Consider a parallel LCR circuit maintained at a temperature T >> ( k b

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1 UW Physics PhD. Qualifying Exam Spring 8, problem Consider a parallel LCR circuit maintained at a temperature T >> ( k b LC ). Use Maxwell-Boltzmann statistics to calculate the mean-square flux threading the inductor and the mean-square charge on the capacitor. Solution : by means of partition function Recall that the energy of a capacitor is E The energy stored in the circuit is E Q C and of an inductor is E L I Φ L. Q C + Φ L. Assert that the relative probability that a circuit element has energy E is well-described by the Boltzmann factor: P(E) e E kt. (Note that this assertion about the system energy is completely independent of what statistics are obeyed by the particles in the system). The partition function is then Z e β i Q i C + what the charge and flux of the ith state are?? Φ i L, where β /kt. How are we to decide To proceed we must convert the sum to an integral over phase space. What variables define phase space? Phase space is defined by generalized coordinates p and momenta q. Here is one recipe for assigning the coordinates and momenta.. Write down the Lagrangian. In the circuit above, note that V C V L, that is Q C L di dt Φ, since Φ LI. Rewrite the energy E C Φ + Φ L. Observe that the energy, written in this fashion, is analogous to the energy of an harmonic oscillator, with Φ taking the role of position. Therefore identify the inductor flux Φ as the generalized coordinate. View the first term as a kinetic energy term, the second as a potential energy term, and write the Lagrangian as L C Φ Φ L.

2 . Find the momentum p that is conjugate to the coordinate. The conjugate momentum is defined to be p L q L Φ C Φ Q. Phase space is thus defined for this problem by the variables Φ, Q. The phase space volume element is therefore dφ dq. Recall the recipe for converting sums over quantum states to integrals (this was discussed in connection with the Bose-Einstein Condensation problem): dpdq t, h where t is the number of sets of (p,q) in the problem (in this case t ). Thus, the correct form for the partition function is Q i Φ i Z e β C + L Z e β dφdq i h h e β Q C + Now we can easily calculate the expectation values that we want: Q CkT, h h Q e β Q C Φ L dq e β dφ Q e β Q C dq C β Qe β Q C β Φ e β dq e L dφ e β Q C dq Φ L Q C β dq e Q C e β C + β Q C dq Φ L e β dφ Q C dq Φ LkT. Solution : by means of equipartition Follow the solution above, until arriving at this expression for the energy: E C Φ + Φ L. Observe this has the same form as the energy of a -D harmonic oscillator. Deduce that this system must have two degrees of freedom, like the harmonic oscillator. Use the classical theorem of equipartition of energy to conclude that there must be! kt associated with each degree of freedom, that is! kt for kinetic energy (first term) and! kt for potential energy (second term). CΦ Thus Q C kt, Φ L kt, and Φ LkT, Q CkT, as before.

3 Solution 3: Johnson Noise (following Nyquist, Phys. Rev. 3 (98), -3). The method of Nyquist involves imagining the resistor is connected across an ideal transmission line, which is terminated at the far end by an identical resistor: The normal modes of the system are then the normal modes of the transmission line, with wavelength λ n l/n, where l is the length of the transmission line. The energy contained in each mode is calculated by using Bose-Einstein statistics in the limit kt>>e n, where E n is the energy of the nth mode E n hν n hc λ n. The result is that the energy contained in each mode is kt, which is identical to result expected from classical equipartition, on the assumption that there are two degrees of freedom per normal mode. One degree of freedom represents a travelling wave carrying energy down the transmission line from resistor to resistor, and the other degree of freedom represents a travelling wave carrying energy from resistor to resistor. Since all energy is contained in travelling waves the power delivered to each resistor per unit time may be equated with the V /R power dissipated in the resistor. This gives the Nyquist result: V 4RkT Δf See Nyquist s (very readable) paper for details. Only circuit elements with non-zero real impedances (ie resistors) serve as voltage sources. Thus a resistor at a finite temperature may be modeled as an ideal resistor in series with a voltage source: The fact that finite-temperature resistors act as voltages sources was noticed experimentally by Johnson (before being explained almost immediately by Nyquist, the theorist in the next office), so the phenomenon is known as Johnson noise. The factor Δf indicates that all frequency bands contribute equally to <V >. Thus Δf may be interpreted equally well as in infinitesimal (in which case the left-hand side ought to be d<v >), or as a bandwidth, or as a maximum frequency on the assumption that the minimum frequency is zero.

4 If we now model the LRC circuit problem like this:, the voltage across the inductor and capacitor in a particular frequency range df is related to the Johnson noise voltage < V > by: d V L d V C d V + R (ωc ωl ) 4RkT df We find the mean-square flux linking the inductor by using V L di dt V L ω I ω Φ, which implies that d Φ ω d V L, + R (ωc ωl ) so Φ 4kTR dω π ω + R ωc. ωl This integral may be solved by the substitution xωc-/ωl, ω ( x + x + 4C L ) C to give πl/r, so that Φ LkT, as before. The total fluctuating voltage is V L V C 4RkT π dω + R ωc ωl kt C The charge on the capacitor is then Q C V CkT. One conclusion that may be drawn from this is that while the mean square charge or flux on an isolated capacitor or an inductor is non-zero, its output impedance (ability to drive current in other circuit elements) is zero; only resistors have non-zero output impedances. Addendum: if the inequality kt >> LC does not apply (high frequencies and/or low temperatures) the Nyquist argument may easily be extended, with the result 4hf RΔf V e hf /kt.

