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Home , Polar coordinate system

Vec!or A dB, 3 dBx dB, + ,E; dBd~, laA A A, ori (ay-- )((ax-ay)~rdl('~xtl~yik~zl

6. V .'B =O From the vector identity Eq. (2.30, if A is a vector field then V.(VxA)=O

This implies that B = V x A.

A = [C exp (-xy) sin"ax1 E;

r 8 r a = I -aY [C exp(-xy)sin ar]- J -ax [C exp(-xy)sin ax]

A = i [-x C exp(-xy)sin ax] - f [a C exp(-xy)cos cru - y C exp(- xy)sin a] = C exp (-xy) sin ar (- x ;+ y j) -i~a exp(- xy) cos ax -. COOmINATE SYSTEMS

Structure 3.1 Introduction Objectives 3.2 Non-Cartesian Coordinate Systems Planc Polar Ooordinate System Cylindrical Spherical Polar&rdinate System 3.3 Expressing a Vector in Polar Coordinates Differential .Element of Vector Length Dlflcrential Element of Vector Gradient of a Scalar Field 3.4 Curvilinear Coordinate System Orthogonality Condition Unit Veclors Vector Differential Operators 3.5 Summary 3.6 Terminal Questions 3.7 Solutions and Answers

INTRODUCTION

In the previous units we restricted ourselves completely to the ca~tesia! coordinate system. This system offers us unique advantage in that the unit vectors i, j and k &reconstant in direction as well as in magnitude. But it is not advisable to use Cartesian coor@inatesin all physical problems. For instance, when we deal with gravitational or electrostatic , use of Cartesian coordinates is usually not convenient. The same is true when we wish to study 'the distribution of gas molecules, flow of a fluid along a pipe, flow of current in a cylindrical conductor, heat flow in a sphere or energy produced in.a reactor. In all these situations, we use non-Cartesian coordinates-- polar, cylindrical or spherical polar coordinates. The best choice for a coordinate system depends on the shape of the system under consideration. FQ~systems with cylindrical or spherical , the use of these coordinates simplifies the algeb~-0.So it is absolutely necessary for you to be familiar with all of them. You will learn about all these coordinate systems in Sec. 3.2. In Sec. 3.3 you will learn to I express a vector in different polar coordinates. The expressions for , surface and volume elements are also developed here. The notCartesian coordinate systems are special cascs of the general orthogonal . In Sec. 3.4, we have introduced curvilinear coordinates by linking them with Cartesian coordinates. The expressions for gradient, divqrgence, curl, and Laplacian operator are derived first in curvilinear and then in pohr coordinates, These expressions will be useful in Unit 4 where you learn to evaluate itltepls involving various scalar and vector fields,

Objectives I After going through this unit, you should be able to relate plane polar, spherical polar and cylindrical polar coordinates to Cartesian coordinates write a vector in terms of polar coordinates differentiate between Cartesian coordinates and curvilinear coordinates write line, surface and volume elements in terms of curvilinear and polar coordinates, and derive expressions for gradient,divergence, curl and loplacian operators in curvilinear and polar coordinates.

3.2 NON-CARTESIAN COORDINATE SYSTEMS

You are familiar with Cartesian coordinates from your and courses. On , the basis of your knowledge, you would agree that Cartesian coordinates are the simplest and often natural. But there are many physical problems where it is advisable to use non-dartesian coordinates. Some familiar examples include the motion of in the solar system, nlotion of charged particles in electric and magnetic fields, flow of current in a conductor, energy produced in a power plant, temperature distribution in a circular plate and so on. Of the various non-Cartesian coordinate systems,-the simplest is the plane polar coordinate system. It is very useful in tlie study of the motion of planets, oscillations of a simple pendulum and calculation of potential due to a dipole. Since you have learnt about it in'your earlier courses (PHE-O1,02), we will discuss it briefly. YA 3.2.1 Plane Polar Coordinate System Refer to Fig. 3.la. It models a flat circular dinner plate. To specify any P on its surface, you have to draw two fixed coordinate axes at right to each other passing through 0, the centre of the plate. This is the familiar Cartesian coordinate system. You can locate the point of interest by giving its from the two axes. This is denoted as P (x,y). You may now ask: Is it the only way to locate this point? The position of point P can also be uniquely defined by measuring its radial distance from the and the Q .between the x-axis and the line joining the point to the origin, as shown in Fig. 3.lb. Point b is called the pole. The distance p is called the vector of the point P and Q is its (a) polar angle. These two coordinates taken together are called plane polar coordinates of P. Now you can specify the position of the point P as P (p ,$). You would now like to know how x and y are related to p and $. 4 To relate the x and y coordinates of P to its p and 4 coordinates, you can easily write referring to Fig. 3.lb

