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Linear Response in Classical Physics
Wouter G. EllenbroekTechnische Universiteit Eindhoven
Notes (section 2) by Fred MacKintosh (Vrije Universiteit),used previously in the 2011 DRSTP course
1 Introduction
The theory of linear response deals with the question how the measurable prop-erties of a physical system change when it is slightly taken away from its equilib-rium state. Traditionally, the term refers to linear response of thermal systemsto slight deviations from thermodynamic equilibrium. In the first part of thiscourse, we will deal with the question how this response is related to the equi-librium thermal fluctuations in the system. The notes for this first part havebeen provided by Fred MacKintosh, who taught a similar course in a DRSTPschool a few years ago.
In the second part, we will consider an athermal system that is in mechan-ical equilibrium, but not in thermodynamic equilibrium, and study its (linear)response to mechanical perturbations.
The course will be followed up with a course on quantum linear responsetheory, taught by Rembert Duine.
2 Fluctuations and Linear Response
2.1 Brownian Motion: Random Walks in 1d
We begin with arguably the simplest problem illustrating the effect of fluctu-ations, e.g., in a fluid: Brownian motion. This term derives from the botanistRobert Brown who observed random motions of pollen grains in a fluid in 1828.This kind of motion is general to sufficiently small particles suspended in a fluid.By observing different kinds of particles, he concluded that the motion was notdependent on the type of particles. We now know that this motion is due to theever-present thermal fluctuations in the fluid itself. In a series of papers begin-ning in 1905, Einstein gave a theoretical basis for Brownian motion. About thesame time, Smoluchowski derived the same results in a somewhat different way.
Here, we derive the properties of what is usually referred to as the randomwalk problem for a single particle. Despite the seeming simplicity, this consti-tutes a rich example that will help to illustrate many of the principles of fluctu-
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ation about equilibrium generally. Examples of such random walks include theBrownian motion of a particle in a fluid, as well as the random conformations oflong polymer molecules under some conditions. In the first of these examples,the essential physical aspects of this problem are: the existence of randomlyfluctuating forces pushing the particle in random directions, the viscous drag ofthe particle in the fluid, and the inertia of the particle. As we shall see later,the the drag of the particle (or equivalently, its mobility when subject to an ex-ternal force) is fundamentally related to the strength of the random forces thatit experiences in the fluid. We’ll see later a statement of this deep connectionbetween fluctuations and dissipation in the form of the so-called Fluctuation-Dissipation Theorem, and it is largely with this in mind that we shall examinethe basic problem of Brownian motion in such great detail.
We first look at Brownian motion in just one spatial dimension. Consider aparticle that can move just left (L) and right (R) along a line by steps of equallength `. Steps to the left and right occur randomly, with equal probabilityp = 1/2. Because the probability to jump to the left and to the right are equal,the average displacement of a particle during the i-th step is zero:
〈∆xi〉 =1
2(+`) +
1
2(−`) = 0 . (1)
However, the mean squared displacement is non zero:
〈∆x2i 〉 =
1
2(+`)2 +
1
2(−`)2 = `2 . (2)
Next consider average displacement after N steps:
〈∆X〉 =
N∑i=1
〈∆xi〉 = 0 . (3)
The mean squared displacement after N steps is:
〈∆X2〉 =
N∑i=1
N∑j=1
〈∆xi∆xj〉 . (4)
The jump directions on different steps are not correlated, therefore
〈∆xi∆xj〉 = 0 for i 6= j . (5)
For i = j, we have 〈∆xi∆xj〉 = `2. If we insert this in the expression for 〈∆X2〉we get
〈∆X2〉 =
N∑i=1
N∑j=1
`2δij . (6)
where δij is a Kronecker delta:
δij ≡ 1 for i = j
= 0 for i 6= j .
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Hence:〈∆X2〉 = N`2 . (7)
If the time interval between successive jumps of the particle is δt, then thenumber of jumps in a time interval t is equal to t/δt and
〈∆X2〉(t) = `2t/δt . (8)
In other words, the mean squared displacement increases linearly with time. Weshould compare this with the ballistic motion of a particle with a velocity v, inwhich case ∆X(t) = vt, and hence 〈∆X2〉(t) = 〈v2〉t2.
For the simple case of one-dimensional Brownian motion we can compute notjust the mean-squared displacement of a particle, but the complete probabilitydistribution to find a particle at a distance x from its original position, after Nsteps. We denote by NL and NR the number of steps that occur to the left andright. The total number of steps is then N = NL + NR, while the net lengthof the walk (to the right) is x = ` (NR −NL). The number of such walks totalsteps N , of which NL steps are to the left and NR to the right is
Ω (x,N) =N !
NL!NR!. (9)
Then, the probability of a such a walk of net length x is
P (x,N) =N !
NL!NR!
(1
2
)N=
N !(N−x/`
2
)!(N+x/`
2
)!
(1
2
)N. (10)
By using Stirling’s approximation n! ∼=√
2πnnne−n, we find that
P (x,N) ∼=NNeNR+NL
√2πN
2NeN(N+x/`
2
)[N+x/`]/2 (N−x/`
2
)[N−x/`]/2√2πNR
√2πNL
=
√2πN
(1 + x/N`)[N+x/`
2 ](1− x/N`)[N−x/`
2 ]√π(N + x/`)
√π(N − x/`)
=
√2
(1− (x/N`)2)N/2(
(1+x/N`)(1−x/N`)
)x/2`√Nπ(1− x2/(N`)2)
Now we make use of the fact that
(1− (x/N`)2)N/2 ≈ exp(−x2/(2N`2))
and that
ln
((1 + x/N`)
(1− x/N`)
)x/2`≈ (x/2`) (x/(N`) + x/(N`)) = x2/(N`2)
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and hence ((1 + x/N`)
(1− x/N`)
)x/2`= exp(x2/(N`2)) .
We can then write
P (x,N) ∼=√
2
πN
1
exp(−x2/(2N`)2) exp(x2/(N`2))√
1− x2/(N`)2
≈√
2
πNexp(−x2/(2N`2)) (11)
Of course, it comes as no surprise that this probability distribution is symmetricand that the most likely walk returns to the origin (x = 0). Thus, the average〈x〉 = 0. The mean-square displacement for such random walks, however, is non-zero. This is most easily calculated by treating P (x,N) above as a continuousdistribution in the variable x. In fact, for small enough x/`, the distributionabove varies slowly over the discrete values of x that are possible for a givenlarge value of N . Thus,
〈x2〉 ∼=∫x2e−x
2/(2N`2)dx∫e−x2/(2N`2)dx
= N`2. (12)
Here, we have assumed that the particle moves with independent random stepsof length `. If we assume further that these steps occur at equally-spaced timeintervals δt, then t = Nδt and
〈x2〉 ∼= t`2/δt. (13)
This behavior x2 ∝ t or x ∝√t is characteristic of diffusion.
Before leaving this simple one-dimensional random walk problem, we canmake another observation. This problem is instructive for other physical sit-uations, such as the conformations that a long polymer chain can make. Theresulting conformations are referred to as Gaussian. We ignore the fact thatthe polymer chains cannot overlap in space. (Although an unphysical micro-scopic assumption, there are real situations where polymer conformations arenevertheless Gaussian.) Then, Ω(x,N) above represents the number of suchconformations, subject to a given end-to-end separation x. Thus, by the funda-mental assumption of statistical mechanics (that microscopic states of the sameenergy are equally likely), the probability of having a given value of x is pro-portional to Ω(x,N), the number of such conformations or states of the system.Since, for this simple problem, there are 2N states in all (N independent steps,with 2 choices at each), the normalized probability is just P (x,N) above.
2.2 Diffusion
One can also study Brownian motion from the point of view of diffusion. Here,we denote the density of (identical) particles suspended in the fluid by a con-centration variable n(~r, t), which is a function of both position ~r and time t.
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We also denote by ~j(~r, t) = n(~r, t)~v(~r, t) the current density of particles at ~rand time t, where ~v is the velocity. In thermal equilibrium, and in the absenceof other forces, the average density 〈n(~r, t)〉 will be constant. Furthermore, bythe Brownian motion of particles in the fluid, an initial inhomogeneity in thedensity will relax over time toward this average, uniform density. Thus, spatialvariations (gradients) in n result in net particle motion, i.e., a particle current.For small variations in n, we expect that the corresponding current ~j is small.Thus, it is natural to assume a linear relationship:
~j = −D~∇n, (14)
which is known as Fick’s law. As we shall see, the minus sign agrees withour physical intuition that Brownian motion will lead over time to a uniformparticle density. A positive sign here would lead to unphysical results. Thus,we see immediately that Brownian motion or diffusion is essentially dissipativeand irreversible. The constant D here is known as the diffusion constant, whichhas units of square length per time.
There is one more physical aspect to Brownian motion that we have neglectedin this treatment so far, namely the fact that particles are conserved. If noparticles are created or destroyed while they fluctuate around in the fluid, thenthe particle current and density are related by conservation
∂n(~r, t)
∂t+ ~∇ ·~j(~r, t) = 0,
which is simply a mathematical expression of the fact that a net particle currentinto (out of) any volume results in an increase (decrease) in the number ofparticles in that region. This, together with Fick’s law leads to the diffusionequation,
∂n(~r, t)
∂t= D∇2n(~r, t).
A particular solution to this equation is
n(~r, t) =N
(4πDt)3/2
e−r2/(4Dt). (15)
This is, in fact, the solution 1 to the initial condition that N particles are placedat the origin at time t = 0, i.e.,
n(~r, 0) = N δ(~r).
This result is for a number N of particles in the fluid. Provided that thesedo not interact with each other, which is valid if the density is not too high, thenwe could have equivalently described things in terms of the probability densityP (~r, t) for a single particle, which is given by Eq. (15) with N = 1. We thenfind that
〈~r(t)〉 = 0; 〈r2(t)〉 =
∫r2P (~r, t)d3r = 6Dt ∝ t,
1See Appendix A.
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which is in agreement with our earlier result for the one-dimensional randomwalk:
〈x(t)〉 = 0; 〈x2(t)〉 = t`2/δt = 2Dt ∝ t,
where D = `2/(2δt). The general result depends on the number of dimensionsd: 〈r2(t)〉 = 2dDt.
Even without solving the diffusion equation explicitly, we can relate 〈x2(t)〉to D. We do this as follows. We start with the one-dimensional diffusionequation for the probability density P (x, t):
∂P (x, t)
∂t= D
∂2P (x, t)
∂x2.
We now multiply both sides of this equation by x2 and integrate over x:∫ ∞−∞
dx x2 ∂P (x, t)
∂t=
∫ ∞−∞
dx x2D∂2P (x, t)
∂x2.
