Twisted One 151’s Weblog

Physics Friday 96

November 6, 2009 by twistedone151

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 $P(x,y,z)=-\rho{g}z+C$ , and the force on an area element is:
$d\mathbf{F}=-P\mathbf{\hat{n}}\,dS=-P\,d\mathbf{S}$ . Thus, the torque on that element is
$d\mathbf{\tau}=\mathbf{d}\times\mathbf{F}=-\mathbf{r}\times{P}\mathbf{\hat{n}}\,dS={P}\mathbf{\hat{n}}\times\mathbf{r}\,dS$ ,
where r is the coordinate vector to the element. Recalling that for vectors v and w and scalar a,
$\mathbf{v}\times(a\mathbf{w})=a(\mathbf{v}\times\mathbf{w})=(a\mathbf{v})\times\mathbf{w}$ , so since P(x,y,z) is a scalar, we see
$d\mathbf{\tau}={P}\mathbf{\hat{n}}\times\mathbf{r}\,dS=\mathbf{\hat{n}}\times(P\mathbf{r})\,dS$
and so the total torque (about the origin) is
$\mathbf{\tau}=\iint_{\partial{V}}\mathbf{\hat{n}}\times(P\mathbf{r})\,dS$ ;
and one form of the divergence theorem tells us that for vector field A,
$\iint_{\partial{V}}\mathbf{\hat{n}}\times\mathbf{A}\,dS=\iiint_{V}\mathbf{\nabla}\times\mathbf{A}\,dV$ .
Thus
$\mathbf{\tau}=\iiint_{V}\mathbf{\nabla}\times(P\mathbf{r})\,dS$
Now, the product rule for the curl of the scalar product of a scalar field ψ and a vector field a is
$\mathbf{\nabla}\times(\psi\mathbf{a})=\mathbf{\nabla}\psi\times\mathbf{a}+\psi\mathbf{\nabla}\times\mathbf{a}$ .
Now, applying that to $\mathbf{\nabla}\times(P\mathbf{r})$ , and noting that $\mathbf{\nabla}\times\mathbf{r}=0$ (as it is a radial vector field), and $\mathbf{\nabla}P=-\rho{g}\mathbf{\hat{z}}$ , we see
$\begin{eqnarray}\mathbf{\tau}&=&\iiint_{V}\mathbf{\nabla}\times(P\mathbf{r})\,dV\\&=&\iiint_{V}\mathbf{\nabla}P\times\mathbf{r}+P\mathbf{\nabla}\times\mathbf{r}\,dV\\&=&-\iiint_{V}\rho{g}\mathbf{\hat{z}}\times\mathbf{r}\,dV\\&=&\rho{g}\iiint_{V}\mathbf{r}\times\mathbf{\hat{z}}\,dV\end{eqnarray}$ .
Now, since $\mathbf{\hat{z}}$ is a constant vector, it can be “factored out” of the integral:
$\begin{eqnarray}\mathbf{\tau}&=&\rho{g}\iiint_{V}\mathbf{r}\times\mathbf{\hat{z}}\,dV\\&=&\rho{g}\left[\iiint_{V}\mathbf{r}\,dV\right]\times\mathbf{\hat{z}}\end{eqnarray}$ .
We recall that the total buoyant force on the object is $\mathbf{\vec{F}}=\rho{g}V\mathbf{\hat{z}}$ ; plugging this into the above, we see
$\begin{eqnarray}\mathbf{\tau}&=&\rho{g}\left[\iiint_{V}\mathbf{r}\,dV\right]\times\mathbf{\hat{z}}\\&=&\left[\iiint_{V}\mathbf{r}\,dV\right]\times\left(\frac{\mathbf{\vec{F}}}{V}\right)\\&=&\frac{1}{V}\left[\iiint_{V}\mathbf{r}\,dV\right]\times\left(\mathbf{\vec{F}}\right)\end{eqnarray}$ .

Now, recall that the center of mass of a region with density function ρ(x,y,z) is $\mathbf{r}_{cm}=\frac{\iiint_{V}\rho(\mathbf{r})(\mathbf{r})\,dV}{\iiint_{V}\rho(\mathbf{r})\,dV}$ . If the density is a constant, it factors out of the integrals, and one gets $\mathbf{r}_{cm}=\frac{\iiint_{V}\mathbf{r}\,dV}{\iiint_{V}\,dV}=\frac{1}{V}\iiint_{V}\mathbf{r}\,dV$ . 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 $\mathbf{r}_{b}=\frac{1}{V}\iiint_{V}\mathbf{r}\,dV$ , and then we see that
$\begin{eqnarray}\mathbf{\tau}&=&\frac{1}{V}\left[\iiint_{V}\mathbf{r}\,dV\right]\times\left(\mathbf{\vec{F}}\right)\\&=&\mathbf{r}_{b}\times\mathbf{\vec{F}}\end{eqnarray}$ ,
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.

