Suppose we have a cube, of edge length L, with mass M symmetrically distributed so that the geometric center of the cube is the center of mass. Suppose we have a fluid with density ρ, such that (the average density of the cube is less than that of the fluid), so that the cube will float in the fluid. If the cube is floating such that the upper face is entirely outside the fluid, and is tipped toward one edge by an angle θ from the horizontal, then what is the torque on the cube about its center?


Let us define the ratio of the densities of the cube and the fluid as xi;, so that . Then 0<xi;<1. Then via Archimedes’ Principle, the weight of fluid displaced equals the weight of the cube, and thus the volume V_d of fluid displaced can be found:
,
or, using M=ξρL³, V_d=ξL³, and xi; is also the fraction of the cube submerged.



Now, if it is tilted at an angle θ, the submerged portion is a trapezoidal prism; if the trapezoid has bases b₁ and b₂, then it has area , and thus the prism has volume , which, equating to the above, tells us
.
Our geometry tells us that ; combining, we see

,
and to keep the upper face “dry,” we have
.

Now, the center of buoyancy is located at the centroid of the trapezoidal prism. By symmetry, this center will be in the plane halfway between the trapezoidal faces of the prism. Using xy coordinates on this plane with the origin at the center of the cube and with the axes along those of the cube, we find the center of buoyancy to be the centroid of the cross-section trapezoid.



Now, using these coordinates, we see that the trapezoid is the region , ; the area of the trapezoid is A=ξL², so performing the double integrals for the centroid, we have

and
.



Now, to convert these to coordinates with horizontal and vertical axes, we rotate coordinates by θ. If x’ is our horizontal axis and y’ our vertical axis, then
, so our center of buoyancy in these coordinates (with origin still at the center of our cube) is


And so the torque is, with positive torque being in the direction to decrease θ
.

A useful polynomial factoring rule is that for positive integer n, the linear term x-y is a factor of . More specifically,
;
expanding the right hand side of the above gives you the left after cancellation of various terms.
For example, for the first few values of n
;
;
;
;
;
and so on.

It is this which gives us the formula for a finite geometric series; letting x=r, y=1, and our power be n+1, then
.

Further, note that if x and y are integers, then is also an integer, and thus the integer x-y is a factor of the integer .

Now, let n be odd. If we let x=a and y=-b, then
becomes

For example,
;
;
;
and so on. And we see that if x and y are integers, then is also an integer, and thus the integer x+y is a factor of the integer when n is odd.

Suppose, as in this post, we have a fluid of density ρ, with some arbitrary three-dimensional object immersed in the fluid, occupying a volume V. This object has surface ∂V. We saw that for any point on the surface, the pressure is , and the force on an area element is:
. Thus, the torque on that element is
,
where r is the coordinate vector to the element. Recalling that for vectors v and w and scalar a,
, so since P(x,y,z) is a scalar, we see

and so the total torque (about the origin) is
;
and one form of the divergence theorem tells us that for vector field A,
.
Thus

Now, the product rule for the curl of the product of a scalar field ψ and a vector field a is
.
Now, applying that to , and noting that (as it is a radial vector field), and , we see
.
Now, since is a constant vector, it can be “factored out” of the integral:
.
We recall that the total buoyant force on the object is ; plugging this into the above, we see
.

Now, recall that the center of mass of a region with density function ρ(x,y,z) is . If the density is a constant, it factors out of the integrals, and one gets . Thus, if we consider the volume as if it were filled with the fluid; that is to say, the volume of fluid displaced, its center of mass would be , and then we see that
,
which is equivalent to the torque if the buoyant force acted entirely on the point r_b; this point is called the center of buoyancy. Just as the force of gravity on an extended object can be treated as if it acts entirely on the center of mass, the buoyant force can be treated as if it acts entirely on the center of buoyancy.