We can write the left hand side of this equation as∫ ∞−∞
dx x2 ∂P (x, t)
∂t=
∂
∂t
∫ ∞−∞
dx x2P (x, t) =∂〈x2(t)〉∂t
The right-hand side can be computed by partial integration (making use of thefact that P (x, t) and its derivative vanish at x = ±∞:∫ ∞
−∞dx x2D
∂2P (x, t)
∂x2= −
∫ ∞−∞
dx 2xD∂P (x, t)
∂x= +2D
∫ ∞−∞
dx P (x, t) .
But as P (x, t) is normalized to one, the result is simply 2D and therefore:
∂〈x2(t)〉∂t
= 2D (16)
This we can integrate to yield
〈x2(t)〉 = 2Dt
where we have used the fact that at t = 0, the probability distribution is aδ-function. We can write similar expressions for 〈y2(t)〉 and 〈z2(t)〉. Addingthese, we obtain
〈r2(t)〉 = 〈x2(t)〉+ 〈y2(t)〉+ 〈z2(t)〉 = 6Dt .
In reality, for the example of a particle suspended in a fluid, we expect that thekicks that the particle experiences in the fluid are random. Even so, since theparticle has a mass, there will be some effect of its inertia. So, we expect that themotion is not truly random, but will be correlated for short times. For such shorttimes, the particle will move ballistically, and we expect that the displacementgrows linearly with time. Thus, we expect that 〈x2〉 ∝ t2, rather than 〈x2〉 ∝ t
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for short enough times. Actually, this must be true. We can already see thatthere is a problem with the Eq. (13), because it would imply that x '
√t,
which leads to an unphysical divergence of the velocity at short times. Thus,only for times long compared with some microscopic correlation time (which wecan be more quantitative about below) will diffusion, characterized by Eq. (13),be valid.
2.3 Velocity correlation function
x(t) is the distance that a particle has traveled in the x-direction in a timeinterval t. If we denote the instantaneous velocity in the x-direction by vx(t),then we can write
x(t) =
∫ t
0
dt′vx(t′) .
We can then write the mean-squared displacement as
〈x2(t)〉 =
∫ t
0
dt′∫ t
0
dt′′〈vx(t′)vx(t′′)〉 .
The quantity 〈vx(t′)vx(t′′)〉 is an example of a time correlation function - in thiscase, the “velocity auto-correlation function” (VACF). This quantity measuresthe degree of correlation between the velocity of a particle at two different timest′ and t′′. When t′ = t′′, the value of the of the velocity correlation function issimply 〈v2〉. At sufficiently long times, when the particle has lost all memory ofits original velocity, 〈vx(t′)vx(t′′)〉=〈vx(t′)〉 × 〈vx(t′′)〉 =0 (because the averagevelocity of a particle in equilibrium is zero). Time correlation functions, such asthe VACF are properties of the system in equilibrium. Their value can thereforenever depend on the times t′ and t′′ individually, but only on the difference. Thisis so because the properties of a system in equilibrium are time invariant, i.e.:
〈vx(t′)vx(t′′)〉 = 〈vx(t′ + τ)vx(t′′ + τ)〉 (∀τ)
If we choose τ = −t′′, we get
〈vx(t′)vx(t′′)〉 = 〈vx(t′ − t′′)vx(0)〉 ,
which shows that the VACF depends only on the difference t′− t′′. Another im-portant property of (classical) time correlation functions is that we can permutevx(t′) and vx(t′′):
〈vx(t′)vx(t′′)〉 = 〈vx(t′′)vx(t′)〉 .
Because of this symmetry, we can write∫ t
0
dt′∫ t
0
dt′′〈vx(t′)vx(t′′)〉 = 2
∫ t
0
dt′∫ t′
0
dt′′〈vx(t′)vx(t′′)〉 .
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We make use of this result in evaluating
∂〈x2(t)〉∂t
= 2∂
∂t
∫ t
0
dt′∫ t′
0
dt′′〈vx(t′)vx(t′′)〉
= 2
∫ t
0
dt′′〈vx(t)vx(t′′)〉
= 2
∫ t
0
dτ〈vx(τ)vx(0)〉 ,
where we have defined τ ≡ t − t′′. If we insert the above result in Eqn. 16, weobtain
D =
∫ t
0
dτ〈vx(τ)vx(0)〉 . (17)
This equation is only valid for times much longer than the decay time of corre-lations in the velocity, and hence we may replace the upper limit in the integralby +∞. The expression that results is
D =
∫ ∞0
dτ〈vx(τ)vx(0)〉 . (18)
2.4 Langevin Theory
In order to see at a deeper level the connection between the random fluctuationsin the fluid and the dissipative nature of the resulting particle motion, we turnnow to an approach suggested by Langevin. This will also solve a problem notedabove, namely that there must be a failure of our prior analysis of Brownianmotion at short times. Our approach here will be to more directly account forthe fluctuating forces acting on the particles by the surrounding fluid. We shallsee how thermal equilibrium (e.g., the expected uniform spatial distribution ofparticles) can be established by these fluctuations.
We begin with the simple equation of motion satisfied by the particle of massm
md~v
dt= F(t),
where F is the force acting on the particle. We might be tempted at this pointto assume that this force is random, and independent of the particle motion.After all, these forces are due to the fluctuating solvent, right? In fact, there aretwo physically distinct parts to this force: (i) the viscous drag on the particle,
which depends on ~v and (ii) the randomly fluctuating forces ~ζ(t). We assumehere that the first of these can be described by the Stokes formula for the dragon a spherical particle in a viscous fluid:
−6πηa~v,
where η is the fluid viscosity and a is the radius of the sphere. This is validfor small particles and small velocities. It is important to note that this force
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is random and has zero mean, only to the extent that the particle velocity israndom with zero mean. Thus, we can say that, in the absence of externalforces, it has a zero time average.
It is important also to distinguish another kind of average that we have beena bit sloppy about so far: the ensemble average. This represents an average overa large number of similarly prepared systems (an ensemble). We can consider,for instance, the motion of a particle with some initial condition, say velocity~v(0) at time t = 0. We expect that this initial velocity will change over time,and that it will become uncorrelated with the velocity at much later times t. Letτ denote this correlation time, after which memory of the initial conditions islost. As a specific example, if this initial velocity were large compared with theexpected thermal velocities, we would expect that it would decay toward zerofor times large compared with τ . But, the precise way in which it will do thisdepends on the details of the fluctuating force ~ζ due to the fluid. We are notinterested in the exact trajectory that an individual particle makes due to eachand every collision with individual fluid molecules. Rather, we want to knowwhat to expect for the way the particle velocity decays in a typical sample of acertain kind of solvent (e.g., with known viscosity η) at a certain temperature.Thus, we want to know how the particle velocity behaves on the average, i.e.,the ensemble average. However, the ensemble average of the random force iszero 〈ζ(t)〉 = 0, while the ensemble average of the particle velocity is not zero, atleast for times smaller than τ . Mathematically, we could say that the ensemblecorresponds to a large set of possible functions ~ζ(t) consistent with a givenfluid at a given temperature. The properties of these functions are yet to bedetermined, but that the average 〈ζ(t)〉 over this set is zero for all times t is clearby construction. At any time t, ensemble average quantities can be obtained byaveraging over the possible functions ~ζ(t).
We rewrite the equation of motion as follows
md~v(t)
dt= −6πηa~v(t) + ~ζ(t). (19)
If we take the average of this over the ensemble, we obtain
md〈~v(t)〉dt
= −6πηa〈~v(t)〉,
which implies that〈~v(t)〉 = ~v(0)e−t/τ ,
where we can now identify τ = m/(6πηa) as the relaxation time for effects ofthe initial condition to die out. Here again the essentially dissipative natureof this result is evident by the fact that this result cannot be reversed in time,although the precise connection of this irreversibility to the fluctuating forces ~ζis still unclear. This is what we want to establish next.
We can take the scalar product of (19) with the instantaneous position ~r(t).In so doing, we note that ~r · d~vdt = d
dt (~r · ~v)− v2. The result is
md
dt〈~r · ~v〉 = −6πηa〈~r · ~v〉+m〈v2〉. (20)
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We have used the fact that 〈~r · ~ζ〉 = 0. This last point is not as obvious as it
may appear at first sight. Although 〈~r · ~ζ〉 = 0, it can be shown that 〈~v · ~ζ〉 6= 0.In the equation above, if we further assume that the Brownian particle has
attained thermal equilibrium, then we can replace 〈v2〉 by the equipartition value3kT/m. (The kinetic energy mv2/2 of the particle involves three independentdegrees of freedom, in which the energy is quadratic. Classical Statistical Me-chanics then predicts that this energy is 3
2kT on average.) The resulting simpledifferential equation can be solved:
〈~r · ~v〉 = Ce−t/τ +kT
2πηa.
But, the constant C can be determined by the initial condition, which we taketo be that ~r = 0 at t = 0:
〈~r · ~v〉 =kT
2πηa
(1− e−t/τ
).
We can integrate this one more time, to find that
〈r2〉 =kT
πηa
[t− τ
(1− e−t/τ
)],
since 〈~r · ~v〉 = 12ddt 〈r
2〉.At long times, for which t τ , we find again that
〈r2〉 = 6Dt.
Furthermore, we can now identify the diffusion coefficient with the dissipativemechanism of the fluid viscosity:
D =kT
6πηa. (21)
We also find, as expected, that for short times, 〈r2〉 ∝ t2, thus resolving theproblem with our earlier analysis for short times.
More generally, there is a fundamental relationship between the particlediffusion constant D and the mobility µ or response of the particle to externalforces. If we apply an external force ~f to the particle, say due to gravity, thenthe resulting drift velocity is 〈~v〉 = µ~f . This defines what we mean by themobility. For our Stokes sphere in the viscous fluid, the full equation of motionbecomes
md~v(t)
dt= −6πηa~v(t) + ~f + ~ζ(t). (22)
Thus, the mobility in this case is µ = 16πηa , and
D = µkT . (23)
This is known as the Einstein relation, and is more general than this particularproblem of a sphere in a viscous fluid. It is also an example of the fundamental
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relationship that exists between a dissipative response and fluctuations. In thiscase, the response is that of a particle to an external field. This response ischaracterized by the mobility µ. The fluctuating quantity here is the particleposition ~r(t).
There is another way to derive the Einstein relation that makes the relationbetween the mobility µ and the diffusion constant D more obvious. Considersolution in a closed volume V . If there is a concentration gradient of the dis-solved species, this will result in a diffusion current ~j. The diffusion is equal tothe number density ρ of the dissolved species, multiplied by the average velocityof these particles.
