Download Lecture Notes on Linear Response Theory

Lecture Notes on Linear Response Theory
Fred MacKintosh
Vrije Universiteit
Department of Physics and Astronomy
Office: T232
Email: [email protected]
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1.1
Fluctuations and Linear Response
Brownian Motion: Random Walks in 1d
We begin with arguably the simplest problem illustrating the effect of fluctuations, e.g., in a fluid: Brownian motion. This term derives from the botanist
Robert 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 not
dependent on the type of particles. We now know that this motion is due to the
ever-present thermal fluctuations in the fluid itself. In a series of papers beginning in 1905, Einstein gave a theoretical basis for Brownian motion. About the
same time, Smoluchowski derived the same results in a somewhat different way.
Here, we derive the properties of what is usually referred to as the random
walk problem for a single particle. Despite the seeming simplicity, this constitutes a rich example that will help to illustrate many of the principles of fluctuation about equilibrium generally. Examples of such random walks include the
Brownian motion of a particle in a fluid, as well as the random conformations of
long polymer molecules under some conditions. In the first of these examples,
the essential physical aspects of this problem are: the existence of randomly
fluctuating forces pushing the particle in random directions, the viscous drag of
the 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 external force) is fundamentally related to the strength of the random forces that
it experiences in the fluid. We’ll see later a statement of this deep connection
between fluctuations and dissipation in the form of the so-called FluctuationDissipation Theorem, and it is largely with this in mind that we shall examine
the basic problem of Brownian motion in such great detail.
We first look at Brownian motion in just one spatial dimension. Consider a
particle that can move just left (L) and right (R) along a line by steps of equal
length `. Steps to the left and right occur randomly, with equal probability
p = 1/2. Because the probability to jump to the left and to the right are equal,
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the average displacement of a particle during the i-th step is zero:
h∆xi i =
1
1
(+`) + (−`) = 0 .
2
2
(1.1)
However, the mean squared displacement is non zero:
h∆x2i i =
1
1
(+`)2 + (−`)2 = `2 .
2
2
(1.2)
Next consider average displacement after N steps:
N
X
h∆Xi =
h∆xi i = 0 .
(1.3)
i=1
The mean squared displacement after N steps is:
h∆X 2 i =
N
N X
X
h∆xi ∆xj i .
(1.4)
i=1 j=1
The jump directions on different steps are not correlated, therefore
h∆xi ∆xj i = 0
i 6= j .
for
(1.5)
For i = j, we have h∆xi ∆xj i = `2 . If we insert this in the expression for h∆X 2 i
we get
N X
N
X
h∆X 2 i =
`2 δij .
(1.6)
i=1 j=1
where δij is a Kronecker delta:
δij
≡ 1
for
i=j
=
for
i 6= j .
0
Hence:
h∆X 2 i = N `2 .
(1.7)
If the time interval between successive jumps of the particle is δt, then the
number of jumps in a time interval t is equal to t/δt and
h∆X 2 i(t) = `2 t/δt .
(1.8)
In other words, the mean squared displacement increases linearly with time. We
should compare this with the ballistic motion of a particle with a velocity v, in
which case ∆X(t) = vt, and hence h∆X 2 i(t) = hv 2 it2 .
For the simple case of one-dimensional Brownian motion we can compute not
just the mean-squared displacement of a particle, but the complete probability
distribution to find a particle at a distance x from its original position, after N
steps. We denote by NL and NR the number of steps that occur to the left and
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right. The total number of steps is then N = NL + NR , while the net length
of the walk (to the right) is x = ` (NR − NL ). The number of such walks total
steps N , of which NL steps are to the left and NR to the right is
N!
.
NL !NR !
Ω (x, N ) =
(1.9)
Then, the probability of a such a walk of net length x is
P (x, N ) =
N!
NL !NR !
N
N
1
N!
1
=
.
N −x/`
N +x/`
2
2
!
!
2
By using Stirling’s approximation n! ∼
=
P (x, N ) ∼
=
=
=
(1.10)
2
√
2πnnn e−n , we find that
√
N N eNR +NL 2πN
[N +x/`]/2 [N −x/`]/2 √
√
N −x/`
2N eN N +x/`
2πNR 2πNL
2
2
√
2πN
p
N −x/` p
N +x/`
]
[
[
(1 + x/N `) 2 (1 − x/N `) 2 ] π(N + x/`) π(N − x/`)
√
2
x/2` p
(1+x/N `)
(1 − (x/N `)2 )N/2 (1−x/N
N π(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 )
and hence
(1 + x/N `)
(1 − x/N `)
x/2`
= exp(x2 /(N `2 )) .
We can then write
P (x, N ) ∼
=
≈
r
2
1
p
πN exp(−x2 /(2N `)2 ) exp(x2 /(N `2 )) 1 − x2 /(N `)2
r
2
exp(−x2 /(2N `2 ))
πN
(1.11)
Of course, it comes as no surprise that this probability distribution is symmetric
and that the most likely walk returns to the origin (x = 0). Thus, the average
hxi = 0. The mean-square displacement for such random walks, however, is nonzero. This is most easily calculated by treating P (x, N ) above as a continuous
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distribution in the variable x. In fact, for small enough x/`, the distribution
above varies slowly over the discrete values of x that are possible for a given
large value of N . Thus,
R 2 −x2 /(2N `2 )
x e
dx
∼
= N `2 .
hx i = R −x2 /(2N `2 )
e
dx
2
(1.12)
Here, we have assumed that the particle moves with independent random steps
of length `. If we assume further that these steps occur at equally-spaced time
intervals δt, then t = N δt and
hx2 i ∼
= t`2 /δt.
(1.13)
√
This behavior x2 ∝ t or x ∝ t is characteristic of diffusion.
Before leaving this simple one-dimensional random walk problem, we can
make another observation. This problem is instructive for other physical situations, such as the conformations that a long polymer chain can make. The
resulting conformations are referred to as Gaussian. We ignore the fact that
the polymer chains cannot overlap in space. (Although an unphysical microscopic assumption, there are real situations where polymer conformations are
nevertheless Gaussian.) Then, Ω(x, N ) above represents the number of such
conformations, subject to a given end-to-end separation x. Thus, by the fundamental assumption of statistical mechanics (that microscopic states of the same
energy are equally likely), the probability of having a given value of x is proportional 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.
1.2
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 concentration variable n(~r, t), which is a function of both position ~r and time t.
We also denote by ~j(~r, t) = n(~r, t)~v (~r, t) the current density of particles at ~r
and time t, where ~v is the velocity. In thermal equilibrium, and in the absence
of other forces, the average density hn(~r, t)i will be constant. Furthermore, by
the Brownian motion of particles in the fluid, an initial inhomogeneity in the
density will relax over time toward this average, uniform density. Thus, spatial
variations (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,
(1.14)
which is known as Fick’s law. As we shall see, the minus sign agrees with
our physical intuition that Brownian motion will lead over time to a uniform
particle density. A positive sign here would lead to unphysical results. Thus,
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we see immediately that Brownian motion or diffusion is essentially dissipative
and irreversible. The constant D here is known as the diffusion constant, which
has units of square length per time.
There is one more physical aspect to Brownian motion that we have neglected
in this treatment so far, namely the fact that particles are conserved. If no
particles are created or destroyed while they fluctuate around in the fluid, then
the particle current and density are related by conservation
∂n(~r, t) ~ ~
+ ∇ · j(~r, t) = 0,
∂t
which is simply a mathematical expression of the fact that a net particle current
into (out of) any volume results in an increase (decrease) in the number of
particles in that region. This, together with Fick’s law leads to the diffusion
equation,
∂n(~r, t)
= D∇2 n(~r, t).
∂t
A particular solution to this equation is
n(~r, t) =
N
3/2
e−r
2
/(4Dt)
.
(1.15)
(4πDt)
This is, in fact, the solution 1 to the initial condition that N particles are placed
at 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 these
do not interact with each other, which is valid if the density is not too high, then
we could have equivalently described things in terms of the probability density
P (~r, t) for a single particle, which is given by Eq. (1.15) with N = 1. We then
find that
Z
h~r(t)i = 0; hr2 (t)i = r2 P (~r, t)d3 r = 6Dt ∝ t,
which is in agreement with our earlier result for the one-dimensional random
walk:
hx(t)i = 0; hx2 (t)i = t`2 /δt = 2Dt ∝ t,
where D = `2 /(2δt). The general result depends on the number of dimensions
d: hr2 (t)i = 2dDt.
Even without solving the diffusion equation explicitly, we can relate hx2 (t)i
to D. We do this as follows. We start with the one-dimensional diffusion
equation for the probability density P (x, t):
∂ 2 P (x, t)
∂P (x, t)
=D
.
∂t
∂x2
1 See
Appendix 1.13.
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We now multiply both sides of this equation by x2 and integrate over x:
Z ∞
Z ∞
∂ 2 P (x, t)
2 ∂P (x, t)
dx x
.
=
dx x2 D
∂t
∂x2
−∞
−∞
We can write the left hand side of this equation as
Z ∞
Z ∞
∂hx2 (t)i
∂P (x, t)
∂
dx x2
dx x2 P (x, t) =
=
∂t
∂t −∞
∂t
−∞
The right-hand side can be computed by partial integration (making use of the
fact that P (x, t) and its derivative vanish at x = ±∞:
Z ∞
Z ∞
Z ∞
∂ 2 P (x, t)
∂P (x, t)
dx x2 D
dx
2xD
dx P (x, t) .
