Part 3 Additional Topics

1. Part 3Additional Topics PHYS 4315 R. S. Rubins, Spring 2011

2. Kinetic Theory PHYS 4315 R. S. Rubins, Fall 2009

3. Kinetic Theory: Introduction • Features of kinetic theory • Kinetic theory goes beyond the limitations of classical thermodynamics by taking into account the structures of materials. However, there is a price to be paid in complexity, since general relationships may be obscured. • Unlike both classical thermo and statistical mechanics, kinetic theory may be used to describe non-equilibrium situations. • Kinetic theory has been particularly useful in describing the properties of dilute gases. • It gives a deeper insight into concepts such as pressure, internal energy and specific heat, and explains transport processes, such as viscosity, heat conduction and diffusion.

4. Classical Dilute Gas Assumptions 1 • Molecules are classical particles with well-defined positions and momenta. • A macroscopic volume contains an enormous number of molecules. • At STP (0oC and 1 atm), 1 kmole of a gas occupies 22.4 m3, which is a density of 3 x 1025 molecules/m3. • Molecular separations are much larger than both their dimensions and the range of intermolecular forces. • At STP, the molecular separation is roughly 3 x 10– 9 m. • The Lennard-Jones or 6-12 potential V = K [(d/r)12 – (d/r)6], where K and d are empirical constants is negligible for molecular separations of 3 x 10– 9 m.

5. Classical Dilute Gas Assumptions 2 • The mean free path – the average distance a molecule moves between collisions is of the order 10– 7 m. • The only interaction between particles occur during collisions so brief that their durations may be neglected. • The container is assumed to have an idealized surface. • The collisions are elastic, so that momentum and KE are conserved. • The molecules are assumed to be distributed uniformly in both position and velocity direction. • Molecular chaos exists; that is, the velocity of a molecule is uncorrelated with its position.

6. Kinetic Theory: Pressure Macroscopic pressure • Macroscopic pressure (classical thermo) Balancing forceF = PA. • Microscopic pressure (kinetic theory) ImpulseFΔt = Δptot, where, for a single particle, Δp= 2mv cosθ. The piston makes an irregular (Brownian) motion about its equilibrium position, because of the random collisions of molecules with the piston. θ θ Microscopic pressure

7. Molecular Flux A x vaveΔt Simplified calculation • Assume that the molecules move equally in the ±x, ±y, and ±z directions; i.e. n/6 move in the +x direction, where n is the number of molecules per unit volume. • In time Δt, a total of AnvaveΔt/6 molecules will strike the wall. • Thus, the molecular flux Φ, the number of molecules striking unit area of the wall in unit time, is Φ = nvave/6. • An exact calculations, including integrations over all directions and speeds givesΦ = nvave/4.

8. Molecular Effusion • Φ = nvave/4, where (from stat. mech.) vave = √(8kT/mπ). • For an ideal gas, PV = NkT, so that n = N/V = P/kT. • Thus,Φ = P/√(2πmkT). • Imagine two containers containing the same ideal gas at (P1,T1) and (P2,T2), connected by a microscopic hole of diameter D << L, where L is the mean free path, which is of the order 10–7 m at STP. • The number of molecules passing through the hole is so small that the pressure and the temperature on each side of the hole is unchanged over the time of the experiment. • The condition for equilibrium is the absence of a net flux; i.e. Φ1 = Φ2 orP1/√T1 = P2/√T2.

9. Experimental Problem Macroscopic hole (D >> L): At equilibrium, P1 = P2, T1 = T2. Microscopic hole (D << L): At equilibrium, P1/√T1 = P2/√T2. • Under effusion conditions, P1/√T1 = P2/√T2 , so that P1 = P2/ √(T1/T2), or P1 ≈ 0.07 P2. T2= 300 K vacuum P2 ΔP Hg Liquid He He vapor at pressure P1 Liquid He at T1= 1.3 K

10. Boltzmann’s Transport Equation • Boltzmann’s H theorem dH/dt ≤ 0, where H(t) = ∫d3v f(p,t) log f(p,t) . • Since H → Hmin, the entropy has the form S = – const. H. • The equilibrium (Maxwell) distribution is given by

11. Superfluidity in Liquid Helium PHYS 4315 R. S. Rubins, Fall 2009

12. Boiling Points • The two helium isotopes have the lowest boiling points of all known substances, 3.2 K for 3He and 4.2 K for 4He. • Both isotopes would apparently remain liquid down to absolute zero; to solidify helium would require a pressure of about 25 atmospheres. • Two factors produce the reluctance of helium to condense: i. the low mass of the atoms; ii. The extremely weak forces between atoms. • The low atomic mass ensures a high zero-point energy, a result that may be deduced from the uncertainty principle.

