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0

I(t t (13.35)

;1 ! + im

;

The problem we now face is to evaluate this integral. For this we will use complex integra-

tion. The integration variable is now called ! rather than z but this does not change the

principles. We will close the contour by adding a semicircle either in the upper half-plane

or in the lower half-plane to the integral (13.35) along the real axis. On a semicircle with

radius R the complex number ! can be written as ! = R exp i'.

Problem d: Show that e;i!(t;t0 ) = exp (;R sin ' (t0 t)). ;

Problem e: The integral along the semicircle should vanish in the limit R Use

0 the contour should be closed in the

! 1.

the result of problem d to show that for t < t

upper half plane and for t > t0 in the lower half-plane, see gure (13.3).

Problem f: Show that the integrand in (13.35) has one pole at the negative imaginary

axis at ! = =m and that the residue at this pole is given by

;i

(t t0 ) :

Res = exp (13.36)

m

; ;

Problem g: Use these results and the theorems derived in section (13.2) that:

( for t < t0

0

0) =

I(t t (13.37)

exp m (t t0 ) for t > t0

;

;i ; ;

Hint treat the cases t < t0 and t > t0 separately.

CHAPTER 13. COMPLEX INTEGRATION

170

tâ€™ > t tâ€™ < t

X X

Figure 13.3: The poles in the complex plane and the closure of contours for t' larger than

t (left) and t' smaller than t (right).

Let us rst consider (13.37). One can see from (13.33) that I(t t0 ) is a function that

describes the e ect of a force acting at time t0 on the velocity at time t. Expression (13.37)

;

tells us that this e ect is zero when t < t0 . In other words, this expression tells us that the

force f(t0 ) has no e ect on the velocity v(t) when t < t0 . This is equivalent to saying that

the force only a ects the velocity at later times. In this sense, equation (13.37) can be

seen as an expression of causality the cause f(t0 ) only in uences the e ect v(t) for later

times.

Problem h: Insert the equations (13.37) and (13.35) in (13.33) to show that

1 Z t exp (t t0 ) f(t0)dt0 :

v(t) = m (13.38)

m

;1

; ;

Pay in particular attention to the limits of integration.

Problem i: Convince yourself that this expression states that the force (the \cause")

only has an in uence on the velocity (the \e ect") for later times.

You may be very happy that for this problem we have managed to give a proof of the

causality principle. However, there is a problem hidden in the analysis. Suppose we switch

o the damping parameter , i.e. we remove the syrup from the problem. One can easily

see that setting = 0 in the nal result (13.38) poses no problem. However, suppose that

we would have set = 0 at the start of the problem.

Problem j: Show that in that case the pole in gure (13.3) is located on the real axis

rather than the negative imaginary axis.

This implies that it is not clear how this pole a ects the response. In particular, it is

not clear whether this pole gives a nonzero contribution for t < t0 (as it would when we

13.4. RESPONSE OF A PARTICLE IN SYRUP 171

consider it to lie in the upper half-plane) or for t > t0 (as it would when we consider it to

lie in the lower half-plane). This is a disconcerting result since it implies that causality

only follows from the analysis when the problem contains some dissipation. This is not

an artifact of the employed analysis using complex integration. What we encounter here

is a manifestation of the problem that in the absence of dissipation the laws of physics

are symmetric for time-reversal, whereas the world around us seems to move in one time

direction only. This is the poorly resolved issue of the \arrow of time."

CHAPTER 13. COMPLEX INTEGRATION

172

Chapter 14

Green's functions, principles

Green's function play a very important role in mathematical physics. The Green's function

plays a similar role as the impulse response for linear lters that was treated in section

(11.7). The general idea is that if one knows the response of a system to a delta-function

input, the response of the system to any input can be reconstructed by superposing the

response to the delta function input in an appropriate manner. However, the use of Green's

functions su ers from the same limitation as the use of the impulse response for linear

lters since the superposition principle underlies the use of Green's functions they are

only useful for systems that are linear. Excellent treatments of Green's functions can be

found in the book of Barton 5] that is completely devoted to Green's functions and in

Butkov 14]. Since the principles of Green's functions are so fundamental, the rst section

of this chapter is not presented as a set of problems.

14.1 The girl on a swing

In order to become familiar with Green's functions let us consider the example of a girl on

a swing that is pushed by her mother, see gure (14.1). When the amplitude of the swing

is not too large, the motion is of the swing is described by the equation of a harmonic

oscillator that is driven by an external force F(t) that is a general function of time:

2

x + !0 x = F(t)=m : (14.1)

The eigen-frequency of the oscillator is denoted by !0 . It is not easy to solve this equation

for a general driving force. For simplicity, we will solve equation (14.1) for the special case

that the mother gives a single push to her daughter. The push is given at time t = 0 and

has duration and the magnitude of the force is denoted by F0 , this means that:

8

>0 for t < 0

<

F(t) = > F0 for 0 t < (14.2)

:0 for t

We will look here for a causal solution. This is another way of saying that we are looking

for a solution where the cause (the driving force) precedes the e ect (the motion of the

oscillator). This means that we require that the oscillator does not move before for t < 0.

For t the driving force vanishes and the solution is given by a linear combination of

173

CHAPTER 14. GREEN'S FUNCTIONS, PRINCIPLES

174

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