Non-linear Equations - Lecture 3

1 Introduction

The human mind has evolved to understand linear dynamics because the world in whichwe live appears to be linear. The mathematics of linear systems are easily treated as thesuperposition principle operates on linear systems. Note that for small fluctuations aroundan equilibrium condition, the force is explained in terms of linear behavior,

F (x) = F (x0) + (x − x0) F ′(x0) +(x − x0)

2

2!F ′ ′ (x0) + · · ·

In the above, F (x0) = 0 for equilibrium and the one assumes that terms higher than the firstpower of (x−x0) can be neglected. This is of course, the reason that the harmonic oscillatorhas such wide applications.

In the 18th century Leibniz stated that if one had sufficient knowledge and sufficient compu-tational power, all physical processes could be predicted - classical determinism. However,this ignores that the state of a physical system can never be completely specified, even clas-sically. A deterministic system assumes that the future is predictable from present initialconditions. A stochastic system cannot specify all conditions precisely so that even a deter-ministic system cannot be fully predicted.

As a simple example consider the motion of a ball in a pinball machine. This is illustratedin Figure 1. Any two trajectories that start reasonable close, diverge exponentially in time.As observed in the figure, a finite, and usually a small number of collisions, results in a wideseparation of the spatial dimensions of the system. The displacement of the balls can berepresented by δx(t) ;

|δx(t)| ≈ eλt |δx(0)|

Obviously it takes δx(0) to know δx(t) but in this case the spatial dimensions diverge expo-nentially. The constant, λ, is called the Lyapunov exponent and there is a Lyapunov timegiven by;

TL ≈ −1λ

ln|δx/L|

The Lyapunov time represents the time, given the accuracy of the initial data, over whichthe dynamics are predictable. Non-linear behavior and randomness must be used to explainthe behavior of complex systems. We observe traffic congestion, weather changes, stockmarket fluctuations, etc. There is also the process of self organization, which is the abilityof a system to self restrict its phase space and maintain this restricted phase space. This is

1

Figure 1: An example of trajectories of the balls in a pinball machine whose starting trajec-tories are similar but diverge after only a few bounces.

inherently a non-linear process.

2 Chaos

Newton’s success in solving the Kepler problem of astronomical orbits, reinforced a determin-istic picture of the natural phenomenon. The orbits of planets were viewed as the pendulumsof precision clocks. Thus it was anticipated that all physical processes could be predictedgiven initial conditions and the power to precisely solve these equations. This did not turnout to be the case even in classical processes because a small non-linear interaction can in-troduce unpredictability.

As an example, consider the Lorentz model of weather. The original model consisted of a12 variable system of non-linear ode equations. The solution was aperiodic and extremelysensitive to the initial conditions. To see the effect, consider a simplified model with 3 vari-ables.

dxdt

= σ(y − x)

dydt

= rx − y − xz

dzdt

= xy − bz

In the model, σ, r, and b are positive parameters and represent a model for the buoyantconvection in a fluid.

x measures the rate of convection

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y measures the horizontal temperature gradient

z measures the vertical temperature gradient

The parameters σ and b depend on the properties of the fluid and the geometry of the con-tainer. Common values are σ ≤ 10 and b = 8/3. The parameter r is chosen to match thetemperature gradient in the system. Starting from initial conditions, the solution for x, y, zcan be found numerically as a function of t. Two solutions for x(t) are shown in Figure 2.The solutions for y, z are similar. The solutions are not remarkable with the exception thatthe lower, second solution has only a slight change in the initial conditions. Lorentz calledthis the butterfly effect as only a butterfly flapping its wings could effect the weather. Thisdemonstrates the extreme sensitivity to the initial conditions.

Classical problems are called chaotically deterministic when their solutions becomes so ex-tremely dependent on the initial conditions that predictability is lost. Chaotic motion indynamical systems can described by first order differential equations if;

1. There are at least 3 variables.

2. One or two non-linear terms couple at least 2 of the variables.

The sensitivity to initial conditions requires that the solution does not converge to a cycleof any length. If such a system is started with a finite amount of information, then aftera given time the system is completely unpredictable. Applications of chaotic descriptionsoccur in many fields of physics, biology, astronomy, and engineering.

3 Deterministic Chaos

A positive Lyapunov exponent does not necessarily lead to chaos. There must be a mixingof the trajectories of many elements in the system. That is, the separate trajectories mustfold back on themselves. To be chaotic, the system must;

1. everywhere be locally unstable (positive Lyapunov constant), and

2. the trajectories must mix (positive entropy)

Such a system is not chaotic in a statistical sense as the trajectories can be calculated. Inthe case of the pinball system discussed in the introduction, there are 2n topologically dis-tinct trajectories for n bounces. This assumes a trajectory has the possibility of moving in

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Figure 2: An example of 2 solutions to the Lorentz model equations which shows the extremesensitivity to initial conditions.

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either of 2 directions. Write this as N(n) ≈ ehN , where h is the topological entropy. Hereh = ln2. If the system never returns to an original condition, then the system is open or isa “repeller”. In the sense we are using chaos, it means a mixing of unstable periodic orbitsand the intertwining of stable and unstable trajectories in phase space. There is usually asuccession of nearly periodic but unstable motion. There is seemingly a periodic patternwhich changes to another pattern seemingly at random. Consider the Lorentz equationsintroduced earlier, but somewhat simplified here.

dxdt

= (y − x)

dydt

= x(r − z) − y

dzdt

= xy − bz

First find all the stable points of the system.

dxdt

=dydt

= dzdt

= 0

A)

Suppose x = y so when x is constant in time (dxdt

= 0) y is constant in time. Then

dzdt

= 0 = x(r − z) − x

z = (r − 1) which is constant.

Thus all fixed points are steady state solutions with y = x.

B)

Now choose x = 0. Then;

dxdt

= y

dydt

= −y = −dxdt

Since x and y are independent they must be constants, so x = y = z = 0 is a fixed pointregardless of the initial conditions.

C)

Suppose x = y = c for c a constant. Then z is a constant.