13. Zero-point Energy 1 The uncertainty in momentum of a particle in a cavity is p  h/R. • It’s zero-point energy is thus E0  (p)2/2m or E0  h2/2mR2. • The large zero-point energy must be added to the potential energy of the liquid to give the liquid’s total energy.

14. Zero-point Energy 2 Because He atoms are so light, the zero-point Energy is comparable to the PE, the minimum of the total energy occurs at a relativity high atomic volume. For other inert gases, the zero-point energy is of negligible magnitude.

15. Phase Diagrams • The large zero-point energy of liquid eliminates the solid-vapor curve present for a normal material. • The λ line occurs only for 4He, and is associated with the λ-point transition to superfluid behavior near 2 K.

16. The λ Specific Heat Transition in Liquid 4He If liquid helium, which liquifies below 4.2 K, is cooled by lowering the pressure above it, bubbles of vapor form within the liquid, which boils vigorously. However, below 2.17 K, the λ point, the liquid becomes very still, as the transition from a Normal fluid (He I) to a superfluid (He II) occurs. In 3He, a transition to a superfluid occurs near 3 mK.

17. Macroscopic Properties of Superfluid He II 1 • Zero viscosity • Measurements showing the zero resistance to flow of He II were made in 1964. • This was done by showing that the flow velocity through channels of widths between 0.1 μm and 4 μm were independent of the pressure gradient along the channel.

18. Two-fluid Model of He II • Zero viscosity 2 • Experiments showed an apparent contradiction, that He II was both viscous and non-viscous at the same time. • This result was the source of the two-fluid model of He II, introduced by Tisza in 1938. • This is a quantum effect; the liquid does not consist of two distinct fluids, one normal and the other superconducting. • In Andronikashvili’s 1946 experiment, a series of equally spaced metal disks, suspended by a torsion fibre, were made to oscillate in liquid He. • The results confirmed that He II consists of a normal viscous fluid of density ρn and a superfluid of density ρs, and allowed the ratios ρs/ρ and ρn/ρ to be measured as functions of temperature, where ρ = ρn + ρs.

19. Andronikashvili’s Experiment

20. Macroscopic Properties of Superfluid He II 2 • Infinite thermal conductivity • This makes it impossible to establish a temperature gradient in a bulk liquid. • In a normal liquid, bubbles are formed when the local temperature in a small region in the body of the liquid is higher than the surface temperature. • Unusually thick adsorption film • The unusual flow properties of He II result in the covering of the exposed surface of a partially immersed object being covered with a film about 30 nm (or 100 atomic layers) thick, near the surface, and decreasing with height.

21. Flow of He II over Beaker Walls • The temperature is the same throughout the system, and the superfluid acts as a siphon, flowing through the film to equalize the levels in the two bulk liquids. • By observing the rate at which the beaker level changes, the superfluid velocity has been found to be about 20 cm/s.

22. Thermomechanical Effect 1 • If a temperature gradient is set up between two bulk volumes connected by a superleak, through which only the superfluid can flow, the superfluid flows to the higher temperature side, in order to reduce the temperature gradient. • This example of the thermomechanical effect, shows that heat transfer and mass transfer cannot be separated in He II.

23. At equilibrium, GA = GB; i.e. ΔG = 0. Now, dG = – S dT + V dP = 0  ΔP = (S/V) ΔT = (s/v) ΔT, where s and v are the values per kg. Now ρ = 1/v, so that, ΔP = s ρΔT. Thermomechanical Effect 2

24. The Fountain Effect • In this celebrated experiment of Allen and Jones (1938), the superleak is heated by a flashlight. • In order to equalize temperatures, the superfluid flows through the superleak with sufficient speed to produce a fountain rising 30 cm or more. • According to Landau’s theory (1941), the normal fluid consists of the excited quantum states. The fine channels in the superleak filter out the excited states

25. Bose-Einstein Condensation PHYS 4315 R. S. Rubins, Fall 2009

26. About BEC • In 1924, Einstein applied Satyendra Bose’s explanation of blackbody radiation to matter, predicting the phenomenon known as Bose-Einstein condensation (BEC). • BEC is a quantum mechanical phase-transition, thought to be responsible for superfluidity in liquid helium. • Not until 1995 was it observed in isolated atoms, in 87Rb (NIST), 23Na (MIT) and 7Li (Rice U.). Since then, BEC has been observed around the world, and 1H (MIT) and 4He France. • Samples typically contain of the order of 105 - 106 atoms, in which several thousand form the condensate, with transition temperatures in the range 300 – 600 nK.

