Chapter 0 Non-equilibrium Stat. Mech.: Introduction There is no well-defined frame work for “nonequilibrium” systems. What is nonequilibrium? Any system that does not conform to the rules of “thermodynamics” or does not obey equilibrium statistical mechanics may be called a nonequilibrium system. Agreeably a negative definition. But what is equilibrium Stat Mech anyway? Our starting point is of course mechanics: classical or quantum. The rules to describe a system may be itemized as • Define degrees of freedom: based on our intuitive ideas of length and time • number of particles: simple counting or may be infinite in number: countable or uncountable • The number of degrees of freedom: 3N+3N or
2dN |{z} . We have pairs of variables (xi , pi )
always even
(angle, angular momentum ) and so on. For large N , the number of d.f. is large. P p2i • Define Hamiltonian, e.g., H = + · · ·. And then the appropriate equations of motion. 2m For statmech, follow the same steps upto H, and then get the appropriate partition function Z, a sum of e−βH type terms. Thermodynamics follows from it. At the end, the system is what described by 2, 4, 6....(T, U or S), (p, V ), (H, M ) (µ, N ). Again they are in pairs but only a are few, not going to infinity as N → ∞! Could we have guessed this? There are extensive variables these? like S, U, V, N which grow with the size, and can be traced (except S) to mechanics. But their conjugate partners are intensive, not changing with size, like T, P, µ. What are these? Only thing we know is that they are determined by external objects like reservoirs or baths. E.g. What is T ? Put a thermometer and let it come to equilibrium with the object −→ then we get T . If T is not fixed, the two are not in equilibrium. An example would be heating. While the temperature is changing rapidly, can we even talk of T ? This gives us a taste of problems we face in nonequilibrium situations. It is the dynamics, how things are changing with time, that becomes important; still, it is not strictly a mechanics problem. “Equilibrium” is not “dead”, there is dynamics going on. Laws of mechanics should be applicable, except the effects of the equivalents of T, p, µ need to be put in. In this scenario, 1
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it is necessary, not just convenient, to switch to a probabilistic description. Statistical mechanics involves probabilities. Our approach will therefore be a mixed one. Often we talk of equations of motion, and often in terms of probabilities. We have to get used to this dichotomy. Also we need to simplify things - build simple models, and generalize the results. We shall consider dynamics in equilibrium or near equilibrium, maintaining close contact with the equilibrium language. The problem is clear here. We know we need “friction” to generate heat, we need resistance in an electrical circuit to generate heat, but equilibrium statistical mechanics is completely devoid of this idea of “friction” in spite of its dealing with “heat”. The reason this works nicely is solely due to the emergence of the concept of “entropy”. A probabilistic description makes entropy a natural choice and this S helps us to make connection with Heat (remember, dQ = T dS). Notwithstanding a probabilistic language, an analogue of S is missing for nonequilibrium systems.... Aside: The definition of Gibbs entropy is S = −kB
P
p ln p
We shall also consider situations far away from equilibrium for which it may not be possible or meaningful to think of any Hamiltonian.... Let us consider a few problems chosen at random. • One standard example we are familiar with is the problem of heat conduction. Take a bar maintained at a constant temperature. There is no flow of heat. That’s the definition of equilibrium. However if a small temperature difference is maintained, then there is a steady heat flow as per the Fourier law: T2 − T1 dQ =κ A, dt L
(1)
where T2 and T1 are the temperatures at the two end of the bar of length L and crosssection A, Q is the heat flow at time t. The constant κ is called the thermal conductivity. It is one of the general properties of matter. Still, equilibrium stat mech is silent about it. Dimensional analysis tells us that dimensionally [κ/kB ] = [LT]−1 . Therefore a time-like quantity is needed to give a meaning to κ. Eq. stat mech (ESM) by definition cannot give any time-like quantity. Similar problem arises to define viscosity. • Think q of sound waves. Any system in equilibrium allows sound waves to propagate, c = Kρ where K is some elastic constant, and ρ density. Dimensionally, c is L/T and still we do not need any time like quantity. The sound velocity can be obtained in ESM. In reality, we know that sound does not propagate to infinity. There is attenuation. Where does the energy go? Moreover, attenuation cannot be an equilibrium phenomenon. Is it possible to have propagation of sound without attenuation? Or, are these two related? The question is outside the realm of ESM. • There are many other transport problems, like electrical conductivity or resistance, thermoelectric effect,... ionic conductivity, etc. 2