=
−
=
+2D
∂x2
∂x
−∞
−∞
−∞
But as P (x, t) is normalized to one, the result is simply 2D and therefore:
∂hx2 (t)i
= 2D
∂t
(1.16)
This we can integrate to yield
hx2 (t)i = 2Dt
where we have used the fact that at t = 0, the probability distribution is a
δ-function. We can write similar expressions for hy 2 (t)i and hz 2 (t)i. Adding
these, we obtain
hr2 (t)i = hx2 (t)i + hy 2 (t)i + hz 2 (t)i = 6Dt .
In reality, for the example of a particle suspended in a fluid, we expect that the
kicks that the particle experiences in the fluid are random. Even so, since the
particle has a mass, there will be some effect of its inertia. So, we expect that the
motion is not truly random, but will be correlated for short times. For such short
times, the particle will move ballistically, and we expect that the displacement
grows linearly with time. Thus, we expect that hx2 i ∝ t2 , rather than hx2 i ∝ t
for short enough times. Actually, this must be true. We can already see that
√
there is a problem with the Eq. (1.13), 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
we can be more quantitative about below) will diffusion, characterized by Eq.
(1.13), be valid.
1.3
Velocity correlation function
x(t) is the distance that a particle has traveled in the x-direction in a time
interval t. If we denote the instantaneous velocity in the x-direction by vx (t),
then we can write
Z t
x(t) =
dt0 vx (t0 ) .
0
6
We can then write the mean-squared displacement as
Z t
Z t
hx2 (t)i =
dt00 hvx (t0 )vx (t00 )i .
dt0
0
0
The quantity hvx (t0 )vx (t00 )i is an example of a time correlation function - in this
case, the “velocity auto-correlation function” (VACF). This quantity measures
the degree of correlation between the velocity of a particle at two different times
t0 and t00 . When t0 = t00 , the value of the of the velocity correlation function is
simply hv 2 i. At sufficiently long times, when the particle has lost all memory of
its original velocity, hvx (t0 )vx (t00 )i=hvx (t0 )i × hvx (t00 )i =0 (because the average
velocity of a particle in equilibrium is zero). Time correlation functions, such as
the VACF are properties of the system in equilibrium. Their value can therefore
never depend on the times t0 and t00 individually, but only on the difference. This
is so because the properties of a system in equilibrium are time invariant, i.e.:
hvx (t0 )vx (t00 )i = hvx (t0 + τ )vx (t00 + τ )i (∀τ )
If we choose τ = −t00 , we get
hvx (t0 )vx (t00 )i = hvx (t0 − t00 )vx (0)i ,
which shows that the VACF depends only on the difference t0 − t00 . Another important property of (classical) time correlation functions is that we can permute
vx (t0 ) and vx (t00 ):
hvx (t0 )vx (t00 )i = hvx (t00 )vx (t0 )i .
Because of this symmetry, we can write
Z t
Z t
Z t
Z
dt0
dt00 hvx (t0 )vx (t00 )i = 2
dt0
0
0
0
t0
dt00 hvx (t0 )vx (t00 )i .
0
We make use of this result in evaluating
Z t
Z t0
∂
∂hx2 (t)i
= 2
dt0
dt00 hvx (t0 )vx (t00 )i
∂t
∂t 0
0
Z t
00
= 2
dt hvx (t)vx (t00 )i
0
Z t
= 2
dτ hvx (τ )vx (0)i ,
0
where we have defined τ ≡ t − t00 . If we insert the above result in Eqn. 1.16, we
obtain
Z
t
dτ hvx (τ )vx (0)i .
D=
(1.17)
0
This equation is only valid for times much longer than the decay time of correlations in the velocity, and hence we may replace the upper limit in the integral
by +∞. The expression that results is
Z ∞
D=
dτ hvx (τ )vx (0)i .
(1.18)
0
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1.4
Langevin Theory
In order to see at a deeper level the connection between the random fluctuations
in the fluid and the dissipative nature of the resulting particle motion, we turn
now to an approach suggested by Langevin. This will also solve a problem noted
above, namely that there must be a failure of our prior analysis of Brownian
motion at short times. Our approach here will be to more directly account for
the fluctuating forces acting on the particles by the surrounding fluid. We shall
see how thermal equilibrium (e.g., the expected uniform spatial distribution of
particles) can be established by these fluctuations.
We begin with the simple equation of motion satisfied by the particle of mass
m
d~v
= F(t),
m
dt
where F is the force acting on the particle. We might be tempted at this point
to 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 are
two 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 assume
here that the first of these can be described by the Stokes formula for the drag
on 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 valid
for small particles and small velocities. It is important to note that this force
is random and has zero mean, only to the extent that the particle velocity is
random with zero mean. Thus, we can say that, in the absence of external
forces, it has a zero time average.
It is important also to distinguish another kind of average that we have been
a bit sloppy about so far: the ensemble average. This represents an average over
a 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. Let
τ denote this correlation time, after which memory of the initial conditions is
lost. As a specific example, if this initial velocity were large compared with the
expected thermal velocities, we would expect that it would decay toward zero
for times large compared with τ . But, the precise way in which it will do this
depends on the details of the fluctuating force ζ~ due to the fluid. We are not
interested in the exact trajectory that an individual particle makes due to each
and every collision with individual fluid molecules. Rather, we want to know
what to expect for the way the particle velocity decays in a typical sample of a
certain 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 is
zero hζ(t)i = 0, while the ensemble average of the particle velocity is not zero, at
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least for times smaller than τ . Mathematically, we could say that the ensemble
~
corresponds to a large set of possible functions ζ(t)
consistent with a given
fluid at a given temperature. The properties of these functions are yet to be
determined, but that the average hζ(t)i over this set is zero for all times t is clear
by construction. At any time t, ensemble average quantities can be obtained by
~
averaging over the possible functions ζ(t).
We rewrite the equation of motion as follows
m
d~v (t)
~
= −6πηa~v (t) + ζ(t).
dt
(1.19)
If we take the average of this over the ensemble, we obtain
m
dh~v (t)i
= −6πηah~v (t)i,
dt
which implies that
h~v (t)i = ~v (0)e−t/τ ,
where we can now identify τ = m/(6πηa) as the relaxation time for effects of
the initial condition to die out. Here again the essentially dissipative nature
of 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 ζ~
is still unclear. This is what we want to establish next.
We can take the scalar product of (1.19) with the instantaneous position
v
d
r · ~v ) − v 2 . The result is
~r(t). In so doing, we note that ~r · d~
dt = dt (~
m
d
h~r · ~v i = −6πηah~r · ~v i + mhv 2 i.
dt
(1.20)
~ = 0. This last point is not as obvious as it
We have used the fact that h~r · ζi
~ = 0, it can be shown that h~v · ζi
~ =
may appear at first sight. Although h~r · ζi
6 0.
In the equation above, if we further assume that the Brownian particle has
attained thermal equilibrium, then we can replace hv 2 i by the equipartition value
3kT /m. (The kinetic energy mv 2 /2 of the particle involves three independent
degrees of freedom, in which the energy is quadratic. Classical Statistical Mechanics then predicts that this energy is 23 kT on average.) The resulting simple
differential equation can be solved:
h~r · ~v i = Ce−t/τ +
kT
.
2πηa
But, the constant C can be determined by the initial condition, which we take
to be that ~r = 0 at t = 0:
kT h~r · ~v i =
1 − e−t/τ .
2πηa
We can integrate this one more time, to find that
i
kT h
hr2 i =
t − τ 1 − e−t/τ ,
πηa
9
d
since h~r · ~v i = 12 dt
hr2 i.
At long times, for which t τ , we find again that
hr2 i = 6Dt.
Furthermore, we can now identify the diffusion coefficient with the dissipative
mechanism of the fluid viscosity:
D=
kT
.
6πηa
(1.21)
We also find, as expected, that for short times, hr2 i ∝ t2 , thus resolving the
problem with our earlier analysis for short times.
More generally, there is a fundamental relationship between the particle
diffusion constant D and the mobility µ or response of the particle to external
forces. If we apply an external force f~ to the particle, say due to gravity, then
the resulting drift velocity is h~v i = µf~. This defines what we mean by the
mobility. For our Stokes sphere in the viscous fluid, the full equation of motion
becomes
d~v (t)
~
= −6πηa~v (t) + f~ + ζ(t).
(1.22)
m
dt
1
Thus, the mobility in this case is µ = 6πηa
, and
D = µkT .
(1.23)
This is known as the Einstein relation, and is more general than this particular
problem of a sphere in a viscous fluid. It is also an example of the fundamental
relationship that exists between a dissipative response and fluctuations. In this
case, the response is that of a particle to an external field. This response is
characterized by the mobility µ. The fluctuating quantity here is the particle
position ~r(t).
There is another way to derive the Einstein relation that makes the relation
between the mobility µ and the diffusion constant D more obvious. Consider
solution in a closed volume V . If there is a concentration gradient of the dissolved species, this will result in a diffusion current ~j. The diffusion is equal to
the number density ρ of the dissolved species, multiplied by the average velocity
of these particles.
~j = ρh~v i .
Now suppose that the dissolve particles are subject to an external potential
U (x). If this potential is not constant, there will be a net force acting on the
particles:
∂U (x)
fx = −
,
∂x
where we have assumed that the potential ia a function of x only. The average
velocity of a particle due to this force is
vx = µfx = −µ
10
∂U (x)
.
∂x
As the particles move under the influence of this force, the density will become
inhomogeneous. But once the density is not constant, there will also be a
diffusion current. When the system reaches equilibrium, the diffusion current is
exactly balanced by the current due to the external force, i.e.:
0 = ρhvx i = −ρ(x)µ
∂ρ(x)
∂U
−D
.
∂x
∂x
(1.24)
But we also know that, in equilibrium, the probability to find a particle at a
position 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. 1.24, we get
0 = ρhvx i = −ρ(x)µ
∂U
D
∂U (x)
+
ρ(x)
.