27. BEC: Scientific Entanglements BEC belongs to atomic physics, condensed matter physics and stat. mech. It could not have been produced without the tools of optics and laser physics, the manipulation of magnetism and fluid dynamics, and the use of new techniques in low temperature physics. BEC is a deep entanglement of fields, giving rise to a totally new field of physics. See Physics Today, December 2006

28. Bosons and Fermions • Identical particles follow either Bose-Einstein or Fermi-Dirac statistics. • Bosons have integer angular momentum quantum numbers (e.g. photons, atoms with an even no. of neutrons.). • They have symmetrical wavefunctions; i.e.; if two particles (1 and 2) are in the states a and b, then Ψsym = ψa(1) ψb(2) + ψa(2) ψb(1)  ψa(1) ψa(2) if a = b. • Fermions have half-integer angular momentum quantum nos. (e.g. electrons, nucleons, atoms with an odd no. of neutrons.). • They have antisymmetrical wavefunctions; i.e.; if two particles (1 and 2) are in the states a and b, then Ψanti = ψa(1) ψb(2) – ψa(2) ψb(1)  0 if a = b.

29. Boson and Fermion Gases Below 1 mK In these Rice University images of atomic clouds, those of 7Li (a boson with 4 neutrons) continue to collapse as the temperature is lowered. Since identical fermions cannot occupy the same space (the Pauli exclusion principle), the atomic cloud of 6Li (a fermion with 3 neutrons) shows a smaller collapse.

30. BEC Photo from Rice University • Cloud of about 70,000 7Li atoms, with about 1200 in the BEC peak at the center, at about 600 nK.

31. BEC: a Phase Transition in an Ideal Gas • Like the ferromagnetic transition at the Curie point of iron (1043 K),BEC is a phase transition,but unlike the ferromagnetic transition, which occurs because of the strong interaction between iron atoms,BEC occurs in an ideal gas, for which interatomic forces are negligible.

32. BEC Atoms: Each in the Same Wave Function The de Broglie wavelengthλdB = h/mv, becomes for a quantum gas λdB =h/(2πmkT)1/2. Thus λdB increases as T is lowered, and a phase transition to a BEC state occurs when λdB reaches the atomic separation.

33. Interference Between BEC Waves • Like the interference patterns that may be produced by the coherent light from lasers,BEC waves show interference phenomena. • However, unlike laser beams, which are in non-equilibrium states,a BEC wave is an equilibrium state.

34. Loading a Magnet Trap for Li7 (Rice U.) • The apparatus is contained in an ultra-high vacuum at room temperature. • Hot Li7 atoms, emitted from an oven at 800 K, form an atomic beam. • The atomic beam is slowed by an oppositely directed laser beam, and deflected by a second laser beam towards a magnetic and optical trap. • Another laser beam collimates the deflected atomic beam, and optically pumps it, so that each atom is in the same magnetic state. • Once in the trap, the atomic beam is contained by a set of six laser beams.

35. Magnetic Trap (Rice U.) If the magnetic moment of an atom is parallel to the magnetic field, it will be attracted to a local minimum of the field, which occurs at the center of the magnet distribution. If the direction of the magnetic moment is reversed, the center of the distribution becomes a local maximum, which causes that atom to leave. The magnetic field at the minimum must not be zero, otherwise the atomic moments may spontaneously reverse their directions. In practice, the field at the minimum was 0.1 T. Atoms in the trap may be lost by collisions in which the moment direction is reversed.

36. Laser Cooling 1 • Laser cooling is achieved by using the Doppler effect to reduce vrms. • Two opposing laser beams of equal intensity are each tuned to the low frequency side of an optical transition. • The beam opposing the atom’s motion is blue-shifted to higher frequencies, so that the force on it is increased. • The beam in the same direction as the atom’s motion is red-shifted to lower frequencies, so that the force on it is decreased.

37. Laser Cooling 2 The net effect of the two opposing laser beams is to reduce the magnitude of the velocity component of each atom along the axis of the two beams. Three orthogonal pairs of lasers are used to slow the motions of atoms moving in all directions. Using laser cooling for Rb87, the NIST group in Boulder, achieved temperatures of 10 μK, which are still ten to a hundred times too high for observing BEC. The effect of reducing vrms on the temperature of the sample may be calculated using the equipartition theorem; i.e. ½ mvrms2 = (3/2)kT.

38. Evaporative Cooling 1 • This method is analogous to the cooling of a hot liquid by evaporation. • The fastest moving atoms move furthest from the minimum, to a position of highest energy (see the upper atom shown in the figure). • Magnetic resonance is used to reverse the moments of the most energetic atoms, causing them to leave the trap, which is now an energy maximum. • Slowly reducing the radio frequency removes progressively cooler atoms. • At the end, only about 1% of the atoms remain in the trap, and the temperature is reduced by a factor of about 100, giving a temperature of the order of 100 nK.