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• Such problems are well-defined in the sense that we know what the Hamiltonian is and how to describe it in equil;ibrium. What distinguishes all of these examples from any equilibrium system is the occurrence of a current, like, the heat current, particle current, charge current and so on. There is no current in an equilibrium state. This current is a hallmark of nonequilibrium systems. • We may go to the extreme where the currents actually vitiate the description through a Hamiltonian. We may be forced to define the problem by some dynamics or rules for changes which need not be strictly from mechanics. (a)
(b)
(c)
(d)
Figure 1: Particles in a box. The blue arrow indicates the direction of the field. (a) Cylinder, (b) Torus, (c) One dimensional road, (d) circular road. Here (a) and (c) have closed boundaries at the ends while (b) and (d) have periodic boundary conditions. Let’s take the example of a gas in a uniform gravitational field. This is the problem of atmospheric density variation with height, assuming uniform temperature throughout. We have a long cylinder closed on all sides, top and bottom. The particles are under the acceleration due to gravity g vertically down. See Fig. 1. The Hamiltonian is H = p2 + mgz, with z measured from the bottom (Fig 1(a).) The answer comes from the 2m probability p(z) ∝ exp(−βmgz) and the density is proportional to this probability. Here β = 1/kB T , T being the temperature and kB the Boltzmann constant. What temperature is doing here is to provide the energy required to go against the force and thereby creating a hydro-static balance. One get the same result by taking a thin horizontal slice of thickness dz at height z with pressures p and p + dp on two surfaces. The upward force due to the pressure difference, with gas law P V = N kB T , balances the weight of the slice. This prevents the motion of all particles settling on the floor. There is no current - this is equilibrium. There is dynamics, the particles are bouncing off the walls, maintaining the equilibrium. All particles settling on the floor, in contrast, would be a dead phase. Fig 1(c) is an equivalent one dimensional situation. Let us make a simple modification. Suppose instead of a cylinder, we use periodic boundary condition to get a torus in Fig. 1(b) or a circle (d). Locally in any neighbourhood, there is a “g” like force and one would be tempted to write the same Hamiltonian. It is the periodic boundary condition that prevents us from doing that. There cannot be a global Hamiltonian. If we use an angular variable θ, then we face a problem defining the potential energy at θ = 0 and θ = 2π. We do see that the absence of the walls prevents bouncing off, and under the local field, the particles would go along the circle. There will be a current eventhough in the transverse direction we would see “equilibrium”. How to describe this system? • Let us consider a biology problem. In a cell some things are produced in one region and need to be transported to some other region. These are done by motor proteins which 3
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walk on tracks. There are thermal fluctuations, viscosity, crowding and so on. A protein while walking may fall off or it may restart depending on the process involved. Therefore probabilities are required to describe the process. It is not a problem of classical or quantum mechanics, nor of equilibrium statistical mechanics. • Another very common problem in biology is enzymatic reaction. Most processes do not occur spontaneously. That’s needed for stability. Whenever something is needed, say something needs to be formed, like DNA needs to opened or unzipped, oxygen needs to be carried from point to point. These are done by enzymes. An object, called substrate, S becomes product P with the help of an enzyme E. First, thers is a transient compound ES and then ES gives P, with the enzyme back in the original state. What one is looking for is an overall coarse description and not a Schrodinger equation involving all the atoms. Such situations occur in many chemical processes too. • Growth problems: Suppose we are dropping particles on a surface. It could be atoms or molecules falling on a surface in an evaporation-deposition process or electroplating, sand grains or rice falling on a surface and making shapes, where we see a growth process evolving with time but characterized by some noise from external sources. Again not a mechanics problem. It requires a probabilistic description involving averages and fluctuations. There are more such examples like earthquakes, avalanches, hysteresis in various physical systems. • What is friction? A question we may have to leave unanswered at the end. These examples are chosen because some of these turn out to be rather typical of a class of non-equilibrium situations. We hope to discuss these in the course as specific examples. These examples are mostly from classical physics because most of the time we shall study classical problems but once in a while we would consider quantum problems too.