~j = ρ〈~v〉 .Now suppose that the dissolve particles are subject to an external potentialU(x). If this potential is not constant, there will be a net force acting on theparticles:
fx = −∂U(x)
∂x,
where we have assumed that the potential ia a function of x only. The averagevelocity of a particle due to this force is
vx = µfx = −µ∂U(x)
∂x.
As the particles move under the influence of this force, the density will becomeinhomogeneous. But once the density is not constant, there will also be adiffusion current. When the system reaches equilibrium, the diffusion current isexactly balanced by the current due to the external force, i.e.:
0 = ρ〈vx〉 = −ρ(x)µ∂U
∂x−D∂ρ(x)
∂x. (24)
But we also know that, in equilibrium, the probability to find a particle at aposition x must be proportional to the Boltzmann factor exp(−U(x)/kT ):
ρ(x) = ρ0 exp(−U(x)/kT ) .
If we insert this expression in Eqn. 24, we get
0 = ρ〈vx〉 = −ρ(x)µ∂U
∂x+
D
kTρ(x)
∂U(x)
∂x. (25)
This equation is satisfied for all x if
D = µkT ,
that is, the Einstein relation.We can actually establish a more detailed connection between dissipation
(specifically, the drag coefficient α = 6πηa here) and the fluctuating force ~ζ(t).To do this, we rewrite the equation of motion, dividing out by the particle mass
d
dt~v(t) = −1
τ~v(t) +
1
m~ζ(t). (26)
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This, we rearrange slightly and multiply by et/τ :
et/τd
dt~v(t) +
1
τet/τ~v(t) =
d
dt
(et/τ~v(t)
)= et/τ
1
m~ζ(t). (27)
We now change t to t′ and integrate from 0 to t, with the result that
~v(t) = ~v(0)e−t/τ + e−t/τ1
m
∫ t
0
et′/τ~ζ(t′)dt′. (28)
Again, we see that if we take the ensemble average of this equation, we obtain
〈~v(t)〉 = ~v(0)e−t/τ ,
as we found before. This is because 〈~ζ(t′)〉 = 0 for any t′. If we square bothsides of (28) and then take the ensemble average, however, we find that
〈v2〉 = v(0)2e−2t/τ + e−2t/τ 1
m2
∫ t
0
∫ t
0
e(t′+t′′)/τ 〈~ζ(t′) · ~ζ(t′′)〉dt′dt′′. (29)
Again, a term had been dropped because 〈~ζ〉 = 0. The correlation function
K(t, t′) = 〈~ζ(t) · ~ζ(t′)〉, however, is non-zero in general. This is where the uglydetails of the fluctuating molecules in the fluid begin to appear directly in ouranalysis. We finally have a measurable quantity that would appear to dependdirectly on the random forces ~ζ.
These appear through the correlation function K(t, t′) = 〈~ζ(t) · ~ζ(t′)〉, whichhas several general properties that we can identify. First of all, so long as weare looking at an equilibrium system that is not evolving in any macroscopicsense in time, then this function can only depend on the time interval s =t− t′: K(t, t′) = K(s). Furthermore, provided that we confine our attention toclassical variables (as opposed to possibly non-commuting quantum mechanical
operators), we can reverse the order of ~ζ(t) and ~ζ(t′), which tells us that K(s)is symmetric, i.e., that it depends only on |s|. On simple physical grounds, wealso expect that K(s) decreases as the time interval s increases, since we expect
the force ~ζ(t) to lose memory of its earlier values. More precisely, we expect
that ~ζ(t) and ~ζ(t′) become decorrelated as |t− t′| grows, i.e., that
K(s) = 〈~ζ(s) · ~ζ(0)〉 → 〈~ζ(s)〉 · 〈~ζ(0)〉 = 0
as s grows. Furthermore, we can see that K(s) attains its maximum value ats = 0. In fact, since
〈[~ζ(s)± ~ζ(0)
]2〉 = 〈ζ2(s)〉+ 〈ζ2(0)〉 ± 2〈~ζ(s) · ~ζ(0)〉 = 2 (K(0)±K(s)) ≥ 0,
the function K(s) is bounded by the limits −K(0) and K(0). In particular,K(s) ≤ K(0).
This all leads us to the inevitable conclusion that K(s) = 〈~ζ(s) · ~ζ(0)〉 is afunction peaked at s = 0, from which it decays rapidly toward zero in a short
12
time τ∗, which is determined by the fast dynamics of the individual molecules ofthe fluid surrounding the Brownian particle. We expect a very short time scalefor this, of order the time between collisions with the individual molecules. Moregenerally, it should be the time scale that characterizes the rate with which thedegrees of freedom responsible for the random forces ~ζ relax to internal equi-librium after a displacement of the particle. In fact, for all practical purposes,this correlation function decays so quickly away from s = 0 that we may takeit to be proportional to the delta function, that is that it can be taken to benon-zero only for s = 0, but with finite integral over s:
C =
∫ ∞−∞
K(s)ds.
Returning to the problem that led us to consider this function K(s) in thefirst place, namely the evaluation of the mean-square velocity, we note that thedouble integral in (29) is appreciably non-zero only for t′ ∼= t′′. Thus,
〈v2〉 ∼= v(0)2e−2t/τ + e−2t/τ C
m2
∫ t
0
e2t′/τdt′
= v(0)2e−2t/τ + e−2t/τ C
m2
τ
2
(e2t/τ − 1
), (30)
where we have made the approximation suggested above:
K(s) ∼= Cδ(s).
In particular, however, Eq. 30 is true for times t large compared with τ , at whichpoint all memory of the initial conditions (e.g., ~v(0)) should have been forgotten.Then, the ensemble average 〈v2〉 = 3kT/m, consistent with the EquipartitionTheorem. The only way this can be true is if
Cτ
2m= 3kT,
i.e., that the drag coefficient α is fundamentally related to the random forces ~ζ:
α =1
6kT
∫ ∞−∞〈~ζ(s) · ~ζ(0)〉ds. (31)
The above result can be obtained in a few lines, using the techniques describedin Appendix B (see, in particular, section B.1). The relation between the dragcoefficient and the random forces acting on a particle is special case of what wewill later learn generally as the Fluctuation-Dissipation theorem. For now, wenote that the drag that a particle experiences in the fluid is due, at a micro-scopic level, to the randomly fluctuating forces that the particle feels at finitetemperature. The random forces are at the origin of macroscopic dissipation.
We began this analysis of the motion of a particle in a dissipative mediumby considering the equation of motion
md~v(t)
dt= −α~v(t) + ~f + ~ζ(t), (32)
13
where, for instance, the drag coefficient α = 6πηa for a sphere of radius a movingin a fluid with viscosity η. Here, ~f represented a possible external force appliedto the particle (e.g., by an electric or gravitational field).
We established a fundamental connection between dissipation (specifically,
the drag coefficient α here) and the fluctuating force ~ζ(t). For this, it wasconvenient to rewrite the equation of motion (26) in the absence of externalforces. This allowed us to solve for the velocity in terms of an initial value ~v(0)
and the fluctuating force ~ζ(t). We thus found a measurable quantity (specifically,
〈v2〉) that depends directly on the random forces ~ζ. This dependence, however,
is only through the correlation function K(t, t′) = 〈~ζ(t)·~ζ(t′)〉, which has severalgeneral properties that identified above.
Finally, we found that the drag coefficient α is fundamentally related to therandom forces ~ζ, via Equation 31, which expresses the fact that the drag that aparticle experiences in the fluid is due, at a microscopic level, to the randomlyfluctuating forces that the particle feels at finite temperature. In other words,if we were to pull the particle through the fluid, the drag force experienced canbe entirely expressed in terms of the forces ~ζ, which are presumably due to therandom equilibrium fluctuations of the fluid! These random forces are at theorigin of macroscopic dissipation.
2.5 Spectral (Fourier) analysis of fluctuations
When dealing with fluctuating quantities such as ~ζ and correlation functionssuch as K(t, t′) it is often most convenient to characterize them in terms offrequency. We have already noted that each specific realization of the randomforce ~ζ(t) (i.e., for each specific member of an ensemble of similar systems) isa very rapidly varying function, with a characteristic time scale of τ∗, whichis very short compared with other time scales of interest, such as that of themotion of the particle (which also fluctuates). ~ζ(t) varies much more rapidly
than does, for instance, ~v(t). Thus, ~ζ is expected to contain very high-frequencycomponents (details) that are not apparent in the motion of our hypotheticalparticle moving around in the fluid. As we shall see, for all practical purposes,we can say that the frequencies characteristic of ~ζ are unbounded, or that it hascomponents that vary arbitrarily rapidly in time.
In order to make these notions more precise, let’s look at the Fourier trans-forms of fluctuating quantities such as ~ζ(t). This can be done by using theorthogonality property of complex exponential functions (see Appendix C),
1
2π
∫ ∞−∞
e−iω(t−t′)dω = δ (t− t′) , (33)
where δ(t − t′) is the Dirac δ function, which is zero for t 6= t′, yet whichintegrates to unity: ∫ ∞
−∞δ(t)dt = 1.
14
Consider some fluctuating quantity f(t) for a particular member of the ensemble.
This could be, for instance, ~ζ itself, but we’ll suppress any vector aspects. Wecan write
f(t) =
∫ ∞−∞
δ(t− t′)f(t′)dt′
=1
2π
∫ ∞−∞
dt′∫ ∞−∞
e−iω(t−t′)f(t′)dω.
In other words,
f(t) =1
2π
∫ ∞−∞
f(ω)e−iωtdω,
where
f(ω) =
∫ ∞−∞
f(t′)eiωt′dt′, (34)
which define the Fourier transform and inverse Fourier transform.Here, f(ω) describes the “frequency content” of f(t). Assuming that f is
real, then because the complex conjugate satisfies
f∗(t) = f(t), (35)
we also have thatf∗(ω) = f(−ω). (36)
It is also worth noting that we have made some serious assumptions aboveconcerning convergence of the various integrals. This will not pose a practicalproblem, but one might worry about this, since the function ζ is presumably non-zero and is fluctuating over the same range for arbitrarily large times. Thus, forinstance, ζ(ω) might not be convergent. One could instead define the transformsfor a modified function ζΘ(t) that agrees with ζ(t) for |t| < Θ, and vanishes for|t| > Θ. One could then consider a limiting procedure in which Θ→∞.
