Second Sound University of California, Santa Cruz September 12, 2006
Contents 0.1 Apparatus.......................................... 2 0.2 Experiment......................................... 3 0.3 Exercise........................................... 5 0.4 Second Sound Equations.................................. 7 In 1908, Kamerlingh Onnes discovered superfluidity in liquid Helium. He was alone in doing experiments on this phenomenon until 1930! He measured quantities such as flow resistance, and other curious effects such as the superfluid fountain. However, he did not observe any wave phenomena. A pulse of heat in a classical fluid should diffuse, with the temperature reducing monotonically. In 1930, Landau predicted a wave phenomenon based on quantum mechanical principles. Once Helium is cooled below the lambda point (about 2.173 K), it can be regarded as consisting of two interpenetrating fluids: one normal and the other superfluid. The fraction of normal fluid is a function of temperature. If a source of heat is pulsed in time in such an environment, the heat does not simply diffuse out from the source as it would in a normal fluid. Instead, it travels as a pulse, with a well-defined speed. This is a phenomenon with the name second sound (there are also 3 rd and 4 th sound phenomena). It is not a pressure wave (as sound is), but it is a wave: a quantum-mechanical temperature wave. As soon as Landau predicted this wave, others besides Onnes jumped into the business of liquid Helium experiments. In this experiment you will learn some basic techniques of cryogenics while measuring parameters of second-sound propagation. Because the apparatus is complicated and easily damaged, and liquid helium and the Dewar system are expensive, the lab manager or instructor will pre-instruct and transfer the helium. Make arrangements with the lab manager or instructor a week in advance to ensure sufficient helium and their availability. Start the cooling process (using liquid nitrogen) early in the morning so that the Helium transfer can take place before lunch. This will give you time to complete the experiment before 6 PM. There are a number of important cautions in this experiment: NEVER INTRODUCE HELIUM INTO THE VACUUM SPACE THAT ISO- LATES THE INNER DEWAR (access through glass valve). After flushing the space, introduce only a small amount of air into that space (as thermal transfer fluid while cooling with liquid nitrogen). The reason this space needs to be flushed is that helium gas will slowly diffuse through glass at room temperature. It must be removed to ensure good thermal isolation at low T. DO NOT RECONNECT THE VACUUM TUBING OR OPEN THE VALVE TO 1
THE VACUUM SPACE OF THE INNER DEWAR. THESE ARE ACTIONS ONLY FOR THE LAB MANAGER. Flush the inner Dewar (where the liquid helium will go) with helium and maintain a slight positive pressure with helium while cooling the Dewar and second sound head. This will reduce the amount of air snow that can be dangerous and obscures vision. The head and atmosphere admittance flanges act as over-pressure relief valves. LEAVE THESE FLANGE BOLTS VERY LOOSE. Ensure that you understand the plumbing circuit and the consequences of operating the valves and pumps under all conditions. VERIFY with the lab manager or the instructor the procedure for shutting down the system, if you plan to continue after they leave for the day. 0.1 Apparatus The apparatus has a heat source (transmitter) and a heat detector (receiver) at the two ends of a cavity that is filled with the superfluid Helium. Both the transmitter and receiver are carbon film resistors which have an electrical resistance that varies with the temperature. The receiver is often called a Bolometer, a device that measures the input power (or energy per pulse) by detecting a small change in temperature. For this experiment we use a temperature-dependent resistor, and observe heat pulses as small changes in the bolometer resistance. By putting a current through the bolometer, a voltage is generated, which you will be watching in this experiment (see section below on the Bolometer Operation). If the temperature remains constant, the voltage remains constant. If a second-sound (heat pulse) signal arrives at the receiving bolometer, the voltage changes as the temperature changes. For example if the second-sound wave is a sine wave, you will see a sine wave in the voltage. If the wave is a pulse, or train of pulses, you will see that in the voltage. The circuit that supplies the transmitter signal starts from a wave-form voltage generator. This is either a sine-wave generator, or a more complicated system involving making a single pulse, or a train of pulses. The single-pulse generator has a button on it to press for creating the single pulse, and it also has knobs to control the width and amplitude of the pulse. If a train of pulses is desired, then an earlier pulse generator provides a sawtooth wave-form that triggers the pulse generator in a regular fashion. It is also possible to trigger the pulse generator with the sine-wave generator. Note on Bolometers Bolometers are devices for detecting small amounts of heat. In the present experiment we use the carbon film resistor as a bolometer. It is sensitive to heat pulses because its resistance R(T ), is strongly dependent on the film temperature. For carbon films, the resistance increases as T is lowered. The resistance varies with T roughly as ( ) R B = R 0 1 + ae T 0 T Where R 0 (1 + a) is approximately the resistance at room temperature and T 0 determines the rate of increase at low T. The transmitter sends out a heat pulse that is received by the receiving bolometer. The receiving bolometer decreases in resistance R B when it is hit by the heat pulse. In order to measure R B, a bias current I B, must be sent through the bolometer, generating a voltage V B that will vary with 2
