note however, that while ord9(10)=1 divides 9-1=8, ord21(10)=6 does not divide 21-1=20. So what is our general rule for all d coprime with 10?
Consider Euler’s totient function φ(n), and note that for p prime, φ(p)=p-1. So, redoing the chart to add φ(d):

d ordd(10) φ(d)
3 1 2
7 6 6
9 1 6
11 2 10
13 6 12
17 16 16
19 18 18
21 6 12

From this, we see that in every case here, ordd(10) divides
φ(d). This is the general rule. However, proving that this is true requires group theory. This rule does limit the valid possibilities for ordd(10). For example, φ(23)=22, so ord23(10) must be 1, 2, 11, or 22. Now, we can rule out 1 and 2, as 9 and 99 are not divisible by 23. Thus ord23(10) must be 11 or 22 (ord23(10)=22).

.

### Physics Friday 84

August 7, 2009

The canonical ensemble is a powerful tool for statistical mechanics. Let us now illustrate another powerful formalism, an extension known as the “grand canonical” formalism.

In the canonical formalism, our system is in contact with a thermal reservoir with which it may exchange energy; the reservir is large enough that it’s temperature be treated as a constant (thus giving it an entropy which varies linearly with energy). In the grand canonical formalism, we add the ability to exchange particles, as well as energy, with the particle reservoir large enough that the chemical potential $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$ is treated as a constant.

If we treat the combination of our system and the reservoir as a closed system with every microstate equally probable, we have for a state where our system has energy Ei and particle number Ni, the fractional occupation is the number of microstates of the reservoir for which its energy is EtEi and it’s particle number is NtNi divided by the total number of microstates for the system and reservoir combination with total energy Et and a total of Nt particles: $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$
However, using the entropy definition $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$, we can rewrite this in terms of the reservoir and total entropies (as functions of energy and particle number): $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$.
As with our canonical formalism, we can expand the entropy as a Taylor series in energy about EtU and in particle number about the average particle number N $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$, (using $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$).
And via additivity of the entropy, $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$.
Thus $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$,
where $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$ is the thermodynamic beta and $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$ is the “grand potential”. As with the canonical ensemble, we normalize this: $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$, where $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$ is the “grand canonical partition function”, and the sum is over all states, with state s having energy Es and Ns particles.

We see that just as $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$, we can solve for the grand potential: $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$.
Let us define $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$. Now, the expectation value for the number of particles is $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$,
or, using the chain rule, $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$.
Similarly, one can show that $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$,
the variance in the particle number.

Now, consider the logarithmic partial derivative of Ƶ with respect to β: $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$,
and we have our internal energy. We can also show by a second differentiation that $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$.

From our equation for the grand potential, we see $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$.

Next, consider the fundamental thermodynamic relation $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$. Soving for the mechanical work term, $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$. Now, if we continue to hold β (and thus T) and μ constant, the right hand side of the previous relation is just –: $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$.
From this, one can obtain the equation of state $S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$.