Spontaneous symmetry breaking in superfluid helium-4

We derive an analytical expression for a critical temperature of spontaneous symmetry breaking in a repulsive hard-core interacting Bose system. We show that the critical temperature of spontaneous symmetry breaking in a hard-core interacting Bose system is determined by the three physical parameters: the density of Bose liquid at absolute zero (p(0)), the mass (in) and the hard sphere diameter (a) of a boson. The formula that we have derived is T, = po pi h(2)sigma/m(2)k(B). We report T, of liquid helium-4 is 2.194 K, which is significantly close to the lambda-temperature of 2.1768 K. The deviation between the predicted and experimental values of the X-temperature is less than 1%. (c) 2007 Elsevier B.V. All rights reserved.