the bolometer resistance. This bias current, however, changes the temperature of the bolometer, in turn changing its resistance. In this experiment, the bias current is set up with a battery (V 0 = 36 volts) connected in series with a resistor R 1 and the bolometer, with resistance R B. The voltage across the bolometer is thus V B = R BV 0 R B + R 1 As R B varies with the impinging heat pulse, the voltage changes such that, to first order, V B V B = R B R B R 1 R 1 + R B It would appear that the larger the current, the larger V B. However the current I B cannot be increased indefinitely; if I B is too large it heats the bolometer and R B decreases also the total input power must be kept below roughly 100 mw to avoid boiling off the helium. The signal V B is a complicated function of temperature, such that the maximum signal is obtained when the resistance R 1 is at some particular empirical value that you will find by varying it. You can measure R B (T ) as you cool down using a low value of current (large value of R 1 ) which assures that the bolometer is not heating up significantly. To do this you will need to measure V B (T ), V 0, and R 1. Show R B = V B R 1 V 0 V B Then at low T you can measure the second sound signal height using the analog scope as a function of bias current by varying R 1 in the bias box to determine the best value of current to use. You may need to repeat this when you get close to T λ perhaps around 1.9K (but not too close or you will not be able to find the signal). We expect to have a constant current source available for the bias current in the near future. In this case I B can be set at a given constant value and V B = R B I B If this is available, vary I B to find the maximum in V B. 0.2 Experiment There are three different methods of looking at the second-sound phenomenon. When the source is single-pulsed, the second sound travels up the cavity to the receiver, reflects back to the transmitter, and continues to travel back and forth, decaying in amplitude due to damping. The signal on the detector shows pulses arriving at regular intervals of time. The time intervals determine the speed of second sound (knowing the length of the cavity), and the decay in the amplitudes of each successive pulse determines the damping information. A second method has the heat source driven with a sine-wave input. As this frequency is varied, resonances in the amplitude are observed when the wavelength of the second sound matches the cavity dimension. The width of the resonance (in frequency) determines the damping information in this method. When the resonance method is used, one can observe not only the fundamental, but also the first, second, etc harmonics. It is important not to confuse a harmonic with the fundamental when estimating the speed of second sound. However, this method continuously inputs heat to the 3
system and can make it more difficult to keep T constant. To minimize the evaporation of liquid Helium do not leave the sine-wave generator on for long periods of time. Also note that the second sound frequency is at twice the sine-generation frequency. The maximum heating occurs when the ac voltage amplitude is at the maximum (that occurs for both positive and negative voltage). Use this in determining the second sound period. Another way to think about the transmitter in this mode is in terms of the instantaneous power applied to it. If V = V 0 cos ωt, P = V 2 R or P = V 2 0 cos2 ωt R = V 0 2 [1 + cos 2ωt] 2R Thus the power oscillates at a frequency 2ω. You will find that as the lambda point is approached, these methods breaks down as noise becomes dominant. Take data, including finding the resonance widths, over as great a range of temperature as you possibly can. However, the speed of second sound does not vary much until you get above 2.0 K, and the lambda point is 2.173 K. You will find it difficult to see the speed change much if you can t get data close to 2.17K. There is a third method that makes it possible to take very good data well above 2.0 K. This is the method in which you transmit a regular train of pulses. If the received pulses at the detector fall right on top of each other in time, then they can constructively interfere, giving a much higher pulse amplitude. Because the pulses are narrow, this only happens if the time between the transmitted pulses corresponds exactly to the time that it takes for a previously generated pulse to travel to the end of the cavity and back. The way to set up this method is to use another generator