n / x=pcos$ (-m

kxand y = p sin $ (-w

and on dividing one by the other, you can write Fig. 3.1 : Specification of a pin\ in (ajcartesii and (b) polar 9 = tan- coordinates. In Literature, the notation (r, 0) is also used frequently V-/ for polar coordinates. We take p to be positive ivhen it is measured from the origin along the line OP. Similarly, $ is taken to be positive in the anticlockwise direction from x-axis. The rang of variation of these coordinates is given by

Now you Hay logically ask: Can we uniquely determine plane polar coordinates for all points? From Eq. (3.2) you will note that only for y = 0 = x, I$ is undefined. This means that but for the origin, the polar coordinates are nniquely determined. You will note that the transformation given'by Eqs. (3.1) and (3.2) gives a one-to-one correspondence between (x, y)and (p, $) coordinates. This means that in a two- dimensional (2-D) geomet;y, instead of labelling P by the'cartesian coordinates, wti can equally well use plane polar coordinates.

Let US now-consider the relationx2 + y2 = 1 . You would recall that this equation defines a . So you can say that in plane polar coordinates, this relation defines a circle of unit 2 radius about the origin. Similarly, x2 + y -: 2 defines another circle of radius \rZ and so on. Thus for different values of p, you would obtain afamily of concentric about the origin. Similarly, you may recall that thc relation y = nu + c defines a straight line with c as intercept on the y-axis. For c = 0, the straight line.wil1 pass through the origin. Does this not mean that y = x tan $, for different values of $, defines a family of concurrent lines passing through the origin? This is illustrated in Fig. 3.2. We know that Cartesian coordinate system is orthogonal, i.e. the x and y-axes meet at right angles at any point in the plane and their directions are fixed. Can you say the same about plane polar coordinate system? To discover the answer to these questions, again refer to Fig. 3.2. You will note that each concurrent line cuts the concentric circles at one point only. When you draw a tangent at that point, you will note that the coordinate axes dcfining the directions of increasing p and increasing 9 are at right angles. That is , the polar coordinates form an orthogonal coordinate system in the plane. But for two different points (p, , @,) Coordinate Smtems

Fi. 3.2 : ~adyof amcurrent lines passing throw origin for diKerent values ol+ and. Pz (pZ, 42), the directions of p and I$ coordinate axes are not the same. From this you may conclude that unlike in the Cartesian system, the directions of plane polar coordinate axes vary from point to point. Let us pause for a minute and summarise our findings about plane polar coordinates.

i) Any point can be represented uniquely by specifying its distance from the origin and the angle radius vector makes with the x-axis. ii) The Cartesian coordinates are uniquely related to plane polar coordinates of a point x=p cos4

y = p sin c$ lii) The polar coordinates form an orthogonal coordinate system in. a plane. , iv) The directions of coordinate axes are not the same for all points.

SAQ 1 A simple pendulum is suspended from the +ling of a lift. It is moving upward with Spend 5 min a. The string of effective length of the pendulum makes angle 4 with the vertical. Choose the point of suspension as origin, x-axis along the horizontal and y-axis vertically downwards. Express x and y in terms of plane polar coordinates at time t after the lift takes off. .- . So far we have confined our discussion to plane polar coordinates. The space is three-dimensional (3-0) and we know of many 3-0 systkms. The ball hit by a cricketer, r veins carrying blood into our heart, pipes carrying water and LPG; and a conductor carrying current are some of the more familiar examples. Some of these involve cylindrical geometry and for them the use of cylindrical polar coordinates appears natural. Let us now discuss these.

