LOG#167. D-dimensional laws(II).

Posted on by August 27, 2015

be BECtraps boseBEC PhaseTrans

Let me begin this article in D=d+1 spacetime. We are going to study quantum gases and their statistics in multidimensional space. Usual notation:

    \[\sum_{\vec{k}}\rightarrow V_d\int\dfrac{d^dk}{(2\pi)^d}=V_d\int\dfrac{d^dk}{(2\pi \hbar)^d}\]

In D=d+1 spacetime, the massless free ideal relativistic gas satisfies, as we will show, certain relations between thermodinamical variables. For instance,

    \[P=\dfrac{\varepsilon}{d}=\dfrac{u}{d}\]

or

    \[P=\dfrac{s\varepsilon}{d}\]

if the dispersion relationship is \varepsilon=ap^s in d-dimensional SPACE. The Bose-Einstein integral reads

(1)   \begin{equation*} I_{BE}=\int_{0}^\infty\dfrac{x^n}{e^x-1}=\Gamma(n+1)\zeta(n+1) \end{equation*}

We define n=\dfrac{N}{V_d} and the energy density

(2)   \begin{equation*} \varepsilon=\rho=\dfrac{E}{N}=\int\dfrac{d^dk}{(2\pi)^d}\dfrac{E}{e^{\beta E}-1} \end{equation*}

Generally, we will work with natural units \hbar=k_B=1 (they will be reintroduced if necessary) and with zero chemical potentical \mu=0 and massless particles. We will explore the issue of massive particle statistics though.

Spherical coordinates (in d-dimensional euclidean space they have d-1 angles (\varphi,\theta_1, \theta_2,\ldots,\theta_{d-2})) introduce a (d-1)dimensional solid angle

    \[d\Omega_{d-1}=2\pi\prod_{n=1}^{d-2}\sin^n\theta_nd\theta_n\]

and where

    \[\Omega_{d-1}=\dfrac{2\pi^{d/2}}{\Gamma\left(\dfrac{d}{2}\right)}=\dfrac{2\left(\Gamma\left(\dfrac{1}{2}\right)\right)^d}{\Gamma\left(\dfrac{d}{2}\right)}\]

 and

    \[\int_0^\pi d\theta\sin^n\theta=\sqrt{\pi}\dfrac{\Gamma\left(\dfrac{n+1}{2}\right)}{\Gamma\left(\dfrac{n+2}{2}\right)}\]

    \[\langle\cos^2\theta_{d-2}\rangle=1-\langle\sin^2\theta_{d-2}\rangle\]

for the solid angle \Omega_{d-1}. Moreover, a trivial calculation

    \[\langle \sin^2\theta_{d-2}\rangle=\dfrac{\Gamma\left(\dfrac{d}{2}\right)\Gamma\left(\dfrac{d+1}{2}\right)}{\Gamma\left(\dfrac{d-1}{2}\right)\Gamma\left(\dfrac{d+2}{2}\right)}=\dfrac{d-1}{d}=\dfrac{D-2}{D-1}\]

    \[\langle\cos^2\theta_{d-2}\rangle=1-\langle\sin^2\theta_{d-2}\rangle=1-\dfrac{d-1}{d}=\dfrac{1}{d}=\dfrac{1}{D-1}\]

For a dispersion relationship \varepsilon=ap in d-dimensional space, we have

    \[\boxed{\varepsilon=dP(T)=\dfrac{2\Gamma\left(d+1\right)\zeta(d+1)\left(k_BT\right)^{d+1}}{\Gamma\left(\dfrac{d}{2}\right)\left(4\pi\right)^{d/2}}=\Omega_{d-1}\dfrac{1}{(2\pi)^d}\Gamma\left(D\right)\zeta\left(D\right)\left(k_BT\right)^D}\]

and where

    \[\Omega_{d-1}=\dfrac{2\pi^{d/2}}{\Gamma\left(\dfrac{d}{2}\right)}\]