Tags: Buoyancy, Center of Buoyancy, Center of Mass, Curl, Divergence Theorem, Friday Physics, physics, Torque
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Monday Math 95

November 2, 2009 by twistedone151

A notable arithmetic function which is neither additive nor multiplicative is the von Mangoldt function, denoted Λ(n). It is defined as
$\Lambda(n)=\left\{\begin{eqnarray}\ln{p}&\;\;&\text{if}\;{n=p^k},\;p\;\text{prime,}\;k\;\text{a positive integer,}\\{0}&\;\;&\text{otherwise.}\end{eqnarray}\right\.$
It has a few notable properties. First, consider $\sum_{d|n}\Lambda(d)$ . Considering the prime factorization $n=p_1^{k_1}p_2^{k_2}\cdots{p_r^{k_r}}$ , we see that the only divisors d of n for which the von Mangoldt function is non-zero are $d=p_i^{k}$ , $1\le{k}\le{k_i}$ , with $\lambda(p_i^{k})=\ln{p_i}$ . There are k_i terms for each p_i, so
$\begin{eqnarray}\sum_{d|n}\Lambda(d)&=&k_1\ln{p_1}+k_2\ln{p_2}+\cdots+k_r\ln{p_r}\\&=&\ln(p_1^{k_1}p_2^{k_2}\cdots{p_r^{k_r}})\\\sum_{d|n}\Lambda(d)&=&\ln{n}\end{eqnarray}$ ,
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 $F(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}$ . Now, let us consider the derivative of the Dirichlet series generating function. Taking the derivative with respect to s,
$F'(s)=\sum_{n=1}^{\infty}\frac{\partial}{\partial{s}}\frac{f(n)}{n^s}=-\sum_{n=1}^{\infty}\frac{f(n)\ln{n}}{n^s}$ .

Next, let us consider the Dirichlet convolution of f^-1(n) and f(n)ln(n). This is
$[f^{-1}*(f\cdot\ln)](n)=\sum_{d|n}f^{-1}(d)f\left(\frac{n}{d}\right)\ln\left(\frac{n}{d}\right)$ .
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 $n=p_1^{k_1}p_2^{k_2}\cdots{p_r^{k_r}}$ , we have $d=p_1^{i_1}p_2^{i_2}\cdots{p_r^{i_r}}$ , with $0\le{i_1}\le{k_1},\;\;0\le{i_2}\le{k_2},\;\ldots,\;\;0\le{i_r}\le{k_r},\;\;$ . I showed here that the Dirichlet inverse f^-1(n) of a completely multiplicative function f(n) is the multiplicative function with
$f^{-1}\left(p^k\right)=\Large\left\{\begin{eqnarray}1&,\;&k={0}\\-f(p)&,\;&k=1\\{0}&,\;&k\gt1\end{eqnarray}\right\.$ .
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 $f^{-1}(1)f(n)\ln(n)=f(n)\ln(n)$ .
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
$\begin{eqnarray}f^{-1}(d)f\left(\frac{n}{d}\right)\ln\left(\frac{n}{d}\right)&=&f^{-1}(p_j)f\left(\frac{n}{p_j}\right)\ln\left(\frac{n}{p_j}\right)\\&=&-f(p_j)f\left(\frac{n}{p_j}\right)(\ln{n}-\ln{p_j})\\&=&f(n)(\ln{p_j}-\ln{n})\end{eqnarray}$ .
Thus, for those cases where n has only one distinct prime factor, $n=p^k$ , then
$[f^{-1}*(f\cdot\ln)]\left(n\right)=(f(n)\ln{n})+f(n)(\ln{p}-\ln{n})=f(n)\ln(p)$ .
Next, for d a product of two distinct prime factors p_j and p_k,
$\begin{eqnarray}f^{-1}(d)f\left(\frac{n}{d}\right)\ln\left(\frac{n}{d}\right)&=&f^{-1}(p_jp_k)f\left(\frac{n}{p_jp_k}\right)\ln\left(\frac{n}{p_jp_k}\right)\\&=&f^{-1}(p_j)f^{-1}(p_k)f\left(\frac{n}{p_jp_k}\right)\ln\left(\frac{n}{p_jp_k}\right)\\&=&(-f(p_j))(-f(p_k))f\left(\frac{n}{p_j}\right)(\ln{n}-\ln{p_j}-\ln{p_k})\\&=&f(n)(\ln{n}-\ln{p_j}-\ln{p_k})\end{eqnarray}$ .
So for n with $r=\omega(n)=2$ , we see
$\begin{eqnarray}\left[f^{-1}*(f\cdot\ln)\right]\left(n\right)&=&(f(n)\ln{n})+f(n)(\ln{p_1}-\ln{n})\\&\:&+f(n)(\ln{p_2}-\ln{n})+f(n)(\ln{n}-\ln{p_1}-\ln{p_2})\\&=&{0}\end{eqnarray}$ .
And when d is the product of l distinct primes (2≤l≤r), $d=p_1p_2\cdots{p_l}$ , we see
$\begin{eqnarray}f^{-1}(d)f\left(\frac{n}{d}\right)\ln\left(\frac{n}{d}\right)&=&f^{-1}(p_1p_2\cdots{p_l})f\left(\frac{n}{p_1p_2\cdots{p_l}}\right)\ln\left(\frac{n}{p_1p_2\cdots{p_l}}\right)\\&=&(-1)^{l}f(n)(\ln{n}-\ln{p_1}-ln{p_2}-\ldots-\ln{p_l})\end{eqnarray}$ .
Summing these, we see that an $f(n)$ can be factored from each term. Cancelling the logarithms, one finds that $[f^{-1}*(f\cdot\ln)]\left(n\right)=0$ whenever $r=\omega(n)\gt1$ , so that the Dirichlet convolution is a function that is nonzero only for $n=p^k$ , p prime and k a positive integer; in that case, $[f^{-1}*(f\cdot\ln)]\left(p^k\right)=f\left(p^k)\ln{p}$ . Thus
$[f^{-1}*(f\cdot\ln)]\left(n\right)=f(n)\Lambda(n)$ .