∂x
kT
∂x
(1.25)
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
1
1~
~v (t) = − ~v (t) + ζ(t).
dt
τ
m
This, we rearrange slightly and multiply by et/τ :
1
d t/τ
d
1~
et/τ ~v (t) + et/τ ~v (t) =
e ~v (t) = et/τ ζ(t).
dt
τ
dt
m
We now change t to t0 and integrate from 0 to t, with the result that
Z t
0
−t/τ
−t/τ 1
~ 0 )dt0 .
~v (t) = ~v (0)e
+e
et /τ ζ(t
m 0
(1.26)
(1.27)
(1.28)
Again, we see that if we take the ensemble average of this equation, we obtain
h~v (t)i = ~v (0)e−t/τ ,
~ 0 )i = 0 for any t0 . If we square both
as we found before. This is because hζ(t
sides of (1.28) and then take the ensemble average, however, we find that
Z tZ t
0
00
1
~ 0 ) · ζ(t
~ 00 )idt0 dt00 . (1.29)
e(t +t )/τ hζ(t
hv 2 i = v(0)2 e−2t/τ + e−2t/τ 2
m 0 0
~ = 0. The correlation function
Again, a term had been dropped because hζi
0
0
~
~
K(t, t ) = hζ(t) · ζ(t )i, however, is non-zero in general. This is where the ugly
11
details of the fluctuating molecules in the fluid begin to appear directly in our
analysis. We finally have a measurable quantity that would appear to depend
~
directly on the random forces ζ.
~ · ζ(t
~ 0 )i, which
These appear through the correlation function K(t, t0 ) = hζ(t)
has several general properties that we can identify. First of all, so long as we
are looking at an equilibrium system that is not evolving in any macroscopic
sense in time, then this function can only depend on the time interval s =
t − t0 : K(t, t0 ) = K(s). Furthermore, provided that we confine our attention to
classical variables (as opposed to possibly non-commuting quantum mechanical
~ and ζ(t
~ 0 ), which tells us that K(s)
operators), we can reverse the order of ζ(t)
is symmetric, i.e., that it depends only on |s|. On simple physical grounds, we
also 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
~
~ 0 ) become decorrelated as |t − t0 | grows, i.e., that
that ζ(t) and ζ(t
~ · ζ(0)i
~
~
~
K(s) = hζ(s)
→ hζ(s)i
· hζ(0)i
=0
as s grows. Furthermore, we can see that K(s) attains its maximum value at
s = 0. In fact, since
h
i2
~ ± ζ(0)
~
~ · ζ(0)i
~
h ζ(s)
i = hζ 2 (s)i + hζ 2 (0)i ± 2hζ(s)
= 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).
~ · ζ(0)i
~
This all leads us to the inevitable conclusion that K(s) = hζ(s)
is a
function peaked at s = 0, from which it decays rapidly toward zero in a short
time τ ∗ , which is determined by the fast dynamics of the individual molecules of
the fluid surrounding the Brownian particle. We expect a very short time scale
for this, of order the time between collisions with the individual molecules. More
generally, it should be the time scale that characterizes the rate with which the
degrees of freedom responsible for the random forces ζ~ relax to internal equilibrium 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 take
it to be proportional to the delta function, that is that it can be taken to be
non-zero only for s = 0, but with finite integral over s:
Z ∞
C=
K(s)ds.
−∞
Returning to the problem that led us to consider this function K(s) in the
first place, namely the evaluation of the mean-square velocity, we note that the
double integral in (1.29) is appreciably non-zero only for t0 ∼
= t00 . Thus,
Z t
0
2
2 −2t/τ
−2t/τ C
∼
e2t /τ dt0
hv i = v(0) e
+e
m2 0
C τ 2t/τ
= v(0)2 e−2t/τ + e−2t/τ 2
e
−1 ,
(1.30)
m 2
12
where we have made the approximation suggested above:
K(s) ∼
= Cδ(s).
In particular, however, Eq. 1.30 is true for times t large compared with τ , at
which point all memory of the initial conditions (e.g., ~v (0)) should have been
forgotten. Then, the ensemble average hv 2 i = 3kT /m, consistent with the
Equipartition Theorem. The only way this can be true is if
Cτ
= 3kT,
2m
~
i.e., that the drag coefficient α is fundamentally related to the random forces ζ:
Z ∞
1
~ · ζ(0)ids.
~
α=
hζ(s)
(1.31)
6kT −∞
The above result can be obtained in a few lines, using the techniques described
in Appendix 1.14 (see, in particular, section 1.14.1). The relation between the
drag coefficient and the random forces acting on a particle is special case of what
we will later learn generally as the Fluctuation-Dissipation theorem. For now,
we note that the drag that a particle experiences in the fluid is due, at a microscopic level, to the randomly fluctuating forces that the particle feels at finite
temperature. The random forces are at the origin of macroscopic dissipation.
We began this analysis of the motion of a particle in a dissipative medium
by considering the equation of motion
m
d~v (t)
~
= −α~v (t) + f~ + ζ(t),
dt
(1.32)
where, for instance, the drag coefficient α = 6πηa for a sphere of radius a moving
in a fluid with viscosity η. Here, f~ represented a possible external force applied
to 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 was
convenient to rewrite the equation of motion (1.26) in the absence of external
forces. 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,
~ This dependence, however,
hv 2 i) that depends directly on the random forces ζ.
0
~
~ 0 )i, which has several
is only through the correlation function K(t, t ) = hζ(t)·
ζ(t
general properties that identified above.
Finally, we found that the drag coefficient α is fundamentally related to the
~ via Equation 1.31, which expresses the fact that the drag that
random forces ζ,
a particle experiences in the fluid is due, at a microscopic level, to the randomly
fluctuating 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 can
~ which are presumably due to the
be entirely expressed in terms of the forces ζ,
random equilibrium fluctuations of the fluid! These random forces are at the
origin of macroscopic dissipation.
13
1.5
Spectral (Fourier) analysis of fluctuations
When dealing with fluctuating quantities such as ζ~ and correlation functions
such as K(t, t0 ) it is often most convenient to characterize them in terms of
frequency. We have already noted that each specific realization of the random
~
force ζ(t)
(i.e., for each specific member of an ensemble of similar systems) is
a very rapidly varying function, with a characteristic time scale of τ ∗ , which
is very short compared with other time scales of interest, such as that of the
~
motion 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-frequency
components (details) that are not apparent in the motion of our hypothetical
particle 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 has
components 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 the
orthogonality property of complex exponential functions (see Appendix 1.15),
Z ∞
0
1
e−iω(t−t ) dω = δ (t − t0 ) ,
(1.33)
2π −∞
where δ(t − t0 ) is the Dirac δ function, which is zero for t 6= t0 , yet which
integrates to unity:
Z ∞
δ(t)dt = 1.
−∞
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. We
can write
Z ∞
f (t) =
δ(t − t0 )f (t0 )dt0
−∞
Z ∞
Z ∞
0
1
=
dt0
e−iω(t−t ) f (t0 )dω.
2π −∞
−∞
In other words,
f (t) =
1
2π
where
f˜(ω) =
Z
∞
f˜(ω)e−iωt dω,
−∞
Z
∞
0
f (t0 )eiωt dt0 ,
(1.34)
−∞
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),
14
(1.35)
we also have that
f˜∗ (ω) = f˜(−ω).
(1.36)
It is also worth noting that we have made some serious assumptions above
concerning convergence of the various integrals. This will not pose a practical
problem, but one might worry about this, since the function ζ is presumably nonzero and is fluctuating over the same range for arbitrarily large times. Thus, for
instance, ζ̃(ω) might not be convergent. One could instead define the transforms
for 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) = hζ(t +
s)ζ(t)i. We call the Fourier transform of this C(ω), and the two are related by
Z ∞
1
C(ω)e−iωs dω
K(s) =
2π −∞
and
Z
∞
0
K(s0 )eiωs ds0 .
C(ω) =
(1.37)
−∞
But, based on simple and general physical considerations, we showed before that
K(t, t0 ) = hζ(t)ζ(t0 )i not only depends just on the time interval s = t − t0 , but
K(s) is also symmetric. It is also, of course, real. It is not hard to show that
this also means that C(ω) is a real function and is symmetric in ω. We also
argued before that K(s) is a function peaked at s = 0 and rapidly decaying
to 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
in (1.37) is small wherever the K(s) is appreciable. This suggests that C(ω) is
nearly 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 for
the motion of our sphere in the fluid. So, we say that the noise spectrum is
white. Just as we argued before that
K(s) ∼
= Cδ(s),
we can say that
C(ω) ∼
= C,
a constant. Note in particular that
Z ∞
C(0) =
K(s0 )ds0 = 6kT α,
∞
and
1
~ · ζ(0)i
~
K(0) = hζ(0)
=
2π
Z
∞
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 ζ̃(ω). First of all, we note that (for most systems) we can
15
replace the ensemble average for a single realization of the fluid or member of
the ensemble in the definition of K by a time average over a wide enough time
interval −Θ < t < Θ:
K(s)
=
∼
=
=
=
hζ(t + s)ζ(t)i
Z Θ
1
ζ(t + s)ζ(t)
2Θ −Θ
Z Θ Z ∞
Z ∞
dω −iω(t+s)
1
dω 0 −iω0 t
dt
e
ζ̃(ω)
e
ζ̃(ω 0 )
2Θ −Θ
−∞ 2π
−∞ 2π
Z ∞
1
dω −iωs
e
ζ̃(ω)ζ̃(−ω).
2Θ −∞ 2π
In other words,
1
|ζ̃(ω)|2 .