We can do the same transform for the correlation function K(s) = 〈ζ(t +s)ζ(t)〉. We call the Fourier transform of this C(ω), and the two are related by
K(s) =1
2π
∫ ∞−∞
C(ω)e−iωsdω
and
C(ω) =
∫ ∞−∞
K(s′)eiωs′ds′. (37)
But, based on simple and general physical considerations, we showed before thatK(t, t′) = 〈ζ(t)ζ(t′)〉 not only depends just on the time interval s = t − t′, butK(s) is also symmetric. It is also, of course, real. It is not hard to show thatthis also means that C(ω) is a real function and is symmetric in ω. We alsoargued before that K(s) is a function peaked at s = 0 and rapidly decayingto zero within a time interval s of order τ∗, a very short molecular time scale.Thus, if |ω| is much smaller than 1/τ∗, then the argument of the exponential
15
in (37) is small wherever the K(s) is appreciable. This suggests that C(ω) isnearly independent of frequency ω for |ω| less than 1/τ∗, a very high frequency.In fact, such (molecular) frequencies are higher than any relevant frequency forthe motion of our sphere in the fluid. So, we say that the noise spectrum iswhite. Just as we argued before that
K(s) ∼= Cδ(s),
we can say thatC(ω) ∼= C,
a constant. Note in particular that
C(0) =
∫ ∞∞
K(s′)ds′ = 6kTα,
and
K(0) = 〈~ζ(0) · ~ζ(0)〉 =1
2π
∫ ∞−∞
C(ω)dω
is of order ∼ C/τ∗.We can also relate the correlation functions K(s) and C(ω) to the Fourier
transform of the noise ζ(ω). First of all, we note that (for most systems) we canreplace the ensemble average for a single realization of the fluid or member ofthe ensemble in the definition of K by a time average over a wide enough timeinterval −Θ < t < Θ:
K(s) = 〈ζ(t+ s)ζ(t)〉
∼=1
2Θ
∫ Θ
−Θ
ζ(t+ s)ζ(t)
=1
2Θ
∫ Θ
−Θ
dt
∫ ∞−∞
dω
2πe−iω(t+s)ζ(ω)
∫ ∞−∞
dω′
2πe−iω
′tζ(ω′)
=1
2Θ
∫ ∞−∞
dω
2πe−iωsζ(ω)ζ(−ω).
In other words,
C(ω) =1
2Θ|ζ(ω)|2.
Here, we have used the orthogonality of the complex exponential functions andthe fact that ζ is a real function. It may seem strange that a measurable quantityC(ω), which we have seen is directly related to the drag coefficient, and thus thediffusion constant, seems to depend inversely on the time over which we observethe system Θ. However, as we have also noted, the Fourier transform ζ(ω) maynot be well defined (e.g., finite). In fact, one should use the modified functionsζΘ(t) described above. These have finite transforms that grow with Θ. But, theexpression relating C(ω) to ζΘ(ω) can be expected to become independent of Θfor large enough Θ (see Appendix D.1).
16
An important historical application of these ideas was to simple electricalcircuits, which satisfy an equation similar to that of a particle in a dissipativemedium:
Ld
dtI(t) = Vext −RI(t) + V ′(t), (38)
where L is the inductance, I is the current, Vext represents a (constant orslowly varying) external applied field (emf), R is the resistance, and V ′ is thefluctuating field related to dissipation. Taking the ensemble average of thisequation yields the familiar law relating the voltage, current, and resistance:
〈I〉 =1
RVext.
The resistance here is analogous to the drag coefficient α above, and a similaranalysis to what was done above leads to
R =1
2kT
∫ ∞∞〈V ′(t+ s)V ′(t)〉 =
CV V (0)
2kT.
Here, the factor of 2 rather than 6 is due to the fact that we are dealing withan inherently one-dimensional problem rather than a three-dimensional one.Furthermore, if the fluctuations of V ′ are very fast (i.e., characterized by ashort correlation time τ∗, here corresponding to the time between collisions ofelectrons in the wires, for instance), as for the fluid, then the spectrum of thesefluctuations is very broad range
CV V (ω) =
∫ ∞−∞〈V ′(t+ s)V ′(t)〉eiωsds ∼= 2kTR
for ω in a range of order −1/τ∗ to 1/τ∗. This general relationship betweenthe voltage fluctuations and the resistance in a circuit is known as Nyquist’stheorem, and is yet another special case of the fluctuation-dissipation theorem.For a step-by-step derivation, see Appendix D.3. In Appendix D.2, we discussanother important noise type, namely shot noise.
We have seen so far that the macroscopic drag coefficient, and thus the mo-bility and diffusion coefficient, of a particle in a simple liquid is related directlyto the spectrum of microscopic fluctuations within the liquid. This also meansthat the physics governing the response of a system that is taken out of equilib-rium (e.g., by dragging a sphere through a liquid) can be described entirely interms if the fluctuations of the system about the equilibrium state. This is thefundamental observation behind the Fluctuation-Dissipation Theorem and thesomewhat earlier Regression Hypothesis made by Lars Onsager.
Before deriving these general relationships between fluctuations and response,however, let’s examine one last aspect of our simple Langevin treatment of par-ticle motion (fluctuations) within liquids. Namely, let us see how the mobilityof the particle (e.g., when subjected to an external driving force) relates to therandom fluctuations of the particle motion rather than the random fluid mo-tion. We’ll do this in just one dimension, and will thus neglect vector notation.
17
Consider the position of the particle, which we assume to start at t = 0 fromthe origin. This satisfies
r(t) =
∫ t
0
v(t′)dt′.
Thus,
〈r2(t)〉 =
∫ t
0
∫ t
0
dt′dt′′〈v(t′)v(t′′)〉. (39)
Here, what we see is a correlation function
Kvv(t′, t′′) ≡ 〈v(t′)v(t′′)〉
that is reminiscent of K we saw before. In fact, Kvv satisfies all the propertiesthat we previously identified for K = Kζζ , except that we can expect to find alonger correlation time of order τ . In particular, Kvv should be a (symmetric)function only of the time interval between t′ and t′′.
We can anticipate a result based on (39). Since we expect Kvv(s) to decaytoward zero for times long compared with the correlation time τ , if we areinterested in times long compared with this time, then we expect that we canalso approximate Kvv(s) by a δ function, resulting in
〈r2(t)〉 ∼= tCvv,
where
Cvv =
∫ ∞−∞
Kvv(s)ds
is some constant. In other words, we expect to find diffusion for long times.We can, however, do better than approximating Kvv(s) by a δ function. We
know from before that
v(t) = v(0)e−t/τ + e−t/τ1
m
∫ t
0
et′/τζ(t′)dt′. (40)
Thus,
〈v(t+ s)v(t)〉 = v(0)2e−(2t+s)/τ +e−(2t+s)/τ
m2
∫ (t+s)
0
∫ t
0
e(t′+t′′)/τ 〈ζ(t′)ζ(t′′)〉dt′dt′′
∼= v(0)2e−(2t+s)/τ +e−(2t+s)/τ
m2Cζζ
∫ (t+s)
0
∫ t
0
e(t′+t′′)/τδ(t′ − t′′)dt′dt′′,
where we have made the approximation
〈ζ(t′)ζ(t′′)〉 = Cζζδ(t′ − t′′)
made before. Continuing, we find that
〈v(t+ s)v(t)〉 ∼= v(0)2e−(2t+s)/τ +e−(2t+s)/τ
m2Cζζ
∫ min(t+s,t)
0
e2t′/τdt′,
= v(0)2e−(2t+s)/τ +e−(2t+s)/τ
m2Cζζ
τ
2
e2t/τ − 1 if s > 0e2(t+s)/τ − 1 otherwise
18
For times t long enough ( τ) so that all memory of the initial value of thevelocity is lost,
〈v(t+ s)v(t)〉 → kT
me−|s|/τ .
(This is a one-dimensional result.) Indeed, for t τ , we find that
〈r2(t)〉 → tCvv,
where
Cvv =
∫ ∞−∞〈v(t+ s)v(t)〉ds =
2kTτ
m,
Which agrees with our prior result for the three-dimensional diffusion, once weaccount for the appropriate factors of 3:
〈r2(t)〉 = 6Dt,
where D = kT/α.Continuing in one dimension, however, we note the complementary expres-
sions for the drag coefficient α and the mobility µ:
α =1
2kT
∫ ∞−∞〈ζ(t+ s)ζ(t)〉ds (41)
and
µ =1
α=
1
2kT
∫ ∞−∞〈v(t+ s)v(t)〉ds. (42)
For the electric circuit problem we introduced before, the analogous results forthe resistance R and conductivity Σ are:
R =1
2kT
∫ ∞−∞〈V (t+ s)V (t)〉ds (43)
and
Σ =1
2kT
∫ ∞−∞〈I(t+ s)I(t)〉ds, (44)
where V is the (fluctuating) voltage and I is the corresponding current. Thesecond of these relationships is usually written in terms of the current densityj and the (local) conductivity σ. This is known as the Kubo formula
σ =1
2kT
∫ ∞−∞〈j(t+ s)j(t)〉ds. (45)
In each of these cases, we have a pair of corresponding quantities: force andvelocity, voltage and current, . . . . In each case, the pair of quantities is relatedin the sense that some generalized force (force or voltage) results in a response(drift velocity or current). In each case, the corresponding correlation functions,e.g. Kζζ and Kvv, are related by an equation of the form(∫ ∞
−∞〈ζ(t+ s)ζ(t)〉ds
)(∫ ∞−∞〈v(t+ s)v(t)〉ds
)= (2kT )
2. (46)
19
2.6 The Fluctuation-Dissipation Theorem and Linear Re-sponse
We have seen in the examples above that one can, at least sometimes, relatequantities that have to do with systems out of equilibrium to the spectrumof thermal fluctuations about equilibrium. These principles are based on theFluctuation-Dissipation Theorem, which was proven in general by Callen andWelton in 1952, although special cases (as we have seen) of this were understoodmuch earlier. The basic idea behind the fluctuation dissipation theorem wasalso captured in the 1930s by the regression hypothesis of Onsager, who arguedthat the physics of macroscopic relaxation of a system back to equilibrium isgoverned by the same physics as the relaxation of spontaneous fluctuations aboutequilibrium. We have seen this, for instance in the relaxation of an initial valueof v(0) of the velocity of a particle in a viscous liquid is described by the sametime constant τ that appears in the correlation function Kvv above.
