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Problem d: In order to reformulate (8.54) as an equation of conservation of momentum,

di erentiate (8.54) with respect to xi . Do this, use the de nition (8.55) and the

relation between force and potential (F = ) to write the result as:

; rV

1 @ (v v ) + h2 @ 1 2@i 1 2 1

2

@t vi + 2 i j j 8m i 2 = m Fi : (8.56)

jr j ; r

The second term on the left hand side does not look very much to the term @j ( vj vi )

in the left hand side of (8.13) To make progress we need to rewrite the term @i (vj vj ) into

a term of the form @j (vj vi ). In general these terms are di erent.

Problem e: Show that for the special case that the velocity is the gradient of a scalar

function (as in expression (8.55)) that:

1 @ (v v ) = @ (v v ) : (8.57)

2i jj j ji

With this step we can rewrite the second term on the left hand side of (8.56). Part of the

third term in (8.56) we will designate as Qi :

1@ 1 2@i 1 2

2

Qi : (8.58)

i2

8

; jr j ; r

Problem f: Using equations (8.6) and (8.56) through (8.58) derive that:

@t ( v) + ( vv) = m F+h2 Q : (8.59)

r

CHAPTER 8. CONSERVATION LAWS

88

Note that this equation is identical with the momentum equation (8.13). This im-

plies that the Schrodinger equation is equivalent with the continuity equation (8.6) and

the momentum equation (8.13) for a classical uid. In section (6.4) we have seen that

microscopic particles behave as waves rather than point-like particles. In this section we

discovered that particles also behave like a uid. This has led to hydrodynamic formu-

lations of quantum mechanics 28]. In general, quantum-mechanical phenomena depend

critically on Planck's constant. Quantum mechanics reduces to classical mechanics in the

limit h 0. The only place where Planck's constant occurs in (8.59) is the additional

!

force Q that multiplied with Planck's constant. This implies that the action of the force

term Q is fundamentally quantum-mechanical, it has no analogue in classical mechanics.

Problem g: Suppose we consider a particle in one dimension that is represented by the

following wave function:

!

x2 exp i (kx !t) :

(x t) = exp (8.60)

L2

; ;

Sketch the corresponding probability density and use (8.58) to deduce that the

quantum force acts to broaden the wave function with time.

This example shows that (at least for this case) the quantum force Q makes the wave

function \spread-out" with time. This re ects the fact that if a particle propagates with

time, its position becomes more and more uncertain.

Chapter 9

Scale analysis

In most situations, the equations that we would like to solve in mathematical physics

are too complicated to solve analytically. One of the reasons for this is often that an

equation contains many di erent terms which make the problem simply too complex to be

manageable. However, many of these terms may in practice be very small. Ignoring these

small terms can simplify the problem to such an extent that it can be solved in closed

form. Moreover, by deleting terms that are small one is able to focus on the terms that

are signi cant and that contain the relevant physics. In this sense, ignoring small terms

can actually give a better physical insight in the processes that really do matter.

Scale analysis is a technique where one estimates the di erent terms in an equation

by considering the scale over which the relevant parameters vary. This is an extremely

powerful too for simplifying problems. A comprehensive overview of this technique with

many applications is given by Kline 33].

Many of the equations that are used in physics are di erential equations. For this

reason it is crucial in scale analysis to be able to estimate the order of magnitude of

derivatives. The estimation of derivatives is therefore treated rst. In subsequent sections

this is then applied to a variety of di erent problems.

9.1 Three ways to estimate a derivative

In this section three di erent ways are derived to estimate the derivative of a function f(x).

The rst way to estimate the derivative is to realize that the derivative is nothing but the

slope of the function f(x). Consider gure 9.1 in which the function f(x) is assumed to

be known in neighboring points x and x + h.

Problem a: Deduce from the geometry of this gure that the slope of the function at x

is approximately given by (f(x + h) f(x)) =h.

;

Since the slope is the derivative this means that the derivative of the function is approxi-

mately given by

df f(x + h) f(x) (9.1)

;

dx h

89

CHAPTER 9. SCALE ANALYSIS

90

f(x+h)

f(x)

x x+h

Figure 9.1: The slope of a function f(x) that is known at positions x and x + h.

The second way to derive the same result is to realize that the derive is de ned by the

following limit:

df lim f(x + h) f(x) : (9.2)

;

dx h!0 h

If we consider the right hand side of this expression without taking the limit, we do not

quite obtain the derivative, but as long as h is su ciently small we obtain the approxima-

tion (9.1).

The problem with estimating the derivative of f(x) in the previous ways is that we

do obtain an estimate of the derivative, but we do not know how good these estimates

are. We do know that if f(x) would be a straight line, which has a constant slope, that

the estimate (9.1) would be exact. Hence is it the deviation of f(x) from a straight line

that makes (9.1) only an approximation. This means that it is the curvature of f(x)

that accounts for the error in the approximation (9.1). The third way of estimating the

derivative provides this error estimate as well.

Problem b: Consider the Taylor series (2.17) of section 2.1. Truncate this series after

the second order term and solve the resulting expression for df=dx to derive that

df = f(x + h) f(x) 1 d2 f h + (9.3)

;

2 dx2

dx h ;

where the dots indicate terms of order h2 .

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