2Θ
Here, we have used the orthogonality of the complex exponential functions and
the fact that ζ is a real function. It may seem strange that a measurable quantity
C(ω), which we have seen is directly related to the drag coefficient, and thus the
diffusion constant, seems to depend inversely on the time over which we observe
the system Θ. However, as we have also noted, the Fourier transform ζ̃(ω) may
not be well defined (e.g., finite). In fact, one should use the modified functions
ζΘ (t) described above. These have finite transforms that grow with Θ. But, the
expression relating C(ω) to ζ̃Θ (ω) can be expected to become independent of Θ
for large enough Θ (see Appendix 1.16.1).
An important historical application of these ideas was to simple electrical
circuits, which satisfy an equation similar to that of a particle in a dissipative
medium:
d
L I(t) = Vext − RI(t) + V 0 (t),
(1.38)
dt
where L is the inductance, I is the current, Vext represents a (constant or
slowly varying) external applied field (emf), R is the resistance, and V 0 is the
fluctuating field related to dissipation. Taking the ensemble average of this
equation yields the familiar law relating the voltage, current, and resistance:
C(ω) =
hIi =
1
Vext .
R
The resistance here is analogous to the drag coefficient α above, and a similar
analysis to what was done above leads to
Z ∞
1
CV V (0)
R=
hV 0 (t + s)V 0 (t)i =
.
2kT ∞
2kT
Here, the factor of 2 rather than 6 is due to the fact that we are dealing with
an inherently one-dimensional problem rather than a three-dimensional one.
Furthermore, if the fluctuations of V 0 are very fast (i.e., characterized by a
16
short correlation time τ ∗ , here corresponding to the time between collisions of
electrons in the wires, for instance), as for the fluid, then the spectrum of these
fluctuations is very broad range
Z ∞
CV V (ω) =
hV 0 (t + s)V 0 (t)ieiωs ds ∼
= 2kT R
−∞
for ω in a range of order −1/τ ∗ to 1/τ ∗ . This general relationship between
the voltage fluctuations and the resistance in a circuit is known as Nyquist’s
theorem, and is yet another special case of the fluctuation-dissipation theorem.
For a step-by-step derivation, see Appendix 1.16.3. In Appendix 1.16.2, we
discuss another important noise type, namely shot noise.
We have seen so far that the macroscopic drag coefficient, and thus the mobility and diffusion coefficient, of a particle in a simple liquid is related directly
to the spectrum of microscopic fluctuations within the liquid. This also means
that the physics governing the response of a system that is taken out of equilibrium (e.g., by dragging a sphere through a liquid) can be described entirely in
terms if the fluctuations of the system about the equilibrium state. This is the
fundamental observation behind the Fluctuation-Dissipation Theorem and the
somewhat 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 particle motion (fluctuations) within liquids. Namely, let us see how the mobility
of the particle (e.g., when subjected to an external driving force) relates to the
random fluctuations of the particle motion rather than the random fluid motion. We’ll do this in just one dimension, and will thus neglect vector notation.
Consider the position of the particle, which we assume to start at t = 0 from
the origin. This satisfies
Z
t
v(t0 )dt0 .
r(t) =
0
Thus,
hr2 (t)i =
Z tZ
0
t
dt0 dt00 hv(t0 )v(t00 )i.
(1.39)
0
Here, what we see is a correlation function
Kvv (t0 , t00 ) ≡ hv(t0 )v(t00 )i
that is reminiscent of K we saw before. In fact, Kvv satisfies all the properties
that we previously identified for K = Kζζ , except that we can expect to find a
longer correlation time of order τ . In particular, Kvv should be a (symmetric)
function only of the time interval between t0 and t00 .
We can anticipate a result based on (1.39). Since we expect Kvv (s) to decay
toward zero for times long compared with the correlation time τ , if we are
interested in times long compared with this time, then we expect that we can
also approximate Kvv (s) by a δ function, resulting in
hr2 (t)i ∼
= tCvv ,
17
where
Z
∞
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
Z
1 t t0 /τ 0 0
e ζ(t )dt .
(1.40)
v(t) = v(0)e−t/τ + e−t/τ
m 0
Thus,
2 −(2t+s)/τ
hv(t + s)v(t)i = v(0) e
∼
= v(0)2 e−(2t+s)/τ
e−(2t+s)/τ
+
m2
Z
(t+s)
t
0
(t+s)
0
e−(2t+s)/τ
Cζζ
+
m2
Z
Z
0
0
00
e(t +t )/τ hζ(t0 )ζ(t00 )idt0 dt00
Z
t
0
00
e(t +t )/τ δ(t0 − t00 )dt0 dt00 ,
0
where we have made the approximation
hζ(t0 )ζ(t00 )i = Cζζ δ(t0 − t00 )
made before. Continuing, we find that
e−(2t+s)/τ
hv(t + s)v(t)i ∼
Cζζ
= v(0)2 e−(2t+s)/τ +
m2
= v(0)2 e−(2t+s)/τ
min(t+s,t)
Z
0
e2t /τ dt0 ,
0
τ
e−(2t+s)/τ
Cζζ
+
2
m
2
e2t/τ − 1
if s > 0
e2(t+s)/τ − 1 otherwise
For times t long enough ( τ ) so that all memory of the initial value of the
velocity is lost,
hv(t + s)v(t)i
→
kT −|s|/τ
e
.
m
(This is a one-dimensional result.) Indeed, for t τ , we find that
hr2 (t)i → tCvv ,
where
Z
∞
hv(t + s)v(t)ids =
Cvv =
−∞
2kT τ
,
m
Which agrees with our prior result for the three-dimensional diffusion, once we
account for the appropriate factors of 3:
hr2 (t)i = 6Dt,
where D = kT /α.
18
Continuing in one dimension, however, we note the complementary expressions for the drag coefficient α and the mobility µ:
Z ∞
1
hζ(t + s)ζ(t)ids
(1.41)
α=
2kT −∞
and
1
1
µ= =
α
2kT
∞
Z
hv(t + s)v(t)ids.
(1.42)
−∞
For the electric circuit problem we introduced before, the analogous results for
the resistance R and conductivity Σ are:
Z ∞
1
R=
hV (t + s)V (t)ids
(1.43)
2kT −∞
and
1
Σ=
2kT
Z
∞
hI(t + s)I(t)ids,
(1.44)
−∞
where V is the (fluctuating) voltage and I is the corresponding current. The
second of these relationships is usually written in terms of the current density
j and the (local) conductivity σ. This is known as the Kubo formula
Z ∞
1
hj(t + s)j(t)ids.
(1.45)
σ=
2kT −∞
In each of these cases, we have a pair of corresponding quantities: force and
velocity, voltage and current, . . . . In each case, the pair of quantities is related
in 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
Z ∞
Z ∞
2
hζ(t + s)ζ(t)ids
hv(t + s)v(t)ids = (2kT ) .
(1.46)
−∞
1.6
−∞
The Fluctuation-Dissipation Theorem and Linear Response
We have seen in the examples above that one can, at least sometimes, relate
quantities that have to do with systems out of equilibrium to the spectrum
of thermal fluctuations about equilibrium. These principles are based on the
Fluctuation-Dissipation Theorem, which was proven in general by Callen and
Welton in 1952, although special cases (as we have seen) of this were understood
much earlier. The basic idea behind the fluctuation dissipation theorem was
also captured in the 1930s by the regression hypothesis of Onsager, who argued
that the physics of macroscopic relaxation of a system back to equilibrium is
governed by the same physics as the relaxation of spontaneous fluctuations about
equilibrium. We have seen this, for instance in the relaxation of an initial value
19
of v(0) of the velocity of a particle in a viscous liquid is described by the same
time constant τ that appears in the correlation function Kvv above.
In order to derive the general relationship between, say the relaxation of some
macroscopic observable and its equilibrium fluctuations, we need to look at the
microscopic evolution of the system. This can either be done in the context of
quantum mechanics, as is done in the book by Chaikin and Lubensky, or in the
context of classical, Hamiltonian mechanics, as we shall do below. We describe
the microscopic state of the system by a point in phase space described by a set
of 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 Hamiltonian
mechanics that we need to recall is the fact that, given some initial point in phase
space (p(0), q(0)) at time t = 0, the state of the system (p(t), q(t)) at time t in
the future is completely determined by the initial condition and the Hamiltonian
H (p(0), q(0)). It is convenient for us to denote this evolution of the system by
some time-evolution operator Tt , defined by (p(t), q(t)) = Tt (p(0), q(0)). In
classical statistical mechanics, the microstates of a system in equilibrium are
distributed according to a probability distribution
P (p, q) =
where
Z
Q=
1 −βH(p,q)
e
,
Q
dp dq e−βH(p,q)
and β = 1/(kT ). The integral is over all of phase space. Again, we use an
abbreviated notation, in which dp dq indicates an infinitesimal volume element
in 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 new
Hamiltonian H0 ). In fact, we shall look at the reverse problem, in which a
perturbing field was applied to the system in the distant past, resulting in a new
equilibrium at time t = 0 characterized by the modified Hamiltonian H0 . At time
t = 0, we turn off this field, and allow the system to relax back to equilibrium
(characterized by H). It is this relaxation of the system back to equilibrium
that we want to examine. Let A(p, q) be some macroscopic observable whose
relaxation we are interested in. We shall consider only linear response theory,
in which we assume the perturbation is sufficiently weak that the Hamiltonian
H0 can be taken to be H0 = H + ∆H, where ∆H = −f A and f is the perturbing
field. (It is not hard to see that with this definition, f and A are conjugate
thermodynamic variables, like pressure and volume, since f = − ∂F
∂A , where F is
the free energy.) The initial state of the system corresponds to distribution
P 0 (p, q) =
where
0
Q =
Z
1 −βH0 (p,q)
e
,
Q0
0
dp dq e−βH (p,q) .