3.2.2 Cylindrical Coordinate System You know that in Cartesian coordinate system, the positioi of? pointP,in space is denoted by P (x, y, z). We denote the cylindrical polar coordinates of this point by P (p, +, z). By referring to Fig. 3.3, you will note that p and z respectively denote the distance of P from the z-axis, and the xy plane while + is the angle which the line joining the origin with the projection of P on xy plane makes with the x-axis. It is called the azimuthal angle. From the preceding section you may recall that the equation x2 + y2 - p2 with p = constant defines a circle in a plane, This suggests that if you add a Cartesian z-axis to the plane polar coordinate system, sb that z denotes the distance from the x y plane, you would obtain a cylinder. That is, cylindrical polar coordinate system extends the plane polar coordinate system to three . So if a point has Cartesian coordinates (x, y, z) and cylindrical polar coordinates (p, 9, z), by analogy with Eq. (3.1) or by referring directly to Fiq. 3.3, you can write Veclor Calculus

Fig. 3.3 : Representation cf a point in Cartesian and cylindrical coordinates. 1

As before, you can invert these relations to write (b.m.-,I 1 4 = tan - l (:I (0 s 4

In case of plane polar coordinates, 4 is undefined at the origin. But in cylindrical coordinates + is undefined for a11 points on the z-axis (x=O=y)

Fig. 3.4 : (a) Contours of constant temperature (b)2 -D plot of counbyide hills. Refer to Fig. 3.4a, which shows contours of constant temperature over different parts of the country on a particular day. (You may have seen it over your TV sets during National News Telecast.) Fig. 3.4b shows a map of countryside containing hills. The map shows contours for area at a given height. These are 2-D representation of a 3-D view. Mathematically speaking, a surface is defined by a relation between the (x, y, z) coordinates, once you fix on6 of these coordinates. For Cartesian coordinates, these are planes and illustrated in Fig. 3.5. You can --visualize these by performing the following activity. Activity Take a 4" x 6" plane paper. Make a cut mark in the shape of plus (t) sign (size 2" x 3") at the centre using a razor blade. Now take another plane paper of size 2" x 6". Make a 4" cut mark at its centre lengthiise. Insert it in the first paper so that the cut is normal to its plane. Now take another paper of size 2" x 4" and insert it so that it is normal to both of them. These three papers define three plane surfaces for the case of cartesian coordinates.-Identify . them by comparing with Fig, 3.5. By looking at your activity you can see that the three .74 planes intersect orthogonally at one point. Coordinate Systems

Fig. 3.5 : Schematic representation of p+es in Cartcsian coordinate system. Now refer to Fig. 3.68, which sh ws a cylinder of radius p and whose axis of symmetry is along the z-axis. You will note t at for any point on the surface of cylinder, p is constant. That is, p = constant defines a cii cular cylindrical surface. (This also gives this coordinate system its name.) This is also called the 4 - z surface. For different values of p, you will obtain co-axial right circular cylinders. Their common axis of symmetry is z-axis.

A p =Iconstant /

I I I I .------I 4 - e 0 Y

X /------.

(a)

Flg. 3.6 : Surfaces defined in cylindrical coordinate system.

SAQ 2 Draw 4 -z surfaces for the cylindrical coordinate system. Spend 5 min '

Can you draw the surface defined by 4 = constant and z = constant? The p z-syrface defined by 4 = constant is half-plane bounded on one edge by thez-axis (Fig. 3.6b). But the p $ - surface given by z = constant is a plane parallel to the xy plane (Fig.3.6~)just as in the Cartesian coordinate system. - . - SAQ 3 - Figs. (3.6a-c) depict surfaces defined by p = constant, r$ - constant and z = constant Spend I0 min separately. To ensure that you have understood this concept,we would like you to draw these surfaces on one figure,

In the above paragraphs you have learnt that once one coordinate out of the three (x, y, z) or (p, $J)defining a coordinate system is fixed, the relation between them manifests as a surface. What will happen ifyou fix two of these? You will observe that intersection of y = constant and z = constant surfaces defines the x-axis. Similarly, intersection of surfaces given by z = constant and x = constant defines the y-axis. What does the intersection of x = constant, and y = constant define? It defines the z-axis. In the case of cylindrical coordinates, intersection of surfaces given by 4 = constant and r = constant defines the p-curye. It is a ray perpendicular to the z-axis. What is the shape of (a) $-curve given by z = constant, p = constant and (b) z-curve given by p = constant, J, = constant? While curve Vector Cakulus is a horizontal circle centred on the z-axis, the z-curve is a line parallel to the z-axis. These are illustrated in Fig. 3.7.