A notable arithmetic function which is neither additive nor multiplicative is the von Mangoldt function, denoted Λ(n). It is defined as

It has a few notable properties. First, consider . Considering the prime factorization , we see that the only divisors d of n for which the von Mangoldt function is non-zero are , , with . There are k_i terms for each p_i, so
,
or in terms of the Dirichlet convolution, (Λ*1)(n)=ln(n). Note that since Λ(1)=0, the von Mangoldt function has no Dirichlet inverse.

Now, let us consider a completely multiplicative function f(n), with Dirichlet inverse f^-1(n), and Dirichlet series generating function . Now, let us consider the derivative of the Dirichlet series generating function. Taking the derivative with respect to s,
.

Next, let us consider the Dirichlet convolution of f^-1(n) and f(n)ln(n). This is
.
For n=1, we note that ln(1)=0, so that the above is also zero in that case. Now, for n>1 with prime factorization , we have , with . I showed here that the Dirichlet inverse f^-1(n) of a completely multiplicative function f(n) is the multiplicative function with
.
Thus, f^-1(d)=0 whenever d is not squarefree, and so our terms are nonzero only when i₁, i₂, …, i_r are each either 0 or 1; this reduces our sum to 2^r terms. When all of the is are zero, so that d=1, the term in the sum is .
Now, suppose that only one of the is is zero, so that d is equal to one of the prime factors p_j of n. Then
.
Thus, for those cases where n has only one distinct prime factor, , then
.
Next, for d a product of two distinct prime factors p_j and p_k,
.
So for n with , we see
.
And when d is the product of l distinct primes (2≤l≤r), , we see
.
Summing these, we see that an can be factored from each term. Cancelling the logarithms, one finds that whenever , so that the Dirichlet convolution is a function that is nonzero only for , p prime and k a positive integer; in that case, . Thus
.

Now, using the relationship between Dirichlet convolution and Dirichlet series, we have that the Dirichlet series generating function for is thus the product of those for f^-1(n) and f(n)ln(n); the former is , and we saw already that the latter is . Thus we see that the logarithmic derivative of F(s) has series
.
Letting the completely multiplicative function f(n) be 1(n), then we see

(for ), displaying the connection between the von Mangoldt function and the Riemann zeta function which is used in the earliest proofs of the prime number theorem.

Now, since Λ(1)=0, we see . Performing termwise integration with respect to s on the latter sum, one can then find
.

Consider a projectile launched from the ground at an initial velocity v₀ at an angle θ above the horizontal, with negligible air resistance. The projectile will thus follow a parabolic trajectory. What is the angle θ such that the total distance the projectile travels in the air is maximized? Note, this is the length of the parabolic trajectory, not the distance along the ground to the impact point, and so differs from the h=0 case of Physics Friday 24.


The only acceleration is due to gravity; so our basic kinematic equations tell us that and .
Solving for t as a function of x and substituting, we get
, and thus
.
The trajectory ranges between the two x values where y=0, which are x=0 and .

Thus, we find the length of the parabola via the arc length formula

Making the substitution , we get
.
(the last step is due to symmetry, as the integrand is an even function; see here.)
Now, by consulting a table of integrals (or via trigonometric substitution), we find that
,
and thus our length becomes
.

To determine the maximum, we find and set it equal to zero.
Now, taking the derivative, we get
.

Since we want 0<θ<π/2, the above will be zero when

.
(Note that this is an angle independent of our initial velocity.) However, the above equation cannot be solved analytically; a numerical solution gives θ≈0.98551≈56.466°.

Consider two arithmetic functions f(n) and g(n), with Dirichlet series generating functions F(s) and G(s), respectively. What, then, can we say about the product F(s)G(s)?

Using different summation indicies a and b, we have
and . The product, then, is
.

Now, suppose we reorder the sum so as to group terms with the same value of ab. We let n=ab; then we have , and there is one term with given n for each a which divides that n, so within each set of terms with the same n, we sum over a|n, and we get:,
and so we see the product F(s)G(s) is the Dirichlet series generating function of the Dirichlet convolution of f(n) and g(n) (compare to the relationship between the convolution and the Fourier transform).

