Question

Consider a two-dimensional (2D) Bose gas at finite but low T confined in a square box potential with side lengths L and area A = L^2.

2. Consider a two-dimensional (2D) Bose gas at finite but low T confined in a square box potential with side lengths L and area A = L2. Using the density of states function as you found above, derive an expression for the 2D phase space density and argue why Bose-Einstein condensation does not occur in 2D at finite temperature. You will find the integral below useful. stats function as you found above, derive an

Answer 1

Therefore the density of state function has been found So the density of state function for 2D: - g (E) = L^2 m/ 2 pi hstrok^2 therefore Hence the number of particles is then equal to: - N = integral^infinity_0 g (E) 1/ e^beta E - 1 dE here E is the energy of the system and beta = 1/KT So N = L^2 M/ 2 pi planck^2 integral^infinity_0 1/e^beta_E - 1 dE let BE = x implies dE = dx/beta So N = L^2 M/2 pi hstrok^2 integral^infinity_0 1/e^x - 1 (dx/ beta) N = L^2 M KT/ 2 pi hstrok integral^infinity_0 1/(e^x - 1) dx therefore from here: - T = 2 pi hstrok^2 N/L^2 M K {integral^infinity_0 1/(e^x - 1) dx}^-1 therefore integral^infinity_0 (1/e^x/z - 1) dx = -ln (1 - z) \r\n therefore The denominator or the term in the bracket diverges So the T rightarrow 0 for the Base Einstein condensation.