In order to derive the general relationship between, say the relaxation of somemacroscopic observable and its equilibrium fluctuations, we need to look at themicroscopic evolution of the system. This can either be done in the context ofquantum mechanics, as is done in the book by Chaikin and Lubensky, or in thecontext of classical, Hamiltonian mechanics, as we shall do below. We describethe microscopic state of the system by a point in phase space described by a setof generalized momenta p1, p2, . . . and corresponding coordinates q1, q2, . . ..Although this is a very high-dimensional space (with of order 1023 coordinates!),we’ll denote is systematically by (p, q). The only essential result of Hamiltonianmechanics that we need to recall is the fact that, given some initial point in phasespace (p(0), q(0)) at time t = 0, the state of the system (p(t), q(t)) at time t inthe future is completely determined by the initial condition and the HamiltonianH (p(0), q(0)). It is convenient for us to denote this evolution of the system bysome time-evolution operator Tt, defined by (p(t), q(t)) = Tt (p(0), q(0)). Inclassical statistical mechanics, the microstates of a system in equilibrium aredistributed according to a probability distribution
P (p, q) =1
Qe−βH(p,q),
where
Q =
∫dp dq e−βH(p,q)
and β = 1/(kT ). The integral is over all of phase space. Again, we use anabbreviated notation, in which dp dq indicates an infinitesimal volume elementin phase space.
Consider what happens if we disturb the system, taking it out of equilibrium(or, more precisely, taking it to a new equilibrium state characterized by a newHamiltonian H′). In fact, we shall look at the reverse problem, in which aperturbing field was applied to the system in the distant past, resulting in a newequilibrium at time t = 0 characterized by the modified HamiltonianH′. At timet = 0, we turn off this field, and allow the system to relax back to equilibrium
20
(characterized by H). It is this relaxation of the system back to equilibriumthat we want to examine. Let A(p, q) be some macroscopic observable whoserelaxation we are interested in. We shall consider only linear response theory,in which we assume the perturbation is sufficiently weak that the HamiltonianH′ can be taken to be H′ = H+∆H, where ∆H = −fA and f is the perturbingfield. (It is not hard to see that with this definition, f and A are conjugatethermodynamic variables, like pressure and volume, since f = −∂F∂A , where F isthe free energy.) The initial state of the system corresponds to distribution
P ′ (p, q) =1
Q′e−βH
′(p,q),
where
Q′ =
∫dp dq e−βH
′(p,q).
The initial (macro)state of the system is one in which
〈A〉 =1
Q′
∫dp dq e−βH
′(p,q)A(p, q).
After the field is turned off, however, the system (i.e., each microstate (p, q))evolves according to the Hamiltonian H(p, q) and the time-evolution operatorT , rather than H′(p, q) and T ′. In other words, at times t > 0,
〈A(t)〉 =1
Q′
∫dp dq e−βH
′(p,q)A (Tt(p, q))
∼=∫dp dq e−βH(p,q) (1− β∆H)A (Tt(p, q))∫
dp dq e−βH(p,q) (1− β∆H).
Expanding, and keeping only terms through linear order gives
〈A(t)〉 ∼=∫dp dq e−βH(p,q)A (Tt(p, q))∫
dp dq e−βH(p,q)
+βf
∫dp dq e−βH(p,q)A(p, q)A (Tt(p, q))∫
dp dq e−βH(p,q)
−βf∫dp dq e−βH(p,q)A(p, q)∫
dp dq e−βH(p,q)
∫dp dq e−βH(p,q)A (Tt(p, q))∫
dp dq e−βH(p,q).
This means that, to linear order,
〈A(t)〉 − 〈A(t)〉0 = βf(〈A(0)A(t)〉0 − 〈A〉20
),
= βf〈δA(0)δA(t)〉0, (47)
where δA(t) = A(t) − 〈A〉0 and 〈〉0 refers to an ensemble average in the un-perturbed system. We have been somewhat sloppy in the notation here. Thefirst term in this equation refers to a evolution to time t of an ensemble of notonly similarly prepared systems, but systems subject to the same perturbation.
21
It does not strictly refer to any ensemble average in the usual sense. Thus, itmight be less confusing to write
∆A(t) = βf〈δA(0)δA(t)〉0, (48)
where this expresses the average deviation A from its equilibrium value. Theaverage is to be understood to mean over the ensemble of systems with the sameperturbation.
This is one form of the Fluctuation-Dissipation theorem, expressing themacroscopic evolution of a system out of equilibrium to the spontaneous fluctu-ations of the system about equilibrium. This is usually expressed, however, interms of the (linear) response of the system. This is implicit in the expressionabove. We have seen how the response ∆A(t) above is linear in the force (or,more generally, the perturbation) f . This perturbation may, however, be timedependent. The most general form of a linear response to a time-dependent fis
∆A(t) =
∫dt′ χ(t, t′)f(t′),
where χ expresses the fact that the response at time t depends on the waythe system was perturbed at other times. In principle, we should allow forarbitrarily influences from distant times. Thus, the integral above should beover all times t′. It would be unphysical, however, if this influence extendedto future times. How, for instance, can the system anticipate the perturbationthat it will experience in the future? This simple, and physically reasonableassumption of causality can be expressed mathematically as χ(t, t′) = 0 for allt′ > t. Note that χ in this sort of description is assumed to be a property of the(equilibrium) system, and not itself dependent on f . Thus, we also expect thatχ should depend only on the time interval between t and t′, so that
χ(t, t′) =
χ(t− t′) if t > t′
0 otherwise.
As noted already, we expect that χ is a property of the (equilibrium) system,and not itself dependent on f . We should, therefore, be able to express it interms if the dynamics of the equilibrium system. We have examined a particularcase above, in which we perturbed the system for times t < 0, then removed theperturbation. This corresponds to
f(t) =
f if t < 00 otherwise
.
Here, we found that
∆A(t) = βf〈δA(0)δA(t)〉0, (49)
But,
∆A(t) =
∫ 0
−∞dt′ χ(t, t′)f
22
in this case, since f = 0 for t′ > 0. By a change of variable of integration, thisbecomes
∆A(t) = f
∫ ∞t
dt′ χ(t′).
But, this only applies for t > 0. Thus,
χ(t) =
−β d
dt 〈δA(0)δA(t)〉 if t > 00 otherwise
.
This expresses the way in which a system responds to a perturbation (taking itout of equilibrium) in terms of the spontaneous fluctuations of the system aboutequilibrium.
Last time, we saw how the response of the system, specifically in terms ofsome observable quantity A, could be described by a linear response functionχ:
∆A(t) =
∫ ∞−∞
dt′ χ(t, t′)f(t′), (50)
where χ expresses the fact that the response at time t depends on the waythe system was perturbed by some generalized force f at other times. Here,∆A(t) ≡ A(t) − 〈A〉0, where A(t) represents the expected value of A at timet (this is actually an ensemble average quantity, but for a particular ensembleof similarly perturbed systems.) Based on the physical principle of causality,however, we expect the response only to depend on the history of the systemand its perturbation. Furthermore, since the response function should be onlyan equilibrium property of the system itself (and not, for instance, dependenton f , at least if the perturbation is weak enough that the response is, indeed,linear). Thus, we expect that
χ(t, t′) =
χ(t− t′) if t > t′
0 otherwise.
In deriving the above, we assumed a very particular form of the perturbedHamiltonian:
H′ = H− fA.
In general, however, we expect if the (generalized) force f is weak, then one cansimply expand this Hamiltonian to linear order, with the result that
H′ = H− fB.
That is, the linear coefficient of f need not be A, but may be some other variableB. When the system is subjected to such a perturbation, we expect to find alinear response of the variable A, as given by (50), where we shall denote theresponse function by χAB(t − t′), since it represents the response of A to aperturbation f that is conjugate to B. Here, f and B are conjugate in the
23
thermodynamic sense, in that
∂F
∂B= −kT ∂
∂BlogQ
= −kT 1
Q
∂
∂B
∫dp dq e−β(H−fB)
= −f,
where
Q =
∫dp dq e−β(H−fB)
is the partition function and
F = −kT logQ
is the free energy. This is just like the relationship between the conjugatevariables volume V and pressure P :(
∂F
∂V
)T,N
= −P.
As before, if we consider a perturbing field f that is turned on in the distantpast, and then switched off at time t = 0, we can write the evolution of (theensemble average of) A as
A(t) =1
Q′
∫dp dq e−βH
′(p,q)A (Tt(p, q))
=
∫dp dq e−β(H(p,q)−fB(p,q))A (Tt(p, q))∫
dp dq e−β(H(p,q)−fB(p,q)).
An alternative way of deriving the linear response that we calculated before isto look at the derivative of this with respect to the perturbation f :
∂A(t)
∂f=
β
Q′
∫dp dq e−βH
′(p,q)B (p, q)A (Tt(p, q))
−∫dp dq e−βH
′(p,q)A (Tt(p, q))(∫dp dq e−βH′(p,q)
)2 β
∫dp dq e−βH
′(p,q)B (p, q) .
Evaluating this derivative for f = 0, and noting that the point (p, q) in phasespace represents the state of the system at time t = 0, we find that
∆A(t) ' βf (〈B(0)A(t)〉0 − 〈A(t)〉0〈B(0)〉0)
= βf〈B(0)δA(t)〉0. (51)
Following the same analysis as before, we also find the response functionχAB(t) in terms of the fluctuating quantities δA and B:
χAB(t) = −β ddt〈B(0)δA(t)〉0. (52)
24
Again, here, we should note that these thermodynamic averages 〈〉0 refer tothe unperturbed (i.e., equilibrium) system. Thus, they represent equilibriumfluctuations of the system, which exhibit some temporal correlations. The mainexample we have been looking at is the motion of a particle in a viscous fluid.Here, the perturbing field can be taken to be an external force f applied tothe particle (say, by some magnetic or gravitational field). The response of thesystem is to develop a drift velocity v. But, the thermodynamically conjugatevariable to f is the displacement x of the particle (in the direction of the force,which we take to be the x direction). Thus,
H′ = H− fx.
So, the ensemble average drift velocity is
v(t) =
∫ t
−∞χvx(t− t′)f(t′)dt′ = f0
∫ ∞t
χvx(t′)dt′,
when, as before, a force f0 is applied in the distant past and switched off attime t = 0. But, in this case, we expect that v(0) = µf0, where µ is the particlemobility. But, we know that
µ =1
2kT
∫ ∞−∞〈v(t)v(0)〉dt =
1
kT
∫ ∞0
〈v(t)v(0)〉dt,
where we have used the symmetry of Kvv(t) = 〈v(t)v(0)〉. This suggests thatχvx(t) = βKvv(t). This says that the dynamics associated with the response χ ofa system to a non-equilibrium perturbation are governed by the same underlyingprinciples as the fluctuations Kvv(t) = 〈v(t)v(0)〉 about equilibrium. This is,indeed, the case, as Onsager suggested in his regression hypothesis. But, let’ssee if we can derive the relationship χvx(t) = βKvv(t).