Now, this implies that the Dirichlet series generating function of the unit function ε(n) is 1, which should be obvious, as
.
And since , we see that if f(n) has Dirichlet series generating function F(s), then its Dirichlet inverse f^-1(n) has Dirichlet series generating function 1/F(s). For example, consider the Dirichlet inverses 1(n) and μ(n), which have Dirichlet series generating functions and , respectively. You may also recall that here I showed that the Dirichlet series generating function for |μ| is . I also demonstrated here that |μ| and the Liouville function \lambda(n) are Dirichlet inverses, so we can see that \lambda(n) has Dirichlet series generating function ; we can confirm this via the Euler product for Dirichlet series generating functions of completely multiplicative functions.

One should also note that if a function f(n) has Dirichlet series generating function F(s), then the function has Dirichlet series generating function F(s-a); since
,
then
.
For example, the Dirichlet series generating function of is thus .

Combining this with our above relationship between Dirichlet convolution and Dirichlet series generating functions, we can develop a number of proofs. For example, the proof that the Dirichlet series generating function of the divisor function σ_x(n) is becomes much more simple than the proof seen here:
Since , we see ; since the Dirichlet series generating function of 1(n) is , and the Dirichlet series generating function of is , the Dirichlet series generating function of σ_x(n) is their product, .
Similarly, we can use the Möbius inversion formula relationship that , found here, to confirm that φ(n) has Dirichlet series generating function , as we found via Euler product here.

A classic problem: A skier starts at rest at the top of a large, hemispherical hill with radius (and thus height), R. If friction is negligible, at what height h above the base of the hill will she become airborne? What is her velocity, both magnitude and angle below the horizontal, at this point?


At any point before the skier becomes airborne, she is acted on by two forces; gravity, and the normal force exerted by the hill. If she is at a point on the hill which is at an angle θ from the top of the hill (as measured at the center of the hill), and moving at speed v, then the normal force N will be at an angle θ from vertical. As she is undergoing non-uniform circular motion, the net force will have both radial and tangential components; however, the radial component must still be centripetal force inward, where m is her mass.



Now, the (inward) radial component of the force is ; equating, we see
.
Now, the skier becomes airborne when the normal force the hill exerts on her becomes zero. The above equation tells us that this occurs when the angle θ and speed v are such that . But, we note from trigonometry that , so that we have
.

Now, we could use the tangential component of the force to find the tangential acceleration as a function of angle, and thus position; this gives a second-order differential equation. However, there is a much easier method to find the velocity as a function of position. Since there is no friction, we can use conservation of energy, with the only potential energy being gravitational. At the top of the hill, the skier has only potential energy . At height h, the potential energy is the smaller , and the kinetic energy is just . Conservation of energy tells us, then, that
; we can cancel out the skier’s mass, which appears in all terms, and then solve for v² to get
.
Plugging this into our relation for the point where the skier goes airborne, we get
.

Thus, we have the height where the skier becomes airborne, which is independent of the skier’s mass (and the acceleration of gravity g). This corresponds to an angle θ of
. As her velocity is still tangential at the point where she becomes airborne, this is also the angle below horizontal of her velocity at that point. That velocity has magnitude
.

[Update 7:15 - Fixed an error in description of tangential acceleration method].

Last Monday, I discussed the Dirichlet inverse. I showed that the Dirichlet inverse can be found by the recursive formula:
,
and for n>0,
.
Further, since the inverse of a multiplicative function is also multiplicative, for a multiplicative function f(n) this formula becomes
, and for p prime and k>0,
.