The pressure P(T) gives us the Stefan-Boltzmann law in higher dimensions

(3)   \begin{equation*} \boxed{P(T)=\dfrac{\varepsilon}{d}=\dfrac{2}{d}\dfrac{\Gamma\left(d+1\right)\zeta\left(d+1\right)\left(k_BT\right)^{d+1}}{\Gamma\left(\dfrac{d}{2}\right)\left(4\pi\right)^{d/2}}} \end{equation*}

since

    \[\dfrac{\Gamma\left(d+1\right)}{\dfrac{d}{2}\Gamma\left(\dfrac{d}{2}\right)\left(4\pi\right)^{d/2}}=\dfrac{\Gamma\left(\dfrac{D}{2}\right)}{\pi^{D/2}}=\dfrac{\Gamma\left(\dfrac{d+1}{2}\right)}{\pi^{(d+1)/2}}\]

We are ready to study the phenomenon of Bose-Einstein condensation (BEC) for ideal massive (relativistic and non-relativistic) bosonic gases. Take the dispersion relation to be now

    \[E_k=\sqrt{c^2\hbar^2k^2+m^2c^4}=\begin{cases}mc^2+\dfrac{\hbar^2k^2}{2m}+\mathcal{O}(k^4),NR(c\hbar k<<mc^2)\\ \hbar ck\left[1+\dfrac{1}{2}\left(\dfrac{mc}{\hbar k}\right)^2+\mathcal{O}(k^{-4})\right],UR(\hbar ck>>mc^2)\end{cases}\]

and where NR denotes non-relativistic, UR denotes ultrarelativistic (massless relativistic or almost massless relativistic). Now, but the ideal bosonic gas in a box of size L, with

    \[\sum_{k}\rightarrow\left(\dfrac{L}{2\pi}\right)\int d^dk=V_d\int d^dk\]

The critical temperature of the BEC, T=T_c is approache when \mu(T_c)=mc^2 and \beta_c=1/k_BT_c. The number density will be then

(4)   \begin{equation*} n=\dfrac{N}{L^d}=\dfrac{N}{V_d}=\dfrac{1}{(2\pi)^d}\int d^dk\dfrac{1}{e^{\beta_c(\vert E_k\vert-mc^2)}-1} \end{equation*}

Define

    \[d^dk=\dfrac{2\pi^{d/2}}{\Gamma\left(\dfrac{d}{2}\right)}k^{d-1}dk\]

    \[z_c=e^{\mu/k_BT_c}\]

where with z\leq 1 the Bose integral diverges. Then, the non-relativistic bosonic BEC temperature reads

(5)   \begin{equation*} \boxed{k_BT_c(NR,B)\equiv \dfrac{2\pi \hbar^2}{m}\left[\dfrac{n}{\zeta\left(\dfrac{d}{2}\right)}\right]^{2/d}} \end{equation*}

In 3d we get the known result

    \[k_BT_c(3d,NR,B)\simeq 3\mbox{.} 31\hbar^2\dfrac{n^{2/3}}{m}\]

where we have used \zeta(3/2)\simeq 2\mbox{.} 612.

In the case of the ultrarelativistic (massless, almost massless) case, we obtain

(6)   \begin{equation*} \boxed{k_BT_c(UR,B)=\left[\dfrac{\hbar^dc^d2^{d-1}\pi^{d/2}\Gamma\left(\dfrac{d}{2}\right)}{\Gamma(d)\zeta(d)}\right]^{1/d}n^{1/d}} \end{equation*}

In the 3d case, we get the very well result

    \[k_BT_c(3d,UR,B)=\hbar c\pi^{2/3}\left(\dfrac{n}{\zeta (3)}\right)^{1/3}\simeq 2\mbox{.}017\hbar c n^{1/3}\]

and where \zeta(3)=1\mbox{.}20206 is the zeta value of 3.

Remark:

The 2d UR case HAS a critical temperature (unlike the 2d non-relativistic case, where BEC does NOT exist) T_c\neq 0. Indeed, you can easily check that

    \[k_BT_c(2d,UR,B)=\hbar c\left(\dfrac{2\pi n}{\zeta (2)}\right)^{1/2}\simeq 1\mbox{.}954\hbar c n^{1/2}\]

where \zeta (2)=\pi^2/6.