4
Chapter 1 Harmonic Oscillator 1.1
Harmonic Oscillator: On time scales
We know how to solve an oscillator problem, but we recapitulate it in a way that will be useful for us.
1.1.1
Pure Oscillator
Let us consider a linear harmonic oscillator. 2
• Newton’s law : m ddt2x = −kx, Here x is displacement, t is time, m is mass, and k is the spring constant (Spring force / Hooke’s law). • Lagrangian : L = 12 mx˙ 2 − 12 kx2 , where the dot denotes derivative with respect to t. The − ∂L =0⇒ potential is zero at the origin but not at ±∞. The equation of motion is dtd ∂L ∂ x˙ ∂x m¨ x + kx = 0. Both the approaches give a second order differential equation in t which is invariant under t ↔ −t. Another consequence is that with two conditions, we have a perfectly predictable (deterministic) motion. 2
of motion x˙ = {H, x}, etc., defined in • Hamiltonian : H = 12 pm + 12 kx2 , with the P equation ∂B ∂A ∂B terms of the Poisson bracket {A, B} = i { ∂A − }. ∂p ∂x ∂x ∂p • Two first order equations in the Hamiltonian formulation— first order equations involving d which is not time reversible. For t → −t, x → x (even under time reversal) but from dt intuition we know p → −p. With this intuitive input, the two formulations are consistent. In particular, x˙ = p/m | {z }
∂H = {H, x} = − ,& ∂p
∂H p˙ = −kx = {H, p} = − . | {z } ∂x
(1.1)
no change
both sides change sign
How many conditions? Each first order equation requires 1 ⇒ total 2. E.g specifying (x, p) at t = 0. 5
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Reactive coupling, time reversibility
We have two variables (x, p),√with opposite time √ reversible behaviour. To write in a symmetric form let us introduce xˆ = kx and pˆ = p/ m. In a matrix form, the equation of motion involves an anti-symmetric matrix: q k 0 d xˆ xˆ m q =M , with M = pˆ k dt pˆ − m 0
(1.2)
or said verbosely, the even variable satisfies an equation with a ’+’ sign while the odd variable satisfies an equation with a ’-’ sign. This is a very crucial point to be generalized later on. This opposite coupling between the odd and the even variables is to be called a ”reactive” coupling. Soon, we shall be acquainted with a different type of coupling, called a “dissipative” coupling. An antisymmetric matrix has purely imaginary eigenvalues. In this case, the eigenvalues, ±iω satisfy r k k =0⇒ ω=± ≡ ω0 , (1.3) −ω 2 + m m with orthogonal eigenvectors
√1 (1, ±i), 2
1 x + iˆ p), a = √ (ˆ 2
or,
1 and a∗ = √ (x − iˆ p), so that H = a∗ a. 2
(1.4)
Under time reversibility a ↔ a∗ . If we assume a solution of the type x, p ∼ exp(iωt), then a, a∗ have frequencies ω = ±ω0 , while x, p will be linear combinations of the eigenvectors involving these two frequencies. Aside:
In QM, H = |{z} ~ω ( |{z} a† a + 21 ), where (a, a† ) →
q
1 2~ω0
(a, a∗ ).
energy dim−less
A comment on scales: In QM, given a frequency ω, we have a scale ~ω for energy, which ~ , may be the width of the wavefunction. In classical gives a scale for distance ξ ∼ √2m~ω mechanics, we do not have any scale for an oscillator. A scale is fixed by the initial conditions because those fix the total energy E, 0 < E < ∞. Therefore, there is nothing a priori to make a, a∗ dimensionless.