Now, what happens when f(n) is completely multiplicative? Then . We have , and via k=1, , we see that
,
,

and so we see that if f(n) is completely multiplicative, then f^-1(n) is the multiplicative function with
.
For f(n)=1(n), this gives
, which gives , as we’ve found before.
Next, for f(n)=id_x(n), we see
, which, with a little examination, we see means that , which reduces to the above for x=0. [Just as the inverse pair of 1(n) and μ(n) gives us the Möbius inversion formula, the above pair of id_x(n) and id_x(n)μ(n) gives a generalized Möbius inversion.]

Lastly, consider the Liouville function . This is completely multiplicative, with λ(p)=-1. Then we have , which we should recognize as .

Continuing from last Friday, we now find the behavior of an ideal Bose gas at a temperature below the condensation temperature , where our previous analysis fails due to the integral approximation neglecting the ground state.
First, let us examine the spacing between the energy levels. For our quantum particle in a box, we have energy levels , where n_x, n_y, and n_z are positive integers. Using , we rewrite as . The ground state is . The first excited state is three-fold degenerate: , and the difference between these states is
. For a volume of one liter (=10^-3 cubic meter), and particles the mass of ⁴He (6.6×10^-27 kg), this energy difference is about 2.5×10^-39 J, which divides by k to get an equivalent temperature in the vicinity of 2×10^-16 K, and so the spacing is close enough that for any reasonable temperature, replacing the sum by an integral makes sense. Thus, our problem is purely the neglect of the ground state.
Let us examine what happens as the chemical potential μ approaches the ground state from below. We name the number of particles in the ground state as n₀; our Bose-Einstein distribution tells us that . If our occupation number of the ground state is large, , then we see that ; we thus expand the exponential in first order:
, so that
. If the total number of particles is Ñ, then we see that the population of the orbital ground state is comparable to this when , and the chemical potential cannot get closer to the ground state than ; the higher states are shielded from Gibbs potential by the ground state.
When the ground state population is significant, which occurs below T_c, we can see that the occupation number of any other individual state is relatively small; compare to . Thus, the integral approximation remains valid for representing all states except the ground state. Thus, we simply add an additional separate term explicitly listing the ground state in the sum of states.
The total number of particles is , where is here the number of particles in the excited states, and thus the reason for the subscript we added before. Writing n₀ in terms of fugacity, and using ε=0 for the ground state, we have
.
Since particles in the ground state have zero energy, the energy of the gas remains , and the reinterpreting of Ñ_e is the key correction.
Correcting the grand canonical potential , we see that our explicit ground state adds a term , and so our fundamental equation is
.

Now, we can use the above to analyze the properties of the Bose gas below the condensation temperature. First, for T<T_c, the maximum number of particles in the excited states is
. As T→em>T_c, we have Ñ_e→Ñ, and so we have , where λ_c is the value of the thermal de Broglie wavelength at the condensation temperature. Dividing these,
; this tells us that the fraction of the gas particles in the ground state as a function of temperature is
.

Now, examine the energy. For T>T_c, we have as before, . For T<T_c, we have
.

From this, we find the heat capacity at constant volume by
.
Below the condensation temperature, we take the derivative of the above, to get
, giving at a value of , significantly higher than the classical value 1.5Ñk, which we approach in the high-temperature classical regime. For temperatures above the condensation temperature, we have to take the derivative with respect to temperature of at constant Ñ, and eliminating using , and consider the temperature dependence of the thermal de Broglie wavelength. Performing that calculus, the net result is that
. This is greater than the classical value, and approaches it as T becomes large.
Both the T<T_c and T>T_c equations give at T=T_c; however, the heat capacity has a cusp at this point; C_V is increasing for T<T_c and decreasing for T>T_c. This unique cusp in heat capacity is one of the signatures of Bose condensation.

Last Monday, I discussed the Dirichlet convolution. We noted that it is commutative, associative, and has an identity element (the “unit function” ). We also noted that the Dirichlet convolution of two multiplicative functions is also multiplicative.