20
The initial (macro)state of the system is one in which
Z
0
1
hAi = 0
dp dq e−βH (p,q) A(p, q).
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 operator
T , rather than H0 (p, q) and T 0 . In other words, at times t > 0,
Z
0
1
dp dq e−βH (p,q) A (Tt (p, q))
hA(t)i =
Q0
R
dp dq e−βH(p,q) (1 − β∆H) A (Tt (p, q))
∼
R
.
=
dp dq e−βH(p,q) (1 − β∆H)
Expanding, and keeping only terms through linear order gives
R
dp dq e−βH(p,q) A (Tt (p, q))
∼
R
hA(t)i =
dp dq e−βH(p,q)
R
dp dq e−βH(p,q) A(p, q)A (Tt (p, q))
R
+βf
dp dq e−βH(p,q)
R
R
dp dq e−βH(p,q) A(p, q) dp dq e−βH(p,q) A (Tt (p, q))
R
R
−βf
.
dp dq e−βH(p,q)
dp dq e−βH(p,q)
This means that, to linear order,
hA(t)i − hA(t)i0
= βf hA(0)A(t)i0 − hAi20 ,
= βf hδA(0)δA(t)i0 ,
(1.47)
where δA(t) = A(t) − hAi0 and hi0 refers to an ensemble average in the unperturbed system. We have been somewhat sloppy in the notation here. The
first term in this equation refers to a evolution to time t of an ensemble of not
only similarly prepared systems, but systems subject to the same perturbation.
It does not strictly refer to any ensemble average in the usual sense. Thus, it
might be less confusing to write
∆A(t)
=
βf hδA(0)δA(t)i0 ,
(1.48)
where this expresses the average deviation A from its equilibrium value. The
average is to be understood to mean over the ensemble of systems with the same
perturbation.
This is one form of the Fluctuation-Dissipation theorem, expressing the
macroscopic evolution of a system out of equilibrium to the spontaneous fluctuations of the system about equilibrium. This is usually expressed, however, in
terms of the (linear) response of the system. This is implicit in the expression
above. 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 time
21
dependent. The most general form of a linear response to a time-dependent f
is
Z
∆A(t) = dt0 χ(t, t0 )f (t0 ),
where χ expresses the fact that the response at time t depends on the way
the system was perturbed at other times. In principle, we should allow for
arbitrarily influences from distant times. Thus, the integral above should be
over all times t0 . It would be unphysical, however, if this influence extended
to future times. How, for instance, can the system anticipate the perturbation
that it will experience in the future? This simple, and physically reasonable
assumption of causality can be expressed mathematically as χ(t, t0 ) = 0 for all
t0 > 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 that
χ should depend only on the time interval between t and t0 , so that
χ(t − t0 ) if t > t0
χ(t, t0 ) =
.
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 in
terms if the dynamics of the equilibrium system. We have examined a particular
case above, in which we perturbed the system for times t < 0, then removed the
perturbation. This corresponds to
f if t < 0
f (t) =
.
0 otherwise
Here, we found that
∆A(t)
=
βf hδA(0)δA(t)i0 ,
(1.49)
But,
Z
0
dt0 χ(t, t0 )f
∆A(t) =
−∞
in this case, since f = 0 for t0 > 0. By a change of variable of integration, this
becomes
Z ∞
∆A(t) = f
dt0 χ(t0 ).
t
But, this only applies for t > 0. Thus,
d
hδA(0)δA(t)i if t > 0
−β dt
.
χ(t) =
0
otherwise
This expresses the way in which a system responds to a perturbation (taking it
out of equilibrium) in terms of the spontaneous fluctuations of the system about
equilibrium.
22
Last time, we saw how the response of the system, specifically in terms of
some observable quantity A, could be described by a linear response function
χ:
Z
∞
dt0 χ(t, t0 )f (t0 ),
∆A(t) =
(1.50)
−∞
where χ expresses the fact that the response at time t depends on the way
the system was perturbed by some generalized force f at other times. Here,
∆A(t) ≡ A(t) − hAi0 , where A(t) represents the expected value of A at time
t (this is actually an ensemble average quantity, but for a particular ensemble
of similarly perturbed systems.) Based on the physical principle of causality,
however, we expect the response only to depend on the history of the system
and its perturbation. Furthermore, since the response function should be only
an equilibrium property of the system itself (and not, for instance, dependent
on f , at least if the perturbation is weak enough that the response is, indeed,
linear). Thus, we expect that
χ(t − t0 ) if t > t0
χ(t, t0 ) =
.
0
otherwise
In deriving the above, we assumed a very particular form of the perturbed
Hamiltonian:
H0 = H − f A.
In general, however, we expect if the (generalized) force f is weak, then one can
simply expand this Hamiltonian to linear order, with the result that
H0 = H − f B.
That is, the linear coefficient of f need not be A, but may be some other variable
B. When the system is subjected to such a perturbation, we expect to find a
linear response of the variable A, as given by (1.50), where we shall denote
the response function by χAB (t − t0 ), since it represents the response of A to
a perturbation f that is conjugate to B. Here, f and B are conjugate in the
thermodynamic sense, in that
∂F
∂B
∂
log Q
∂B Z
1 ∂
= −kT
dp dq e−β(H−f B)
Q ∂B
= −f,
= −kT
where
Z
Q=
dp dq e−β(H−f B)
is the partition function and
F = −kT log Q
23
is the free energy. This is just like the relationship between the conjugate
variables volume V and pressure P :
∂F
= −P.
∂V T,N
As before, if we consider a perturbing field f that is turned on in the distant
past, and then switched off at time t = 0, we can write the evolution of (the
ensemble average of) A as
Z
0
1
dp dq e−βH (p,q) A (Tt (p, q))
A(t) =
0
Q
R
dp dq e−β(H(p,q)−f B(p,q)) A (Tt (p, q))
R
=
.
dp dq e−β(H(p,q)−f B(p,q))
An alternative way of deriving the linear response that we calculated before is
to look at the derivative of this with respect to the perturbation f :
Z
0
∂A(t)
β
=
dp dq e−βH (p,q) B (p, q) A (Tt (p, q))
0
∂f
Q
R
Z
0
0
dp dq e−βH (p,q) A (Tt (p, q))
−
β
dp dq e−βH (p,q) B (p, q) .
2
R
0
dp dq e−βH (p,q)
Evaluating this derivative for f = 0, and noting that the point (p, q) in phase
space represents the state of the system at time t = 0, we find that
∆A(t) '
=
βf (hB(0)A(t)i0 − hA(t)i0 hB(0)i0 )
βf hB(0)δA(t)i0 .
(1.51)
Following the same analysis as before, we also find the response function
χAB (t) in terms of the fluctuating quantities δA and B:
χAB (t) = −β
d
hB(0)δA(t)i0 .
dt
(1.52)
Again, here, we should note that these thermodynamic averages hi0 refer to
the unperturbed (i.e., equilibrium) system. Thus, they represent equilibrium
fluctuations of the system, which exhibit some temporal correlations. The main
example 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 to
the particle (say, by some magnetic or gravitational field). The response of the
system is to develop a drift velocity v. But, the thermodynamically conjugate
variable to f is the displacement x of the particle (in the direction of the force,
which we take to be the x direction). Thus,
H0 = H − f x.
24
So, the ensemble average drift velocity is
Z t
Z
χvx (t − t0 )f (t0 )dt0 = f0
v(t) =
−∞
∞
χvx (t0 )dt0 ,
t
when, as before, a force f0 is applied in the distant past and switched off at
time t = 0. But, in this case, we expect that v(0) = µf0 , where µ is the particle
mobility. But, we know that
Z ∞
Z ∞
1
1
µ=
hv(t)v(0)idt =
hv(t)v(0)idt,
2kT −∞
kT 0
where we have used the symmetry of Kvv (t) = hv(t)v(0)i. This suggests that
χvx (t) = βKvv (t). This says that the dynamics associated with the response χ of
a system to a non-equilibrium perturbation are governed by the same underlying
principles as the fluctuations Kvv (t) = hv(t)v(0)i about equilibrium. This is,
indeed, the case, as Onsager suggested in his regression hypothesis. But, let’s
see if we can derive the relationship χvx (t) = βKvv (t).
We see from (1.52) that
χvx (t) = −β
d
hx(0)v(t)i0 .
dt
(1.53)
But,
hx(0)v(t)i0 = hx(t0 )v(t + t0 )i0 ,
which cannot depend on t0 . Hence,
0
=
=
=
d
hx(t0 )v(t + t0 )i0
dt0
d
d
h 0 x(t0 )v(t + t0 )i0 + hx(t0 ) 0 v(t + t0 )i0
dt
dt
hv(t0 )v(t + t0 )i0 + hx(t0 )v̇(t + t0 )i0 ,
and
d
hx(t0 )v(t + t0 )i0 = hx(t0 )v̇(t + t0 )i0 = −hv(t0 )v(t + t0 )i0 = −Kvv (t).
dt
This shows that
1 −t/τ
e
,
m
for our particle in the fluid. We see that this represents what is sometimes
referred to as a memory function. It shows how the response at one time depends
on the history of disturbances, but with vanishing influence from distant past.
That is, the memory fades over time. (As we all know too well!)