1.1.3
Modes
The purpose of this discussion is to identify and count modes. Our mode counting would be: one mode per first order equation. In this case, there are 2 modes. The normal modes are the eigen vectors described by a and a∗ and each one has its own frequency (±ω0 ). For the oscillator, the reactive coupling gave us real frequencies, meaning oscillations. In the Newtonian or the Lagrangian formulation, x is a linear superposition of the two timereversed parts. In the Hamiltonian case it is a linear combination of exp(iω0 t) from one mode 6
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of frequency ω and exp(−iω0 t) from the other mode of a different freq. −ω0 . Since x is real, the two coefficients have to be complex conjugate, not independent. We may write x(t) = A exp(iω0 t) + c.c..
(1.5)
However, a complex number needs two real numbers and so still two conditions needed. Modes : We define modes which would diagonalize the Hamiltonian : these normal modes have individual frequencies. We started with 2 d.f. ⇒ linear combination gave us two modes. A reactive coupling, that respects time reversibility, gives oscillations or real frequencies, with conservation of energy. Aside: you may recognize a 2-d vector space. Well, space for what?
1.2
Force
If we apply a steady force F on the oscillator, it comes to an equilibrium where the forces balance, F = −kxeq . This may be stated as the equilibrium length of the spring under a force F is 1 (1.6) xeq = χ F, χ = , k the stronger the spring, the less is the extension. Under an oscillatory force F = Fω exp(iωt), there is a forced oscillation of frequency ω and amplitude x(ω) = χ(ω) F (ω), χ(ω) =
1 . m(−ω 2 + ω02 )
(1.7)
See Fig. 1.1. The zero frequency (Ω → 0) limit of χ(Ω) gives the static behaviour of Eq. (1.6). There is a divergence, a giant response when the external frequency matches the natural frequency set by the reactive coupling.
χ(ω)
ic stat ω −ω0
0
ω0
Figure 1.1: Plot of χ(Ω) vs Ω for a linear oscillator.
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1.3
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Damped Oscillator
Damping = friction = loss of energy = dissipation Friction decreases the velocity, never ever increasing it. Newton’s law gives
t→−t:
m¨ x2 + |{z} αx˙ + mω02 = |{z} |{z} even
odd
even
f |{z}
.
(1.8)
external force
Energy is not conserved. To test that, for zero force, multiply by x˙ and rewrite Eq. (1.8) in terms of energy E = 21 mx˙ 2 + 12 kx2 , as dE = −αx˙ 2 < 0, ifα < 0, dt
(1.9)
(as a square of a real number, x˙ 2 ≥ 0). The energy is conserved if and only if α = 0. Here, the total energy is decreasing with time. This loss is dissipation. The equation of motion can be expressed in the form of Eq. (1.2) with q k 0 m q , M= k − m −γ
(1.10)
which does not have any definite symmetry. Consequently it need not have pure imaginary eigen values. A coupling that destroys the anti-symmetric nature of M is called a dissipative coupling. Another way of defining a dissipative coupling is that it connects two quantities of opposite time reversal behaviour. dp/dt is even while p is odd and γ connects these two. Take x = xω e−iωt . For a non-zero solution we require, p iγ ± −γ 2 + 4ω02 α −mω 2 − iαω + mω02 = 0, ⇒ ω = , γ= , 2 m
(1.11)
Depending on the nature of the roots (real or complex), two cases are possible. There is a critical case that is left as a problem.
1.3.1
Underdamped case
If ω0 > γ/2, it is the under-damped case q ω± = ± ω02 − γ 2 /4 + iγ/2 = ±ω1 + iγ/2.
(1.12)
The solution is x = Aeiω1 t−γt/2 + c.c. = |A|e−γt/2 cos(ω1 t + φ).