Now, suppose we have an arithmetic function f(n). Is it possible to find another arithmetic function, g(n), such that f*g=ε? First, note that ε(1)=1. Thus, we have 1=(f*g)(1)=f(1)g(1), and thus, we see that we can find , so long as f(1)≠0. Assuming that this is true, we now can examine higher n; using the fact that ε(n)=0 for n>0, we note that
.
Similarly,
,
,
and so on; more specifically, for n>0,
.
Thus, we can find the value of g(n) for each positive integer n recursively, in terms of the values of g for the proper divisors of n, so long as f(1)≠0, and we see that for a given f(n), g(n) is unique. This function, which we shall denote f^-1(n), is called the Dirichlet inverse of f(n). Note that the Dirichlet inverse of the Dirichlet inverse of a function is that function, as we expect from inverse elements.
Writing the above recursive procedure succinctly:
,
and for n>0,
.

An important property of the Dirichlet inverse is that the Dirichlet inverse of a multiplicative function is also multiplicative. Thus, since the set of multiplicative functions is closed under the Dirichlet convolution, the Dirichlet convolution is commutative and associative, has an identity, and every multiplicative function has a multiplicative function as an inverse, we see that the multiplicative functions form an Abelian group under Dirichlet convolution.

Now, since the Dirichlet inverse of a multiplicative function is also multiplicative, we can limit the recursive procedure to the powers of primes; that is, we use

and then use , p prime and k>0, to get
,

For example, let’s consider the Möbius function μ(n). We have μ^-1(1)=1, and, using
along with
,
we see

,
and, since for k>1, we see that all terms in the sum are zero except the i=k-1 term,

and thus, via this recursion, for any prime p and non-negative integer k; and thus
μ^-1(n)=1(n), the constant function that returns 1 for all arguements.

This fact allows us to prove a relation known as the Möbius inversion formula, which states that for arithmetic functions f(n) and g(n), with
for every positive integer n, then
.
Note that in the notation of Dirichlet convolutions, the first relation is just g=f*1, and the second is f=μ*g. Now, if g=f*1, then
,
and we have proven the formula.

To show it in use, suppose that g(n) is the identity function id(n)=n, and we want to find f; that is, we want to find the function f(n) such that for every positive integer n. The Möbius inversion formula tells us that f(n)=(id*μ)(n). Using the fact that both id and μ are multiplicative, we see f is multiplicative, so for p prime and k>0,
,
and we should thus recognize that f(n)=(id*μ)(n)=φ(n), the totient function, and so .

We can also see that, since we showed last time that , the Möbius inversion formula then tells us that
.

The polylogarithms are the functions
;
the sum converges in the complex plane over the open unit disk (and can be extended to the whole plane via analytic continuation).
The specific cases Li₂(z) and Li₃(z) are known respectively as the dilogarithm and trilogarithm. Note that from the Mercator series , we see that . Similarly, .

Taking the derivative, we see:
.
Combining this with the above, we see when the parameter s is a negative integer, the polylogarithm is a rational function:

,
and so on. Similarly,

,
and so on.

Examining the series, we see right away that and that . Further, for s>1

,
where is the Riemann zeta function and is the Dirichlet eta function.
Combining these with the derivative relation,

.

For s≥0, Li_s(z) is monotonically increasing (with z<1 for convergence of the series).

With regards to Fermi-Dirac and Bose-Einstein statistics, I show here that
and
.
[More generally, we have a pair of integral definitions:
.]

Using all the above, we see that our functions and used in the Bose gas analysis have the following properties:

• and for small ξ.
• and .
• and , which diverges (infinite slope).
• , with equality only at ξ=0
• Expanding to first order in ξ, .
• and are both monotonically increasing functions for |ξ|≤1

Last week, we began modeling the ideal Bose gas, including showing that it possesses the same classical limit as the Fermi gas, and that the mean occupation number for a given state in the Bose gas (Bose-Einstein distribution), , differs from that in the Fermi gas (Fermi-Dirac distribution), , by a change in sign in the denominator. After using the particle-in-a-box “density of states” to approximate sums over states as integrals over energy, we found several parameters in terms of polylogarithms; with fugacity , we had
,
,

(the subscript placed here on the particle number will become important for our notation later).
We also found previously that due to the divergence of at , the Gibbs potential must always be negative. Since β is always positive, then the fugacity must be less than unity: 0<ξ<1.