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y(r − z) − y = 0

y2 = bz

y = x = ±√

b(r − 1)

z = r − 1

Then if z = −1 dxdt

= −dydt

D)

Finally a fixed point must have x = Q(bz) and this is only possible if r > 1

dxdt

= (y − x) = 0

dydt

= x(r − z) − y = 0

dzdt

= xy − bz = 0

This gives y = x and

x(r − 1) = xz → z = r − 1

x2 = bz so r > 1 for x to be real.

Therefore we expect 3 fixed points. Two points for (x, y) are attractors, the point x = 0 y =0, z = 0 is a repeller. Figure 3 shows a trajectory in 3-D space and is said to lie on anattractor. The motion is a type of aperiodic orbit as the solutions orbit one of the fixedpoints and flips to another. The two solutions move rapidly apart no matter how initiallyclose. The center of the attractor mixes trajectories, some to the left and some to the right.Particles with almost the same initial conditions are eventually separated. The 3 points inthe figure are the steady states (fixed points) of the Lorentz model.

4 Mathematical details

In the example of the last section there is a function f(x) = 0, and we attempt for solvefor the values of x that satisfy it. We first attempt a solution by Newton’s method. Thefunction is;

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Figure 3: An example of aperiodic motion moving away from a repeller to an attractor

f(xi) = 0

Suppose xi is the ith root of the equation. Expand the function about this root. Thus write,x1 ≈ xi + δxi for δx < 1;

f(xi) = f(x1 − δx) = f(x1) − dfdx

|x=x1δx + · · ·

Keep only terms to 1st order and by definition f(xi) = 0.

δx =f(x1)f ′(x1)

Thus we obtain a prediction for the root to lie at x2

x2 = x1 − δx = x1 − f(x1)f ′(x1)

The steps are iterated to convergence. When the method converges it does so very rapidly,however, it does not always converge. To assure convergence one needs a beginning guesssufficiently close to the answer. This procedure is illustrated in Figure 4

Now for the multi-variable problem, set the unknown to a vector ~x, ([x1, x2, · · · xj ]). Wewant to find the roots of the vector function;

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f(x ) = xN N

xN

f(x)

Attractor

Iteration Path

Figure 4: An example of the iteration procedure to find a fixed point using Newton’s method

~f(~x) = [f1(~x), f2(~x), · · · fN (~x)]

Assume that the error in the 1st approximation with respect to the root ~x∗ is;

δ~x = ~x1 − ~x∗

Perform a Taylor expansion about ~x∗ ;

~f(~x∗) = ~f(~x1 − δ~x) = ~f(~x1) − δx ~D(~x1)

In the above, ~D(~x) is defined by;

Dij =∂fj(~x)∂xi

Then to 1st order;

δ~x = ~f(~x1)D−1(~x1)

5 Maps

Consider the motion of a system as a trajectory in multi-dimensional phase space. Thereis an assumed hyperspace called a Poincare section such that the trajectory intercepts thissection. The value at the intersection provides a measure of the flow of the system. Supposewe observe the trajectory at a series of time points, ti Then write x1(t1) → x1 for the phasespace vector, ~x = (x1, x2, · · · xn). As an example, consider the ideal pendulum. The motion

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M = (x, p)2d

x

p

Figure 5: An example of phase space for the ideal pendulum. The Poincare sections are thecrossings of the x axis

is;

x = x0 sin(ωt)

v = ω x0 cos(ωt)

p = mv

Figure 5 shows an example of a Poincare section in a phase space, M2d = (x, p). Eachtime the trajectory crosses the x axis the same point is generated. If the pendulum has aresistance, then the motion is damped to zero and the trajectory spirals to the origin. ThePoincare sections approach the origin as the pendulum comes to rest.

In a multi-dimensional phase space a Poincare section is useful in reducing the trajectorymotion in a multi-dimensions to a few. Thus we have in effect a projection of the trajectoryonto a subplane. This type of mapping is useful in looking at flows in multi-dimensionalphase space.

Suppose a vector in Ms = (x1, x2 · · · xd). The motion is described as;

dxidt

= vi(x, t)

Then;

dxkdt

= Vk = dxkdx1

dx1dt

= V1dxkdx1

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dxkx1

= VkV1

This gives all the xk at a time ∆t from the time when xi = xoi . Proceed for each time interval

and obtain a map of an ∞ set of discrete points.

The motivation was that we anticipate a large number of problems to exhibit the same be-havior. Thus we have a set of points, M to which we apply a map;

F (M1) → M2

The trajectory is measured by the set of points, fn(M).

x, f 1(x), f 2(x) · · · fn(x)

Then suppose the logistic map;

xn+1 = µ xn(1 − xn)

xnǫ[0, 1] 1 < µ < 4

This map comes from the differential equation;

dxdt

= µx(1 − x)

For xk+1 as determined by xk for a time τ

xk+1 − xk = µtk+τ∫

tk

dt xk(1 − xk)

xk+1 = xk + µG(tk + τ) − µG(tk)

In the above, dGdt

= x(1 − x) . This is now a map that takes xk into xk+1 The solution isshown in Figure 6. There is one fixed point which is an attractor and the region betweenthe origin and this fixed point is the basin of the attractor. the fixed point (attractor) isstable provided the slope at f(x∗) < 1. This point occurs when;

f(x∗) = µx∗(1 − x∗) = x∗

x∗ = 1 − 1/µ

For an initial, x0 when 0 < x0 < 1/µ then the xi converge to the attractor. For µ > 1 andx0 < 0 or x0 > 1 then xi → −∞

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f(x)

xx x x

x

1 2 3

i

Figure 6: An illustration of a solution using Newton’s method to solve the equations

f ′(x∗) = µ(1 − 2x∗) = 2 − µ = −1

Use x∗ = 1 − 1/µ and two fixed points occur. Thus the stable attractor bifurcates into 2fixed points. In this case;

x∗

i = f(f(x∗

2)) = µ(µx∗

2(1 − x∗

2))[1 − µx∗

2(1 − x∗

2)]

This continues until an ∞ numbers of bifurcations occur. The increasing period doubling isa route to chaos and is characterized by a universal constant δ (Feigenbaum number). theratio of the spacings of the bifurcations is given by;

limn→∞

µn − µn−1µn+1 − µn

= δ = 4.6692161 · · ·

All this is illustrated in Figures 7 and 8. For the equation xn+1 = α(1−xn) the Lyapunovcoefficient is ;

λ = limn→∞ (1/n)n−1∑

i=0

ln|df(xi)dx

|

The Lyapunov coefficient is shown in Figure 9 and a bifurcation diagram in Figure 10.