(1.13)
We need both the solutions and two conditions. So far as time is concerned, we see two different scales, one scale 1/ω1 ∼ 1/ω0 and the other scale is the decay time 1/γ. For small γ there is a clear separation of scales. To understand this separation, we set γ very small to write ω1 ≈ ω0 −γ 2 /8ω0 . A dimensionless time is e t = γt so that in one time period 2π/ω0 , the change in e t is γ2π/ω0 1. In other words, 8
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in one oscillation we won’t see any significant change in the amplitude but there will be a decay over a long time. An explicit way to express this is to take x(t) = A(e t) exp(iω0 t) + c.c.,
(1.14)
as an ansatz where the oscillation part is explicitly taken into account by the exp term. This is like viewing the motion under a stroboscope of angular frequency ω0 . For the undamped case, the amplitude would be constant. The normal expectation for small γ is that the time variation introduced by the “small” damping term is a slow modulation in the amplitude. From the solution in Eq. (1.13) we do see the slow behaviour: A(T ) = A exp(−e t/2) exp(−iγ e t/8ω0 ). A direct approach would be to get the equation for A by substituting Eq. (1.14) in Eq. (1.8) whose solution is of the above quoted form. To leading order order in γ one gets (see problem below) d 1 γ A(e t) = − A(e t) − i A(e t) 2 8ω de t
(1.15)
iγA00 (e t) iγA0 (e t) Problem: From Eqs. (1.14) and (1.8), show that A satisfies − − + 2ω 2ω A(e t) A0 (e t)+ = 0, where a prime denotes a derivative with respect to e t. Use the derivative 2 rule dA/dt = γdA/de t. Then, by successive iteration, recover Eq. (1.15).
A being complex, we still have two modes, two slow modes described in a rapidly oscillating frame. Our discussion is not to redo the mechanics problem. The equation for A is instructive. There are two types of terms on the rhs of Eq. (1.15). The term with an imaginary coefficient gives us oscillations. This is a reactive term. The term with real coefficient leads to decay and is responsible for dissipation or loss of energy. This is an example of a “dissipative” term, to be generalized to dissipative coupling.
Figure 1.2: Underdamped and over-damped case. x vs t plot.
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1.3.2
Lecture 1
Overdamped: time scales
On the other hand if ω0 < γ2 , then the characteristic “frequencies” are imaginary, r ω=i ±
γ γ2 − ω02 + 4 2
with decay times τf =
!
√ γ(1+
√ √ γ γ 2 2 2 2 so that x = Ae− 2 (1+ 1−4ω0 /γ )t + Be− 2 (1− 1−4ω0 /γ )t ,(1.16)
2 1−4ω02 /γ 2 )
and τs =
√ γ(1−
2 . 1−4ω02 /γ 2 )
See Fig. 1.2. Obviously, the τf
mode decays faster. So for large t, x = Be−t/τs . At long times we are left with the slow modes. ⇒ One mode is lost. Problem: Take the two cases of Fig. 1.2. The functions plotted are given there. Determine the values of ω0 and γ for each case.
Let’s look at the two time scales in the large friction limit s α γ 2 ω0 γ 4ω , ωf = 1 + 1 − 20 2−→ γ = 2 γ m s 2 γ 4ω0 γ2 ω0 ω02 k ωs = 1+ 1− 2 = . −→ 2 γ γ α
(1.17)
(1.18)
In the large friction limit, the faster scale is controlled by K.E. and friction. Dimensionally 2 hαi ω0 1 ≡ [ω0 ] ≡ ≡ , (T is the dimension of time). (1.19) m γ T The comparison of the friction and the inertial term is like a free motion with friction and particle has not yet felt the action of the spring. On the other hand, the comparison of friction and the oscillator term is like comparing the two forces on the particle. This gives two time scales. When the first mode, more like a free particle, dies out, we may ignore inertial part. This the like taking m → 0.