In the following work, I will use various properties of the polylogarithm functions. A separate post exploring the mathematics involved can be found here.

Using the thermal de Broglie wavelength , the above expressions for energy and particle number simplify to:
.
and
.
[Note that when the fugacity is small, these are approximated to first order in fugacity by and , which we found in the classical approximation here].
Now, dividing these to cancel the terms, we get
,
which differs from the classical by a factor of , which equals 1 at zero fugacity, and decreases with increasing fugacity (see the addendum). At the ξ→1 limit (μ→0 from below), we have

and
.

Now, we have a way to analyze a given Bose gas. Suppose we know Ñ_e, V, and T. Then
tells us that , and this can be solved numerically for the fugacity, which in turn allows us to evaluate the various thermodynamic functions. Note that the dependence of the fugacity upon volume and particle number can be expressed as dependence on particle density : .

However, we should note that diverges for ξ>1, so the maximum value of the polylogarithm in question is . However, one can be given values of Ñ_e, V, and T with the temperature low enough or density high enough that . In this situation, our proceeding analysis gives no solution for the fugacity; our analysis breaks down in this strong quantum limit. Where is the error?

The error is in how we considered the ground state. Recall that at μ=0, the occupation number n₀ becomes infinite; as the Gibbs potential rises toward zero, an increasing number of particles are found in the ground state. However, recall that in approximating our sum over states as an integral, we used a “density of states” , which for zero energy gives zero state density; our approximation neglects the ground state. This wasn’t a problem for our Fermi gas, as the ground state there can only hold a small number (2s+1) of particles.

Examining this “breakdown” point, we set , and seek the temperature:
,
where T_c is called the Bose condensation temperature. Above this temperature, our integral approximation still holds; below it, a phenomenon called Bose condensation happens, where the population of the ground state becomes significant. For example, ⁴He has atomic mass m=4.0026 amu=6.6465×10^-27 kg, and liquid ⁴He has a density of approximately 125 kg/m³, which gives number density n≈1.88×10²⁸ m^-3. Using these, we get T_c≈2.8 K, which is fairly close to the 2.17 K temperature where liquid helium-4 becomes superfluid.

Next week, I will explore how we correct the above analysis for temperatures below the condensation temperature.

An important binary operation on arithmetic functions is the Dirichlet convolution: for two artithmetic functions f(n) and g(n), the Dirichlet convolution gives a new arithmetic function, denoted f*g, defined by
.

Noting, first of all, that if d is a divisor of n, then n/d is also a divisor, we see that
,
and so the Dirichlet convolution is commutative.
Another property (but more complicated to prove, so I present without proof) is that the Dirichlet convolution is also associative: f*(g*h)=(f*g)*h.

Now, suppose that f and g are both multiplicative functions. Then let us consider (f*g)(mn), where m and n are coprime:
. Let d₁ be any divisor of m, and d₂ be any divisor of n. Then d₁d₁ divides mn. Further, since gcd(m,n)=1, any divisor d of mn may be uniquely expressed as such a product d₁d₂. This allows us to rewrite the sum in the Dirichlet convolution above:
,
and, since gcd(d₁,d₂)=1, and thus (since respective factors of coprime integers must also be coprime), we apply the fact that f and g are both multiplicative:
,
and so the Dirichlet convolution of multiplicative functions is also multiplicative. Since a multiplicative function can be defined entirely by its values on the powers of primes, this makes finding the Dirichlet convolution of multiplicative functions much easier.

