Question

What is Bose-Einstein condensation? Show that Bose-Einstein condensation temperature is given by Tc=h^2/2πmkB x [N / 2.612V]^2/3

Answer 1

Answer on Question #50642-Physics-Molecular Physics-Thermodynamics

What is Bose-Einstein condensation? Show that Bose-Einstein condensation temperature is given by

$$T _ {c} = \frac {h ^ {2}}{2 \pi m k _ {B}} \cdot \left[ \frac {N}{2 . 6 1 2 V} \right] ^ {\frac {2}{3}}$$

Answer

A Bose-Einstein condensate is a rare state (or phase) of matter in which a large percentage of bosons collapse into their lowest quantum state, allowing quantum effects to be observed on a macroscopic scale. The bosons collapse into this state in circumstances of extremely low temperature, near the value of absolute zero.

The BEC critical temperature for a uniform 3D system is given by the condition that all the particles accommodated in excited single particles state (except for a ground state) when $\mu = u_0 = 0$ are equal to the total number of particles in the system:

$$N = \int_ {0} ^ {\infty} g (u) \overline {{n _ {u}}} d u.$$

where $g(u)du$ is the density of states in terms of the energy density:

$$g (u) d u = \frac {V}{4 \pi^ {2}} \left(\frac {2 m}{h ^ {2}}\right) ^ {\frac {3}{2}} \sqrt {u} d u.$$

Thus

$$N = \frac {V}{4 \pi^ {2}} \left(\frac {2 m}{h ^ {2}}\right) ^ {\frac {3}{2}} \int_ {0} ^ {\infty} \sqrt {u} \frac {d u}{e ^ {\frac {u}{k _ {B} T}} - 1}.$$

We can evaluate the energy integral using the relation $\frac{1}{e^x - 1} = \sum_{n=1}^{\infty} e^{-nx}$, where $x = \frac{u}{k_B T}$:

$$\int_ {0} ^ {\infty} \sqrt {u} \frac {d u}{e ^ {\frac {u}{k _ {B} T}} - 1} = (k _ {B} T) ^ {\frac {3}{2}} \int_ {0} ^ {\infty} \sum_ {n = 1} ^ {\infty} e ^ {- n x} \sqrt {x} d x = (k _ {B} T) ^ {\frac {3}{2}} \sum_ {n = 1} ^ {\infty} n ^ {- \frac {3}{2}} \int_ {0} ^ {\infty} e ^ {- t} \sqrt {t} d t = (k _ {B} T) ^ {\frac {3}{2}} \zeta \left(\frac {3}{2}\right) \Gamma \left(\frac {3}{2}\right)$$

where $\zeta\left(\frac{3}{2}\right) = \sum_{n=1}^{\infty} n^{-\frac{3}{2}}$ and $\Gamma\left(\frac{3}{2}\right) = \int_{0}^{\infty} e^{-t} \sqrt{t} \, dt$.

Finally, we obtain the BEC critical density for a uniform 3D system:

$$n _ {c} = \frac {N _ {c}}{V} = \frac {1}{4 \pi^ {2}} \left(\frac {2 m k _ {B} T}{h ^ {2}}\right) ^ {\frac {3}{2}} \zeta \left(\frac {3}{2}\right) \Gamma \left(\frac {3}{2}\right).$$

We know that

$$\zeta \left(\frac {3}{2}\right) \cong 2. 6 1 2; \Gamma \left(\frac {3}{2}\right) = \frac {\sqrt {\pi}}{2}.$$

For a given density $\frac{N}{V}$, the previous relation, written in terms of the temperature, defines the critical temperature $T_{c}$:

$$T _ {c} = \frac {h ^ {2}}{2 \pi m k _ {B}} \cdot \left[ \frac {N}{2 . 6 1 2 V} \right] ^ {\frac {2}{3}}.$$

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