By now you must have realised that the choice of a coordinate system is determined by the geometry of the system under consideration. Suppose you wish to study the flow of heat in the interior of a sphere or calculate potential at a point due to a uniformly charged sphere. Will the use of Cartesian or cylindrical coordinates' look natural? Spherical symmetry suggests that it will be more convenient to define a polar angle. This brings us to spherical polar coordinates. Let us now discuss it.

3.2.3 Spherical Polar Coordinate System In spherical polar coordinate system, the position of a point is specified by the radial distance r, the polar angle 8 and the azimuthal angle 4 as shown in Fig. 3.8. You can see that while 8 is measured in the clockwise direction from the z-axis, 4 is measured in the anticlockwise direction from the x-axis.

Ng. 3.8 : Representation of a point in Cartesian and spherical polar coordinate system. .

Referring to this figure you can see that the projection of .OP onto the x y plane, OP' = r sin8 while its projection on z-axis is r cos 8. The components of OP' alongx arid y-axes are OP' cos 4 and OF' sin 4. Hence, for a point P having Cartesian coordinates (x, y, z ) and spherical polar coordinate. (r, 0, $) , you can write Thc z-axis ii also rcferred to as the x-OP'cos~=rsin0cos~ (x1 axis because of the analogy hclween [he Earth's surface ant1 thc surrucc of a sphcrc. z=OPcos8=rcos8 (-a

Thesurfi~cesdefinedby r = constant, 8 = constant and $ - constant are illustrated in Fig. 3.9. It is instructive to note that in spherical coordinates, $ = constant is the half plane as in cylindrical coordinates.

4 lane z +=constant

cone z 0 = constant r = constant

Y

b X Y

Fig. 3.9 : Surfaces defined in spherical polar amrdinate system.

SAQ 4 Draw a picture of the solid defined by the set of points (r, 8,9) such that Spend 5 min

Before you proceed, we would like you to go through the following example.

Example 1 The Cartesian coordinates of a point are (2, - 2,3)in arbitrary units. Compute its (i) cylindrical &d (ii) spherical polar coordinates. Solution i) Cylindrical coordinates From Eq. (3.4), we have p=GG7=.\/22+(-2)2 =2n

= tan- (- 1) - 3n/4 or 7d4. But x is positive and y is negative. So

Hence 9 7n/4. and z ;. 3

ii) Spherical polar coordinates From Eq. (3.6), we have

and $ = tan- (:) = tan- (- 1) = 7rd4 Vector Calculus In physics, we make good use of vectors. You may recall that velocity and acceleration are the most fundamental quantities used to describe motion. It is often necessary to express th~ components of a vector expressed in one coordinate system in teims of another. So we must know to express a vector in polar coordinates.

3.3' EXPRESSING A VECTOR IN PO COORDINATES

From Unit 1 you would recall that we can resolve a vector A in Cartesian coordinates as

AA A where i, j and k are unit vectors along the x, y, and c%x:s respectively. This means that to express A in polar coordinates, we must first relate i, j, k with the unit vectors associated with the system of interest. Let us first consider the cylindrical coordinate system. Cylindrical coordinate system A A For cylindrical coordinate system, we define the unit vectoFs ep, el and 2; at 8 given point P :s :s follows : Let R be the point on the z-axis with the same z-coordinate as P. Then we define e,, to be the at P which is normal to the cylindrical surface p = constant through A P. So e, will be in the direction of RP,i.e, along the direction of increasing p , as shown in A Fig. 3.10a. Similarly, we define ee to be the unit vector normal to the half plane 4 = constant -through P in the direction of increasing 4. The unit vector izis defined normal to the plane t = constant through Pin the direction of increasingz. Now refer to Fig. 3.10b. You will note A n x that ep makes an angle $ with the x-axis, - 4 with the y-axis and - with the z-axis. So the 2- - 2 AA A components of ;p along the directions of i, j and k are, respectively

