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depends on a large number of constant. In addition, the scaling of the heat equation has

led in a natural way to the key role of the Rayleigh number in the mode of heat transport

in a uid.

Problem e: Use (9.41) to show that convective heat transport dominates over conductive

heat transport when Ra 1.

CHAPTER 9. SCALE ANALYSIS

102

Problem f: Suppose that this condition is satis ed and that heat conduction plays a0

negligible role. Show that the characteristic time-scale of the dimensionless time t

is much less than unity. Give a physical interpretation of this result.

Transforming dimensional equations to dimensionless equations is often used to derive

the relevant dimensionless physical constants of the system as well as for setting up al-

gorithms for solving systems numerically. The basic rationale behind this approach is

that the physical units that are used are completely arbitrary. It is immaterial whether

we express length in meters or in inches, but of course the numerical values of a given

length changes when we change form meters to inches. Making the system dimensionless

removes all physical units from the system because all the resulting terms in the equation

are dimensionless.

Chapter 10

Linear algebra

In this chapter several elements of linear algebra are treated that have important appli-

cations in (geo)physics or that serve to illustrate methodologies used in other areas of

mathematical physics

10.1 Projections and the completeness relation

In mathematical physics, projections play an extremely important role. This is not only

in linear algebra, but also in the analysis of linear systems such as linear lters in data

processing (see section 11.10) and the analysis of vibrating systems such as the normal

modes of the earth. Let us consider a vector v that we want to project along a unit vector

n, see gure (10.1). In the examples of this section we will work in a three-dimensional

^

space, but the arguments presented here can be generalized to any number of dimensions.

v

Pv

v Ï•

^

n

Figure 10.1: De nition of the geometric variables for the projection of a vector.

We will denote the projection of v along n as Pv, where P stands for the projection

^

operator. In a three-dimensional space this operator can be represented by a 3 3 matrix.

It is our goal to nd the operator P in terms of the unit vector n as well as the matrix

^

103

CHAPTER 10. LINEAR ALGEBRA

104

form of this operator. By de nition the projection of v is directed along n, hence:

^

Pv =C^ :

n (10.1)

This means that we know the projection operator once the constant C is known.

Problem a: Express the length of the vector Pv in the length of the vector v and the

angle ' of gure (10.1) and express the angle ' in the inner product of the vectors

v and n to show that: C = (^ v).

^ n

Inserting this expression for the constant C in (10.1) leads to an expression for the pro-

jection Pv:

Pv = n (^ v) :

^n (10.2)

Problem b: Show that the component v? perpendicular to n as de ned in gure (10.1)

^

is given by:

v? = v n (^ v) :

^n (10.3)

;

Problem c: As an example, consider the projection along the unit vector along the x-

axis: n = x. Show using the equations (10.2) and (10.3) that in that case:

^^

01 01

vx C

B 0 A and v? = B v0y C :

Pv = @ @A

vz

0

Problem d: When we project the projected vector Pv once more along the same unit

vector n the vector will not change. We therefore expect that P(Pv) = Pv. Show

^

using expression (10.2) that this is indeed the case. Since this property holds for

any vector v we can also write it as:

P2 = P : (10.4)

Problem e: If P would be a scalar the expression above would imply that P is the

identity operator I. Can you explain why (10.4) does not imply that P is the

identity operator?

In expression (10.2) we derived the action of the projection operator on a vector v. Since

this expression holds for any vector v it can be used to derive an explicit form of the

projection operator:

P = nnT :

^^ (10.5)

This expression should not be confused withe the inner product (^ n), instead it denotes

n^

the dyad of the vector n and itself. The superscript T denotes the transpose of a vector or

^

matrix. The transpose of a vector (or matrix) is found by interchanging rows and columns.

For example, the transpose AT of a matrix A is de ned by:

AT = Aji (10.6)

ij

10.1. PROJECTIONS AND THE COMPLETENESS RELATION 105

and the transpose of the vector u is de ned by:

01

B ux C

uT = (ux uy uz ) u = @ uy A

when (10.7)

uz

i.e. taking the transpose converts a column vector into a row vector. The projection

operator P is written in (10.5) as a dyad. In general the dyad T of two vectors u and v

is de ned as

T = uvT : (10.8)

This is an abstract way to de ne a dyad, it simply means that the components Tij of the

dyad are de ned by

Tij = ui vj (10.9)

where ui is the i-component of u and vj is the j-component of v.

In the literature you will nd di erent notations for the inner-product of two vectors.

The inner product of the vectors u and v is sometimes written as

(u v) = uT v : (10.10)

Problem f: Considering the vector v as a 1 3 matrix and the vector vT as a 3 1

matrix, show that the notation used in the right hand sides of (10.10) and (10.8) is

consistent with the normal rules for matrix multiplication.

Equation (10.5) relates the projection operator P to the unit vector n. From this the

^

representation of the projection operator as a 3 3-matrix can be found by computing

the dyad nnT .

^^

01

B1C

Problem g: Show that the operator for the projection along the unit vector n = p114 @ 2 A

^

3

is given by 0 1

123

P = 14 B 2 4 6 C :

1

@ A

369

Verify explicitly that for this example P^ = n, and explain this result.

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