For example, consider the Dirichlet convolution of the Möbius function μ(n), and the multiplicative function 2^ω(n) appearing in this post. Since both are multiplicative, μ*2^ω will be multiplicative, so we need only examine its values over the primes.
Now, from the definition of the Dirichlet convolution,
.
Now, for p prime, the divisors of p^k are just 1, p, p², …, p^k. Thus,
.
Now, we recall that for powers of primes,

and
.
Thus, we see that for k=0, we get
,
as we expect of a multiplicative function; for k=1:
,
Now, for k≥2, we see that any terms in the sum with i≥2 will be zero, as the Möbius function for pⁱ is zero for i≥2. Similarly, for k≥2, both and both equal 2, since both k and k-1 are greater than zero.
Thus, we find that for k≥2,

Thus, μ*2^ω is the multiplicative function with
,
and so we see that


Similarly, we can see that the Dirichlet convolution of the constant function 1(n) with itself is
,
the divisor function, and the Dirichlet convolution of the constant function 1(n) with the identity function id(n) is
,
the sum-of-divisors function.

Lastly, consider the “unit function” ε(n) defined by
. We see that this function is completely multiplicative; for any positive integers m and n.
Now, consider the Dirichlet convolution of this function with any arithmetic function f(n); all terms of the sum will be zero except that where d=1, and thus n/d=n:
.
Thus, we see that the unit function serves as the identity element for Dirichlet convolution.

When we modeled the ideal Fermi fluid (previous posts here, here, here, here, and here), we used the fact that the grand canonical partition function can be factored over the individual states:
,
where the individual state partition is
,
where the sum is over all possible values of n, the number of particles in the state. For fermions, the Pauli exclusion principle meant that n=0 and n=1 were the only allowed values, and the above series had only two terms. Bosons, however, can be placed in the same quantum state in any number, so for an ideal Bose gas, the above sum becomes an infinite (geometric) series:
.

Next, let us find the occupancy (mean occupation number) f_k,ms:
.
Now, we note that for |x|<1,
.
This tells us that the sum in the numerator is thus , and so
.
Compare this to the occupancy for a fermion fluid ; the difference is a change of sign in the denominator. Note that unlike the fermion case, for a boson fluid, the occupancy can be greater than one.

For a boson fluid, we generally choose our energy scale so that our zero energy is the ground state. Now, note that the occupancy has a singularity at , where it goes to infinity. This means that for a Bose gas with a finite number of particles, the Gibbs potential μ must always be less that all our energy values; thus, with the above choice of energy scale, we see that μ must always be negative.

We can, as in the fermion case, use the quantum particle in a cubical box to develop the concept of the “density of states” D(ε):
,
which allows us to approximate the sum over all states as an integral over energy:

(where g₀ is the number of different spin states: g₀=2s+1, where s is the particle spin, here an integer). Using the same integration by parts as in the fermion case, we can change this to:
.
Similarly,
,
and
.
Note the difference in signs in the denominators in the integrals from the fermion case, and that still holds.
Using the math from this addendum, we can rewrite the above integrals in terms of polylogarithms:

and
.

We also note from our equation for Ñ, that if the number of particles is held constant, we see that the integral must be constant, and so as the temperature increases, the GIbbs potential μ must decrease (as it also does for a Fermi gas).

When examining the ideal Fermi fluid in the classical limit, we noted that when the fugacity is small (), we see that , and so the fermion occupancy can be approximated via
.
Now, note that means that for the boson occupancy,
,
the exact same approximation. Thus, the results from applying this approximation is the same (except for the different value of g₀), and thus, the ideal Bose gas has the same condition for classical behavior, and, as expected, also approaches the classical ideal gas. Thus, the difference between Bose gases and Fermi gases is only significant in the quantum regime.

Next, let us find the occupancy (mean occupation number) f_k,ms:
.
Now, we note that for |x|<1,
.
This tells us that the sum in the numerator is thus , and so
.
Compare this to the occupancy for a fermion fluid ; the difference is a change of sign in the denominator. Note that unlike the fermion case, for a boson fluid, the occupancy can be greater than one.