A A e, . i = cos $

'4 ip On combining these results, you can write

A A A ep= cos$ i + sin4 j (3.9a) Fig. 3.10 A Another method to derive thw Again, by looking at Fig. 3.10(b), you will note that e9 makes an angle relations isto note that X h A Re x-axis, 4 with the y-axis and - with the z-axis. Can you now compute the components of el e p=- 2 A JRP I A A along the directions of i, j and k following the procedures outlined for ep?The result is - 2 P+Y 3 A A A P e, 5 - sin4 i + cos4 j (3.9b) n subtituling and Eq* Since e, parallel to z-axis, its projection along the x and y-axes will be zero. Hence, (3.31, you Will get h A A A PcOS+P+psin+ f e p= e, = k (3.9~) P =ma+f+.sin9 f These relations cap readily be inverted. To this end, you have to multiply Eq. (3.9a) by cos + A and Eq. (3.9b) by sin 4. Add the resulting expressions and use the identity Since e+ makes an angle cos2 4 + sin" = 1. On simplification you will get A A A (5++)*hbher-ub i 5 cos 4 ep- sin 4 eg (3.10a) and an angle + with thepaxig, YOU YOU can similarly'write can write .f=~in~~~+cos~~ (3. lob) td=ma (:t~)?+als~f A A and k=e, (3.1 OC) =- sin @ f +cos @ f AAA 78 We know that directions of the Cartesian unjt vectors (i, j, k) are uniquely fixed. Is the same AA A true for unit vectors e,, e+ and e,? The d~rectionsof these unit vectors vary from point to Coordinate Systems point. But computing their dot and cross products, you can easily check that these vectors , are normal to each other and form a right-handed system. We will now like you to study the following example carefully. * Example 2 In Cartesian coordinates, the position vector is given by

AAA r=xi+yj+zk Fxpies; it in terms of cylindrical coordinates (p, I$, z) and the associated unit vectors ep, e+,e, . Solution The position vector is.given by

AAA r=xi+yj+zk On substituting forx , y and z from Eq. (3.3) and (i,AAA j, k) from Eqs.(3.l0a, b, c), we get

A. A. A A r = p cos4 (cos4 lP- sin4 eS+ -Q sin4 (sin+ e, + kos4 e ,+) + ze, A A A A h On collecting the coefficients of e;, e$ and e, wcget r = p ep + z el

In this example, you have'zee: that tke position vector can be resolved into components parallel to the unit vectors ep, ee and e, . In fact you can do so for any vector and write

A A=A,~+A+;,,,+A,~, (3.11) where A;=A, cos$ +A, sin4 (3.11a)

and A+ = -A, sin@+A, cos4 (3.11b)

Example 3 A particle is moving through space. Compute the components of its velocity in cylindrical coordinates. Solution From Example 2 we recall that position vector of a particle moving through space is given by A A r (t)= p ep + z e, On substituting for Gp from Eq. (3.9a), we hale

r (t)= (ihcos4 +;sin$) + E; z

AA since e, = k.

c, To compute the velocity of this particle, we have to keep in mind that p as well as 4 change Y with time as r changes. So on differentiating (i) with respect to time, we get

v a t=p(i^cos+ +j^sin+)i + (-?ki?$ +j^cos$)f~4 + -;pp+G,,,p4+ii (ii) where dot over r, p, I$ and z denotes their respective first time . On comparing this expression with the equation A A A v-v e +v,+e,,,+v,e, P P we find that v, - P, V+ = p 4, and vZ = (iii)

Spherical Coordinate System

1 For spherical polar coordinates, we define the unit vectors as follows: At the point 1 Vcclor Calculus A P (r, 0,4), the unit vector iris normal to the surface r = constant, e, is normal to the surface A 8 = constant and e4is normal to the surface defined by I$ = constant through the given point. Their directions are along increasing r, 0 and 4 respectively as shown in Fig. 3.11a. By AAA considering the components of the unit vectors e,, e,, eg in the directions of unit vectors

AAA i,jrk (Fig. 3.1lb) we would like you to show that A e, = sin8 cm+ itsine sin$i+ case e, = cos0 cos6 i + cos0 sin4. .j- - sin0 k A" e+,=-sin+:+ we;

Conversely A ;= cos&, + cose cos6 :,- sin6 e A+ j= sin0 sin+ B + c0s0 sin+ Go+ CO@I e+ A . k = cose Cr - sine e,

Fig. 3.11

In case you cannot establish these results now, you will get another opportunity in the next section . So you should not feel depressed.