χvx (t) = βKvv (t) =
25
1.7
The Damped Harmonic Oscillator
The harmonic oscillator illustrates nicely many of the ideas we have been developing. It also turns out to be a very practical application of the Langevin
analysis. There are many experimental situations of current interest where one
has a harmonic potential for particles that would otherwise undergo brownian 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. Our
equation of motion becomes
mẍ(t) + αẋ(t) + Kx(t) = f (t),
where f is the force acting on the particle. We’ll consider the case where this
is some external driving force. Of course, there will also be our familiar random brownian forces. But, we shall treat ensemble quantities below. Fourier
transforming the equation of motion leads to
K − mω 2 − 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 linear
response:
1
fω ,
xω =
(K − mω 2 − iαω)
where
1
K − mω 2 − iαω
is the Fourier transform of the response function defined by
Z ∞
x(t) =
dt0 χ(t − t0 )f (t0 ).
χ̃(ω) =
−∞
It is tempting at this point to anticipate the frequency-dependent version of
our expression
χxx (t) = −β
d
d
hx(t)x(0)i = −β Kxx (t),
dt
dt
(1.54)
which would read
χ̃(ω) = iωβhxω x−ω i.
(incorrect)
But, this would be wrong. Recall that the left-hand side of (1.54) is non-zero
only for t > 0, while the right-hand side is a symmetric function of time. Thus,
the equation is valid only for positive time. In fact, given the properties of
Kxx , the right-hand side of the equation is an odd function. We have noted
before that the 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 would be odd and purely imaginary. We can write (1.54) in a way that
26
is valid for all time. 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
(χ(t) + χ(−t))
2
and
1
(χ(t) − χ(−t)) .
2
The Fourier transform χ̃E (ω) is thus real and symmetric. We call it χ0 (ω).
Similarly, the Fourier transform χ̃O (ω) is imaginary and odd in ω. We call it
iχ00 (ω). Note that both χ0 and χ00 are real, and that
χO (t) =
χ̃(ω) = χ̃0 (ω) + iχ̃00 (ω).
Now, returning to the expression (1.54) above, we can write this in a way
valid for all times:
2χO (t) = −β
d
d
hx(t)x(0)i = −β Kxx (t).
dt
dt
(1.55)
Thus,
ωβ
ωβ
hxω x−ω i =
h|xω |2 i,
(1.56)
2
2
which is a common way to write the fluctuation dissipation theorem. The imaginary part of the (Fourier transform of the) response function is directly related
to the power spectral density of the equilibrium fluctuations.
χ00 (ω) =
1.8
Kramers-Kronig relations
There are very important and far-reaching implications of the physical observation of causality in the response function χ(t), i.e., the fact that it vanishes
for t < 0. Consider the Laplace transform of this as a function of a complex
frequency z:
Z
Z
∞
∞
eizt χ(t)dt =
χ̃(z) =
−∞
eizt χ(t)dt.
(1.57)
0
(Here, the notation is going to go down-hill even further, but I hope the context
makes the various quantities clear enough.) Again, the function χ(t) is zero for
negative times. Furthermore, on simple physical grounds, we also expect that
χ is bounded at long times. In fact, it will usually be the case that it vanishes
at 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 the
function χ̃(z) of the complex variable z is actually analytic in the upper half
plane (=z > 0). Thus, we can apply the Cauchy theorem for complex functions:
I
dz 0 χ̃(z 0 )
,
χ̃(z) =
2πi z 0 − z
27
for any contour in the upper half plane containing the point z. Of course, as
the complex variable z approaches some point ω on the real axis, χ̃(z) → χ̃(ω).
Let z = ω + i, which lies in the upper half plane. Since the contour must
contain z but can otherwise be any contour in the upper half plane, we can
take the contour defined by a line very close to the real axis together with a
large semicircle defined by |z| equal to some very large radius R. The only
contributions will then come from the portion of the contour defined by z 0 =
ω 0 + i0 , where we will take 0 to zero. Thus,
Z ∞
χ̃(ω 0 )
1
dω 0 0
.
χ̃(ω + i) =
2πi −∞
ω − ω − i
Separating the real and imaginary parts of singular part of the integrand,
we find
1
ω0 − ω
=
+i
2
2
0
2
0
ω 0 − ω − i
(ω − ω) + (ω − ω) + 2
The imaginary part of this is an increasingly highly peaked function as → 0.
In fact, in this limit, it approximates a δ function:
(ω 0
2
− ω) + 2
→ πδ(ω 0 − ω).
Thus, for small enough ,
1
2πi
χ̃(ω) '
Z
∞
dω 0 χ̃(ω 0 )
−∞
ω0 − ω
(ω 0
2
− ω) +
2
1
+ χ̃(ω).
2
So that
χ̃(ω)
=
=
χ0 (ω) + iχ00 (ω)
Z ∞
1
ω0 − ω
lim
dω 0
(χ0 (ω 0 ) + iχ00 (ω 0 )) .
2
πi →0 −∞
(ω 0 − ω) + 2
Comparing real and imaginary parts, we find that
Z ∞
Z ∞
1
ω0 − ω
1
χ00 (ω 0 )
0
0
00
0
lim
dω
dω 0 0
χ (ω) =
χ (ω ) = P
2
π →0 −∞
π
ω −ω
(ω 0 − ω) + 2
−∞
and
χ00 (ω)
=
−
1
lim
π →0
Z
∞
−∞
dω 0
ω0 − ω
1
χ0 (ω 0 ) = − P
2
0
2
π
(ω − ω) + Z
∞
−∞
dω 0
χ0 (ω 0 )
.
ω0 − ω
These are the Kramer-Kronig relations, which express the fact that the functions
χ0 and χ00 are not independent of each other. In fact, one can completely recover
one from the other via a sort of transformation involving the full frequency
information.
28
1.9
General Properties of Response Functions
Last time, for the harmonic oscillator, we found an expression,
ωβ
h|xω |2 i,
2
χ00 (ω) =
relating the response function to the the power spectral density of thermal
fluctuations of the position. The more general result when we have a perturbed
Hamiltonian H0 = H − f A is
χ00AA (ω) =
ωβ
h|Aω |2 i.
2
(1.58)
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 this
expression as follows,
χ00 (ω)
h|Aω |2 i = 2kT AA
.
(1.59)
ω
The left-hand-side is clearly a real, symmetric, and strictly positive quantity.
Thus, we learn quite generally that χ00AA (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 harmonic
oscillator, although the results can easily be generalized. Then,
Z t
Z t
0
0
0
x(t) =
χ(t − t )f (t )dt = f0
χ(t)dt0 .
−∞
−∞
If this is bounded, as is, for instance, the displacement x for the harmonic
oscillator, then
x(t) → χ0 f0 ,
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,
Z ∞
χ0 =
χ(t)dt = χ0 (0).
−∞
Using a Kramers-Kronig relation, we discover a thermodynamic sum rule concerning χ00 :
Z
1 ∞ χ00 (ω 0 ) 0
χ0 =
dω .
(1.60)
π −∞ ω 0
Here, we have used the fact that χ00 is an odd function, which means that the
integrand is non-singular.
29
1.10
Dissipation
Among the many general properties of response functions is the connection
between χ00 and macroscopic dissipation. Non only is this an important fundamental observation, but there are also many experimental situations in which
dissipative processes such as absorption are measured. Here, we show that such
measurements can be used to determine the response function. As above, we
shall look at this connection between χ00 and dissipation within the context of
the damped harmonic oscillator, although this is only meant to illustrate the
more general results. The power dissipated can be expressed here in terms of the
force and the velocity. The work done in displacing a particle a small amount
dx with a force f is dW = f dx. Thus, the power is
dW
= f ẋ(t).
dt
Consider what happens when we drive the system with a monochromatic
force
1
f (t) = f0 cos (ω0 t) = f0 e−iω0 t + eiω0 t .
2
Thus, the Fourier transform of this is
fω = πf0 [δ(ω − ω0 ) + δ(ω + ω0 )] .
The displacement is then given by
Z ∞
1
x(t) =
χ̃ (ω) fω e−iωt d ω
2π −∞
Z
f0 ∞ χ̃(ω0 )e−iω0 t + χ̃(−ω0 )eiω0 t
=
2 −∞
=
f0 [χ0 (ω0 ) cos(ω0 t) + χ00 (ω0 ) sin(ω0 t)] .
The average power dissipated during one period T = 2π/ω0 is then
1
T
Z
T
f (t)ẋ(t)dt =
0
1 2
f ω0 χ00 (ω0 ).
2 0
Thus, the power dissipated is determined by χ00 alone. Since this is also directly
related to the power spectral density of fluctuations, hence the name fluctuationdissipation theorem. We also find, that
ωχ00 (ω) > 0,
since, physically, the power dissipated must be positive. We previously found
that χ00 (ω)/ω was strictly positive.
30
1.11
Example: the Harmonic Oscillator, Again
We can see many examples of the general properties of linear response theory
in the damped harmonic oscillator problem discussed last time. Previously,
we analyzed the response of the damped harmonic oscillator, described by the
equation of motion
mẍ(t) + αẋ(t) + Kx(t) = f (t),
where K is the harmonic spring constant. We found that the response function
can be written as
1
.
χ̃(ω) =
K − mω 2 − iαω
We noted that χ̃(ω), thought of as a complex function, should be an analytic
function of a complex variable ω in the upper half-plane, i.e., for positive imaginary part of ω. This might seem surprising, since the response function χ̃ above
clearly has some singularities in ω. Nevertheless, it is analytic in the upper
half-plane, as can be seen by rewriting the response function as follows
χ̃(ω) =
1
−1
,
m (ω − Ω− ) (ω + Ω+ )
where
Ω± = Ω ± iγ/2,
Ω2 = ω02 − γ 2 /4,
p
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 in
the lower half-plane. Thus, there are no singularities in the upper half-plane.
