(10.4) Evaporative cooling A cloud of atoms has a Boltzmann energy distri- bution N(E)= Ae-E, where 1/3 = kBT and the normalisation constant A is found from Af -BE dE = The cloud has a total energy given by Etotal = A Ee-BE DE AS Ntotal = A 3 = NtotalkBT. Hence each atom has a mean energy E = kBT. In an evaporative cooling step all atoms with energy greater than e escape. (a) Calculate the fraction of atoms lost AN/Ntotal. (b) Calculate the fractional change in the mean energy per atom. (c) Evaluate your expressions for cuts with 3 = 3 and 6. Compare the ratio of energy lost and the number of atoms removed in the two cases and comment on the implications for evaporative cooling. (d) The collision rate between atoms in the cloud is Rcoll= nuo. Assuming that the collision cross-section is independent of the energy, show that Reoll x Ntotal/Etotal in a harmonic trapping potential. Show that the collision rate increases during evaporation in such a potential. 10.4) Initial Boltzmann distribution: N(E)= Ae-E Normalization condition: No = A fedE = A/B Total energy: E = Af Ee-dE = A/ = NokT Mean energy: E=1/3=kT For a distribution truncated at energy e N = A fedE = No (1-e) where 3= kBT E = A EedE = (1-e-c-ce-) Therefore the mean energy after truncation is: E =kBT x 3 6 Fractional changes (in terms of r= c/kBT): (a) = , (b) and (c) exp(-x) = | 16% e-6-0.25% 1.5% Bee-le 1-e-Be 6.0 Cutting less deeply is more efficient, i.e. it de- creases the temperature more for a given loss I atoms. In this example 20 small cuts with Be= 6 would give about the same loss of atoms sa single cut with e= 3, but the many small cuts reduce the energy by twice as much as the single large cut. (d) AE/F ANIN Density n~ N/4r and M = TE therefore n x N/E/2, Speed x E/2, So the collision rate Reall na x AR = AR R N EREN Write this as R=AN/E where A = constant. A (AN-EN) -45 AN AE N E N = E The collision rate increases, e.g. using the val ues from part (c), a cut with z=3 gives =-5% - (- 16 %) = +11%