A particle is moving in space. Express its position vector and components of its velocity in spherical coordinates. Solution The position vector of a particle moving in space is written as

hAA r=xi+yjczk

AA A Substituting for x, y, z from Eq. (3.5) and i, j, k from Eq. (3.13) you will get

A r= r sine COS$(sine cosb 8 + COS~as4 e, - sin+ t r sine sin+ (sine sin+ $ + cose sin( &, + cos+ :+)

On collecting the coefficients of$, Go, and 4,this expression gives us therequired result:

+ to(r sin0 as2+cos0 + r sin0 cos0 sin2$ - r cose sine) In spherical polar coordinates, the position vector of a particle moving in space is given by r=rerA

= r (sin0 cos+ i'+sine sin+ j'+ mse Ci) Differentiating it with respect to time, you will get v = ;6 + fr (cod cos$ (3 - sin6 sin+ 4)+ r f(cosB sin4 6 + sine cos4 6)- r k sine 6

- i- Gr + r 6 (cos0 cos+ ;+ cos0 sin@$- & sine) + r I$ (- ;sine sin@+ ;sin0 cos+) . A + A =;$+r9eO+rsin84eg

You must have realised that while evaluating the'velocity components in non-Cartesian coordinates, we first expressed polar unit vectors in terms of Cartesian unit yecttrs; These expressions were differentiated and the result was re-expressed in terms of (ep,e,, er) and (G,, 4,4) . This involves straightfonvard algebra and we would like you to understand it rather well. You are therefore advised to carefully go through the following example where we have computed acceleration in cylindrical coordinates.

Example 5 Compute the components of acceleration in cylindrical coordinates for a particle moving in space. Solution From Example 3 you would recall that v=pep+p$e.A .A+ +ze, .A (i) On differentiating this with respect to time, we get

, ..A .A 7 A .. A .A ..A .A a-v=pep+pei,+p4e+ +p+eg+pi$eg+ze,+ze, (ii)

A A A To evaluate time derivatives of e,,, e+, e; ; we recallsthat

A A ep - i cos$ +;sin+

A A A e,--isin++ j co*

Hence A n e,, - (- [sin$ + ica@) ) - e+ 4

A A A 8 e, - (- fms$ - j sin+) )( - - e,, ( A e, - 0 Using these results in (ii), we get 6 .. A a-[p-P(()2]~+(p;b+2~~)e~+~ez

1 d .. '! -[6:-~(4)~l~~+~~(~~be,+~e,(iii)

I So the cylindrical components of acceleration of a particle moving in space are a,=p-p(+)2,a,=p++2p+ a,=z (iv)

A special case of cylindrical coordinates is plane polar coordinate system in which the z-coordinate is not present. So expressions for velocity and acceleration in the plane polar coordinate system are respectively given by (i) and (iii) with z = 0. - Vector Calculus In the preceding examples you have learnt to express the position vector as well as its first and second time derivatives (velocity and acceleration) in polar coordinates. Now consider a F(t) acting on a body. Suppose it displaces the body through d I. (In mathematical language, d 1 is a differential element of vector length. ) Then, the work done by this force on the body is given by 6W - F. d I. Similarly, thenmagnetic flux coming out of a surface of area dA is given by the relation,@= B . dA, where n is unit normal in the outward direction. You must, therefore, know to express differential elements of vector length, in terms of non-Cartesian polar coordinates. Another important quantity is differential volume element, which is a scalar . You will use these expressions extensively in the next unit. So we would like you to master the basic technique rather than remembering1 these formulae by heart.