Now, using the relationship between Dirichlet convolution and Dirichlet series, we have that the Dirichlet series generating function for $f(n)\Lambda(n)$ is thus the product of those for f^-1(n) and f(n)ln(n); the former is $\frac{1}{F(s)}$ , and we saw already that the latter is $-F'(s)$ . Thus we see that the logarithmic derivative of F(s) has series
$\frac{F'(s)}{F(s)}=-\sum_{n=1}^{\infty}\frac{f(n)\Lambda(n)}{n^s}$ .
Letting the completely multiplicative function f(n) be 1(n), then we see
$\frac{\zeta'(s)}{\zeta(s)}=-\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^s}$
(for $\Re(s)\gt1$ ), 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 $\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^s}=\sum_{n=2}^{\infty}\frac{\Lambda(n)}{n^s}$ . Performing termwise integration with respect to s on the latter sum, one can then find
$\ln\zeta(s)=\sum_{n=2}^{\infty}\frac{\Lambda(n)}{\ln{n}}\frac{1}{n^s}$ .

Tags: Dirichlet Convolution, DIrichlet Inverse, Dirichlet Series Generating Function, Logarithmic Derivative, Math, Monday Math, Riemann Zeta Function, Von Mangoldt Function
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Physics Friday 95

October 30, 2009 by twistedone151

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.
Solution

Tags: Friday Physics, Kinematics, physics, Projectile
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Monday Math 94

October 26, 2009 by twistedone151

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
$F(s)=\sum_{a=1}^{\infty}\frac{f(a)}{a^s}$ and $G(s)=\sum_{b=1}^{\infty}\frac{g(b)}{b^s}$ . The product, then, is
$\begin{eqnarray}F(s)G(s)&=&\left(\sum_{a=1}^{\infty}\frac{f(a)}{a^s}\right)\left(\sum_{b=1}^{\infty}\frac{g(b)}{b^s}\right)\\&=&\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\frac{f(a)g(b)}{(ab)^s}\end{eqnarray}$ .

Now, suppose we reorder the sum so as to group terms with the same value of ab. We let n=ab; then we have $b=\frac{n}{a}$ , 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: $\begin{eqnarray}F(s)G(s)&=&\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\frac{f(a)g(b)}{(ab)^s}\\&=&\sum_{n=1}^{\infty}\sum_{a|n}\frac{f(a)g\left(\frac{n}{a}\right)}{n^s}\\&=&\sum_{n=1}^{\infty}\frac{\left(\sum_{a|n}f(a)g\left(\frac{n}{a}\right)\right)}{n^s}\\F(s)G(s)&=&\sum_{n=1}^{\infty}\frac{(f*g)(n)}{n^s}\end{eqnarray}$ ,
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
$\sum_{n=1}^{\infty}\frac{\epsilon(n)}{n^s}=\frac{1}{1^s}+\frac{0}{2^s}+\frac{0}{3^s}+\cdots=1$ .
And since $f*f^{-1}=\epsilon$ , 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 $\zeta(s)$ and $\frac{1}{\zeta(s)}$ , respectively. You may also recall that here I showed that the Dirichlet series generating function for |μ| is $\frac{\zeta(s)}{\zeta(2s)}$ . 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 $\frac{\zeta(2s)}{\zeta(s)}$ ; 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 $id_a\cdot{f}$ has Dirichlet series generating function F(s-a); since
$F(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}$ ,
then
$\sum_{n=1}^{\infty}\frac{f(n)\cdot{n^a}}{n^s}=\sum_{n=1}^{\infty}\frac{f(n)}{n^(s-a)}=F(s-a)$ .
For example, the Dirichlet series generating function of $id_x(n)=1\cdot{id_x}(n)$ is thus $\zeta(s-x)$ .

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 $\zeta(s)\zeta(s-x)$ becomes much more simple than the proof seen here:
Since $\sigma_x(n)=\sum{d|n}d^x$ , we see $\sigma_x=1*id_x$ ; since the Dirichlet series generating function of 1(n) is $\zeta(s)$ , and the Dirichlet series generating function of $id_x(n)=1\cdot{id_x}(n)$ is $\zeta(s-a)$ , the Dirichlet series generating function of σ_x(n) is their product, $\zeta(s)\zeta(s-x)$ .
Similarly, we can use the Möbius inversion formula relationship that $id*\mu=\phi$ , found here, to confirm that φ(n) has Dirichlet series generating function $\frac{\zeta(s-1)}{\zeta(s)}$ , as we found via Euler product here.

Tags: Dirichlet Convolution, DIrichlet Inverse, Dirichlet Series Generating Function, Divisor Functions, Euler Product, Math, Möbius Function, Möbius Inversion Formula, Monday Math, Riemann Zeta Function, Totient
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Physics Friday 94

October 23, 2009 by twistedone151

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?
Solution:

Tags: Centripetal Force, Conservation of Energy, Friday Physics, Kinematics, Non-Uniform Circular Motion, physics
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Monday Math 93

October 19, 2009 by twistedone151

Last Monday, I discussed the Dirichlet inverse. I showed that the Dirichlet inverse can be found by the recursive formula:
$f^{-1}(1)=\frac{1}{f(1)}$ ,
and for n>0,
$f^{-1}(n)=-\frac{1}{f(1)}\sum_{d|n,\,d\lt{n}}f\left(\frac{n}{d}\right)f^{-1}(d)$ .
Further, since the inverse of a multiplicative function is also multiplicative, for a multiplicative function f(n) this formula becomes
$f^{-1}(1)=1$ , and for p prime and k>0,
$f^{-1}\left(p^k\right)=-\sum_{i=0}^{k-1}f\left(p^{k-i}\right)f^{-1}\left(p^i\right)$ .