6 Flows

Above we discussed a Poincare section and its use in discussing flows in phase space. Theexample is a stream of representation points moving in time as a bundle that distorts as itmoves in space. This is shown in Figure 11. Consider the transport of the velocity vector~v to a new position.

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Figure 7: An illustration of fixed points in a set of trajectories

Figure 8: The solution using Newton’s method to find the fixed points in a trajectory.

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Figure 9: The Lypanov exponent as a function of α for the example in the text.

Figure 10: Bifurcation diagram for the example in the text

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Figure 11: An illustration of vector flow through phase space of a bundle of velocities

~v(x(t)) = Jt(x0)~v(x0)

In the above Jt is the Jacobian of the transformation between the coordinate frames.

Jij(x0) = ∂xi∂xj

|x=x0

Now linear fields lead to linear ode and the equations can be solved analytically. The phasespace is ;

~x = ~v(x) = A~x

In the above, A is a matrix of variations. This leads to stable motion. A Hamiltonian systemhas stability because of linearity about an eigenvalue. Non-linear behavior, as seen from theprevious examples, can result in instability.

7 The Rossler System

The Rossler system is one of the simplest 3-dimensional, non-linear systems which exhibitchaotic behavior. The system consists of the equations;

dxdt

= −(y + z)

dydt

= x + ay

dzdt

= b + z(x − c)

In the above, a, b, c are the system parameters and x, y, z define the system states. We at-tempt to find the fixed points of this system by setting the derivatives equal to zero. Lookingat the equations, the fixed points must have z = −y = x/a. The value of x is determinedby;

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Rössler attractor with a = 0.2, b = 0.2, c = 5.7

illustration of the repulsing plane generated

. The magenta

– it

illustrates the behavior of points that become

that the general appearance and behavior of

Figure 12: Numerical solution for the Rossler system of equations for the parameters shownin the figure

x2 − cx + ab = 0

Thus x = [c ±√

c2 − 4ab]/2 and a solution exists (real x) when c2 − 4ab) > 0 and a 6= 0.The solution when solved numerically, is shown in Figure 12. The initial conditions areshown in the figure. The orbit follows an outward spiral around an unstable fixed point closeto the (x, y) plane. The outward motion then becomes influenced by the second fixed pointcausing the trajectory to rise and twist in the z direction. These equations are found to beuseful in modeling equilibrium in chemical reactions.

A poincare map can be constructed by obtaining the value of the trajectory as it passesthrough a plane. As an example, suppose one plots the (y, z) plane at all points when x = 0as the trajectories move upward through the plane. Crossing this plane always occurs whenz = 0 so there is little to plot. However for increasing x, plot the crossings in the planes, x =constants, with initial constants a = 0.1, b = 0.1, c = 14. This plot is shown in Figure 13where the up-swing of the trajectory is obvious. The behavior of the solution is dependenton the choice of parameters. For example, bifurcation occurs as shown in Figure 14 as theparameter c is changed.

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Poincaré map for Rössler attractor with a = 0.1, b = 0.1, c = 14

value every time it passes

due to the nature

increases, as is to be expected

Figure 13: A Poincare plot of the R’ossler system as the plane x = constant is varied

Bifurcation diagram for the Rössler attractor for varying c

eriodicity interspersed with periods of chaos,

s higher-period orbits as

values illustrates the

general behavior seen for all of these parameter

Figure 14: Bifurcation of the R’ossler solution as the parameter c is varied

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8 Traffic Flow

The equations of motion for this problem are;

∂ρ∂t

= −~∇ · ~F (ρ)

This is basically the diffusion equation. Simplify the example to one spatial dimension.

∂ρ∂t

= − ∂∂x

F (ρ)

In the above ρ represents any one of the conserved quantities of mass density, momentumdensity, or energy density. The function, F , is the mass flux, momentum flux, or energy flux.Use here the mass flux which is the mass density times the fluid velocity.

∂ρ∂t

= − ∂∂x

(ρv)

Obviously this represents the conservation of mass (equation of continuity). We presumethat the fluid velocity is a function of density;

v(x, t) = v(ρ) = vm(1 − ρ/ρm)

This essentially represents an expansion about a maximum density, ρm, and the equationrepresents some of the characteristics of traffic flow. The parameter vm represents the speedlimit and the max density occurs when the cars are bumper to bumper. The equation to besolved is;

∂ρ∂t

= −[ ddρ

(ρv)]∂ρ∂x

Let c(ρ) =d(ρv)dρ

so that;

∂ρ∂t

= c(ρ)∂ρ∂x

c(ρ) = vm[1 − 2ρ/ρm]

The function c(ρ) is the speed of waves in the traffic flow. The waves can move in eitherdirection but never faster than vm. Now apply the initial condition that;

ρ(x, 0) =

[

ρm x < 00 x > 0

]

This represents a line of cars stopped at a traffic light, as illustrated in Figure 15. Theanalytic solution occurs in the form of ρ(x, t) = ρ(x − ct)

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ρ( )x, 0

x

t = 0

Figure 15: The mass density of cars stopped behind a red light as the initial condition forthe traffic problem

x

t = 0t = 1t = 2t = 3

ρ( )x, t

Figure 16: An illustration showing the change in the density as a function of time after thelight turns green

ρ(x, t) =

ρm x ≤ −vmt1/2[1 − x/(vmt)]ρm −vmt < x < vmt0 x ≥ vmt

Note the density decreases linearly with time. The result is illustrated in Figure 16.

