Physics 301 01-Oct-2004 10-1Johnson NoiseThis is another application of the thermal equilibrium of electromagnetic modes. Con-sider an ideal transmission line, like a long piece of lossless coaxial cable. Suppose itslength is L and suppose it is shorted out at each end. Then any wave that travels alongthe line is reflected at each end and t o have an appreciable amplitude, the length of theline must contain an integral number of half wavelengths. In other words this is just aone-dimensional cavity of length L. There are modes of the electromagnetic fields, mostconveniently represented by the potential difference between t he inner and outer conduc-tors. Vn= Vn,0sin(ωt) sin(nπx/L), where n is any positive integer. Vnand Vn,0representthe potential difference and the amplitude of the p otential difference. The fields mustsatisfy Maxwell’s eq uat ions, so ω = nπc/L. Actually, if the coax is filled with a dielectric,the speed of propagation can be different from c, let’s a ssume it’s filled with vacuum. Ifthis line is in thermal equilibrium at temperature τ, each mode a cts like an oscillator andhas average energy ¯hω/(e¯hω/τ− 1). Let’s consider the low frequency limit so the averageenergy in each mode is just τ . The number of modes per integer n is just 1. Then thenumber of modes per unit frequency isn(ω) dω =Lπcdω .The energy per unit length per unit frequency is thenuω=τπc,at low frequencies.As you may know, all transmission lines have a characteristic impedance, R. If aresistor R is connected across the end of the line, then a wave traveling down the line iscompletely absorbed by the resistor. So, let’s take a resistor, in equil ibrium at temperatureτ, and connect it to the end of the line. Since the resistor and the line are at the sametemperature, they are already in thermal equilibrium and no net energy transfer takesplace. Each mo de in the line is a standing wave compo sed equally of traveling wavesheaded in both directions. The waves traveling towards the resistor will be completelyabsorbed by the resistor. This means that the resistor must emit waves with equal powerin order that there be no net transfer of energy. The energy in the frequency band dω perunit length headed towards the resistor is τ dω/2πc. This is traveling a t speed c, so thepower incident on t he resistor is τ dω/2π which is also t he power emitted by the resistor,What we’ve established so far is that the line feeds power τ dω/2π into the resistorand vice-versa. This means that a voltage must appear across the resistor. This willbe a fluctuating voltage with mean 0 since it’s a random t hermal voltage. However, it smean square val ue wil l not be zero. Let’s see if we can calculate this. As an equivalentcircuit, we have a resistor R, a voltage generator (the thermally induced voltage source),Copyrightc 2004, Princeton University Physics Department, Edward J. GrothPhysics 301 01-Oct-2004 10-2a filter (to limit the frequencies to dω), and another resistor of resistance R representingthe transmission line. Then the current is I = V/2R . The average power delivered to theresistor is then hI2iR = hV2i/4R = τ d ω/2π. In the la b, one measures frequencies in Hertzrather than radians per second, ν = ω/2π. FinallyhV2i = 4Rτ dν .This relates the mean square noise voltag e which appears across a resistor to the temper-ature, resistance, and bandwidth (dν) . Of course, this voltage results from fluctuationsin the motions of electrons inside the resistor, but we calculated it by considering electro-magnetic modes in a one-dimensional cavity, a much simpler system! This thermal noisevoltage is called Johnson noise.Debye Theory of Lattice VibrationsA little thought will show that sound waves in a solid are not all that different fromelectromagnetic waves in a cavity. Further thought will show that there are some importantdifferences that we must take into account.The theory of lattice vibrations that we’ll discuss below applies to the ion lat tice ina conductor. In addition, one needs to account for the thermal effects of the conductionelectrons which behave in many respects like a gas. We’ll consider the electron gas laterin the term. For now, we imagine that we’re dealing with an insulator.We will treat crystalli ne solids. This is mai nly for conceptual convenience, but alsobecause we want reasonably well defined vibrational modes.As a model, suppose the atoms in a solid are arra nged in a regular cubic lattice. Eachatom vibrates around its equilibrium position. The equilibrium and the characteristicsof the vibrations are determined by interactions with the neighboring atoms. We canimagine that each a tom is connected to its six nearest neighbors by springs. At firstsight, this seems silly. But, the equilibrium position is determined by a minimum inthe potential energy, and the potential energy almost surely increases quadratically withdisplacement from equilibrium. This gives a linear restoring force which is exactly whathapp ens with a spring. So our solid is a large number of coupled oscillato rs. In general,the motion of a system of coupled oscil lators is very complex. You probably know fromyour classical mechanics course, that the motion of a system of coupled oscillators can beresolved into a superposition of normal modes with the motion of each mode being simpleharmonic in t ime. So, we can describe the motion with the N vectors riwhich representthe displacement of each atom from its equilibrium position, or we can describe the motionwith 3N normal mode amplitudes. For those of you that know about Fourier transforms,the normal modes are just the Fourier transforms of the position coordinates.Copyrightc 2004, Princeton University Physics Department, Edward J. GrothPhysics 301 01-Oct-2004 10-3These normal modes represent elastic vibrations of our solid. They are standing elasticwaves, or standing sound waves. In this respect, t hey are similar to the cavity modes wediscussed earlier. There are two main differences. First, there are three polarizations:there are two transversely polarized waves (as we had in t he electroma g netic case) andone longitudinally polarized wave (absent in the electromagnetic case). Second, there is alimit to the number of modes. If our solid contains N atoms, there are 3N modes. In theelectromagnetic case, t here is no upper limit to the frequency of a mode. High frequencymodes with ¯hω ≫ τ are not excited,