Remark (II):

BEC does depend not only on the number of dimensions but also on the density of states, i.e., it is highly dependent on the dispersion relationship \varepsilon=\varepsilon(p) we use!

An additional important issue is the following. If we allow k_BT>>mc^2, then pairs boson-antiboson B\overline{B} can be created. In particular, we have that

    \[E_k^2=\hbar^2c^2k^2+m^2c^4\]

for each particle. Moreover, you have

    \[E_k=\pm E_p\]

with + for bosons and – for antibosons, so

    \[(N_k-\overline{N}_k)V_d=\sum_kn_k-\overline{n}_k\]

and then

    \[N_k-\overline{N}_k=\sum_k\left(\dfrac{1}{e^{\beta(E_k-\mu)}-1}-\dfrac{1}{e^{\beta(E_k+\mu)}-1}\right)\]

with -mc^2<\mu<mc^2. Thus, we get

    \[n=\dfrac{N-\overline{N}}{V_d}=\dfrac{N-\overline{N}}{L_d}=\dfrac{2\pi^{d/2}}{\Gamma\left(\dfrac{d}{2}\right)\left(2\pi\right)^d}\int_0^\infty dk \dfrac{k^{d-1}\sinh(\beta_cmc^2)}{\cosh\left(\beta_c\sqrt{\hbar^2c^2k^2-m^2c^4}\right)}-\cosh\left(\beta_cmc^2\right)\]

Case 1. Low T, with k_BT<<mc^2. Then T_C(B\overline{B},NR)=T_c(NR,B), as we would expect.

Case 2. High T, with k_BT>>mc^2. Then, the critical temperature DOES change to take into account the boson-antiboson pair creation. It yields

(7)   \begin{equation*} \boxed{k_BT_c(UR,B\overline{B})=\left[\dfrac{\hbar^dc^{d-2}\Gamma\left(\dfrac{d}{2}\right)(2\pi)^d}{4m\pi^{d/2}\Gamma(d)\zeta(d-1)}\right]^{1/(d-1)}n^{1/(d-1)}} \end{equation*}

Note that the UR boson-antiboson massive case is not equal to the UR boson (massless) case, even in 3d! Indeed, you find that

    \[k_BT(UR,B\overline{B},3d,m)=\left(\dfrac{3\hbar^3c}{m}\right)^{1/2}n^{1/2}\]

In fact, with no antiboson (massless), the case provides the following thermodynamical quantities:

    \[n=n_0+\dfrac{\Omega_d}{(2\pi \hbar)^d}\int_{0^+}^\infty \dfrac{1}{e^{\beta(E_p-\mu)}-1}p^{d-1}dp\]

    \[n_0\equiv\dfrac{1}{V\left[e^{\beta(mc^2-\mu)}-1\right]}=\dfrac{N_0}{L^d}\]

    \[\dfrac{U^B}{V}=u_B=nmc^2+\dfrac{\Omega_d}{(2\pi \hbar)^d}\int_{0^+}^\infty\dfrac{E_p-mc^2}{e^{\beta(E_p-\mu)}-1}p^{d-1}dp\]

    \[\dfrac{F(T,V,n)}{V}=\mathcal{F}=nmc^2+k_BT\dfrac{\Omega_d}{(2\pi \hbar)^d}\int_{0^+}^\infty \ln\left[1- e^{\beta(E_p-\mu)}\right]p^{d-1}dp\]

    \[\dfrac{S^B}{V}=s_B=\dfrac{k_B\Omega_d}{(2\pi \hbar)^d}\int_{0^+}^\infty p^{d-1}dp\dfrac{(E_p-mc^2)\beta}{e^{\beta(E_p-\mu)}-1}-\ln\left(1-e^{\beta(\mu-E_p)}\right)\]