9 Navier-Stokes equation

The Navier stokes equation describes the flow of a fluid. It is written in terms of rela-tions between the rates of the variables of interest (velocity, pressure, etc.) The equationis non-linear and applies not only to fluid (liquid or gas) flows but also applies to weather,economics, and other problems. In order to derive the equations, note that a time derivativeof a system moving with velocity, ~v(t, ~x), is written as (convective derivative);

ddt

= ∂∂t

+∑

i

∂xi∂t

∂∂xi

= ∂∂t

+ ~v · ~∇

The term on the right hand side is the derivative with respect to a fixed point and the terms

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dσ

dx = v dt

dΩ

Figure 17: An illustration of the expression representing the outflow of L through the surfaced~σ

on the left are the changes with respect to the moving fluid. Now apply conservation ofmass, energy, momentum, and angular momentum. Suppose a variable of a system, L(t, ~x).In a small volume, dΩ, the quantity that is lost is due to the flow across the boundaries andthe sinks or sources in the volume.

[∂L∂t

] dΩ = −L(~V · n) dσ + Q dΩ

This equation represents the outward flow through the surface dσ n enclosing the volume dΩhaving sources (or sinks) Q. Consider the flow out of the surface dσ of a small volume, dΩas shown in Figure 17.

Use Gauss’s theorem to write;

∂L∂t

+ ~∇ · (L~v) = Q

If L = ρ the equation represents the mass density. For an incompressible fluid, ρ is constantin time and space. Therefore;

~∇ · ~v = 0

To include conservation of momentum, L = ρ~v. This results in the equation;

∂∂t

[ρ~v] + ~∇[ρ~v]~v = ρ~F

One can write the expression ~v ~v as a tensor of rank 2. Thus using Newton’s law, ~F = m~a;

ρ d~vdt

= ρ~F − ~∇P

In the above, ~F is the external force on a small volume of fluid, ρ is the mass density, andP is the pressure, ie a potential energy term. This becomes;

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∂~V∂t

+ ~V · ~∇~v = ~F − (1/ρ)~∇P

Now assume P is given as a tensor, ie a stress tensor where the diagonal components are apressure.

P =

σxx τxy τxz

τyx σyy τyz

τzx τzy σzz

= −

P 0 00 P 00 0 P

+ T

In the above σxx + σxx + σxx = 3P and T is traceless. This leads to the equation;

ρ ∂~v∂t

+ ρ[~V · ~∇]~v + ~∇P − ~∇ · T − ρ~F = 0

A solution requires the properties of the fluid which determines the tensor T

10 Example

We use a plasma as an example of the application of the Navier-Stokes equation. To begin,the plasma equations must obey Maxwell’s equations.

~∇ · ~E = q(n − n0)/ǫ n and n0 number densities of ± charge

~∇ · ~B = 0

~∇ × ~E = −∂ ~B∂t

~∇ × ~H = ~J + ∂ ~E∂t

~D = ǫ ~E

~B = µ ~H

The Lorentz force is ~F = q[ ~E + ~V × ~B]. Now rewrite these equations in terms of charge

and current densities; q → ρ, q~V → ~J . The force density is then;

F = ρ~E + ~J × ~B = ǫ(~∇ · ~E) ~E + [(1/µ)~∇ × ~B − ǫ∂~E

∂t] × ~B

F = ǫ(~∇ · ~E) ~E − ~B × [~∇ × ~B] − ǫ ∂∂t

[ ~E × ~B] − ~E × [~∇ × ~E]

We now apply a 1-fluid model of a cold plasma. The plasma remains neutral so that only

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currents are generated. This means that the frequencies are low so that ions and electronsmove to cancel any electric field. Thus ~E = 0. Assume a scalar pressure, P . The forcedensity is then;

σ d~Vdt

= −~∇P + ~J × ~B

These are the magnetohydrodynamic equations. Maxwell’s equations are needed with ∂ ~E∂t

=0 as well as the equation of continuity.

∂σ∂t

+ ~∇ · (σ~V ) = 0

Substitution into the fluid equation yields;

σ ∂~V∂t

= −~∇P − (1/µ) ~B × [~∇ × ~B]

The equations are non-linear so to solve them we expand about an equilibrium point, keepingonly linear terms. Thus let;

~B ≈ ~B0 + ǫB~B1

~V ≈ ~V0 + ǫV~V1

σ ≈ σ0 + ǫσ σ1 σ1

P ≈ P0 + ǫP P1

The ǫ’s are ≪ 1. Substitution into the plasma equations gives to first order;

Equation of Continuity

∂σ1∂t

+ ~∇ · (σ0~V1 + σ1

~V0) = 0

Lorentz Force (perfectly conducting)

F = ρ~E + ρ ~V × ~B = 0

Faraday’s Law

∂ ~B1∂t

= ~∇ × [(~V1 × ~B0) + (~V0 × ~B1)]

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Momentum Conservation (1-fluid model)

σ1d~V0dt

+ σ0d~V1dt

= −~∇P1 − (1/µ)[ ~B0 × (~∇ × ~B1) + ~B1 × (~∇ × ~B0)]

The equilibrium values are constant and we expand the pressure in terms of the density;

P (σ) = P0(σ0) + (σ − σ0)∂P∂σ

Identify ǫp/ǫσ = ∂P∂σ

= s2 and let ~V0 = 0 and P0 = 0. Then collect terms;

Fluid Model

σ0∂~V1∂t

+ s2~∇σ1 = − ~B0/µ × (~∇ × ~B1)

Equation of continuity

∂σ1∂t

+ σ0~∇ · ~V1 = 0

Faraday’s law

∂ ~B1∂t

= ( ~B0 · ~∇)~V1 − (~∇ · ~v1) ~B0

Finally by substitution;

∂2~V1

∂t2− s2~∇(~∇ · ~V1) + ~VA × (~∇ × [~∇ × (~V1 × ~VA)]) = 0

~VA =~B0√µσ0

Assume a harmonic time dependence to obtain the dispersion relation.

~V1 = ~V ei(~k·~x−ωt)

−ω2~V1 + (s2 + V 2A)(~k · ~V1) + i~VA × |~∇ × [(~VA · ~k)~V1 − (~k · ~V1)~VA]| = 0

Apply the curl operations which inserts ~∇ → i~k

−ω2~V1 + (s2 + V 2A)(~k · ~V1) + (~VA · ~k)[(~VA · ~k)~V1 − (~VA · ~V1)~k − (~k · ~V1)~VA] = 0

Consider the dispersion relation above by making several choices of the directions of thewave oscillations.