Analyticity of χ̃ was shown to be a necessary condition for causality—i.e. the
fact that χ(t) vanishes for t < 0. This is also a sufficient condition, as can be
seen by evaluating χ(t) for the damped oscillator. For t < 0, the integral
Z ∞
1
χ(t) =
e−iωt χ̃(ω)dω
2π −∞
can be evaluated by extending the path of integration to a closed contour in
the upper half-plane, for which the result will be zero because the integrand is
analytic there. In contrast, for positive t, we can extend the integration to a
contour in the lower half-plane, resulting in
I
−1
e−izt
χ(t) =
2πm
(z − Ω− ) (z + Ω+ )
i −iΩ−
=
e
− eiΩ+
2Ωm
sin Ωt −γt/2
=
e
.
mΩ
As we have noted, the harmonic oscillator constitutes a very practical application of the Langevin analysis. There are many experimental situations of
31
current interest where one has a harmonic potential for particles that would otherwise undergo brownian motion in a fluid. For many of these cases, the relevant
limit is that of a (strongly) overdamped oscillator. As we can see above, there
are at least two qualitatively different regimes of behavior: (a) the underdamped
case, for which ω02 > γ 2 /4; and (b) the overdamped case, for which ω02 < γ 2 /4.
The strongly overdamped case can be analyzed by ignoring the inertial term in
the equation of motion. Here, the response function can be obtained from
χ̃(ω) =
1
.
K − iαω
Here, too, the only singularity of this is in the lower half-plane. So, this also is
consistent with a causal response.
The imaginary part is given by
αω
.
χ00 (ω) = 2
K + α2 ω 2
There are several things to note about this. It is an odd function, as we expect.
Also, χ00 (ω)/ω is finite (i.e., non-singular) and strictly positive, as is ωχ00 (ω).
The spectrum of the fluctuations in position, h|xω |2 i can also be evaluated with
the aid of the fluctuation-dissipation theorem:
α
h|xω |2 i = 2kT 2
.
K + α2 ω 2
The behavior of this for large ω should be that of brownian motion (provided
the frequency is small compared with τ −1 , of course). This is because high
enough frequencies correspond to times so short that the particle does not feel
the harmonic potential. We see this in the fact that
h|xω |2 i →
2kT
,
αω 2
and thus, that
2kT
,
α
since vω = −iωxω . This last expression is equivalent to the relation
Z ∞
1
1
µ= =
hv(t + s)v(t)ids.
α
2kT −∞
h|vω |2 i →
(1.61)
found earlier.
We can also write χ00 as follows
χ00 (ω) = χ0
τ0 ω
2,
1 + (τ0 ω)
where τ0−1 is the characteristic frequency or relaxation rate for this system (not
to be confused with the resonance frequency ω0 of the underdamped harmonic
oscillator. Here, χ0 = 1/K is the static response function of a spring with spring
constant K, since x = f /K for a constant force f . This satisfies the sum rule
(1.60) above.
32
1.12
The Generalized Langevin Equation
Way back in our first lecture, we looked at the Langevin equation, in which we
assume that the the fluid (or, more generally, the heat “bath”) imparts random
forces to the particle whose motion we were interested in. At the same time, we
assumed that we could use a single drag coefficient to describe the resistance
to motion of the particle through the fluid in addition to the random forces
from the bath.Here, we want to examine this in greater detail. We do so within
the 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. The
surrounding medium is treated as having a large number of degrees of freedom
y1 , 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) − f x + Hb (y1 , . . .),
where
H0 =
mẍ(t)
+ V (x).
2
Here, V represents some conservative potential, and − dV
dx is the force. We
assume a simple, linear relationship between the yi of the bath and the force f
that the particle feels dues to the bath:
X
f=
ci yj .
i
In keeping with our previous notation, we call the correlation function for these
forces
X
Kb (s) = hδf (s)δf (0)ib =
ci cj hδyi (s)δyj (0)i,
i,j
where b denotes the bath. This means that we will express things in terms of
the 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 linear
response, in particular for the force f :
Z ∞
f (t) = fb (t) +
dt0 χb (t − t0 )x(t0 ),
−∞
where
d
Kb (t)
dt
for t > 0. The equation of motion for the particle is now
Z t
dV
mẍ(t) = −
+ fb (t) +
dt0 χb (t − t0 )x(t0 ).
dx
−∞
χb (t) = −β
33
Here, it is worth noting that χb represents the equilibrium fluctuations of the
bath alone. Thus, we expect the correlations in the bath, and hence memory, to extend over a short time such as τ ∗ in our earlier notation. Given the
relationship between χb and Kb , we have that
Z t
d
dV
+ fb (t) − β
dt0 Kb (t − t0 )x(t0 )
mẍ = −
dx
dt
−∞
Z t
dV
d
= −
+ fb (t) + β
dt0 0 Kb (t − t0 )x(t0 )
dx
dt
−∞
Z t
dṼ
= −
dt0 Kb (t − t0 )ẋ(t0 ),
(1.62)
+ δfb (t) − β
dx
0
where we have integrated by parts, and defined
Ṽ (x) = V (x) − βKb (0)x2 ,
and
δfb (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, we
again assume that K approximates a δ function, which leads us to the following
approximate result for the drag force:
Z ∞
−β ẋ(t)
Kb (s)ds = −αẋ,
0
where
α=
1
2kT
Z
∞
Kb (s),
∞
as before. More generally, however, we have a more complete expression for
the effect of the drag on the particle due to the bath, which includes memory
effects. In other words, we see how the relaxation time of the medium itself
affects the locality in time of the drag. The equation (1.62) is often referred to
as the generalized Langevin equation. The effective potential Ṽ is also called
the potential of mean force, which yields the bath-averaged force on the particle.
34
1.13
Appendix: Solving the diffusion equation
As an illustration, we solve the one-dimensional diffusion equation in an infinite
system.
∂ 2 P (x, t)
∂P (x, t)
.
(1.63)
=D
∂t
∂x2
We take as initial condition P (x, 0) = δ(x). We now take the spatial Fourier
transform of this equation:
Z ∞
Z ∞
∂P (x, t)
∂ 2 P (x, t)
dx eikx
.
(1.64)
=D
dx eikx
∂t
∂x2
−∞
−∞
Partial integration of the right-hand side yields:
Z ∞
Z ∞
∂P (x, t)
dx eikx
=D
dx eikx (−k 2 )P (x, t).
∂t
−∞
−∞
(1.65)
or,
∂ P̃ (k, t)
= −Dk 2 P̃ (k, t).
∂t
This equation can be solved to yield:
P̃ (k, t) = P̃ (k, 0)e−k
But we can easily compute P̃ (k, 0):
Z
P̃ (k, 0) =
2
Dt
.
(1.66)
(1.67)
∞
dx eikx δ(x) = 1
−∞
The inverse Fourier transform then yields P (x, t):
Z ∞
1
dke−ikx P̃ (k, t)
P (x, t) =
2π −∞
Z ∞
2
1
=
dke−ikx e−k Dt
2π −∞
Z ∞
2
2
2 2
2
2 2
1
=
dke−Dt(k +2ikx/(2Dt)−x /(4D t )+x /(4D t ))
2π −∞
Z ∞
2
2
2 2
1
=
dke−Dt((k+ix/(2Dt)) +x /(4D t ))
2π −∞
Z ∞
2
2
1
=
dke−Dt(k+ix/(2Dt)) −x /(4Dt)
2π −∞
r
1
π −x2 /(4Dt)
=
e
2π Dt
r
1 −x2 /(4Dt)
=
e
(1.68)
4πDt
35
1.14
Appendix: Relation between memory function and
random force
The Langevin equation makes the assumption that the friction force depends
only on the instantaneous velocity:
~ ,
m~v˙ (t) = −mγ~v (t0 ) + ζ(t)
(1.69)
where mγ = α is the friction coefficient. It is often necessary to consider the
more general case where the friction force depends also on the velocity at earlier
times:
Z t
~
m~v˙ (t) = −
k(t − t0 )~v (t0 ) + ζ(t)
(1.70)
0
~
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:
~
h~v (0) · ζ(t)i
=0
∀t.
(1.71)
Any part of the friction force that is correlated with the initial velocity will be
absorbed in the memory function. It turns out the the memory function k(τ ) is
~ · ζ(τ
~ )i.
closely related to the auto-correlation function of the random force, hζ(0)
This can be shown quite easily by making use of Laplace transforms. The
Laplace transform L(f ) ≡ F (s) of a function f (t) is defined as:
Z ∞
F (s) ≡ L(f (t)) =
dt exp(−st)f (t)
(1.72)
0
(Note: in Eqn. 1.57 we already used a Laplace transform with s = −iz). We will
use two properties of Laplace transforms: First of all, the Laplace transform of
the time derivative of f (t) is given by
L(f˙) = sF (s) − f (0)
(1.73)
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 Laplace
transforms. The convolution f ∗ g is defined as
Z t
f (t) ∗ g(t) ≡
f (t − t0 )g(t0 ) dt0 .
(1.74)
0
Then it is easy to show that
L(f ∗ g) = F (s)G(s)
(1.75)
If we multiply Eqn. 1.70 with v(0) and average, we get:
mh~v (0) · ~v˙ (t)i = −
t
Z
k(t − t0 )h~v (0) · ~v (t0 )i
0
36
(1.76)
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 Laplace
transforms listed above). We denote the Laplace transform of the velocity ACF
by V (s) ≡ L(h~v (0) · ~v (t)i). With this notation, we get
m(sV (s) − hv(0)2 i) = −K(s)V (s)
or
V (s) =
hv(0)2 i
s + K(s)/m
(1.77)
(1.78)
First of all, we note that it follows from Eqn. 1.70
~
m~v˙ (0) = ζ(0)
(1.79)
Let us now multiply Eqn. 1.70 with m~v˙ (0) and average:
m2 h~v˙ (0) · ~v˙ (t)i = −m
Z
t
~ · ζ(t)i
~
k(t − t0 )h~v˙ (0) · ~v (t0 )i + hζ(0)
(1.80)
0
~
~
~
Note that, in the above expression, we have replaced mh~v˙ (0)· ζ(t)i
by hζ(0)·
ζ(t)i.
