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@ (3.14)

^

z

^

'

with the matrix M given by

0 1

sin cos ' sin sin ' cos

M = B cos cos ' cos sin ' sin C :

@ A (3.15)

;

sin ' cos ' 0

;

Of course expression (3.14) can only be considered to be a shorthand notation for the

equations (3.13) since the entries in (3.14) are vectors rather than single components.

However, expression (3.14) is a convenient shorthand notation.

The relation between the spherical components (ur u u' ) and the Cartesian compo-

nents (ux uy uz ) of the vector u can be obtained by inserting the expressions (3.13) for

the spherical coordinate unit vectors in the relation u =ur^+u ^ + u' '.

r ^

Problem b: Do this and collect all terms multiplying the unit vectors x, y and ^ to show

^^ z

that expression (3.5) for the vector u is equivalent with:

u = (ur sin cos ' + u cos cos ' u' sin ') x ^

;

+ (ur sin sin ' + u cos sin ' + u' cos ') y ^ (3.16)

z

+ (ur cos u sin )^

;

Problem c: Show that this relation can also be written as:

01 01

ux C u

B u y A = MT B u r C :

@ @A (3.17)

uz u'

3.3. THE ACCELERATION IN SPHERICAL COORDINATES 21

In this expression, MT is the transpose of the matrix M: Mij = Mji , i.e. it is the

T

matrix obtained by interchanging rows and columns of the matrix M given in (3.15).

We have not reached with equation (3.17) our goal yet of expressing the spherical coor-

dinate components (ur u u' ) of the vector u in the Cartesian components (ux uy uz ).

;1

This is most easily achieved by multiplying (3.17) with the inverse matrix MT , which

gives: 01 01

B ur C = MT ;1 B ux C :

u

@A @ uy A (3.18)

u' uz

However, now we have only shifted the problem because we don't know the inverse

MT ;1 . One could of course painstakingly compute this inverse, but this would be a

laborious process that we can avoid. It follows by inspection of (3.15) that all the columns

of M are of unit length and that the columns are orthogonal. This implies that M is an

orthogonal matrix. Orthogonal matrices have the useful property that the transpose of

the matrix is identical to the inverse of the matrix: M;1 = MT .

Problem d: The property M;1 = MT can be veri ed explicitly by showing that MMT

and MT M are equal to the identity matrix, do this!

Note that we have obtained the inverse of the matrix by making a guess and by verifying

that this guess indeed solves our problem. This approach is often very useful in solving

mathematical problems, there is nothing wrong with making a guess (as long as you check

afterwards that your guess is indeed a solution to your problem). Since we know that

M;1 = MT , it follows that MT ;1 = ;M;1 ;1 = M.

Problem e: Use these results to show that the spherical coordinate components of u are

related to the Cartesian coordinates by the following transformation rule:

0 10 10 1

ur C B sin cos ' sin sin ' cos C B ux C

B u A = @ cos cos ' cos sin ' sin A @ uy A

@ (3.19)

;

u' sin ' cos ' uz

0

;

3.3 The acceleration in spherical coordinates

You may wonder whether we really need all these transformation rules between a Cartesian

coordinate system and a system of spherical coordinates. The answer is yes! An important

example can be found in meteorology where air moves along a sphere. The velocity v of

the air can be expressed in spherical coordinates:

v =vr^+v ^ + v'' :

r ^ (3.20)

The motion of the air is governed by Newton's law, but when the velocity v and the force F

are both expressed in spherical coordinates it would be wrong to express the -component

of Newton's law as: dv =dt = F . The reason is that the basis vectors of the spherical

coordinate system depend on the position. When a particle moves, the direction of the

CHAPTER 3. SPHERICAL AND CYLINDRICAL COORDINATES

22

basis vector change as well. This is a di erent way of saying that the spherical coordinate

system is not an inertial system. When computing the acceleration in such a system

additional terms appear that account for the fact that the coordinate system is not an

inertial system. The results of the section (3.1) contains all the ingredients we need.

Let us follow a particle or air particle moving over a sphere, the position vector r has

an obvious expansion in spherical coordinates:

r =r^ :

r (3.21)

The velocity is obtained by taking the time-derivative of this expression. However, the unit

vector ^ is a function of the angles and ', see equation (3.7). This means that when we

r

take the time-derivative of (3.21) to obtain the velocity we need to di erentiate ^ as well

r

with time. Note that this is not the case with the Cartesian expression r =x^+y^+z^

xyz

because the unit vectors x, y and ^ are constant, hence they do not change when the

^^ z

particle moves and they thus have a vanishing time-derivative.

An as example, let us compute the time derivative of ^. This vector is a function of

r

and ', these angles both change with time as the particle moves. Using the chain rule it

thus follows that:

rr r r

d^ = d^( ') = d @^ + d' @^ : (3.22)

dt dt dt @ dt @'

r r

The derivatives @^=@ and @^=@' can be eliminated with (3.12).

Problem a: Use the expressions (3.12) to eliminate the derivatives @^=@ and @^=@' and

r r

^ and ' to

^

carry out a similar analysis for the time-derivatives of the unit vectors

show that:

r

d^ = _ ^+ sin ' '_^

dt

d^ = _ ^+ cos ' '

r _^ (3.23)

dt ;

d^ = sin ' ^ cos ' ^ :

'

_r _

dt ; ;

In this expressions and other expressions in this section a dot is used to denote the time-

_

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