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Figure 18: Plasma waves about an equilibrium point

23

~k perpendicular to ~VA

This means that ~k · ~V A = 0 and

−ω2~V1 + (s2 + V 2A)(~k · ~V1)~k = 0

This equation has a solution for ~V1 along ~k with phase velocity, Vp;

Vp =√

s2 + V 2A

This is a compressional Alfvin wave. The compressional Alfvin wave is an oscillation betweenthe inertia of the plasma and the compression of the magnetic field lines

~k parallel to ~VA

After reduction of the dispersion relation this results in the equation;

(V 2k2 − ω2)~V1 + (s2/V 2A)(~VA · ~V1)k

2VA − k2(~VA · ~V1)~VA = 0

There are then 2 choices;

xs 1) ~V1 · ~VA = 0

This results in a transverse wave with phase velocity, Vp = VA and is a shear Alfvin wave.

The shear Alfvin wave is incompressible (~∇P = 0) and is a balance between inertia and thefield line tension.

2) ~V1 parallel to ~VA

This results in the dispersion relation;

k2V 2A − ω2 + s2k2 − k2V 2

A = 0

The phase velocity is Vp = s which is an Alfvin sound wave. In the sound wave B is constant.

These waves are illustrated in the Figure 18. The solutions are valid at low frequencies giventhe initial approximations. Hydrodynamic waves appear only in the presence of a magneticfield below the cyclotron resonance frequency. In these waves the ion movement provides theinertia while the restoring forces are magnetic, arising from ~J × ~B. The waves are regardedas oscillations of the lines of force. In general, a wave in a plasma is a superposition of all ofthese wave types.

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11 Radiofrequency Quadrupole (RFQ)

Space charge forces are strong on intense, low velocity ion beams. In a particle accelerator,one wants high injection energies and currents, and small phase space. Magnetic focusingis ineffective at low velocities, so electrostatics must be used. To see how this works out,suppose a current due to a beam of charges moving with velocity, V . This provides a currentdensity given by J = ρV , where ρ is the charge density. Model the beam as a uniform linecharge of circular dimensions parallel to the velocity. The electric field inside the beam isobtained by Gauss’s law to be;

~E =Q

2πǫL r

In the above, Q, is the charge enclosed within a cylindrical Gaussian surface with radius Rand length, L.

Q = ρ (L πR2)

The electric field at position R is;

~E =ρLRπǫ r

The magnetic field at this position is obtained from Ampere’s law applying the line integralaround a circular loop centered on the beam.

~B =µI

2πR φ

The enclosed current I = ρR2 V

~B =µ ρV R

π φ

Then the force on a charge q at the position R in the beam is;

~F =qρRπǫ [1 − µǫ] φ =

qρRπγ2 φ

An RFQ is designed to use electrostatic forces to both focus and accelerate the beam. Recallthat in an electromagnetic wave guide, the wave is a superposition of a wave transverse tothe longitudinal axis of the guide and one that is parallel to this axis. The RFQ is designedto use both wave types, the transverse wave to accelerate the beam and the transverse waveto keep the bean focused. It takes advantage of the ability to accurately calculate the fieldsusing computers, and manufacturing techniques to produce the complicated shapes of thewave guide poles.

25

Equi−Potential

FieldQ

Q

−Q−Q

Figure 19: The production of a quadrupole electric field through the position of ± electriccharges. An example of an equipotential line and a field line is shown.

Take a simple model of an RFQ, and treat the fields in a quasi-static approximation, valid ifa ≪ c/ω = λ. In this expression 2a is the distance between electrodes and ω the frequency.To electrostaticly focus a beam, we want a quadrupole field as shown in Figure 19. Thisfield has the mathematical form;

Ex = kx

Ey = −ky

The field is obtained by placing ± charges at the positions x, y = ±a as shown in Figure19. Then to first order, the finite size of the charge does not matter. Note the equi-potential

surfaces given by P = −E02a (x2 − y2).

Thus we want a static charge distribution to have the above form. Then for the dynamiccase assume we create the fields;

Ex(~r, t) = E0a x sin(ωt)

Ey(~r, t) = −E0a y sin(ωt)

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These fields are only approximate, but are presumed accurate near the z axis. The force inthe x direction is then;

md2xdt2

= (qE0/a) x sin(ωt)

A full solution may be obtained by Mathieu functions. However, here assume the period ofthe transverse orbit is long compared to 2π/ω. In this limit the solution has two motions;

x(t) = X0 sin(Ωt) + X1 sin(ωt)

In the above X1 ≪ X0; Ω ≪ ω; Ω2X0 ≪ ω2X1. Substitute into the ode and keep only thelargest terms.

−ω2X1 ≈ qE0X0ma sin(Ωt)

This results in;

X(t) = X0 sin(Ωt) − qE0X0

ω2masin(ωt) sin(Ωt)

The second term is small. The force averaged over the fast oscillation period, ω, is;

−Ω2X0 sin(Ωt) ≈ −(1/2)(qE0ωma)2 X0 sin(Ωt)

with Ω =qE0√2ωma

. The low frequency motion is oscillatory so the time varying quadrupole

fields provide a net containment of the beam particles. A numerical solution for ω = 3Ω isshown in Figure 20. Solutions of y are similar.

While the transverse wave captures the charge, it does not accelerate it. To accelerate thecharge, we will need an E field along the z axis. However, note that electromagentic wavesare transverse, so a wave traveling in the z direction has the E vector in the (x, y) directions.Thus we modulate with a spatial period, D, the electrodes which produce the quadrupolefield. However, the modulation in the x and y electrodes is 90 out of phase. These electrodesare shown in Figure 21 and produce fields of the form;

Ex = (E0/a)[1 + ǫ sin(2πz/D)] sin(ωt)

Ey = −(E0/a)[1 − ǫ sin(2πz/D)] sin(ωt)

In the above, ǫ is the amplitude of the variation in the electrostatic surfaces. The focusingfields in the transverse direction are suppressed in these equations. Again use the quasi-staticapproximation. These forms are valid only if they satisfy the Laplace equation (ie only if the

27

Figure 20: A numerical solution for the time averaged position of a charge in the transversequadrupole field as a function of time. The motion shows that the charge is captured byoscillations in the field.