Now, what happens when f(n) is completely multiplicative? Then $f\left(p^{k-i}\right)=(f(p))^{k-i}$ . We have $f^{-1}(1)=1$ , and via k=1, $f^{-1}(p)=-f(p)$ , we see that
$f^{-1}(p^2)=-[f(p)f^{-1}(1)+f(1)f^{-1}(p)]=-[f(p)-f(p)]=0$ ,
$f^{-1}(p^3)=-[f(p^2)f^{-1}(1)+f(p)f^{-1}(p)+f(1)f^{-1}(p^2)]=-[(f(p))^2-f(p)f(p)+0]=0$ ,
$f^{-1}(p^k)=-[f(p^{k-1})f^{-1}(1)+f(p^{k-2})f^{-1}(p)+\cdots+f(1)f^{-1}(p^{k-1})]=-[(f(p))^{k-1}-f(p)(f(p))^{k-2}+0+\cdots+0]=0$
and so we see that if f(n) is completely multiplicative, then f^-1(n) is the multiplicative function with
$f^{-1}\left(p^k\right)=\Large\left\{\begin{eqnarray}1&,\;&k={0}\\-f(p)&,\;&k=1\\{0}&,\;&k\gt1\end{eqnarray}\right\.$ .
For f(n)=1(n), this gives
$1^{-1}\left(p^k\right)=\Large\left\{\begin{eqnarray}1&,\;&k={0}\\-1&,\;&k=1\\{0}&,\;&k\gt1\end{eqnarray}\right\.$ , which gives $1^{-1}(n)=\mu(n)$ , as we’ve found before.
Next, for f(n)=id_x(n), we see
$id_x^{-1}\left(p^k\right)=\Large\left\{\begin{eqnarray}1&,\;&k={0}\\-p^x&,\;&k=1\\{0}&,\;&k\gt1\end{eqnarray}\right\.$ , which, with a little examination, we see means that $id_x^{-1}(n)=id_x(n)\mu(n)$ , 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 $\lambda(n)=(-1)^{\Omega(n)}$ . This is completely multiplicative, with λ(p)=-1. Then we have $\lambda^{-1}\left(p^k\right)=\Large\left\{\begin{eqnarray}1&,\;&k={0}\\1&,\;&k=1\\{0}&,\;&k\gt1\end{eqnarray}\right\.$ , which we should recognize as $|\mu|$ .

Tags: DIrichlet Inverse, Math, Monday Math, Multiplicative Function
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Physics Friday 93

October 16, 2009 by twistedone151

Continuing from last Friday, we now find the behavior of an ideal Bose gas at a temperature below the condensation temperature $\T_c=\frac{2\pi\hbar^2}{mk}\left[\frac{\tilde{N}_e}{\zeta\left(\frac{3}{2}\right)g_0V}\right]^{2/3}$ , 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 $E_{n_x,n_y,n_z}=\frac{\pi^2\hbar^2}{2mL^2}(n_x^2+n_y^2+n_z^2)$ , where n_x, n_y, and n_z are positive integers. Using $L=V^{1/3}$ , we rewrite as $E_{n_x,n_y,n_z}=\frac{\pi^2\hbar^2}{2mV^{2/3}}(n_x^2+n_y^2+n_z^2)$ . The ground state is $E_{1,1,1}=\frac{3\pi^2\hbar^2}{2mV^{2/3}}$ . The first excited state is three-fold degenerate: $E_{2,1,1}=E_{1,2,1}=E_{1,1,2}=\frac{3\pi^2\hbar^2}{mV^{2/3}}$ , and the difference between these states is
$E_{2,1,1}-E_{1,1,1}=\frac{3\pi^2\hbar^2}{2mV^{2/3}}$ . 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 $E_{1,1,1}$ from below. We name the number of particles in the ground state as n₀; our Bose-Einstein distribution tells us that $n_0=\frac{1}{e^{\beta\left(E_{1,1,1}-\mu\right)}-1}$ . If our occupation number of the ground state is large, $n_0\gg1$ , then we see that $\beta\left(E_{1,1,1}-\mu\right)\ll1$ ; we thus expand the exponential in first order:
$e^{\beta\left(E_{1,1,1}-\mu\right)}\approx{1+\frac{E_{1,1,1}-\mu}{kT}}$ , so that
$n_0\approx\frac{kT}{E_{1,1,1}-\mu}$ . If the total number of particles is Ñ, then we see that the population of the orbital ground state is comparable to this when $\beta\left(E_{1,1,1}-\mu\right)\sim\frac{1}{\tilde{N}}$ , and the chemical potential cannot get closer to the ground state than $E_{1,1,1}-\mu\approx\frac{kT}{\tilde{N}}$ ; 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 $n_0=\frac{1}{e^{\beta\left(E_{1,1,1}-\mu\right)}-1}$ to $n_{2,1,1}=\frac{1}{e^{\beta\left(E_{2,1,1}-\mu\right)}-1}$ . 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 $\tilde{N}=n_0+\tilde{N}_e$ , where $\tilde{N}_e=\frac{g_0V}{\Lambda_T^3}\text{Li}_{3/2}(\xi)$ 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
$n_0=\frac{1}{e^{\beta\left(-\mu\right)}-1}=\frac{e^{\beta\mu}}{1-e^{\beta\mu}}=\frac{\xi}{1-\xi}$ .
Since particles in the ground state have zero energy, the energy of the gas remains $U=\frac{3}{2}kT\frac{g_0V}{\Lambda_T^3}\text{Li}_{5/2}(\xi)$ , and the reinterpreting of Ñ_e is the key correction.
Correcting the grand canonical potential $\Psi=-kT\ln\mathcal{Z}$ , we see that our explicit ground state adds a term $-kT\ln\left[\frac{1}{\left(1-e^{\beta\mu}\right)^{g_0}}\right]=g_0kT\ln(1-\xi)$ , and so our fundamental equation is
$\Psi=g_0kT\ln(1-\xi)-g_0kT\frac{V}{\Lambda_T^3}\text{Li}_{5/2}(\xi)$ .