When antibosons are present, these integrals become nastier and more complicated:

    \[n=n_0+\dfrac{\Omega_d}{(2\pi \hbar)^d}\int_{0^+}^\infty dpp^{d-1} \dfrac{1}{e^{\beta(E_p-\mu)}-1}+\dfrac{1}{e^{\beta(E_p+\mu)}-1}\]

    \[\dfrac{U^{B\overline{B}}}{V}=u_{B\overline{B}}=nmc^2+\dfrac{\Omega_d}{(2\pi \hbar)^d}\int_{0^+}^\infty dpp^{d-1} \dfrac{E_p-mc^2}{e^{\beta(E_p-\mu)}-1}+\dfrac{E_p+mc^2}{e^{\beta(E_p+\mu)}-1}\]

    \[\dfrac{F(T,V,n)_{B\overline{B}}}{V}=\mathcal{F}=nmc^2+\dfrac{k_BT\Omega_d}{(2\pi \hbar)^d}\int_{0^+}^\infty dpp^{d-1}\{ \ln\left[1- e^{\beta(mc^2-E)}\right]+\ln\left[1- e^{-\beta(mc^2+E)}\right]\}\]

    \[\begin{split}\dfrac{S^{B\overline{B}}}{V}=s_{B\overline{B}}=\dfrac{k_B\Omega_d}{(2\pi \hbar)^d}\int_{0^+}^\infty dpp^{d-1}\{\dfrac{(E_p-mc^2)\beta}{e^{\beta(E_p-mc^2)}-1}+\dfrac{(E_p+mc^2)\beta}{e^{\beta(E_p+mc^2)}-1}+\\-\ln\left(1-e^{\beta(mc^2-E_p)}\right)-\ln\left(1-e^{-\beta(mc^2+E_p)}\right)\}\end{split}\]

Challenge for eager readers:

Take the density of states

    \[\rho(\varepsilon)=\dfrac{2\pi^{d/2}}{(2\pi)^{d/2}\Gamma\left(\dfrac{d}{2}\right)}\varepsilon\left(\varepsilon^2-m^2\right)^{(d-2)/2}\]

and take the UR limit to get and prove

    \[\rho(\varepsilon)=\dfrac{2\pi^{d/2}\varepsilon^{d-1}}{(2\pi)^d\Gamma\left(\dfrac{d}{2}\right)}\]

and calculate/check the critical temperature

    \[T_c=\left[\dfrac{(2\pi)^d\Gamma\left(\dfrac{d}{2}\right)q}{4\pi^{d/2}\Gamma(d)\zeta(d-1)m}\right]^{1/(d-1)}\]

obtaining the value of the “constant” q above.

By the other hand, we can also study the fermionic gas in D-dimensional spacetime, d-dimensional space. In order to simplify the discussion, we are going to study only the non-relativistic (NR) ideal gas. I will study the relativistic Fermi gas in a future post because it is important in extremely degenerate systems, as some particular kind of stars. The ideal non-relativistic Fermi gas in d-dimensional space has the following interesting features:

    \[\varepsilon(ideal)=\dfrac{U}{V_d}=\dfrac{2d\varepsilon_f k_f^d}{2^d\pi^{d/2}(d+2)\Gamma\left(\dfrac{d}{2}+1\right)}\]

Moreover, you have

    \[P(ideal)=\dfrac{4\varepsilon_f k_f^d}{2^d\pi^{d/2}(d+2)\Gamma\left(\dfrac{d}{2}+1\right)}\]

    \[n=\dfrac{2k_f^d}{2^d\pi^{d/2}\Gamma\left(\dfrac{d}{2}+1\right)}\]

and the Fermi energy reads (\hbar=c=1):

    \[E_f=\sqrt{k_f^2+m^2}=\sqrt{p_f^2+m^2}\]

The dimensional Fermi weights are, for D=d+1 dimensional spacetime and the NR and UR case

    \[f_{sw}(NR)=\dfrac{d+2}{d}=\dfrac{D+1}{D-1}\]

    \[f_{sw}(UR)=\dfrac{d+1}{d}=\dfrac{D}{D-1}\]

