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Physics 301 28-Oct-1999 13-1 Superfluid Helium As We Mentioned Last Time, the Transition of 4He at 2.17 K at 1 Atm Is Believed T

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Physics 301 28-Oct-1999 13-1 Superfluid Helium As We Mentioned Last Time, the Transition of 4He at 2.17 K at 1 Atm Is Believed T Physics 301 28-Oct-1999 13-1 Superfluid Helium As we mentioned last time, the transition of 4He at 2.17 K at 1 atm is believed to be the condensation of most of the helium atoms into the ground state—a Bose-Einstein condensation. That this does not occur at the calculated temperature of 3.1 K is believed to be due to the fact that there are interactions among helium atoms so that helium cannot really be described as a non-interacting boson gas! Above the transition temperature, helium is refered to as He I and below the transition, it’s called He II. K&K present several reasons why thinking of liquid helium as a non-interacting gas is not totally off the wall. You should read them and also study the phase diagrams for both 4He and 3He (K&K figures 7.14 and 7.15). The fact that something happens at 2.17 K is shown by the heat capacity versus tem- perature curve (figure 7.12 in K&K) which is similar to the heat capacity curve for a phase transition and also not all that different from the curve you’re going to calculate for the Bose-Einstein condensate in the homework problem (despite what the textbook problem actually says about a marked difference in the curves!). In addition to the heat capacity, the mechanical properties of 4He are markedly different below the transition temperature. The liquid helium becomes a superfluid which means it flows without viscosity (that is, friction). The existence of a Bose-Einstein condensate does not necessarily imply the existence of superfluidity. To understand this we need to examine the mechanics of friction on a microscopic level. Without worrying about the details, friction must be caused by molecular collisions which transfer energy from the average motion (represented by the bulk velocity) to the microscopic motion (internal energy). If our Bose-Einstein condensate were really formed from a gas of non-interacting particles, then it would be possible to excite any molecule in the condensate out of the ground state and into the first excited state simply by providing the requisite energy (and momentum). In the last lecture, we calculated that under typical conditions, the energy 31 difference between the ground state and the first excited state was about 10− erg, an incredibly small amount of energy that would be very easy to provide given that thermal 16 energies are about 10− erg. In order to have superfluid behavior, it must be that there are interactions among the molecules such that it’s not possible to excite just one molecule out of the condensate and into the first excited state. A better way to say this is that due to the molecular interactions, the single particle states are not discrete energy states of the superfluid. We need to consider the normal modes of the fluid—the longitudinal oscillations or the sound waves. In particular, we can consider travelling sound waves of wave vector k Copyright c 1999, Edward J. Groth Physics 301 28-Oct-1999 13-2 and frequency ω. (Rather than the standing waves which carry no net momentum.) A travelling wave carries energy in units of ¯hω and momentum in units ¯hk.Thenumberof units is determined by the number of phonons or quasiparticles in the wave. Now imagine an object of mass M moving though a stationary superfluid with velocity Vi. In order for there to be a force on the object, there must be a momentum transfer to the superfluid. In order to do this, the object must create excitations in the fluid which contain momentum (the quasiparticles in the travelling waves we just discussed). Of course, if quasiparticles already exist, the object could “collide” with a quasiparticle and scatter it to a new state of energy and momentum. (This can also be viewed as the absorption of one quasiparticle and the emission of another.) We will assume that there are not very many existing quasiparticles and consider only the creation (emission) of a quasiparticle. So, let’s consider this emission process. Before the event, the object has velocity Vi and afterwards it has velocity Vf . We must conserve both energy and momentum, 1 1 MV 2 = MV 2 +¯hω , 2 i 2 f and MVi = MVf +¯hk . We can go through some algebra with the goal of solving for Vi k. The momentum equation can be rewritten as · MV ¯hk = MV , i − f squared and divided by 2M, 1 1 1 MV 2 ¯hV k + ¯h2k2 = MV 2 . 