We see from (52) that
χvx(t) = −β ddt〈x(0)v(t)〉0. (53)
But,〈x(0)v(t)〉0 = 〈x(t′)v(t+ t′)〉0,
which cannot depend on t′. Hence,
0 =d
dt′〈x(t′)v(t+ t′)〉0
= 〈 ddt′x(t′)v(t+ t′)〉0 + 〈x(t′)
d
dt′v(t+ t′)〉0
= 〈v(t′)v(t+ t′)〉0 + 〈x(t′)v(t+ t′)〉0,
and
d
dt〈x(t′)v(t+ t′)〉0 = 〈x(t′)v(t+ t′)〉0 = −〈v(t′)v(t+ t′)〉0 = −Kvv(t).
25
This shows that
χvx(t) = βKvv(t) =1
me−t/τ ,
for our particle in the fluid. We see that this represents what is sometimesreferred to as a memory function. It shows how the response at one time dependson the history of disturbances, but with vanishing influence from distant past.That is, the memory fades over time. (As we all know too well!)
2.7 The Damped Harmonic Oscillator
The harmonic oscillator illustrates nicely many of the ideas we have been de-veloping. It also turns out to be a very practical application of the Langevinanalysis. There are many experimental situations of current interest where onehas a harmonic potential for particles that would otherwise undergo brown-ian motion in a fluid. So, to our brownian particle above we add a harmonic,“trapping” potential Kx2, which tends to localize the particle near x = 0. Ourequation of motion becomes
mx(t) + αx(t) +Kx(t) = f(t),
where f is the force acting on the particle. We’ll consider the case where thisis some external driving force. Of course, there will also be our familiar ran-dom brownian forces. But, we shall treat ensemble quantities below. Fouriertransforming the equation of motion leads to(
K −mω2 − iαω)xω = fω,
where we shall tend to employ a shorthand notation for transforms xω of x(t),and similarly for the force. This can also be written in a way suggestive of linearresponse:
xω =1
(K −mω2 − iαω)fω,
where
χ(ω) =1
K −mω2 − iαωis the Fourier transform of the response function defined by
x(t) =
∫ ∞−∞
dt′ χ(t− t′)f(t′).
It is tempting at this point to anticipate the frequency-dependent version ofour expression
χxx(t) = −β ddt〈x(t)x(0)〉 = −β d
dtKxx(t), (54)
which would read
χ(ω) = iωβ〈xωx−ω〉. (incorrect)
26
But, this would be wrong. Recall that the left-hand side of (54) is non-zero onlyfor t > 0, while the right-hand side is a symmetric function of time. Thus, theequation is valid only for positive time. In fact, given the properties of Kxx, theright-hand side of the equation is an odd function. We have noted before thatthe Fourier transform of a real, symmetric function is also real and symmetric.What about a real function that is odd? The transform of such a function wouldbe odd and purely imaginary. We can write (54) in a way that is valid for alltime. We decompose the function χ(t) (which is non-zero only for positive t)into a even and odd parts:
χ(t) = χE(t) + χO(t),
where
χE(t) =1
2(χ(t) + χ(−t))
and
χO(t) =1
2(χ(t)− χ(−t)) .
The Fourier transform χE(ω) is thus real and symmetric. We call it χ′(ω).Similarly, the Fourier transform χO(ω) is imaginary and odd in ω. We call itiχ′′(ω). Note that both χ′ and χ′′ are real, and that
χ(ω) = χ′(ω) + iχ′′(ω).
Now, returning to the expression (54) above, we can write this in a way validfor all times:
2χO(t) = −β ddt〈x(t)x(0)〉 = −β d
dtKxx(t). (55)
Thus,
χ′′(ω) =ωβ
2〈xωx−ω〉 =
ωβ
2〈|xω|2〉, (56)
which is a common way to write the fluctuation dissipation theorem. The imag-inary part of the (Fourier transform of the) response function is directly relatedto the power spectral density of the equilibrium fluctuations.
2.8 Kramers-Kronig relations
There are very important and far-reaching implications of the physical obser-vation of causality in the response function χ(t), i.e., the fact that it vanishesfor t < 0. Consider the Laplace transform of this as a function of a complexfrequency z:
χ(z) =
∫ ∞−∞
eiztχ(t)dt =
∫ ∞0
eiztχ(t)dt. (57)
(Here, the notation is going to go down-hill even further, but I hope the contextmakes the various quantities clear enough.) Again, the function χ(t) is zero fornegative times. Furthermore, on simple physical grounds, we also expect that
27
χ is bounded at long times. In fact, it will usually be the case that it vanishesat long times, i.e, memory tends to be lost with time. We also, of course,assume that the function χ(t) is otherwise well-behaved. This means that thefunction χ(z) of the complex variable z is actually analytic in the upper halfplane (=z > 0). Thus, we can apply the Cauchy theorem for complex functions:
χ(z) =
∮dz′
2πi
χ(z′)
z′ − z,
for any contour in the upper half plane containing the point z. Of course, asthe complex variable z approaches some point ω on the real axis, χ(z)→ χ(ω).Let z = ω + iε, which lies in the upper half plane. Since the contour mustcontain z but can otherwise be any contour in the upper half plane, we cantake the contour defined by a line very close to the real axis together with alarge semicircle defined by |z| equal to some very large radius R. The onlycontributions will then come from the portion of the contour defined by z′ =ω′ + iε′, where we will take ε′ to zero. Thus,
χ(ω + iε) =1
2πi
∫ ∞−∞
dω′χ(ω′)
ω′ − ω − iε.
Separating the real and imaginary parts of singular part of the integrand,we find
1
ω′ − ω − iε=
ω′ − ω(ω′ − ω)
2+ ε2
+ iε
(ω′ − ω)2
+ ε2
The imaginary part of this is an increasingly highly peaked function as ε → 0.In fact, in this limit, it approximates a δ function:
ε
(ω′ − ω)2
+ ε2→ πδ(ω′ − ω).
Thus, for small enough ε,
χ(ω) ' 1
2πi
∫ ∞−∞
dω′χ(ω′)ω′ − ω
(ω′ − ω)2
+ ε2+
1
2χ(ω).
So that
χ(ω) = χ′(ω) + iχ′′(ω)
=1
πilimε→0
∫ ∞−∞
dω′ω′ − ω
(ω′ − ω)2
+ ε2(χ′(ω′) + iχ′′(ω′)) .
Comparing real and imaginary parts, we find that
χ′(ω) =1
πlimε→0
∫ ∞−∞
dω′ω′ − ω
(ω′ − ω)2
+ ε2χ′′(ω′) =
1
πP∫ ∞−∞
dω′χ′′(ω′)
ω′ − ω
and
χ′′(ω) = − 1
πlimε→0
∫ ∞−∞
dω′ω′ − ω
(ω′ − ω)2
+ ε2χ′(ω′) = − 1
πP∫ ∞−∞
dω′χ′(ω′)
ω′ − ω.
28
These are the Kramer-Kronig relations, which express the fact that the functionsχ′ and χ′′ are not independent of each other. In fact, one can completely recoverone from the other via a sort of transformation involving the full frequencyinformation.
2.9 General Properties of Response Functions
Last time, for the harmonic oscillator, we found an expression,
χ′′(ω) =ωβ
2〈|xω|2〉,
relating the response function to the the power spectral density of thermalfluctuations of the position. The more general result when we have a perturbedHamiltonian H′ = H− fA is
χ′′AA(ω) =ωβ
2〈|Aω|2〉. (58)
This is the general statement of the fluctuation-dissipation theorem.We can make several observations concerning this more general result (which
can, of course, be checked for the damped oscillator). First, if we rewrite thisexpression as follows,
〈|Aω|2〉 = 2kTχ′′AA(ω)
ω. (59)
The left-hand-side is clearly a real, symmetric, and strictly positive quantity.Thus, we learn quite generally that χ′′AA(ω)/ω must be positive.
Consider the response of the system to a constant applied force f . That is,let f(t) = f0 for all t. Here, as below, we’ll revert to the notation of harmonicoscillator, although the results can easily be generalized. Then,
x(t) =
∫ t
−∞χ(t− t′)f(t′)dt′ = f0
∫ t
−∞χ(t)dt′.
If this is bounded, as is, for instance, the displacement x for the harmonicoscillator, then
x(t)→ χ0f0,
as t→∞ where χ0 is the static response function. For the harmonic oscillator,for example, this is just the inverse of the spring constant: χ0 = 1/K. Thus,
χ0 =
∫ ∞−∞
χ(t)dt = χ′(0).
Using a Kramers-Kronig relation, we discover a thermodynamic sum rule con-cerning χ′′:
χ0 =1
π
∫ ∞−∞
χ′′(ω′)
ω′dω′. (60)
Here, we have used the fact that χ′′ is an odd function, which means that theintegrand is non-singular.
29
2.10 Dissipation
Among the many general properties of response functions is the connectionbetween χ′′ and macroscopic dissipation. Non only is this an important funda-mental observation, but there are also many experimental situations in whichdissipative processes such as absorption are measured. Here, we show that suchmeasurements can be used to determine the response function. As above, weshall look at this connection between χ′′ and dissipation within the context ofthe damped harmonic oscillator, although this is only meant to illustrate themore general results. The power dissipated can be expressed here in terms of theforce and the velocity. The work done in displacing a particle a small amountdx with a force f is dW = fdx. Thus, the power is
dW
dt= fx(t).
Consider what happens when we drive the system with a monochromaticforce
f(t) = f0 cos (ω0t) =1
2f0
(e−iω0t + eiω0t
).
Thus, the Fourier transform of this is
fω = πf0 [δ(ω − ω0) + δ(ω + ω0)] .
The displacement is then given by
x(t) =1
2π
∫ ∞−∞
χ (ω) fωe−iωtdω
=f0
2
∫ ∞−∞
[χ(ω0)e−iω0t + χ(−ω0)eiω0t
]= f0 [χ′(ω0) cos(ω0t) + χ′′(ω0) sin(ω0t)] .
The average power dissipated during one period T = 2π/ω0 is then
1
T
∫ T
0
f(t)x(t)dt =1
2f2
0ω0χ′′(ω0).
Thus, the power dissipated is determined by χ′′ alone. Since this is also directlyrelated to the power spectral density of fluctuations, hence the name fluctuation-dissipation theorem. We also find, that
ωχ′′(ω) > 0,
since, physically, the power dissipated must be positive. We previously foundthat χ′′(ω)/ω was strictly positive.