5 Solution 3b: a slightly different circuit model It turns out that if one replaces the resistor R with a complex impedance ZR+iX, the Nyquist result still holds. In this case <V >4R kt Δf. Thus our parallel RLC circuit could be modeled like this Then the resistive part of the impedance is: R +R (ωc ωl ) It is now no longer true that all frequency bands contribute equally to <V >. In particular, high frequencies don t contribute at all, because the capacitor becomes a short circuit at high frequencies, and the resistive impedance goes to zero. In this case it makes sense to calculate the total <V > as < V > 4kT R df ktr π dω + R ωc ωl This is the same integral as before, and yields the same answers. The mean-square charge and flux do not depend on the details of the circuit.

6 Method 4: diffusion Consider the difference between electron movement in a perfect conductor and electron movement in a resisitor: --in a perfect conductor, the electrons never collide with anything, and their number density is determined by the local value of the potential --in a resistor, the electrons collide with other particles ( the lattice ), and their number density is determined not only by the local value of the potential but also by diffusion. In the absence of an external electric field, electron movement through a resistor is entirely due to diffusion. The net current is zero, but this may be thought of as the sum of a leftward-going current and a rightward-going current that are each non-zero. Consider a resistor with no external voltage: Red Blue Suppose that electrons that pass through the resistor going to the right reach the blue end, turn blue, and head back through the resistor going to the left. Blue electrons that reach the red end of the resistor turn red, and head back through the resistor going to the right. A diffusive process is may be thought of as a process in which each particle makes a random walk through the system. It may be characterized by a time τ and a length λ. Every τ the particle takes a step of length λ in a random direction. In a onedimensional diffusive process, the direction can be either left or right. Let us assume that λ<<l, where L is the length of the resistor. Then it is not true that all particles going to the right are necessarily red. We expect each electron to change direction multiple times before passing through the resistor. n n n red n blue L x L x The total density of electrons at all points is n. Diffusive processes are described by the equation: Γ D n, where Γ is the particle flux, n is the density, and D is the diffusion constant (dimensions of length /time), which can be approximated as λ /τ.

7 Consider the red and blue electrons separately. In the -d case n has units of particles per length and Γ has units of particles per time. Specifically: Γ Dn qdn, I qγ; I L L Einstein showed that (with certain assumptions) D for charged particles in the absence of an electric field was related to the mobility µ those same charged particles would have in the presence of an electric field by D µ kt q, where the mobility is defined by v µ E, v being the terminal velocity the particles attain in an electric field E. Thus the current becomes I µ kt n L. In this model the completion of the passage of an electron through the resistor may be treated as an event subject to Poisson statistics. Choose a time interval dt such that the number N of electrons that complete the passage through the resistor during dt is N > or so. Then the Poisson distribution simplifies to the Gaussian distribution, and we conclude the probability P(N) that exactly N electrons will complete the passage is P(N) e (N N ) /N, from which we write N N ; σ N N ; I q N qn ;σ I q N q Idt /q. dt dt dt dt Using the previous result for I the variance of the current becomes σ I qi qµ kt n. dt L dt The quantity qµn may be recognized as the conductivity σ (defined by JσE), which has as its inverse the resistivity η/σ, (defined by EηJ). In our -d model the resistance RηL, so σ I kt Rdt. It is customary to write the averages of fluctuating quantities in terms not of dt but of df, an infinitesimal frequency. The formal way to relate dt to df involves taking a Fourier transform of the current from the time interval to dt, and then to calculate the spectral intensity from the Fourier components. Informally, we note that the number of electrons which arrive in time dt will be affected by fluctuations with time scales much greater than dt, but will not be affected by fluctuations with time scales much less than dt. Thus dt and df are related somehow. The results of the formal calculation (involving spectral intensity) are that /dt should be replaced by df. Thus we have: kt df σ I (true for red or blue electrons) R

8 This type of noise is called shot noise. In this case shot noise is entirely due to electron diffusion. The total current (sum of red and blue currents) is zero, but the current fluctuations add in quadrature (this is an example of the Central Limit Theorem), so: I total ; I 4kT df σ I σ red +σ blue R Therefore V R I 4kT Rdf This is the same result for Johnson Noise (solution 3, above), so the calculations for <Q > and Φ will give the same result as solution 3. Comments on solution 4. Solution 4 depends on the Einstein relation DµkT/q. Under what conditions does this hold? Einstein s relation certainly holds if the electrons follow a Maxwell-Boltzmann distribution. However, the electrons in a metal certainly do not follow a Maxwell-Boltzmann distribution. Electrons are Fermi-Dirac particles, and the chemical potential for a typical metal is a few ev (compare with room temperature ~ /4 ev). (Note that the chemical potential µ (not the Fermi energy E f, which is the highest occupied state at T) appears in the Fermi-Dirac distribution function, n i e ( E i µ) /kt + ; note that the chemical potential adjusts itself as a function of temperature to keep the total number of particles N n a constant; and finally note that at T, µe i f while at i higher temperatures µ increasingly deviates from E f, becoming smaller than E f.) The only way to recover Maxwell-Boltzmann statistics from Fermi-Dirac statistics is to limit one s interest to electrons with energy considerably larger than the chemical potential, ie (E i -µ) >> kt. A more qualitative model for why Fermi-Dirac particles might obey the Einstein relation may be found in Guidoni and Aldao, Eur. J. Phys. 3 () 395-4, On Diffusion, Drift, and the Einstein Relation. Also see the discussion in section 5. in Robinson, Noise and Fluctuations, Oxford (974); TK R6.

PHYS 352. Noise. Noise. fluctuations in voltage (or current) about zero average voltage = 0 average V 2 is not zero

PHYS 352. Noise. Noise. fluctuations in voltage (or current) about zero average voltage = 0 average V 2 is not zero PHYS 352 Noise Noise fluctuations in voltage (or current) about zero average voltage = 0 average V 2 is not zero so, we talk about rms voltage for noise V=0 1 Sources of Intrinsic Noise sometimes noise