2 i − i · 2M 2 f Subtract from the energy equation to get 1 ¯hV k =¯hω + ¯h2k2 , i · 2M or k ¯hω ¯hk V = + . i · k ¯hk 2M What we really want to do is place a lower limit on the magnitude of Vi. This means we can drop the term containing M on the right hand side. The smallest value will occur when Vi is parallel to k/k, the unit vector in the k direction. This corresponds to emission of the quasiparticle in the forward direction. Thus ¯hω V > . i ¯hk Copyright c 1999, Edward J. Groth Physics 301 28-Oct-1999 13-3 I’ve left the ¯h’s there in order to emphasize that the right hand side is the ratio of the energy to the momentum of an excitation, Suppose the excitations are single particle states with momentumh ¯k and energy ¯h2k2/2m. This is the travelling wave analog to the standing wave particle in a box states we’ve discussed many times. Then the right hand side becomes ¯hk/2m which goes to zero as k 0. Thus, an object moving with an arbitrarily small velocity can produce an excitation→ and feel a drag force—there is no superfluid in this case. (Note: k must be bigger than 1/L,whereL is the size of the box containing the superfluid, but as we’ve already seen∼ the energies corresponding to this k are tiny compared to thermal energies.) Suppose the excitations are sound waves (as we’ve been assuming) and the phase velocity is independent of k.Then ω V > = v , i k s where vs is the phase velocity of sound in the fluid. This means that if an object flows through the fluid at less than the velocity of sound, the flow is without drag! That is, the fluid is a superfluid. In fact, the vs is not independent of k and what sets the limit is the minimum phase velocity of any excitation that can be created by the object moving through the fluid. 1 Figure 7.17 in K&K shows that this minimum is about 5000 cm s− for the low lying excitations in 4He. K&K point out that Helium ions have been observed to travel without 1 drag through He II at speeds up to about 5000 cm s− ! Comment 1: It appears that we’ve shown that any fluid should be a superfluid as long as we don’t move things through it faster than its speed of sound. In our derivation, we made an assumption that doesn’t apply in most cases. Can you figure out which assumption it was? Comment 2: As a general rule, superfluidity or superconductivity requires the conden- sation of many particles into a single (ground) state and a threshold for creating excitations. The minimum velocity required to create an excitation is the threshold for non-viscous flow. Copyright c 1999, Edward J. Groth Physics 301 28-Oct-1999 13-4 Heat and Work Now we want to discuss the material covered in chapter 8 of K&K. This material might be considered to have a more classical thermodynamics rather than statistical mechanics flavor. We’ve already discussed a lot of this material in bits and pieces throughout the term, so we will try to focus on the material not yet covered and just hit the highlights of the remaining material. Heat and work occur during processes. They are energy transfers. Work is an energy transfer by macroscopic means and heat is an energy transfer by microscopic means. We’ve discussed reversible processes several times and we’ll assume reversible processes unless we explicitly state otherwise. When work is done to or by a system, the macroscopic parameters of the system are changed—for example changing the volume causes pdV work to be performed. Performing work changes the energy, U, of a system. But work does not change the entropy. Heat transfer changes the entropy as well as the energy: dU =¯dQ = τdσ. A very important activity in any modern society is the conversion of heat to work. This is why we have power plants and engines, etc. Basically all forms of mechanical or electrical energy that we use involve heat to work conversion. Not all of them involve fossil fuels, and in some cases it may be hard to see where the heat enters. For example, what about hydro-electric power? This is the storage of water behind a dam and then releasing the gravitational potential energy of the water to run an electric generator. Where is the heat supplied? Heat is supplied in the form of sunlight which keeps the weather going which provides water in the atmosphere to make rain to fill the lake behind the dam. Of course, the economics of this process are quite different from the economics of an oil fired electrical generating plant. It was the steam engine (conversion of heat, obtained by burning coal, into work) that allowed the industrial revolution to proceed.
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