the the the the
For m → 0, τf → 0 (very rapid event) : no “inertial” effect. The equation then is γ x˙ + ω02 x = f /m ⇒ x = Be−t/τ ,
τ=
w0 . γ
In the overdamped limit k 1 1 x˙ = − x + f = α x α
∂H 1 − , with H = kx2 − f.x . |{z} ∂x 2
(1.20)
work done
This is written in a generic form x˙ = −Γ 10
∂H , i.e. ∂x
(1.21)
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where the “force” is −∂H/∂x and Γ a dissipative constant. There is no kinetic energy, no Poisson Bracket relation. Γ is called a kinetic coefficient that relates the force with the velocity, ddt variable = (kinetic coefficient) × force. This equation of motion is not in the standard form x¨ = −
1 ∂H . m ∂x
as dictated by Sri Newton (or Galileo?). Eq. (1.21) is the Aristotle force law. We instead have m = 0. Dimensionally Γ−1 is not like a mass. The first order equation is to be taken as a phenomenological equation. It is phenomenological because we assumed the frictional force to be αx˙ with α a constant. Frictional force could be more complicated than this. The complications could go in the dissipative constant Γ; even it may be very large, ∞, as in systems near a critical point. But, this is the simplest possible choice that does not have time reversibility, thereby giving us a dissipation of energy. Isn’t it ironic that for nonequilibrium Statmech we are going from the Newton/Galileo viewpoint to the older Aristotelian view? Where does the energy go? That’s heat - but we postpone that issue for the time being. All said and done, in this extreme overdamped limit, we still have two modes, governed by two separate equations, x˙ = −Γ
∂H , ∂x
and p˙ = −
α ∂H p = −Γp m ∂p
(fast mode),
(1.22)
with dissipative couplings Γ, Γp . Compare the three cases (boxed equations) discussed so far. For the overdamped oscillator, the equations may be put in way similar to Eq. (1.2)as d dt
k x x −α =M , with M = p p 0
0 α m
(1.23)
Aside: To see how the p equation comes about, multiply Eq. (1.8) by m/α2 and use e t to obtain x00 +x0 +m(k/α2 )x = mf /α2 , where, as before, the prime denotes a derivative with respect to e t. Now by takimg m → 0, we are left with the first two terms which can be recast in the form of Eq. (1.22), with p = mx. ˙
1.4
Lessons from oscillator: summary
The main purpose of the recapitulations is to define or emphasize the importancereactive coupling and dissipative coupling, and link them to time reversibility. The discussion is the framework of first order differential equations. • A reactive coupling for a pair of variables of opposite time-reversal behaviour (like x and p), could be off-diagonal type with a sign difference (see Eq. (1.1)) or equivalently by an antisymmetric matrix as in Eq. (1.2). Otherwise a reactive coupling may occur with an imaginary coupling as in Eq. (1.15). Reactive couplings tend to conserve energy. 11
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Dissipative coupling do not involve any time reversal symmetry, as in Eq. (1.22). They lead to loss of energy. • The idea of modes, countable at this point. Much like the normal modes in small oscillations in mechanics, these modes may have characteristic oscillation frequency (from reactive coupling) and/or characteristic decay time (from dissipative coupling). Aside: A mathematical issue: In the long time limit, we have a solution which involves only one constant, not 2. One solution, fixed by one condition, is like a solution of a first order differential equation. Eq. (1.8)is a 2nd order differential equation with the requirement of two conditions, say x(t = 0) and x(t ˙ = 0), no matter how small m is. But in the limit m → 0, the solution xm (t) for mass m goes over to xm=0 (t) which is the solution of αx˙ + kx = 0. The solution of the first order equation cannot in general satisfy the two given initial conditions. Although there is a continuity in the equation and the solution with respect to m, there is no continuity in the set of initial conditions. The fast time scale actually plays a special role. No matter how small this scale is, for any m, there is always a short time period where this fast motion takes place. May be, we discuss this issue later on. This is called a boundary layer problem or a singular perturbation theory.
Problem: We discussed the underdamped and the overdamped cases. There is still another one — the boundary of the two. This is left as a problem. Discuss the time evolution for the critical case when ω has degenerate roots, i.e., when γ = 2ω0 .
12
Chapter 2 Response function I 2.1
How do we study a system: Response function
Any system is probed by its response to external perturbations. E.g: We apply a force or a stress to see how a system changes its size and shape, we apply a magnetic field to explore the magnetic property of any substance and so on. If we apply a perturbation f , there is a response x. It is perfectly normal to expect that a small cause produces a small change. So x is a function of f and for f → 0, x → 0. It has to be because there is no effect if ”cause” does not exist. With small quantities and nothing spectacular happening, a Taylor series expansion gives x(f ) = χ f +
1 1 χ2 f 2 + χ3 f 3 + ..... 2 3!