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
$\tilde{N}_e=\frac{g_0V}{\Lambda_T^3}\text{Li}_{3/2}(1)=\frac{g_0V}{\Lambda_T^3}\zeta\left(\frac{3}{2}\right)$ . As T→em>T_c, we have Ñ_e→Ñ, and so we have $\tilde{N}=\frac{g_0V}{\Lambda_c^3}\zeta\left(\frac{3}{2}\right)$ , where λ_c is the value of the thermal de Broglie wavelength at the condensation temperature. Dividing these,
$\frac{\tilde{N}_e}{\tilde{N}}=\frac{\frac{g_0V}{\Lambda_T^3}\zeta\left(\frac{3}{2}\right)}{\frac{g_0V}{\Lambda_c^3}\zeta\left(\frac{3}{2}\right)}=\left(\frac{\lambda_c}{\lambda_T}\right)^3=\left(\frac{T}{T_c}\right)^{3/2}$ ; this tells us that the fraction of the gas particles in the ground state as a function of temperature is
$\frac{n_0}{\tilde{N}}=1-\frac{\tilde{N}_e}{\tilde{N}}=1-\left(\frac{T}{T_c}\right)^{3/2}$ .

Now, examine the energy. For T>T_c, we have as before, $U=\frac{3}{2}kT\frac{g_0V}{\Lambda_T^3}\text{Li}_{5/2}(\xi)=\frac{3}{2}\tilde{N}_ekT\frac{\text{Li}_{5/2}(\xi)}{\text{Li}_{3/2}(\xi)}$ . For T<T_c, we have
$\begin{eqnarray}U&=&\frac{3}{2}kT\frac{g_0V}{\Lambda_T^3}\text{Li}_{5/2}(1)\\&=&\frac{3}{2}kT\frac{\tilde{N}_e}{\zeta\left(\frac{3}{2}\right)}\zeta\left(\frac{5}{2}\right)\\&=&\frac{3}{2}\tilde{N}kT\frac{\zeta\left(\frac{5}{2}\right)}{\zeta\left(\frac{3}{2}\right)}\frac{\tilde{N}_e}{\tilde{N}}\\&=&\frac{3}{2}\tilde{N}kT\frac{\zeta\left(\frac{5}{2}\right)}{\zeta\left(\frac{3}{2}\right)}\left(\frac{T}{T_c}\right)^{3/2}\\U&=&\frac{3\zeta\left(\frac{5}{2}\right)}{2\zeta\left(\frac{3}{2}\right)}\tilde{N}kT_c\left(\frac{T}{T_c}\right)^{5/2}\\U&\approx&0.77\tilde{N}kT_c\left(\frac{T}{T_c}\right)^{5/2}\end{eqnarray}$ .

From this, we find the heat capacity at constant volume by
$C_V=\left(\frac{\partial{U}}{\partial{T}}\right)_V$ .
Below the condensation temperature, we take the derivative of the above, to get
$C_V=\frac{15\zeta\left(\frac{5}{2}\right)}{4\zeta\left(\frac{3}{2}\right)}\tilde{N}k\left(\frac{T}{T_c}\right)^{3/2}\approx{1.9}\tilde{N}k\left(\frac{T}{T_c}\right)^{3/2}$ , giving at $T=T_c$ a value of $C_V\approx{1.9}\tilde{N}k$ , 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 $U=\frac{3}{2}\tilde{N}_ekT\frac{\text{Li}_{5/2}(\xi)}{\text{Li}_{3/2}(\xi)}$ at constant Ñ, and eliminating $\left(\frac{d\xi}{dT}\right)_{\tilde{N}}$ using $\tilde{N}_e=\frac{g_0V}{\Lambda_T^3}\text{Li}_{3/2}(\xi)$ , and consider the temperature dependence of the thermal de Broglie wavelength. Performing that calculus, the net result is that
$\begin{eqnarray}C_V&=&\frac{3}{2}\tilde{N}k\left[\frac{5}{2}\frac{\text{Li}_{5/2}(\xi)}{\text{Li}_{3/2}(\xi)}-\frac{3}{2}\frac{\text{Li}_{5/2}'(\xi)}{\text{Li}_{3/2}'(\xi)}\right]\\&=&\frac{3}{2}\tilde{N}k\left[\frac{5}{2}\frac{\text{Li}_{5/2}(\xi)}{\text{Li}_{3/2}(\xi)}-\frac{3}{2}\frac{\text{Li}_{3/2}(\xi)}{\text{Li}_{1/2}(\xi)}\right]\end{eqnarray}$ . 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 $C_V\approx{1.9}\tilde{N}k$ 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.

Tags: Bose Condensation, Bose Gas, Friday Physics, physics, Quantum Mechanics, Statistical Mechanics, Thermal de Broglie Wavelength, Thermodynamics
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Monday Math 92