30
2.11 Example: the Harmonic Oscillator, Again
We can see many examples of the general properties of linear response theoryin the damped harmonic oscillator problem discussed last time. Previously,we analyzed the response of the damped harmonic oscillator, described by theequation of motion
mx(t) + αx(t) +Kx(t) = f(t),
where K is the harmonic spring constant. We found that the response functioncan be written as
χ(ω) =1
K −mω2 − iαω.
We noted that χ(ω), thought of as a complex function, should be an analyticfunction of a complex variable ω in the upper half-plane, i.e., for positive imagi-nary part of ω. This might seem surprising, since the response function χ aboveclearly has some singularities in ω. Nevertheless, it is analytic in the upperhalf-plane, as can be seen by rewriting the response function as follows
χ(ω) =−1
m
1
(ω − Ω−) (ω + Ω+),
whereΩ± = Ω± iγ/2,
Ω2 = ω20 − γ2/4,
γ = α/m, and ω0 =√K/m is the natural resonance frequency of the oscillator.
Thus, there are singularities (simple poles) at ω = ±Ω∓, both of which lie inthe lower half-plane. Thus, there are no singularities in the upper half-plane.Analyticity of χ was shown to be a necessary condition for causality—i.e. thefact that χ(t) vanishes for t < 0. This is also a sufficient condition, as can beseen by evaluating χ(t) for the damped oscillator. For t < 0, the integral
χ(t) =1
2π
∫ ∞−∞
e−iωtχ(ω)dω
can be evaluated by extending the path of integration to a closed contour inthe upper half-plane, for which the result will be zero because the integrand isanalytic there. In contrast, for positive t, we can extend the integration to acontour in the lower half-plane, resulting in
χ(t) =−1
2πm
∮e−izt
(z − Ω−) (z + Ω+)
=i
2Ωm
[e−iΩ− − eiΩ+
]=
sin Ωt
mΩe−γt/2.
As we have noted, the harmonic oscillator constitutes a very practical ap-plication of the Langevin analysis. There are many experimental situations of
31
current interest where one has a harmonic potential for particles that would oth-erwise undergo brownian motion in a fluid. For many of these cases, the relevantlimit is that of a (strongly) overdamped oscillator. As we can see above, thereare at least two qualitatively different regimes of behavior: (a) the underdampedcase, for which ω2
0 > γ2/4; and (b) the overdamped case, for which ω20 < γ2/4.
The strongly overdamped case can be analyzed by ignoring the inertial term inthe equation of motion. Here, the response function can be obtained from
χ(ω) =1
K − iαω.
Here, too, the only singularity of this is in the lower half-plane. So, this also isconsistent with a causal response.
The imaginary part is given by
χ′′(ω) =αω
K2 + α2ω2.
There are several things to note about this. It is an odd function, as we expect.Also, χ′′(ω)/ω is finite (i.e., non-singular) and strictly positive, as is ωχ′′(ω).The spectrum of the fluctuations in position, 〈|xω|2〉 can also be evaluated withthe aid of the fluctuation-dissipation theorem:
〈|xω|2〉 = 2kTα
K2 + α2ω2.
The behavior of this for large ω should be that of brownian motion (providedthe frequency is small compared with τ−1, of course). This is because highenough frequencies correspond to times so short that the particle does not feelthe harmonic potential. We see this in the fact that
〈|xω|2〉 →2kT
αω2,
and thus, that
〈|vω|2〉 →2kT
α,
since vω = −iωxω. This last expression is equivalent to the relation
µ =1
α=
1
2kT
∫ ∞−∞〈v(t+ s)v(t)〉ds. (61)
found earlier.We can also write χ′′ as follows
χ′′(ω) = χ0τ0ω
1 + (τ0ω)2 ,
where τ−10 is the characteristic frequency or relaxation rate for this system (not
to be confused with the resonance frequency ω0 of the underdamped harmonicoscillator. Here, χ0 = 1/K is the static response function of a spring with springconstant K, since x = f/K for a constant force f . This satisfies the sum rule(60) above.
32
2.12 The Generalized Langevin Equation
Way back in our first lecture, we looked at the Langevin equation, in which weassume that the the fluid (or, more generally, the heat “bath”) imparts randomforces to the particle whose motion we were interested in. At the same time, weassumed that we could use a single drag coefficient to describe the resistanceto motion of the particle through the fluid in addition to the random forcesfrom the bath.Here, we want to examine this in greater detail. We do so withinthe context of our original problem that we used the Langevin description for,although the analysis and principles are more general.
Consider a single degree of freedom such as the position of our particle. Thesurrounding medium is treated as having a large number of degrees of freedomy1, y2, . . . , to which the particle described by the single coordinate x is coupled.We assume a linear coupling in the Hamiltonian for the combined system, i.e.,
H = H0(x)− fx+Hb(y1, . . .),
where
H0 =mx(t)
2+ V (x).
Here, V represents some conservative potential, and −dVdx is the force. Weassume a simple, linear relationship between the yi of the bath and the force fthat the particle feels dues to the bath:
f =∑i
ciyj .
In keeping with our previous notation, we call the correlation function for theseforces
Kb(s) = 〈δf(s)δf(0)〉b =∑i,j
cicj〈δyi(s)δyj(0)〉,
where b denotes the bath. This means that we will express things in terms ofthe bath, unperturbed by the presence of the particle, i.e., the variable x.
We expect to be able to describe the effect of x on the bath by a linearresponse, in particular for the force f :
f(t) = fb(t) +
∫ ∞−∞
dt′ χb(t− t′)x(t′),
where
χb(t) = −β ddtKb(t)
for t > 0. The equation of motion for the particle is now
mx(t) = −dVdx
+ fb(t) +
∫ t
−∞dt′ χb(t− t′)x(t′).
33
Here, it is worth noting that χb represents the equilibrium fluctuations of thebath alone. Thus, we expect the correlations in the bath, and hence mem-ory, to extend over a short time such as τ∗ in our earlier notation. Given therelationship between χb and Kb, we have that
mx = −dVdx
+ fb(t)− β∫ t
−∞dt′
d
dtKb(t− t′)x(t′)
= −dVdx
+ fb(t) + β
∫ t
−∞dt′
d
dt′Kb(t− t′)x(t′)
= −dVdx
+ δfb(t)− β∫ t
0
dt′Kb(t− t′)x(t′), (62)
where we have integrated by parts, and defined
V (x) = V (x)− βKb(0)x2,
andδfb(t) = fb(t)− βKb(t)x(0).
We now return to our original problem, in which no potential is applied,and for which we assume the bath degrees of freedom yi relax very quickly(at least compared with the motion of our particle in the medium). Then, weagain assume that K approximates a δ function, which leads us to the followingapproximate result for the drag force:
−βx(t)
∫ ∞0
Kb(s)ds = −αx,
where
α =1
2kT
∫ ∞∞
Kb(s),
as before. More generally, however, we have a more complete expression forthe effect of the drag on the particle due to the bath, which includes memoryeffects. In other words, we see how the relaxation time of the medium itselfaffects the locality in time of the drag. The equation (62) is often referred to asthe generalized Langevin equation. The effective potential V is also called thepotential of mean force, which yields the bath-averaged force on the particle.
34
A Solving the diffusion equation
As an illustration, we solve the one-dimensional diffusion equation in an infinitesystem.
∂P (x, t)
∂t= D
∂2P (x, t)
∂x2. (63)
We take as initial condition P (x, 0) = δ(x). We now take the spatial Fouriertransform of this equation:∫ ∞
−∞dx eikx
∂P (x, t)
∂t= D
∫ ∞−∞
dx eikx∂2P (x, t)
∂x2. (64)
Partial integration of the right-hand side yields:∫ ∞−∞
dx eikx∂P (x, t)
∂t= D
∫ ∞−∞
dx eikx(−k2)P (x, t). (65)
or,∂P (k, t)
∂t= −Dk2P (k, t). (66)
This equation can be solved to yield:
P (k, t) = P (k, 0)e−k2Dt. (67)
But we can easily compute P (k, 0):
P (k, 0) =
∫ ∞−∞
dx eikxδ(x) = 1
The inverse Fourier transform then yields P (x, t):
P (x, t) =1
2π
∫ ∞−∞
dke−ikxP (k, t)
=1
2π
∫ ∞−∞
dke−ikxe−k2Dt
=1
2π
∫ ∞−∞
dke−Dt(k2+2ikx/(2Dt)−x2/(4D2t2)+x2/(4D2t2))
=1
2π
∫ ∞−∞
dke−Dt((k+ix/(2Dt))2+x2/(4D2t2))
=1
2π
∫ ∞−∞
dke−Dt(k+ix/(2Dt))2−x2/(4Dt)
=1
2π
√π
Dte−x
2/(4Dt)
=
√1
4πDte−x
2/(4Dt) (68)
35
B Relation between memory function and ran-dom force
The Langevin equation makes the assumption that the friction force dependsonly on the instantaneous velocity:
m~v(t) = −mγ~v(t′) + ~ζ(t) , (69)
where mγ = α is the friction coefficient. It is often necessary to consider themore general case where the friction force depends also on the velocity at earliertimes:
m~v(t) = −∫ t
0
k(t− t′)~v(t′) + ~ζ(t) (70)
This equation defines the “memory function” k(τ) and the random force ~ζ(t).The force is random in the sense that it is uncorrelated with the initial velocity:
〈~v(0) · ~ζ(t)〉 = 0 ∀t. (71)
Any part of the friction force that is correlated with the initial velocity will beabsorbed in the memory function. It turns out the the memory function k(τ) is
closely related to the auto-correlation function of the random force, 〈~ζ(0) ·~ζ(τ)〉.This can be shown quite easily by making use of Laplace transforms. TheLaplace transform L(f) ≡ F (s) of a function f(t) is defined as:
F (s) ≡ L(f(t)) =
∫ ∞0
dt exp(−st)f(t) (72)
(Note: in Eqn. 57 we already used a Laplace transform with s = −iz). We willuse two properties of Laplace transforms: First of all, the Laplace transform ofthe time derivative of f(t) is given by
L(f) = sF (s)− f(0) (73)
as can be shown by partial integration. Secondly, the Laplace transform of the“convolution” of two function f and g id the product of the individual Laplacetransforms. The convolution f ∗ g is defined as
f(t) ∗ g(t) ≡∫ t
0
f(t− t′)g(t′) dt′ . (74)
Then it is easy to show that
L(f ∗ g) = F (s)G(s) (75)
If we multiply Eqn. 70 with v(0) and average, we get:
m〈~v(0) · ~v(t)〉 = −∫ t
0
k(t− t′)〈~v(0) · ~v(t′)〉 (76)
36
where we have used the fact that the random force is not correlated with ~v(0).We now Laplace transform this equation (making use of the properties of Laplacetransforms listed above). We denote the Laplace transform of the velocity ACFby V (s) ≡ L(〈~v(0) · ~v(t)〉). With this notation, we get
m(sV (s)− 〈v(0)2〉) = −K(s)V (s) (77)
or
V (s) =〈v(0)2〉
s+K(s)/m(78)
First of all, we note that it follows from Eqn. 70
m~v(0) = ~ζ(0) (79)
Let us now multiply Eqn. 70 with m~v(0) and average:
m2〈~v(0) · ~v(t)〉 = −m∫ t
0
k(t− t′)〈~v(0) · ~v(t′)〉+ 〈~ζ(0) · ~ζ(t)〉 (80)
Note that, in the above expression, we have replaced m〈~v(0) · ~ζ(t)〉 by 〈~ζ(0) ·~ζ(t)〉. Next, we take the Laplace transform of Eqn. 80. We denote the Laplace
transform of the random force ACF by Z(s) ≡ L(〈~ζ(0) · ~ζ(t)〉). Using theproperties of the Laplace transforms of derivatives and convolutions, and usingthe fact that 〈A(0)B(t)〉 = −〈A(0)B(t)〉, we get:
−m2(s2V (s)− s〈v(0)2〉 − 〈~v(0) · ~v(0)〉
)= m
(sV (s)− 〈v(0)2〉
)K(s) + Z(s)
(81)
We note that 〈~v(0) · ~v(0)〉 = 0 (because 〈~v(0) ·~v(t)〉 is an even function of time).Rearranging the terms in Eqn. 81, we get
−m2(s2V (s) + sV (s)K(s)/m− s〈v(0)2〉
)= −m〈v(0)2〉K(s) + Z(s) (82)
Using Eqn. 78, we se that the left-hand side of Eqn. 82 vanishes. Therefore
−m〈v(0)2〉K(s) + Z(s) = 0 (83)
orZ(s) = m〈v(0)2〉K(s) (84)
This relation between the Laplace transforms implies that
〈~ζ(0) · ~ζ(t)〉 = m〈v(0)2〉k(t) (85)
In other words: the memory function is proportional to the autocorrelationfunction of the random force.