of variable frequency to trigger the transmitter pulse generator. Start below 2.0 K where the second sound speed is easily measured by any method. Find the frequency on the generator for which the train of pulses is constructive. Note that if the individual pulses are narrow, this resonance condition can be very sharp. Some of the pulses will have bounced back and forth 5-10 times; for them to contribute requires a precise separation between the transmitted pulses. When using this technique note that you will find resonances when you overlap with any of the second sound pulses from the previous heat pulse, but the effect is largest when you overlap with the first pulse. Also note that if you pulse too fast e.g. twice as fast as the second sound pulses, you will again find a resonance, but now the spacing between the pulses will be half as long. You have added a second set of pulses half-way between the previous pulses. You can use this to determine the fundamental pulse frequency. Make sure the spacing between pulses is not too short (ie less than the period for a train of second sound pulses). Use the resonance pulse method as you approach the lambda point because the signal decreases rapidly as the fraction of superfluid decreases. You should be able to get data very close to the lambda point - perhaps within 2-4 mk; however, you will need to follow a given resonance (i.e. continually adjust the pulse frequency to remain in the resonance condition as T is raised) because the resonance frequency changes rapidly with T as you approach T λ. If you do not follow a given resonance, you may end up on another resonance (i.e. another harmonic) and it will not be clear why. This will give good sound-speed measurements in the region in which the sound speed is changing rapidly. You should be able to reduce the speed of second sound by a factor of 4-5. Measurements of Q You will be able to measure Q with the first two methods, but not very close to the lambda point. When using the pulse method, make sure the transmitter pulses are widely 4
spread, so that each pulse train has died into the noise before the next pulse train is generated (see discussion in file folder). Your measurements, unfortunately, are not of some fundamental damping in second sound, but rather are of the losses incurred in bouncing off the bolometers at both ends. (Therefore don t try to measure Q near the lambda point.) You may need to review the concept of Q. Look at mechanics books that discuss damped harmonic oscillators or at chapters in E&M books that deal with LRC circuits. You should be able to find the equation that describes Q for a decaying series of pulses, as well as (for the ac method) Q in terms of the half height width; remember the detector is measuring the power in the wave. Hints: Start each run at the lowest temperature (pressure) 1 and verify equilibrium of pumping speed and pressure. As always, graph your data as you obtain it. There may be insufficient time to transfer additional helium later. If you lose the resonance (near the lambda point), it is usually easier to pump down until you find the resonance and then repeat some of the data. Atkins book, Liquid Helium Ch.5, Section 4.4 presents the simplified two-fluid equations of motion that serve as a starting point. 0.3 Exercise Discuss the boundary conditions for second sound in our cavity, and derive the relationship between resonant frequencies and the velocity of a second sound wave. Calculate ρ S /ρ N at several temperatures from your measured values of second sound velocity and values of specific entropy and heat capacity taken from the tables provided below. (Also, see Donnelly s book, Experimental Superfluidity.) Atkins book treats other aspects of superfluidity as does a more recent book by Wilks, The Properties of Liquid and Solid Helium. Vapor Pressure of Helium vs. Temperature Table 1 is excerpted from the 1958 4 He Scale of Temperatures, National Bureau of Standards Monogram 10 (1960). The pressure P is given in Torr (mm of Hg), and the temperature T in degrees Kelvin. This is a short table - use the more extensive table in the lab folder. Table 2 gives quantities for liquid helium four between 1 and 2 K. Remember the lambda point is T λ = 2.173 K. The heat capacity and entropy apply to the normal component, and C 2 is the speed of second sound. The equation that gives the proportion of normal fluid as a function of temperature is the following: Of course, then the fraction of normal fluid is ρ n ρ = ρ s = CC2 2 ρ n T S 2 r 1 (1) r 1 + r Table 3 gives both r and this fraction as a function of temperature. (2) 1 I have found that after a few hours, when much of the He has evaporated, one can obtain a much lower temperature. 5