October 12, 2009 by twistedone151

Last Monday, I discussed the Dirichlet convolution. We noted that it is commutative, associative, and has an identity element (the “unit function” $\epsilon(n)=\Large\left\{\begin{eqnarray}1&,\;&n=1\\{0}&,\;&n\gt1\end{eqnarray}\right\.$ ). 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 $g(1)=\frac{1}{f(1)}$ , 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
$\begin{eqnarray}\epsilon(2)&=&(f*g)(2)\\{0}&=&f(1)g(2)+f(2)g(1)\\{0}&=&f(1)g(2)+\frac{f(2)}{f(1)}\\f(1)g(2)&=&-\frac{f(2)}{f(1)}\\g(2)&=&-\frac{f(2)}{\left(f(1)\right)^2}\end{eqnarray}$ .
Similarly,
$\begin{eqnarray}\epsilon(3)&=&(f*g)(3)\\{0}&=&f(1)g(3)+f(3)g(1)\\{0}&=&f(1)g(3)+\frac{f(3)}{f(1)}\\f(1)g(3)&=&-\frac{f(3)}{f(1)}\\g(3)&=&-\frac{f(3)}{\left(f(1)\right)^2}\end{eqnarray}$ ,
$\begin{eqnarray}\epsilon(4)&=&(f*g)(4)\\{0}&=&f(1)g(4)+f(2)g(2)+f(4)g(1)\\{0}&=&f(1)g(4)-\frac{\left(f(2)\right)^2}{\left(f(1)\right)^2}+\frac{f(4)}{f(1)}\\f(1)g(4)&=&\frac{\left(f(2)\right)^2}{\left(f(1)\right)^2}-\frac{f(4)}{f(1)}\\g(4)&=&\frac{\left(f(2)\right)^2}{\left(f(1)\right)^3}-\frac{f(4)}{\left(f(1)\right)^2}\end{eqnarray}$ ,
and so on; more specifically, for n>0,
$\begin{eqnarray}\epsilon(n)&=&(f*g)(n)\\{0}&=&\sum_{d|n}f\left(\frac{n}{d}\right)g(d)\\{0}&=&f(1)g(n)+\sum_{d|n,\,d\lt{n}}f\left(\frac{n}{d}\right)g(d)\\f(1)g(n)&=&-\sum_{d|n,\,d\lt{n}}f\left(\frac{n}{d}\right)g(d)\\g(n)&=&-\frac{1}{f(1)}\sum_{d|n,\,d\lt{n}}f\left(\frac{n}{d}\right)g(d)\end{eqnarray}$ .
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:
$f^{-1}(1)=\frac{1}{f(1)}$ ,
and for n>0,
$f^{-1}(n)=-\frac{1}{f(1)}\sum_{d|n,\,d\lt{n}}f\left(\frac{n}{d}\right)f^{-1}(d)$ .

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
$f^{-1}(1)=\frac{1}{f(1)}$
and then use $n=p^k$ , p prime and k>0, to get
$\begin{eqnarray}f^{-1}\left(p^k\right)&=&-\frac{1}{f(1)}\sum_{d|p^k,\,d\lt{p^k}}f\left(\frac{p^k}{d}\right)f^{-1}(d)\\&=&-\sum_{i=0}^{k-1}f\left(p^{k-i}\right)f^{-1}\left(p^i\right)\end{eqnarray}$ ,

For example, let’s consider the Möbius function μ(n). We have μ^-1(1)=1, and, using
$f^{-1}\left(p^k\right)=-\sum_{i=0}^{k-1}f\left(p^{k-i}\right)f^{-1}\left(p^i\right)$ along with
$\mu\left(p^k\right)=\Large\left\{\begin{eqnarray}1&,\;&k={0}\\-1&,\;&k=1\\{0}&,\;&k\gt1\end{eqnarray}\right\.$ ,
we see
$\mu^{-1}(p)=-\mu(p)\mu^{-1}(1)=-\mu(p)=1$
$\mu^{-1}\left(p^2\right)=-\left[\mu\left(p^2\right)\mu^{-1}(1)+\mu(p)\mu^{-1}(p)\right]=-[{0}\cdot{1}+(-1)\cdot{1}]=1$ ,
and, since $\mu\left(p^k\right)=0$ for k>1, we see that all terms in the sum $\sum_{i=0}^{k-1}\mu\left(p^{k-i}\right)\mu^{-1}\left(p^i\right)$ are zero except the i=k-1 term,
$\mu^{-1}\left(p^k\right)=-\mu(p)\mu^{-1}\left(p^{k-1}\right)=\mu^{-1}\left(p^{k-1}\right)$
and thus, via this recursion, $\mu^{-1}\left(p^k\right)=1$ 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
$g(n)=\sum_{d|n}f(d)$ for every positive integer n, then
$f(n)=\sum_{d|n}\mu(d)g\left(\frac{n}{d}\right)$ .
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
$\begin{eqnarray}\mu*g&=&\mu*(f*1)\\&=&\mu*(1*f)\\&=&(\mu*1)*f\\&=&\epsilon*f\\\mu*g&=&f\end{eqnarray}$ ,
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 $\sum_{d|n}f(d)=n$ 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,
$\begin{eqnarray}\left(\mu*id\right)\left(p^k\right)&=&\sum_{d|p^k}\mu(d)id\left(\frac{p^k}{d}\right)\\&=&\sum_{i=0}^{k}\mu\left(p^i\right)p^{k-i}\\&=&\mu(1)p^{k-0}+\mu(p)p^{k-1}\\&=&p^k-p^{k-1}\\\left(\mu*id\right)\left(p^k\right)&=&p^{k-1}(p-1)\end{eqnarray}$ ,
and we should thus recognize that f(n)=(id*μ)(n)=φ(n), the totient function, and so $\sum_{d|n}\phi(d)=n$ .

We can also see that, since we showed last time that $\mu*2^{\omega}=|\mu|$ , the Möbius inversion formula then tells us that
$(1*|\mu|)(n)=\sum_{d|n}|\mu(d)|=2^{\omega(n)}$ .