Restrict to the first order term. That’s called a “linear” response. We are aware of such behaviour in the context of the general properties of matter, e.g., 1 stress = ) strain K M = χ h (χ : magnetic susceptibility) dQ = C dT C : Specific heat P = χe E χe : (electrical susceptibility).
strain = K stress
(elastic modulus =
(2.1) (2.2) (2.3) (2.4)
These are called “response functions”: In most cases, the aim in theory or experiment is to study the nature of various response functions of a system. In these cases, the “cause” is an intensive variable in the thermodynamic sense while the right hand side is an extensive quantity. For generality, we use the nomeclature, “cause” ↔ “generalized” force in short force “effect” ↔ “generalized” displacement in short displacement dropping, without any implication, the adjective “generalized”. Given a system described by a Hamiltonian H in zero force, the external force yields a new source of energy. The Hamiltonian then needs to be augmented by the “work done” term as Hf = H − f x. For a magnet in a field Hh = H − h · M . The linear occurence of f in the 13
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Hamiltonian does not necessarily guarantee a linear form at the end, say after averaging in statistical mechanics or after solving equations of motion in mechanics. Harmonic oscillator is of course a very special case. The examples quoted above refer to static situations. The extension of the idea to time dependent situations, mainly under an oscillatory drive, would not pose any problem, except an important difference. Be ready for complex-valued response function.
2.2
Definition of response functions
How do we define a response function? If for a force f the response is x, then the response function (“generalized susceptibility” or susceptibility in short) is defined as ∂x , χ= ∂f f =0
(2.5)
The condition f = 0 in the definition can be relaxed if we want the response at a non-zero force. The linear response form would be for small deviations, like ∆x = χ(f )∆f, with χ(f ) = ∂x/∂f evaluated at force f . Most of the time, the zero-force response function suffices. If f is not ∂2x small, then higher order response functions are needed, e.g. χ2 = ∂f 2 , and so on. f =0
Aside: Trouble starts with this way of studying a system if it is very ”soft”. Soft means a small external perturbation produces a huge change. One may go to the limit of χ → ∞. In this situation, the Taylor, series expansion fails. It might then help us to consider the inverse function, f as a function of x and write f = χ−1 x + C2 x2 + C3 x3 + .... |{z}
(2.6)
→0 if χ→∞
We then see the leading order behaviour f ∝ xδ or x ∼ f 1/δ : a highly non-linear behaviour, with δ > 1 in general. An example would an anharmonic spring with potential energy 41 k4 x4 (not quadratic). Under a steady force, the equilibrium position is x3eq = k14 f which is of the above form with δ = 3. This is to be compared with the linear form in Eq. (1.6) for the harmonic spring. The nonlinear behaviour is a characteristic of a soft material or a “soft” system. We know of many soft systems. But, by this definition, a magnet at its curie point with infinite susceptibility (remember Curie Weiss law χ ∝ | T − T c |−1 ) is “magnetically” soft though a rigid solid (elastically). Interestingly, at the Curie point, there is a nonlinear law M ∼ h1/δ with δ = 3 as found for the quartic spring. Linear susceptibiity is ∞.
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Oscillator response
Let’s come back to our oscillator problem. To probe a damped oscillator we put it under a f , then gives for x = xω e−iωt periodic force f = fω e−iωt . The equation of motion x¨ + γ x˙ + ω02 x = m χ(ω) =
2.3.1
1 xω 1 = . 2 fω m −ω + ω02 − iωγ
(2.7)
General features
The special features are: • χ(ω) is complex, with poles in the complex ω-plane, at the frequencies associated with the “modes” (ω± for the underdamped case or ωf,s for the overdamped case. • By going over to the complex ω-plane we can treat both the oscillatory and the decaying modes in the same way. See Fig. 2.1
complex ω plane undamped
Re ω ❑ ▲
over−damped
▲
under−damped
❑
Im ω Figure 2.1: Poles of χ(ω) in the complex ω-plane (convention adopted to define ω is exp(iωt). Poles on the real axis represent undamped modes (real frequencies). Poles on the imaginary axis represent overdamped modes (time scales for decay). General complex poles represent under-damped, oscillatory modes with decay.