The non-relativistic (NR) fermionic (or fermi) gas has a dispersion relationship

    \[E=\dfrac{p^2_d}{2m}=\dfrac{p_1^2+\cdots+p_d^2}{2m}\]

and the energy in terms of Fermi quantities reads

    \[\boxed{\varepsilon=-\dfrac{2}{2^d\pi^{d/2}}\dfrac{\varepsilon_fk_f^d}{\Gamma\left(\dfrac{d}{2}+1\right)d}}\]

and the density of states

    \[g(\varepsilon)=\dfrac{g_sV2\pi^{d/2}p^{d-1}_r(\varepsilon)}{(2\pi\hbar)^d}\left(\dfrac{\partial \varepsilon}{\partial p_r}\right)^{-1}\]

Thus, the energy density will be

    \[\varepsilon=\dfrac{E}{V_d}=\dfrac{g_s}{\lambda_T^2}\dfrac{\beta^{d/2}E^{d/2}-1}{\Gamma\left(\dfrac{d}{2}\right)}\]

with the thermal wavelength

    \[\lambda_T=\dfrac{2\pi \hbar}{\sqrt{2m\pi \beta^{-1}}}\]

Furthermore

    \[N=\dfrac{g_sV}{\lambda_T^d}f_{d/2}(z)\]

where z=\exp(\mu \beta) is the fugacity. The Fermi function reads

    \[f_{d/2}(z)=\sum_{n=1}^\infty(-1)^{n+1}\dfrac{z^n}{k^{d/2}}=-Li_{d/2}(-z)\]

and it uses the polylogarithm as well!!! Wonderful, isn’t it? The average energy per fermion in d-space is

    \[\langle E\rangle_d=\dfrac{\langle E\rangle_d}{fermion}=\dfrac{\int_0^{E_F}g_d(E)F(E)dE}{\int_0^{E_f}g_d(E)F(E)dE}=\dfrac{d}{d+2}E_F\]

You get \langle E\rangle=E_F/2 if d=2, \langle E\rangle=3E_F/5 if d=3 and you also have

    \[\lim_{n\rightarrow\infty}\langle E\rangle_d=E_F\]

with

    \[g_d(E)dE=C(m,V)E^{(d-2)/2}dE\]

    \[E=\dfrac{p_1^2+\cdots+p_d^2}{2m}\]

    \[d^dp=\left(\dfrac{d}{2}\right)\dfrac{\pi^{d/2}(2m)^{d/2}\varepsilon^{(d-2)/2}d\varepsilon}{\Gamma\left(\dfrac{d}{2}+1\right)}\]

and the number density

    \[n=\dfrac{2\pi^{d/2}(2m)^{d/2}\varepsilon_f^{d/2}}{h^d\Gamma\left(\dfrac{d}{2}+1\right)}\]

The massless spinless bosonic particles in D=d+1 dimensions have a free energy

    \[\boxed{\mathcal{F}=-\dfrac{1}{\beta^{d+1}}\dfrac{\Gamma\left(d+1\right)\zeta\left(d+1\right)}{2^{d-1}\pi^{d/2}\Gamma\left(\dfrac{d}{2}\right)d}}\]

The Casimir energy of such a bosonic field requires

    \[\left(\sum_{n\in Z}\vert n\vert^d\right)_{reg}=2\zeta(-d)\]

and the regularized energy in vacuum has to be

    \[\varepsilon_0(reg)=-\dfrac{1}{\beta^{d+1}}\pi^{d/2}\Gamma\left(-\dfrac{d}{2}\right)\zeta\left(-d\right)\]

and it shows that the Riemann zeta functional equation \xi(\nu)=\xi(1-\nu) holds iff

    \[\varepsilon_0=\mathcal{F}\]

This striking consequence and relationship between the vacuum structure and pure mathematics is fascinating and not yet completely understood. But this will be a topic for a future discussion here.

See you in my next blog post!

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