Figure 21: A proto-typical design of an RFQ showing the quadrupole vanes with theirmodulation

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fields can be generated by electrostatic surfaces). We also have that ω/k = Vp = Dω/2π.A particle which keeps this velocity, experiences an axial force if it enters the apparatus withphase, t = 0 + φ/ω at z = 0 and keeps this phase such that 0 < φ < π.

Fz = eEz(t = 0 + φ/π) = eǫE0D2πa sin(φ)

The phase φ is assumed constant so that Fz > 0 and a particle experiences acceleration overa wide range of wave amplitudes, 0 < φ < π/2. The equation of motion of a synchronousparticle is;

mdVdt

= eǫE0D2πa sin(φ)

If V = Dω/2π, then by substitution;

dDdz

= 2πω (1/V ) eǫE0D

2πam sin(φ) =2πeǫE0 sin(φ)

mω2a

Thus if ǫ is constant then the modulation length in the electrodes should increase linearlyto keep the phase constant. Next return to the transverse motion. The equation so far hasthe form;

d2xdt2

= eE0xma [1 + ǫ sin(2πz/D)] sin(ωt − φ)

When t = φ/ω then z = 0. This results in ;

d2xdt2

= eE0Xma sin(ωt − φ) + eE0X

2ma [cos(eπz/D − ωt + φ) − sin(2πz/D + ωt − φ)]

Select only synchronous particles. These are the only ones which accelerate.

X = X ′ + Aeαt + Be−αt

Here α = [eǫE02ma cos(φ)]1/2 Finally, obtain a Poncare section for a plane in the phase space

of the accelerated particles. This is shown in Figure 22.

12 Soliton

A soliton is a non-linear wave which occurs when the non-linearities cancel the naturaldispersion in a propagating wave. Soliton waves occur in a number of physical systems,from fluid flow, to electronics, tsunamis, to fluxions in Josephson junctions. The wave am-plitude is essentially self-reinforcing and occurs, as examples, in the mathematical equations;

29

Figure 22: A Poincare section of the phase space for beam in an RFQ

• Korteweg-de Vries equation

• Non-linear Schrodinger equation

• Sine-Gordon equation

• Navier-Stokes equation

To illustrate using a simple example, consider the propagation of an acoustic wave in aplasma. Previously we considered a cold plasma as a stationary, isothermal, unchargedfluid. The plasma was “cold” so that frequencies were low and electrons move to cancel theelectric fields in the plasma. At higher frequencies the ions and electrons move indepen-dently, and a 1-fluid model is not realistic. For this case, consider each particle type, ionor electron, as a different fluid. There is then the possibility of both electric and magneticfields in the plasma, and charges move in circular orbits about the magnetic field lines withfrequencies;

ω = eBmc

However, ions are much heavier than electrons so the electron frequencies are much higherthat those of the ions. Assume an isothermal electron fluid and a cold ion fluid. The 2-fluidequations both have the form;

30

ρd~Vdt

= −~∇P + ρ~F + (1/3)µ~∇(~∇ · ~V ) + µ∇2~V

The terms are obvious, and similar to the 1-fluid model of the Navire-Stokes equation devel-oped earlier. Write the 1-D fluid equations for the ions as;

Fluid Equation

∂V∂t

+ V ∂V∂x

= (Ze/m)E

Poisson Equation

∂E∂x

= e[ZN − n]/ǫ

Contiunity

∂N∂t

+ NV∂x

= 0

For the electrons assume a Boltzmann number density.

n = n0e−eφ/kT

∂n∂x

= −(en0/kT )∂φ∂x

e−eφ/kT = −neE/kT

In the above N is the Ion number density and n is the electron number density. In a framewith the velocity of the plasma, we have boundary conditions as x → ∞;

V = 0; n = n0; N = n0/Z

Set Z = 1 to reduce complexity. If the system is linearlized the dispersion relation is;

( ωωi

)2 =k2λ2

D

1 + k2λ2D

Where ωi is the ion plasma frequency, en0m , k is the wave number, and Λ2

D is the Debye

length, kBTnoe

2 . Then expanding for small kλD ;

ω/ωi ≈ kλD − (kλd)3/2

kλD = ω/ωi + (1/2)(ω/ωi)3

The plane wave phase is,

31

(kx − ωt) = ζ + η/2

ζ =√

ǫ(x/λD − ωit)

ǫ = (kλD)2

η = ǫ3/2ωit

In the above ζ is the dimentionless distance measured in a frame moving to the right withvelocity, ωiλD ( a sound wave) and η is the dimensionless time. Clearly the wave disperses.Now proceed to the non-linear problem.

Consider a traveling wave with small amplitude so that expansions can be made in powersof ǫ. The solution is valid for wavelengths ≪ λD and ω ≪ ωi. Keeping terms to second order;

η = n0 + ǫn1 + ǫ2n2

N = n0 + ǫN1 + ǫ2N2

V = V1 + ǫ2V2

E ′ = ǫE ′

1 + ǫ2E ′

2

Where E =√

ǫE ′. To first order in ǫ;

n1 = N1 = (n0/Cs)V1

E ′

1 = −(ǫn0KBT )1/2

Cs

∂V1∂ζ

Cs =√

ǫkbT/M = ωpλD (sound speed)

To second order in ǫ, the Korteweg-de Vries equation for the dimensionless fluid velocitydevelops.

∂U∂η

+ U ∂U∂ζ

+ (1/2)∂3U

∂ζ3 = 0

U = V1/Cs

The soliton solution is;

U = U0 sech2[√

U0/6(ζ − ζ0 − (U0/3)(η − η0))]

32

Figure 23: The shape of a soliton solution of the Kdv equation

After substitution for the variables, this results in;

U = U0 sech2[x − x0 − Cs(1 + ǫU0/3)(t− t0)

λD∆]

With ∆ =√

6/ǫU0. Thus we find that;

1. limx→±∞U → 0

2.√

ǫU0∆ = constant

3. The propagation speed is Cs(1 + ǫU0/3)

There are also periodic solutions to the KDV equation of the form;

U = U0 Cn2( xλD∆

).