39. Photographing the Condensate (NIST) 1 False color images show the velocity distribution just before the appearance of BEC (right), just after it (center), and for a nearly pure condensate (right). To increase the sample size, the magnetic trap is turned off. The excited- state (thermal) atoms move out faster, leaving the condensate near the center of the trap. These photographs were taken after the atoms had moved for about 0.05 s. The thermal cloud is almost circular, while the condensate cloud is elliptical.

40. Photographing the Condensate (NIST) 2 • The right frame has a horizontal dimension of 40 – 50 μm, equivalent to about 1500 atoms forming a single wave. • The shape of the peak is related to the elliptical shape of the trap, giving a vivid demonstration of the uncertainty principle pxx  ħ. • The temperature within the condensate may be of the order of 1 nK.

41. Information Theory PHYS 4315 R. S. Rubins, Fall 2009

42. Lack of Information • Entropy S is a measure of the randomness (or disorder) of a system. • A quantum system in its single lowest state is in a state of perfect order: S=0 (3rd law). • A system at higher temperatures may be in one of many quantum states, so that there is a lack of information about the exact state of the system. • The greater the lack of information, the greater is the disorder. • Thus, a disordered system is one about which we lack complete information. • Information theory (Shannon, 1948) provides a mathematical measure of the lack of information, which may be linked to the entropy.

43. Missing Information H • For an experiment with n possible outcomes p1, p2,… pn, Shannon introduced a function H(p1, p2,… pn), which quantitatively measures the missing information associated with the set of probabilities. • Three conditions are needed to specify H to within a constant factor. • 1. H is a continuous function of pi. • 2. If all the pi are equal, then pi = 1/n, and H is a monotonically increasing function of n, since the number of possibilities increases with n. • 3. If the possible outcomes of an experiment depend on the outcomes of n subsidiary experiments, then H is the sum of the uncertainties of the subsidiary experiments. • With these assumptions, H was found to be proportional to the entropy S = – k r pr ln pr.

44. Example of Sum of Uncertainties Single experiment, using H = – K(pr lnpr). H(1/2,1/3,1/6) = – K[(1/2) ln(1/2) + (1/3) ln(1/3) + (1/6) ln(1/6)] = K[(1/2)ln2 + (1/3)ln3 + (1/6)ln6] = K[(2/3)ln2 + (1/2)ln3] = 1.01 K. Two successive experiments H(1/2,1/2) = – K[(1/2) ln(1/2) + (1/2) ln(1/2)] = K ln2. (1/2)H(2/3,1/3) = K[(– (1/3)ln2 + (1/3)ln3 + (1/6)ln3] Thus, H(1/2,1/2) + (1/2)H(2/3,1/3) = K{[1 – (1/3)]ln2 + [(1/3) + (1/6)]ln3} = K[(2/3)ln2 + (1/2)ln3] = 1.01 K.

45. Shannon’s Calculation 1 The simplest choice of continuous function (Condition 1)is For simplicity, consider the case of equal probabilities; i.e. pi = 1/n. Since H is a monotonically increasing function of n (Condition 2), For two successive experiments (Condition 3)

46. Shannon’s Calculation 2 Since H(1/n,…1/n) = n f(1/r), . Letting R = 1/r and S =1/s,

47. Shannon’s Calculation 3 Since g(R) + g(S) = g(R,S) and , so that Thus, so that Since R = 1/r, .

48. Shannon’s Calculation 4 For r =1, the result is certain, so that H(r) = f(1/r) = 0. Thus, C =0, so that Now d/dn(– A ln n) must be positive (Condition 2), so that A must be negative. Letting K = – A, and p = pi =1/n, Thus, the missing information function H is given by . With K replaced by Boltzmann’s constant k, H equals S (entropy).

49. Black-Hole Thermodynamics PHYS 4315 R. S. Rubins, Fall 2009

50. Quantum Fluctuations of the Vacuum • The uncertainty principle applied to electromagnetic fields indicates that it is impossible to find both E and B fields to be zero at the same time. • The quantum fluctuations of the vacuum so produced cannot be detected by normal instruments, because they carry no energy. • However, they may be detected by an accelerating detector, which provides a source of energy. • The accelerating observer would measure a temperature of the vacuum (the Unruh temperature), given by TU = aħ/2πc. Notes i. For an acceleration of 1019 m/s2, TU ~ 1 K. ii. TU = 0 if either ħ =0 or c = ∞, which is the classical result.