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Problem b: Use the rst line of (3.23) and the de nition v =dr=dt to show that in

spherical coordinates:

v = r^+r _^ + r sin ' ' :

_r _^ (3.24)

In spherical coordinates the components of the velocity are thus given by:

vr = r_

v =r_ (3.25)

v' = r sin ' _

This result can be interpreted geometrically. As an example, let us consider the radial

component of the velocity, see gure (3.2). To obtain the radial component of the velocity

we keep the angles and ' xed and let the radius r(t) change to r(t + t) over a time

3.3. THE ACCELERATION IN SPHERICAL COORDINATES 23

z-axis

r(t + âˆ† t)

} dr

âˆ†r = âˆ† t = vr âˆ† t

r(t) dt

y-axis

x-axis

Figure 3.2: De nition of the geometric variables used to derive the radial component of

the velocity.

t. The particle has moved a distance r(t + t) r(t) = dr=dt t in a time t, so that

;

the radial component of the velocity is given by vr = dr=dt = r. This is the result given

_

by the rst line of (3.25).

Problem c: Use similar geometric arguments to explain the form of the velocity compo-

nents v and v' given in (3.25).

Problem d: We are now in the position to compute the acceleration is spherical coordi-

nates. To do this di erentiate (3.24) with respect to time and use expression (3.23)

to eliminate the time-derivatives of the basis vectors. Use this to show that the

acceleration a is given by:

sin 'v' ^+ v + _vr cos 'v' ^+ (v' + sin 'vr + cos 'v ) ' :

a = vr _v _r_ _^

_ _ _ _

; ; ;

(3.26)

Problem e: This expression is not quite satisfactory because it contains both the compo-

nents of the velocity as well as the time-derivatives _ and ' of the angles. Eliminate

_

the time-derivatives with respect to the angles in favor of the components of the ve-

locity using the expressions (3.25) to show that the components of the acceleration

in spherical coordinates are given by:

v 2 + v'

2

ar = vr

_ r

;

CHAPTER 3. SPHERICAL AND CYLINDRICAL COORDINATES

24

2

v'

vr v

a = v + r r tan

_ (3.27)

;

a' = v' + vrrv' + rvtan'

v

_

It thus follows that the components of the acceleration in a spherical coordinate system are

not simply the time-derivative of the components of the velocity in that system. The reason

for this is that the spherical coordinate system uses basis vectors that change when the

particle moves. Expression (3.27) plays a crucial role in meteorology and oceanography

where one describes the motion of the atmosphere or ocean 30]. Of course, in that

application one should account for the Earth's rotation as well so that terms accounting

for the Coriolis force and the centrifugal force need to be added, see section (10.4). It also

should be added that the analysis of this section has been oversimpli ed when applied to

the ocean or atmosphere because the advective terms (v r)v have not been taken into

account. A complete treatment is given by Holton 30].

3.4 Volume integration in spherical coordinates

Carrying out a volume integral in Cartesian coordinates involves multiplying the function

to be integrated by an RRR nitesimal volume element dxdydz and integrating over all volume

RRR FdV = in F(x y z)dxdydz. Although this seems to be a simple procedure,

elements:

it can be quite complex when the function F depends in a complex way on the coordinates

(x y z) or when the limits of integration are not simple functions of x, y and z.

Problem a: Compute the volume of a sphere of radius R by taking F = 1 and integrating

the volume integral in Cartesian coordinates over the volume of the sphere. Show

rst that in Cartesian coordinates the volume of the sphere can be written as

Z R Z pR2 ;x2 Z R2 ;x2 ;y2

p

volume = dzdydx (3.28)

;R ;pR2 ;x2 ; R2 ;x2 ;y2

p

and carry out the integrations next.

After carrying out this exercise you probably have become convinced that using Cartesian

coordinates is not the most e cient way to derive that the volume of a sphere with radius

R is given by 4 R3 =3. Using spherical coordinates appears to be the way to go, but

for this one needs to be able to express an in nitesimal volume element dV in spherical

coordinates. In doing this we will use that the volume spanned by three vectors a, b and

c is given by

ax bx cx

volume = det( a b c ) = ay by cy : (3.29)

az bz cz

If we change the spherical coordinate with an increment d , the position vector will

change from r(r ') to r(r + d '), this corresponds to a change r(r + d ') ;

r(r ') = @r=@ d in the position vector. Using the same reasoning for the variation of

3.4. VOLUME INTEGRATION IN SPHERICAL COORDINATES 25

the position vector with r and ' it follows that the in nitesimal volume dV corresponding

to changes increments dr, d and d' is given by

@r @r @r d' ) :

dV = det( @r dr @ d (3.30)

@'

Problem b: Show that this can be written as:

@x @x @x

@r @ @'

@y @y @y

dV = drd d' = Jdrd d' : (3.31)

@r @ @'

@z @z @z

@r @ @'

| {z }

J

The determinant J is called the Jacobian, the Jacobian is sometimes also written as:

J = @ (x y ')

z) (3.32)

@ (r

but is should be kept in mind that this is nothing more than a new notation for the

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