Figure 23 shows the form of a soliton wave solution to the Kdv equation.

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13 Additional considerations on solitons

Consider a linear, dispersionless wave equation;

∂2U∂t2

− C20∂2U∂x2 = 0

Make a change of variable;

η = x − C0t ζ = x + C0t

The solution is U = f(η) + g(ζ). Choose the functional form for the functions, f and g as ;

f = e−η2

g = e−ζ2

At time t = −∞ the pulses are separated, at t = 0 they are colliding and at t = ∞ they areseparated again. They remain coherent without change in shape. Then we have a localizedwave form, U(x, t) at t = 0. Determine the components of a Fourier expansion;

A(k) = (1/√

2π)∞∫

−∞

dxU(x, 0)eikx

Assume that initially, U(x, 0) = (1/√

2π) e−σ2k2/2

By substitution this gives;

U(x, t) = (1/2π)∞∫

−∞

dk A(k) ei(kx−ωt)

To perform the integral obtain an expansion form for the dispersion relation;

ω(k) = ω0 + dωdk

(k − k0) + (1/2)d2ω

dk2 (k − k0)2 + · · ·

dωdk

= Vgd2ωdk2 = a2

Completing the integral

U(x, t) = 1(√

σ2 + ia2t)√

2πe−(x−vgt)2/(2σ2+ia2t)

The Gaussian shape remains but the width increases with time and the peak height de-creases. The wave centroid travels with a group velocity, Vg.

Now suppose we have the inhomogeneous equation;

34

∂2U∂t2

− C20∂2U∂x2 + m2sin(U) = 0

This equation can be used to describe phonons in a solid and massive particles in relativisticfield theory. Determine the dispersion relation by substitution of the form;

U(x, t) = Aei(kx−ωt)

The dispersion relation is;

−ω2 + c20k

2 + m2 = 0ω(k) = ±

√

m2 + c20k

2

Thus the wave disperses as shown above. The group velocity is;

Vg = ∂ω∂k

= ± c0k√

c20k

2 + m2

Vp = ω/k = ±c0

√

1 + (m/c0k)2

The equation is the Sine-Gordon equation and should be compared to the Klein-Gordonequation above. In the limit θ → 0 the Klein-Gordon equation is recovered. The Sine-Gordon equation can produce solitons. A simple soliton solution after much algebra andallowing m = 1 is;

U = 4 tan−1[exp(x − βt

√

1 − β2)]

In the above β =√

α2 − 1/α where α is a constant of integration.

14 General Relativity

The postulates of special relativity are that space-time is homogeneous, there are specialframes of reference called inertial frames, and that the velocity of light is independent ofthe inertial reference frame. These postulates have problems when extended to gravitatingsystems. However, because of the equivalence principle that inertial mass and gravitationalmass cannot be distinguished, there can be no global inertial frames, only approximate in-ertial frames moving near freely falling bodies. Thus the straight lines in spacetime thatdescribe a gravity free inertial frame must be curved (because of acceleration) in generalrelativistic geometry. The equations governing general relativity must take into account thiscurvature. This introduces non-linear pde which are difficult to solve. An understanding ofdifferential forms in a curved coordinate system is required.

35

14.1 Derivatives of a tensor field

The ordinary derivatives of a scalar field are components of a covarient vector field. Ingeneral, however, the derivatives of a tensor field will NOT form a new tensor field. Thederivative of a vector compares its value to the value at another point, slightly displacedpoint. These two vectors do not transform with exactly the same coefficients, so the derivativeof the coefficients differs. Thus to obtain a tensor derivative which remains a tensor, we mustcompare a vectors which are displaced parallel to each other. If this is done then the lineartransformation between coordinate frames is preserved.

Previously when obtaining vector operators, we found that changes in a vector were duenot only to the changes in magnitude, but to changes in direction. Thus when a vector istranslated to a neighboring point without changing its Cartesian coordinates, it has beenparallel displaced. In Cartesian this is easily done, but becomes complex in general curvi-linear coordinates. Assume a contravarient vector, Ai.

∂ ~A∂ηj

=∑

n

an ∂hnfn

∂ηj+

∑

n

hnfn ∂(an)∂ηj

This can be written as;

∂ ~A∂ηj

=∑

n

an [∂hnfn

∂ηj+

∑

m

hmfm Γnmj ]

The change in an individual component is;

∂An

∂ηj= [

∂hnfn

∂ηj+

∑

m

hnfm Γnmj ]

The form, Γnmj , is a Christoffel symbol of the 2nd kind.

∂ai

∂ηj= Γk

ij ak

Γkij = ak · ∂ai

∂ηj

In the previous lecture, we demonstrated that the change in the unit vectors ai, can beexpressed as;

∂ai

∂ηj= (aj/hi)

∂hj

∂ηi

This shows that for orthogonal coordinates;

Γiik = 0 if i, j, k are all different

Γiii = (1/hi)∂hi

∂ηi

36

Γiij = Γi

ji(1/hi)∂hi

∂ηj

Γjii = −(hi/hj 2)∂hi

∂ηj

Γiii = (1/hi)∂hi

∂ηi

The above was developed for a contravarient vector, but a similar development provides thetransformation for a covarient vector. At times the Christoffel symbol is written in the form;

Γnmj →

nm j

A space that has parallel displacement defined is an “affine” space and the Γnmj are the

components of an affine connection. Note that the Christoffel symbols are not tensors andneither are the derivatives of the vector function. However, the combination of the twothrough the above equation does produce a tensor. A Christoffel symbol of the 1st kind isdefined as;

[ij, k] = (1/2)[∂gij

∂ηj+

∂gjk

∂ηi− ∂gij

∂ηk]

Γlij = glk[ij, k]

In the case of Cartesian coordinates, the Christoffel symbols vanish.