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B.1 Langevin equation
A special case results when k(t) decays very rapidly and can be approximatedas
k(t) = 2mγδ(t) = 2αδ(t)
There is a now slight subtlety with the evaluation of∫ t
0k(t′)dt′. But this can
be resolved by writing∫ t
0
dt′ k(t′) =1
2
∫ t
−tdt′ k(t′) =
1
2(2mγ) = mγ = α
If we insert this expression in Eqn. 70, we obtain the original Langevinequation
m~v(t) = −mγ~v(t′) + ~ζ(t) .
From Eqn. 85 it follows that
〈~ζ(0) · ~ζ(t)〉 = m〈v(0)2〉2αδ(t) = 6kTαδ(t) (86)
C Fourier representation of the δ function
To derive the relation
1
2π
∫ ∞−∞
e−iω(t−t′)dω = δ (t− t′) ,
we first consider the Fourier transform
1
2π
∫ ∞−∞
e−iωte−ε|ω|dω ,
with ε > 0. After that, we consider the limit ε → 0. As exp(−ε|ω|) is an evenfunction of ω, its Fourier transform is real. We can therefore write∫ ∞
−∞e−iωte−ε|ω|dω = 2Re
∫ ∞0
e−iωte−εωdω .
We then use ∫ ∞0
e−iωte−εωdω =1
ε+ it=
ε− itε2 + t2
Therefore,
Re∫ ∞
0
e−iωte−εωdω =ε
ε2 + t2
Now the integral ∫ ∞−∞
dtε
ε2 + t2= π
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As ε becomes smaller, the integrand becomes increasingly sharply peaked, butthe value of integral is independent of ε. In fact,
limε→0
ε
ε2 + t2= πδ(t)
Therefore1
2πlimε→0
∫ ∞−∞
e−iωte−ε|ω|dω =1
2π2πδ(t) = δ(t)
D Noise
D.1 Power spectral density
Consider a fluctuating quantity f(t) - for instance voltage noise. Experimentally,we can measure he noise power in a certain frequency interval by passing thefluctuating signal through a frequency filter that transmits only those frequencycomponents that have an angular frequency between ω and ω+ ∆ω. The meansquare value per unit bandwidth of this ”filtered” signal is the noise powerdensity, Pff (ω).
Pff (ω) = 〈f2〉(ω)/∆ν
where ∆ν = ∆ω/(2π). Let us consider the signal f(t) over a time interval from−Θ/2 to Θ/2. We can decompose the signal in this time interval in discreteFourier components:
f(t) =
∞∑m=−∞
am exp(2πimt/Θ)
Component m has an angular frequency ω = 2πm/Θ and a frequency ν =m/Θ. After passing through the frequency filter, only those Fourier componentsremain that have their (absolute) frequency between ν and ν+∆ν. The numberof Fourier modes that passes through this filter is
Nm = 2∆νΘ
The factor 2 results from the fact that both positive and negative frequenciesare transmitted. The mean square signal is than:
〈f2〉 =1
Θ
∫ Θ/2
−Θ/2
dt
(mH∑
m=mL
(am exp(2πimt/Θ) + a−m exp(−2πimt/Θ))
)2
Due to the orthogonality of different Fourier modes, this can be written as:
〈f2〉 = 2
mH∑m=mL
〈ama−m〉
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The number of modes 2∆νΘ, and therefore:
〈f2〉 = 2∆νΘ〈ama−m〉
or〈f2〉∆ν
= 2Θ〈ama−m〉
We can express the am as
am =1
Θ
∫ Θ/2
−Θ/2
dtf(t) exp(−2πimt/Θ)
or, with ω = 2πm/Θ
a(ω) =1
Θ
∫ Θ/2
−Θ/2
dtf(t) exp(−iωt)
The Fourier transform of f(t) over the interval −Θ/2,Θ/2 is
fΘ(ω) =
∫ Θ/2
−Θ/2
dt eiωtf(t) = Θa(ω)
Therefore,
Pff (ν) =〈f2〉∆ν
=2
Θ〈|fΘ(ω)|2〉
We now consider the limit Θ→∞:
Pff (ν) = limΘ→∞
2
Θ〈|fΘ(ω)|2〉
We can write this as
Pff (ω) = limΘ→∞
2
Θ
∫ Θ/2
−Θ/2
∫ Θ/2
−Θ/2
dt dt′ eiω(t−t′)〈f(t′)f(t)〉
If the correlation function 〈f(0)f(τ)〉 decays on a timescale that is much shorterthan Θ, the we can write
Pff (ω) = limΘ→∞
2
Θ
∫ Θ/2
−Θ/2
dt
∫ ∞−∞
dτ eiωτ 〈f(0)f(τ)〉 .
It then follows that
Pff (ω) =
∫ ∞−∞
dτ eiωτ 〈f(0)f(τ)〉 = 2Cff (ω) ,
where Cff (ω) is the Fourier transform of the correlation function of f(t).
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D.2 Shot noise
As an application, consider the noise due to an electric current that is due tothe transport of individual charge carriers with charge e. Every time ti when acharge e crosses the conductor, we have a contribution eδ(t− ti) to the current.The total current during an interval Θ is
I(t) =
N∑i=1
eδ(t− ti)
where N denotes the total number of charges that have crossed in time Θ. Wecan now compute the Fourier transform of this current:
IΘ(ω) =
N∑i=1
e exp(iωti)
The power spectral density of this current is:
PII(ω) = limΘ→∞
2
Θ〈|IΘ(ω)|2〉
This can be written as
PII(ω) = limΘ→∞
2
Θ
N∑i=1
N∑j=1
e2〈exp(iω(ti − tj))〉
If we assume that different charge crossings are uncorrelated, we get 〈exp(iω(ti−tj))〉 = δij and hence
PII(ω) = limΘ→∞
2
Θ
N∑i=1
e2 = limΘ→∞
2N
Θe2
But the average current I = Ne/Θ. Therefore
PII(ω) = 2eI
The interesting feature of shot noise is that, by measuring the power spectrumof the current noise, we can determine the charge e of the charge carriers.
D.3 Thermal Noise
Consider two resistors R1 and R2 connected in a loop by loss-free wires. Due tothe thermal motion in the resistors, there will be voltage fluctuation generatedover both resistors. We denote the power spectra of these two noise sources byP1(ω) and P2(ω) respectively.
P1(ω) =2
T|V1(ω)|2
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and similarly for P2(ω). These fluctuating voltages set up currents in the loop.For example:
I1(ω) = V1(ω)/(R1 +R2)
The power dissipated per unit bandwidth by I1 in resistor 2 is
p12 = |I1(ω)|2 ∗R2 = |V1(ω)|2 R2
(R1 +R2)2
Conversely
p21 = |I2(ω)|2 ∗R1 = |V2(ω)|2 R1
(R1 +R2)2
In thermal equilibrium, p12 = p21. Hence
|V2(ω)|2R1 = |V1(ω)|2R2
This implies that|V1(ω)|2/R1 = |V2(ω)|2/R2
In other words|V1(ω)|2 = R1f(ω, T )
where f(ω, T ) is an, as yet, unknown function of frequency and temperature(but NOT of the resistance itself).
To derive an explicit expression for the thermal noise due to a resistor,consider a closed circuit contain a resistor R and a coil with induction L, againconnected by loss-less wires. The energy of a induction coil with current I isequal to 1
2LI2. In thermal equilibrium, we must have:
LI2
2=kT
2
or
〈I2〉 =kT
L
In a closed circuit, the sum of all potential differences must vanish. In thepresent case, there are three contributions: the voltage over the inductance dueto the change in current
VI = −L∂I∂t
the voltage drop over the resistor
VR = −IR
and the thermal noise VR. As the sum of these three terms must vanish, wehave for all t
L∂I(t)
∂t= −RI(t) + VR(t)
This equation looks exactly like the Langevin equation
mvx(t) = −αvx(t′) + ζx(t) .
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But, in Eqn. 86 we showed that
〈ζx(0)ζx(t)〉 = m〈vx(0)2〉2αδ(t)
By analogy:〈VR(0)VR(t)〉 = L〈I2〉2Rδ(t) = 2RkTδ(t)
By Fourier transforming, we get∫ ∞−∞
dt〈VR(0)VR(t)〉eiωt = 2RkT
But
2
∫ ∞−∞
dt〈VR(0)VR(t)〉eiωt = PV V (ω) ,
the power spectral density of the noise voltage. And hence:
PV V (ω) = 4RkT
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