Next, we take the Laplace transform of Eqn. 1.80. We denote the Laplace
~
~
transform of the random force ACF by Z(s) ≡ L(hζ(0)
· ζ(t)i).
Using the
properties of the Laplace transforms of derivatives and convolutions, and using
the fact that hȦ(0)B(t)i = −hA(0)Ḃ(t)i, we get:
−m2 s2 V (s) − shv(0)2 i − h~v (0) · ~v˙ (0)i = m sV (s) − hv(0)2 i K(s) + Z(s)
(1.81)
We note that h~v (0) · ~v˙ (0)i = 0 (because h~v (0) · ~v (t)i is an even function of time).
Rearranging the terms in Eqn. 1.81, we get
−m2 s2 V (s) + sV (s)K(s)/m − shv(0)2 i = −mhv(0)2 iK(s) + Z(s) (1.82)
Using Eqn. 1.78, we se that the left-hand side of Eqn. 1.82 vanishes. Therefore
−mhv(0)2 iK(s) + Z(s) = 0
(1.83)
Z(s) = mhv(0)2 iK(s)
(1.84)
or
This relation between the Laplace transforms implies that
~ · ζ(t)i
~
hζ(0)
= mhv(0)2 ik(t)
(1.85)
In other words: the memory function is proportional to the autocorrelation
function of the random force.
37
1.14.1
Langevin equation
A special case results when k(t) decays very rapidly and can be approximated
as
k(t) = 2mγδ(t) = 2αδ(t)
Rt
There is a now slight subtlety with the evaluation of 0 k(t0 )dt0 . But this can
be resolved by writing
Z t
Z
1 t 0
1
0
0
dt k(t ) =
dt k(t0 ) = (2mγ) = mγ = α
2
2
0
−t
If we insert this expression in Eqn. 1.70, we obtain the original Langevin
equation
~ .
m~v˙ (t) = −mγ~v (t0 ) + ζ(t)
From Eqn. 1.85 it follows that
~ · ζ(t)i
~
hζ(0)
= mhv(0)2 i2αδ(t) = 6kT αδ(t)
1.15
(1.86)
Appendix: Fourier representation of the δ function
To derive the relation
1
2π
Z
∞
0
e−iω(t−t ) dω = δ (t − t0 ) ,
−∞
we first consider the Fourier transform
Z ∞
1
e−iωt e−|ω| dω ,
2π −∞
with > 0. After that, we consider the limit → 0. As exp(−|ω|) is an even
function of ω, its Fourier transform is real. We can therefore write
Z ∞
Z ∞
e−iωt e−|ω| dω = 2Re
e−iωt e−ω dω .
−∞
We then use
0
Z
∞
e−iωt e−ω dω =
0
Therefore,
Z
− it
1
= 2
+ it
+ t2
∞
e−iωt e−ω dω =
Re
0
Now the integral
Z
∞
dt
−∞
2
2
=π
+ t2
38
+ t2
As becomes smaller, the integrand becomes increasingly sharply peaked, but
the value of integral is independent of . In fact,
lim
→0 2
Therefore
1.16
1
lim
2π →0
Z
= πδ(t)
+ t2
∞
e−iωt e−|ω| dω =
−∞
1
2πδ(t) = δ(t)
2π
Appendix: Noise
1.16.1
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 the
fluctuating signal through a frequency filter that transmits only those frequency
components that have an angular frequency between ω and ω + ∆ω. The mean
square value per unit bandwidth of this ”filtered” signal is the noise power
density, Pf f (ω).
Pf f (ω) = hf 2 i(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 discrete
Fourier components:
f (t) =
∞
X
am exp(2πimt/Θ)
m=−∞
Component m has an angular frequency ω = 2πm/Θ and a frequency ν =
m/Θ. After passing through the frequency filter, only those Fourier components
remain that have their (absolute) frequency between ν and ν +∆ν. The number
of Fourier modes that passes through this filter is
Nm = 2∆νΘ
The factor 2 results from the fact that both positive and negative frequencies
are transmitted. The mean square signal is than:
1
hf i =
Θ
2
Z
Θ/2
dt
−Θ/2
mH
X
!2
(am exp(2πimt/Θ) + a−m exp(−2πimt/Θ))
m=mL
Due to the orthogonality of different Fourier modes, this can be written as:
hf 2 i = 2
mH
X
m=mL
39
ham a−m i
The number of modes 2∆νΘ, and therefore:
hf 2 i = 2∆νΘham a−m i
or
hf 2 i
= 2Θham a−m i
∆ν
We can express the am as
am
1
=
Θ
Θ/2
Z
dtf (t) exp(−2πimt/Θ)
−Θ/2
or, with ω = 2πm/Θ
1
Θ
a(ω) =
Z
Θ/2
dtf (t) exp(−iωt)
−Θ/2
The Fourier transform of f (t) over the interval {−Θ/2, Θ/2} is
f˜Θ (ω) =
Θ/2
Z
dt eiωt f (t) = Θa(ω)
−Θ/2
Therefore,
hf 2 i
2
= h|f˜Θ (ω)|2 i
∆ν
Θ
We now consider the limit Θ → ∞:
Pf f (ν) =
Pf f (ν) = lim
Θ→∞
2 ˜
h|fΘ (ω)|2 i
Θ
We can write this as
2
Pf f (ω) = lim
Θ→∞ Θ
Z
Θ/2
Z
−Θ/2
Θ/2
0
dt dt0 eiω(t−t ) hf (t0 )f (t)i
−Θ/2
If the correlation function hf (0)f (τ )i decays on a timescale that is much shorter
than Θ, the we can write
2
Θ→∞ Θ
Z
Θ/2
Z
∞
dt
Pf f (ω) = lim
−Θ/2
dτ eiωτ hf (0)f (τ )i .
−∞
It then follows that
Z
∞
Pf f (ω) =
dτ eiωτ hf (0)f (τ )i = 2Cf f (ω) ,
−∞
where Cf f (ω) is the Fourier transform of the correlation function of f (t).
40
1.16.2
Shot noise
As an application, consider the noise due to an electric current that is due to
the transport of individual charge carriers with charge e. Every time ti when a
charge e crosses the conductor, we have a contribution eδ(t − ti ) to the current.
The total current during an interval Θ is
N
X
I(t) =
eδ(t − ti )
i=1
where N denotes the total number of charges that have crossed in time Θ. We
can now compute the Fourier transform of this current:
I˜Θ (ω) =
N
X
e exp(iωti )
i=1
The power spectral density of this current is:
PII (ω) = lim
Θ→∞
2 ˜
h|IΘ (ω)|2 i
Θ
This can be written as
N
N
2 XX 2
e hexp(iω(ti − tj ))i
Θ→∞ Θ
i=1 j=1
PII (ω) = lim
If we assume that different charge crossings are uncorrelated, we get hexp(iω(ti −
tj ))i = δij and hence
N
2 X 2
2N 2
e
e = lim
Θ→∞ Θ
Θ→∞ Θ
i=1
PII (ω) = lim
But the average current I = N e/Θ. Therefore
PII (ω) = 2eI
The interesting feature of shot noise is that, by measuring the power spectrum
of the current noise, we can determine the charge e of the charge carriers.
1.16.3
Thermal Noise
Consider two resistors R1 and R2 connected in a loop by loss-free wires. Due to
the thermal motion in the resistors, there will be voltage fluctuation generated
over both resistors. We denote the power spectra of these two noise sources by
P1 (ω) and P2 (ω) respectively.
P1 (ω) =
2
|V1 (ω)|2
T
41
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
p21 = |I2 (ω)|2 ∗ R1 = |V2 (ω)|2
R1
(R1 + R2 )2
Conversely
In thermal equilibrium, p12 = p21 . Hence
|V2 (ω)|2 R1 = |V1 (ω)|2 R2
This implies that
|V1 (ω)|2 /R1 = |V2 (ω)|2 /R2
In other words
|V1 (ω)|2 = R1 f (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, again
connected by loss-less wires. The energy of a induction coil with current I is
equal to 21 LI 2 . In thermal equilibrium, we must have:
LI 2
kT
=
2
2
or
kT
L
In a closed circuit, the sum of all potential differences must vanish. In the
present case, there are three contributions: the voltage over the inductance due
to the change in current
∂I
VI = −L
∂t
the voltage drop over the resistor
hI 2 i =
VR = −IR
and the thermal noise VR . As the sum of these three terms must vanish, we
have for all t
∂I(t)
L
= −RI(t) + VR (t)
∂t
This equation looks exactly like the Langevin equation
mvx (t) = −αvx (t0 ) + ζx (t) .
42
But, in Eqn. 1.86 we showed that
hζx (0)ζx (t)i = mhvx (0)2 i2αδ(t)
By analogy:
hVR (0)VR (t)i = LhI 2 i2Rδ(t) = 2RkT δ(t)
By Fourier transforming, we get
Z ∞
dthVR (0)VR (t)ieiωt = 2RkT
−∞
But
Z
∞
2
dthVR (0)VR (t)ieiωt = PV V (ω) ,
−∞
the power spectral density of the noise voltage. And hence:
PV V (ω) = 4RkT
43