14.2 Covarient derivative

The covarient derivative assures that a vector is independent of its description in an arbi-trary coordinate frame. That is, the vector has a magnitude and points is a specific directionindependent of the frame of reference. The covarient derivative removes the change in thevector due the curvature of the coordinate frame. Therefore the components of a covarientvector representing the rate of change of an ordinary vector, ~A, with respect to the ηi axisis;

Aji = ∂Aj

∂ηi+

∑

k

AkΓjki

The comma indicates that this is the covarient derivative. The Ai, j are the components of

a mixed tensor, covarient with respect to the index j and contravarient with respect to theindex i. Covarient differentiation can also be extended to tensors of higher order.

The contracted tensor,∑

n

An, n, represents the Div of the vector ~A;

37

∑

n

An, n =

∑

n

∂An

ηn+

∑

n,m

AnΓmnm =

∑

n

1h1h2h3

∂∂ηn

(Anh1h2h3)

Identify An = F n/hn. Substitution in the above yields the Div ~A.

Div ~A = 1h1h2h3

∑

n

∂∂ηn

h1h2h3 An

hn

The curl of vector ~B, can also be developed for orthogonal coordinates by consideration of

the component Ai = −( 1h1h2h3

)[Bj ,k − Bk ,j]. The scalar Laplacian can also be written as

a covarient derivative,∑

n

( 1h2

n

∂φ∂ηn

), n

Thus the use of the covarient derivative allows one to express an equation in the same formfor any coordinate system.

14.3 Geodesic lines

In a general curvilinear coordinate system (Riemann space) there is a unique shortest linethat connects two points. As you know, in a spherical system this line is a great circle. Theselines are called geodesics. The length is given by;

s =b∫

a

ds =b∫

a

dt [gijdxi

dtdxj

dt]1/2

In the above t is the parametric parameter. To find the minimum we apply the calculus ofvariations. This will be discussed later in the course. Basically one varies all paths betweena and b that go through the end points a and b choosing the one which is stationary, ie thisis similar to the derivative. This reduces to the Euler-Lagrange equations and finally thedifferential equation;

d2xl

ds2 + Γlik

dxids

dxk

ds= 0

In a Cartesian system the Chrsitoffel symbol vanishes which leads to a linear relation betweenthe length and the coordinates

14.4 Riemann curvature tensor

Take the integral of a vector in parallel transport around a closed loop. There is a non-zerochange in the vector when arriving at its original position.

38

∆vi = 1/2Rjklvj0S

lk

The factor Slk is the loop area, vj0 is the original vector, and Rjkl is the Riemann curvature

tensor. It is a non-linear, second order differential form in the metric which is a function ofposition.

Rijlk = ( ∂

∂ql Γijk + Γi

lm Γmjk) − ( ∂

∂qk Γijl + Γi

km Γmjl )

The Ricci tensor is the contraction of the Riemann curvature tensor and represents thechange in the volume of a geodesic ball differs from a ball in Euclidian space. The scalarcurvature of space is determined by the contraction of the Ricci tensor. Thus

Ricci Tensor Rij = Rkikj

Scalar curvature R = Rii = RkiRki

The Einstein tensor is defined in terms of these tensors and the metric of the space, gij.

Gij = Rij − (1/2)gijR

The covarient divergence of the Einstein tensor vanishes. This is important in its use in theGeneral relativistic equations.

14.5 Einstein field equations

The assumption is that the Einstein tensor is proportional to the energy-momentum tensor,T . Thus;

Gij = κT ij

The constant κ is determined through measurements. This tensor equation is a relationbetween physical quantities and holds locally at each point in space time. It provides adifferential equation for the metric which is determined by the energy-momentum. Thesolution determines the associated curvature of spacetime. In special relativity the energy-momentum tensor is divergenceless which means it is conserved. In curved spacetime thecovarient derivatives of each quantity are conserved, and G is constructed to maintain thisrelationship. Experimentally κ = 8πG/c4 where G is the gravitational constant.

The interpretation is that action of gravity corresponds to changes in spacetime whichchanges the possible paths that an object moves. The curvature is caused bu the energy-momentum of matter. “Spacetime tells matter how to move and matter tells spacetime how

39

to curve” (Wheeler). As examples we find that the Lorentz contraction and time dilationare the same as in special relativity, but their interpretation is different. In general relativitythe effects arise from the acceleration of the frames rather than their relative velocity.

Simple analytic exposition of solutions to the equations does not exist. There are a fewanalytic solutions in the weak field case based on perturbations, but these are still to com-plicated to develop here. If interested, try several elementary books in General relativitysuch as B. F. Schultz “A first course in General Relativity” (second edition), CambridgeUniversity Press, 2009.

15 Self Organization

Physicsal phenomenon are rich wiht examples of self organization. The issue here is theability of a system to restrict and maintain its phase space through internal and not externalinteractions. Scientific study usually attempts to understand phenomenon by reductionism,that is by breaking a system into components and assume that the system can be explainedby through the sum of its individual parts. There is also the possibility of looking at thesystem globally, looking for general properties and laws which govern the behavior of a classof phenomena. Thus one looks to find general rules which determine the behavior of systems.Thus a system started from a particular state evolves until it arrives and maintains itself inanother state.

Measuring organization quantitavely is a problem because the definition of self orga-nization is not precise. Perhaps the most useful way to measure self organization is byinformation content, although entropy was originally used. In any event, entropy and infor-mation content of a system are related. Thus it is proposed to define statiscal complexityby the amount of information required to describe the behavior of the system at a point,and this is developed through the log of the probability of that action. This thought will bedeveloped further in later half of the course.

15.1 Boolean Network

A Boolean network is a collection of N logic gates (AND, XOR, NOT) with K inputs con-nected together. This network with certain connections can self organize. This is applied inparticular to represent various biological systems with N the number of genes and K theirinterdependencies.

15.2 Entropy

Entropy is defined in the second law of thermodynamics. However, thermodynamics isusually based on statistical systems in equilibrium. In open systems the flow of matter and

40

energy through the system allows the system to self organize, and self organization onlyoccurs far away from thermodynamic equilibrium. A closed system can gain macroscopicorder while increasing overall entropy by becoming more ordered macroscopically at theexpense of microscopic disorder. For example, self assembly of large molecules is compensatedby the disorder in small molecules, especially water. A laser is another example of a systemfroced out of equilibrium where a higher state is more populated than a lower one. Thisleads to coherent growth (amplification) of light output.

41