Tags: Abelian Group, Arithmetic Function, Dirichlet Convolution, DIrichlet Inverse, Math, Möbius Function, Möbius Inversion Formula, Monday Math, Multiplicative Function
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Mathematical Addendum to Physics Friday 92

October 9, 2009 by twistedone151

The polylogarithms are the functions
$\text{Li}_{s}(z)=\sum_{k=1}^{\infty}\frac{z^k}{k^s}$ ;
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 $\ln(1+x)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}z^k}{k}$ , we see that $\text{Li}_1(z)=\sum_{k=1}^{\infty}\frac{z^k}{k}=-\ln(1-z)$ . Similarly, $\text{Li}_0(z)=\sum_{k=1}^{\infty}z^k=\frac{z}{1-z}$ .

Taking the derivative, we see:
$\begin{eqnarray}\frac{d}{dz}\text{Li}_s(z)&=&\frac{d}{dz}\sum_{k=1}^{\infty}\frac{z^k}{k^s}\\&=&\sum_{k=1}^{\infty}\frac{kz^{k-1}}{k^s}\\&=&\frac{1}{z}\sum_{k=1}^{\infty}\frac{z^{k-1}}{k^{s-1}}\\\frac{d}{dz}\text{Li}_s(z)&=&\frac{1}{z}\text{Li}_{s-1}(z)\end{eqnarray}$ .
Combining this with the above, we see when the parameter s is a negative integer, the polylogarithm is a rational function:
$\text{Li}_{-1}(z)=z\frac{d}{dz}\text{Li}_0(z)=z\frac{d}{dz}\frac{z}{1-z}=\frac{z}{(1-z)^2}$
$\text{Li}_{-2}(z)=z\frac{d}{dz}\text{Li}_1(z)=\frac{z(1+z)}{(1-z)^3}$ ,
and so on. Similarly,
$\text{Li}_2(z)=\int_0^z\frac{\text{Li}_1(u)}{u}\,du=-\int_0^z\frac{ln(1-u)}{u}\,du$
$\text{Li}_3(z)=\int_0^z\frac{\text{Li}_2(u)}{u}\,du$ ,
and so on.

Examining the series, we see right away that $\text{Li}_s(0)=0$ and that $\text{Li}_s'(0)=1$ . Further, for s>1
$\text{Li}_{s}(1)=\sum_{k=1}^{\infty}\frac{1}{k^s}=\zeta(s)$
$\text{Li}_{s}(-1)=\sum_{k=1}^{\infty}\frac{(-1)^k}{k^s}=-\eta(s)$ ,
where $\zeta(s)$ is the Riemann zeta function and $\eta(s)$ is the Dirichlet eta function.
Combining these with the derivative relation,
$\text{Li}_{s}'(1)=\frac{1}{1}\text{Li}_{s-1}(1)=\zeta(s-1)$
$\text{Li}_{s}'(-1)=\frac{1}{-1}\text{Li}_{s-1}(-1)=\eta(s-1)$ .

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 $\operatorname{\Gamma}(s)\text{Li}_s(-e^a)=-\int_0^{\infty}\frac{x^{s-1}}{e^{x-a}+1}\,dx$
and
$\operatorname{\Gamma}(s)\text{Li}_s(e^a)=\int_0^{\infty}\frac{x^{s-1}}{e^{x-a}-1}\,dx$ .
[More generally, we have a pair of integral definitions:
$\begin{eqnarray}\text{Li}_s(z)&=&\frac{1}{\operatorname{\Gamma}(s)}\int_0^{\infty}\frac{x^{s-1}}{\frac{e^x}{z}-1}\,dx\\\\\text{Li}_s(-z)&=&-\frac{1}{\operatorname{\Gamma}(s)}\int_0^{\infty}\frac{x^{s-1}}{\frac{e^x}{z}+1}\,dx\end{eqnarray}$ .]

Using all the above, we see that our functions $\text{Li}_{3/2}(\xi)$ and $\text{Li}_{5/2}(\xi)$ used in the Bose gas analysis have the following properties:

$\text{Li}_{3/2}(\xi)\approx\xi$ and $\text{Li}_{5/2}(\xi)\approx\xi$ for small ξ.
$\text{Li}_{3/2}(1)=\zeta\left(\frac{3}{2}\right)\approx{2.612375}$ and $\text{Li}_{5/2}(1)=\zeta\left(\frac{5}{2}\right)\approx{1.341487}$ .
$\text{Li}'_{5/2}(1)=\text{Li}_{3/2}(1)=\zeta\left(\frac{3}{2}\right)\approx{2.612375}$ and $\text{Li}'_{3/2}(1)=\text{Li}_{1/2}(1)$ , which diverges (infinite slope).
$\text{Li}_{3/2}(\xi)\ge\text{Li}_{5/2}(\xi)$ , with equality only at ξ=0
Expanding to first order in ξ, $\frac{\text{Li}_{5/2}(\xi)}{\text{Li}_{3/2}(\xi)}\approx1-\frac{\xi}{2^{5/2}}$ .
$\text{Li}_{3/2}(\xi)$ and $\text{Li}_{5/2}(\xi)$ are both monotonically increasing functions for |ξ|≤1

Tags: Dirichlet Eta Function, Math, Mercator Series, Polylogarithm, Riemann Zeta Function
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Physics Friday 92