Table 1: Usual atmospheric pressure in the lab is about 741 mmhg corresponding to a Helium vapor temperature of Helium of 4.19 K. Vapor Pressure of Helium P T P T P T P T 4 1.523 26 2.031 50 2.291 350 3.49 6 1.612 27 2.044 60 2.375 400 3.60 8 1.682 28 2.057 70 2.449 450 3.70 10 1.740 29 2.071 80 2.516 500 3.80 11 1.765 30 2.083 90 2.578 550 3.89 12 1.789 31 2.095 100 2.635 600 3.97 13 1.812 32 2.107 150 2.87 650 4.05 14 1.833 33 2.119 200 3.06 700 4.13 15 1.854 34 2.130 250 3.22 750 4.20 16 1.873 35 2.141 300 3.36 17 1.891 36 2.152 18 1.909 37 2.163 19 1.926 38 2.174 20 1.943 39 2.185 21 1.959 40 2.195 22 1.974 23 1.989 24 2.003 25 2.017 Table 2: Table of quantities related to second sound for liquid helium four. From J. Maynard (see Second Sound file for additional details). T (K) C (J kg 1 K 1 ) S (J kg 1 K 1 ) C 2 (ms 1 ) 1.00 100 17 19.40 1.10 190 30 18.78 1.20 318 51.5 18.78 1.30 515 84.3 19.03 1.40 786 131.9 19.58 1.50 1142 197.80 20.07 1.60 1598 285.5 20.37 1.70 2174 399 20.36 1.80 2896 542.4 19.89 1.90 3804 722.8 18.78 2.00 4990 946.5 16.69 2.10 6972 1231.6 12.39 2.15 8723 1416.12 7.99 6
Table 3: Table of the fraction of normal fluid as a function of temperature in liquid helium four. 0.4 Second Sound Equations ρ nρ T r 1.0 0.008 0.008 1.1 0.015 0.015 1.2 0.029 0.0283 1.3 0.050 0.0472 1.4 0.081 0.0748 1.5 0.128 0.1131 1.6 0.197 0.1643 1.7 0.300 0.2310 1.8 0.463 0.3166 1.9 0.740 0.4253 2.0 1.289 0.5632 2.1 2.975 0.7484 2.15 7.741 0.8865 by Prof. Stanley Flatté The equations for second sound depend on the fact that heat in the superfluid state is not a diffusive phenomena but has more coherent properties. In the superfluid state we may approximate behavior as having no dissipation. There is a one-to-one correspondence between the amount of heat in a gram of liquid and the fraction of superfluid in that gram. Likewise the temperature is in one-to-one correspondence with the fraction of superfluid, and so is the entropy. If heat is input to some part of the fluid, the fraction of superfluid in that part must immediately go down, and therefore there must be a flow of superfluid OUT OF (and correspondingly a flow of normal fluid IN TO) that place in the fluid. Since all the entropy and heat of the state is carried by the normal fluid, these flows can be described in terms of those quantities also. Now, let us give the equations. Most texts give the five equations that have both ordinary and second sound within them. Here I want to give the equations that have second sound alone. I can do this easily because ordinary sound is a pressure wave. I will give the equations that result from setting the pressure to a constant, thus disallowing ordinary sound. Second sound, with its changing entropy, heat, and fraction of superfluid does NOT require pressure to change! (One good assumption here is that the coefficient of expansion of helium is quite small and can be neglected.) The first equation is the conservation of entropy! Thus, even though entropy is changing, it is in fact flowing from one place to another, not appearing or disappearing: δ t {ρ n S} = (ρ n Sv n ) (3) where S is the entropy per unit mass, ρ n is the density of normal fluid, and v n is the velocity of the normal fluid. (It will be easier if we regard our example as one dimensional, so that the velocity has only an x component.) The second equation is the relation between the amount of normal fluid and the gradient of temperature. If there is a gradient of temperature, it must be that there is a time dependence on 7
the fraction of normal fluid, and that means that there must be a flux of normal fluid flowing from the high-temperature location into the low-temperature location: δ t v n = ρ s ρ n S T (4) Now here is the key to second sound being a wave phenomenon. There is no dissipation in these equations. If the normal fluid rushes from the high-temperature part to the low-temperature location, then eventually it raises the temperature of the low- temperature location up past zero. The flow of normal fluid then reverses. This is the oscillatory behavior that is required for a wave phenomenon. In order to continue with our derivation, it turns out to be important to distinguish between those quantities that will remain constant to first order, and those quantities that fluctuate, thus providing the oscillatory wave phenomenon. In order to make this clear, we define the following: S = S 0 + S (5) T = T 0 + T (6) where the subscripted zero quantities are the average quantities in the absence of the wave, and the primed quantities are the fluctuating (small) parts that represent the wave. Now we use an equation of thermodynamics to relate some of these quantities: S = U T 0 = CT T 0 (7) where U is the internal energy and C is the heat capacity per unit mass of fluid. Finally, we will re-express our two fundamental equations in terms of these more explicitly defined quantities: δ t {ρ n S } = (ρ n S 0 v n ) (8) δ t v n = ρ s ρ n S 0 T (9) Note that where there exists a v n, we take other factors to be constant, while in terms without a v n, we have to decide which factor is varying in time. Now we use our thermodynamic relation on the equation above with S in it; and our other equation is still the same: ρ n C T 0 δ t T = ρ n S 0 v n (10) δ t v n = ρ s ρ n S 0 T (11) We now take the time derivative of Equation 11 and the gradient of Equation 10 and set the resulting cross products of T equal to produce our desired wave equation! δ 2 t v n = c 2 2 2 v n (12) 8
where the speed of the wave, c 2 is given by: c 2 2 = ρ s T 0 S0 2 ρ n C This should provide you with the physical background for this very interesting phenomenon of heat waves! (13) 9