• Take ω = 0. This is the case of a static force. It gives kx = f or χ = 1/k. This agrees with the ω → 0 limit of χ(ω). • Let us write χ(ω) = χ0 (ω) + iχ00 (ω) (real and imaginary part), so that χ0 (ω) =
ω02 − ω 2 1 m (ω02 − ω 2 )2 + ω 2 γ 2
1 For large ω, χ0 (ω) ∼ − mω 2.
& χ00 (ω) =
1 ωγ . 2 m (ω0 − ω 2 )2 + ω 2 γ 2
Why negative?
Let us discuss the underdamped and the overdamped cases separately. 15
SMB/NESM
2.3.2
Lecture 1
Underdamped case
• Imaginary part: χ00 (ω) is odd in ω 1 1 1 1 1 1 1 00 χ (ω) = − + − m 2ω1 2 ω − ω1 − i γ2 ω + ω1 − i γ2 ω − ω1 + i γ2 ω + ω1 + i γ2 " # 1 1 γ/2 γ/2 = 2 − 2 γ 2 m 2ω1 (ω − ω1 ) + (ω + ω1 )2 + γ 4
γ→0
−→
(2.8) (2.9)
4
1 [δ(ω − ω0 ) − δ(ω + ω0 )] 2mω0
(2.10)
• Real part χ0 (ω).
0 at ω = ω0 , but → ∞ as γ → 0, 1/k at ω = 0, χ (ω) = 2 −1/mω for ω → ±∞, 0
(2.11)
and it is symmetric around ω = 0. Fig.XXX The solution is x(t) = Re χ(ω) fω e−ωt = |χ(ω)| f cos(ωt − φ), with tan φ =
χ00 (ω) ωγ = 2 . 0 χ (ω) ω0 − ω 2
(2.12)
Here, χ(ω) = |χ(ω)|eiφ . There is a phase shift in response. What is a phase shift? When the force is maximum, the displacement is not. This lag is coming from the imaginary part. The phase shift is completely characterized by χ(ω). There is a phase shift because there is friction and φ → 0 as γ → 0. Problem: Plot |χ(ω)| and φ vs ω.
2.3.3
Overdamped case
For the overdamped case, γ is large, ωf,s = −iτf−1 , iτs−1 , so that 1 1 −1 m −(ω + iτf )(ω + iτs−1 ) 1 τf = m −(ωτf + i)(ω + iτs−1 )
χ(ω) =
τf ≤ τs
(2.13) (2.14)
If ωτf 1 i.e. the time periodicity is much larger than τf , we may ignore it to get χ(ω) =
16
1 1 1 τf τs 1 = =χ 2 m 1 − iωτs 1 − iωτs +mω0 (1 − iωτs ) | {z }
(2.15)
Lecture 1
SMB/NESM
This gives τs τs−1 χ00 (ω) =χ =χ 2 , ω 1 + ω 2 τs2 ω + τs−2 so that for large ω,
χ00 (ω) ω
(2.16)
→ 0, as 1/ω 2 .
We are talking of complex functions and complex functions need to satisfy some special conditions: Cauchy-Riemann Cond. What’s that? ⇒ Kramers-Kr¨onig relation. Problem: 1. Plot χ0 (ω) and χ00 (ω) for both the underdamped and the overdamped cases. 2. For the oscillator, both undamped and damped, derive the first order equations of motion for the normal modes a and a∗ . From these, devise the rules t identify reactive and dissipative couplings for complex variables. 3. Suppose a set of coupled oscillators is described by equations of the type: x¨n + ω 2 xn + γ x˙ n + D(xn+1 + xn−1 − 2xn ). Identify the reactive and dissipative coupling. 4. Represent the amplitude and phase of the oscillators by a complex number zn = rn eiθn . Suppose the equation of motion for a set of oscillators is z˙n = iωzn − γzn + (K + iΓ)(zn+1 + zn−1 − 2zn ) = 0. Identify the reactive and dissipative couplings. 5. The equation for a damped harmonic oscillator resembles the equation for an LCR circuit. Identify the reactive and dissipative couplings for an LCR circuit.
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