October 9, 2009 by twistedone151

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), $f_{\mathbf{k},m_s}=\frac{1}{e^{\beta\left(\epsilon_{\mathbf{k}}-\mu\right)}-1}$ , differs from that in the Fermi gas (Fermi-Dirac distribution), $f_{\mathbf{k},m_s}=\frac{1}{e^{\beta\left(\epsilon_{\mathbf{k}}-\mu\right)}+1}$ , 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 $\xi\equiv{e}^{\beta\mu}=e^{\frac{\mu}{kT}}$ , we had
$U= \frac{g_0m^{3/2}V}{2^{1/2}\pi^2\hbar^3}\int_0^{\infty}\frac{\epsilon^{3/2}}{e^{\beta(\epsilon-\mu)}-1}\,d\epsilon=\frac{3g_0m^{3/2}V}{2^{5/2}\pi^{3/2}\hbar^3\beta^{5/2}}\text{Li}_{5/2}(\xi)$ ,
$\tilde{N}_e=\frac{g_0m^{3/2}V}{2^{1/2}\pi^2\hbar^3}\int_0^{\infty}\frac{\epsilon^{1/2}}{e^{\beta(\epsilon-\mu)}-1}\,d\epsilon=\frac{g_0m^{3/2}V}{2^{3/2}\pi^{3/2}\hbar^3\beta^{3/2}}\text{Li}_{3/2}(\xi)$ ,
$\Psi=-\frac{2^{1/2}g_0m^{3/2}V}{3\pi^2\hbar^3}\int_0^{\infty}\frac{\epsilon^{3/2}}{e^{\beta(\epsilon-\mu)}-1}\,d\epsilon=-\frac{2}{3}U$
(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 $f_{\mathbf{k},m_s}=\frac{1}{e^{\beta\left(\epsilon_{\mathbf{k}}-\mu\right)}-1}$ at $\mu=\epsilon_{\mathbf{k}}$ , 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 $\Lambda_T=\sqrt{\frac{2\pi\beta\hbar^2}{m}}$ , the above expressions for energy and particle number simplify to:
$U=\frac{3g_0V}{2\beta\Lambda_T^3}\text{Li}_{5/2}(\xi)=\frac{3}{2}kT\frac{g_0V}{\Lambda_T^3}\text{Li}_{5/2}(\xi)$ .
and
$\tilde{N}_e=\frac{g_0V}{\Lambda_T^3}\text{Li}_{3/2}(\xi)$ .
[Note that when the fugacity is small, these are approximated to first order in fugacity by $U\approx\frac{3}{2}kT\frac{gV}{\Lambda_T^3}\xi$ and $\tilde{N}_e\approx\frac{gV}{\Lambda_T^3}\xi$ , which we found in the classical approximation here].
Now, dividing these to cancel the $\frac{g_0V}{\Lambda_T^3}$ terms, we get
$\begin{eqnarray}\frac{U}{\tilde{N}_e}&=&\frac{\frac{3}{2}kT\frac{g_0V}{\Lambda_T^3}\text{Li}_{5/2}(\xi)}{\frac{g_0V}{\Lambda_T^3}\text{Li}_{3/2}(\xi)}\\\\\frac{U}{\tilde{N}_e}&=&\frac{3}{2}kT\frac{\text{Li}_{5/2}(\xi)}{\text{Li}_{3/2}(\xi)}\\U&=&\frac{3}{2}\tilde{N}_ekT\frac{\text{Li}_{5/2}(\xi)}{\text{Li}_{3/2}(\xi)}\end{eqnarray}$ ,
which differs from the classical $U=\frac{3}{2}\tilde{N}_ekT$ by a factor of $\frac{\text{Li}_{5/2}(\xi)}{\text{Li}_{3/2}(\xi)}$ , 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
$\text{Li}_{3/2}(1)=\zeta\left(\frac{3}{2}\right)\approx{2.612375}$
and
$\text{Li}_{5/2}(1)=\zeta\left(\frac{5}{2}\right)\approx{1.341487}$ .

Now, we have a way to analyze a given Bose gas. Suppose we know Ñ_e, V, and T. Then
$\tilde{N}_e=\frac{g_0V}{\Lambda_T^3}\text{Li}_{3/2}(\xi)$ tells us that $\text{Li}_{3/2}(\xi)=\frac{\Lambda_T^3\tilde{N}_e}{g_0V}$ , 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 $n=\frac{\tilde{N}_e}{V}$ : $\text{Li}_{3/2}(\xi)=\frac{\Lambda_T^3\tilde{N}_e}{g_0V}=\frac{\Lambda_T^3}{g_0}n$ .

However, we should note that $\text{Li}_{3/2}(\xi)=\sum_{k=1}^{\infty}\frac{\xi^k}{k^{3/2}}$ diverges for ξ>1, so the maximum value of the polylogarithm in question is $\text{Li}_{3/2}(1)=\zeta\left(\frac{3}{2}\right)\approx{2.612375}$ . However, one can be given values of Ñ_e, V, and T with the temperature low enough or density high enough that $\frac{\Lambda_T^3\tilde{N}_e}{g_0V}\gt\zeta\left(\frac{3}{2}\right)$ . 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” $D(\epsilon)\propto\sqrt{\epsilon}$ , 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 $\frac{\Lambda_T^3\tilde{N}_e}{g_0V}=\zeta\left(\frac{3}{2}\right)$ , and seek the temperature:
$\begin{eqnarray}\frac{\Lambda_T^3\tilde{N}_e}{g_0V}&=&\zeta\left(\frac{3}{2}\right)\\\\\Lambda_T^3&=&\frac{g_0V}{\tilde{N}_e}\zeta\left(\frac{3}{2}\right)\\\\\Lambda_T&=&\left[\frac{g_0V}{\tilde{N}_e}\zeta\left(\frac{3}{2}\right)\right]^{1/3}\\\\\sqrt{\frac{2\pi\hbar^2}{mkT}}&=&\left[\frac{g_0V}{\tilde{N}_e}\zeta\left(\frac{3}{2}\right)\right]^{1/3}\\kT_c&=&\frac{2\pi\hbar^2}{m}\left[\frac{\tilde{N}_e}{\zeta\left(\frac{3}{2}\right)g_0V}\right]^{2/3}\end{eqnarray}$ ,
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.