3.3.1 Differential Element of Vector Length From Example 4 you would recall that in spherical polar coordinates, the position vector of a point moving through space is given by '

On substituting for $. from Eq. (3.12), we have

To computed r, you would recall that r, 0 and $ will change as r changes. Hence, you can write

dr -. dr ([sine cos+ + isin0 sing + k cos0) + r (fcosQcos+ d0 - ;sin0 sin41 d+ + fsinQ cos+ d$ + icos0 sing d0 - k sin8 dB) In spherical coordinates, small line elements along a, ioand 4 are given by dryrd0 and r sin9 d+, respectively. On collecting coefficients of dr, rd0 and rsin0 d+, this expression becomes

Combining this with Eq. (3.12), we find that differential element of vector length in spherical polar coordinates can be written as dc-~dr+jrd0+&rsin0d0 (3.14)

SAQ 5 Spend 5 min In cylindrical polar coordinates, the position vector of a point moving through space is given by 'LA r - p e,, t,ze, Show that a small change d'r in the position vector can be written as - - dr-6 dp+ipd$+g& (3.15)

where dp, pdg and & denote small in the increasing directions of p, 4 and z, respectively.

3.3.2 Differential Element of Vector Area You now know to express the line element in terms of polar cooq.linates. On re-examining Eq. (3.14) you can say that in spherical coordinates dr, rd0 and r sin0 @ are respectively, analogous to dx, dy, and dz in Cartesian coordinate system. Similarly from Eq. (3.15) you

I can identify that in cylindrical coordinates dp, pd$ and dz are analogousYo ah, dy and dz, respectively. You can use this analogy to obtain expressions for components of differential element of vector area. In Cartesian coordinate system, the area of the face normal to x-axis is given by 82 AX-dydz

I You can use the same definition for non-Cartesian coordjnates as well. In terms of Coordinate Systems cylindrical coordinates, you can write

Similarly

and dt4,=pdpd4 1; On combining Eqs. (3.16a, b, c), you can express the differential element of vector area in - - I cylindrical coordinates as I

You can follow the same procedure and show that differential element of vector area in spherical coordinates can be expressed as

Show that the volume element in cylindrical and spherical polar coordinates can be written Spend J.min as and

These are illustrated in Fig. 3.12.

dV= (r sini3, d4)(rde) (dr) = rZ sin9 drd0 d+

Y

X X (a)

Fig. 3.12 Let us pause for a minute and ask: What have we achieved sb far? In this section you have learnt to express a vector in terms of polar coordinates . In physics, V is a very special vector operator. In Unit 2 you learnt that it can operate on real-valued functions, i.e. scalar fields. ! 1 Veaor Calculus While operatingan a scalar field, it gives us a vector field. For instance, the negative I gradient of electrostatic potential defines electric field and gradient of temperature is a measure of the heat flux vector. So whenever you wish to calculate electric field due to a uniformly charged conducting sphere or a cylindrical transmission cable, you must know to express the gradient operator in terms of polar coordinates. Let us now learn to do so.

3.3.3 Gradient of a Scalar Field To express gradieni of a scalar field f in cylindrical coordinates, let us first write V f in terms of its components :

where g,,, gQand g, are unknown functions and we have to determine them. To this end, let A A n n us compute the dot of ep and V f. Since ep, e9 and e, are orthogonal, we get

n That is, g, is the spatial rate of change off in the direction of unit vector ep. At the point P (p, $,z), gpsignifies the rate of change off in the direction of increasing p (with I$ and z held constant ). Refer to Fig, 3.13a. You will observe that the distance between the polnts (p, $, z) and (p + Ap, 9,z) is Ap. So when these points are close together, you can write

where d denotes partial with respect to p .

Fig. 3.1 3 Tbus, the component of V f in the :, direction is

I Similarly, gQis the component of V f in the 4direction, i.e. g$ = 4. V f.But . V f i denotes the rate of change off in the direction of the unit vector 4,which is equal to

This form of the term in the denominator of this limit arises because the distance between the points (p, I$, z) and (p, (I + A$, z) is p A 4 (Fig 3.13b). Hence

13 (3.20b)

Finally, you can write the component of V f in the,$= direction, g, (= e,A . Vf) as (Fig. 3.13~): 3 (3.20~) gz er a~ On combining Eqs. (3.19) and (3.20a-c), you wili obtain

Vf=- af; +la;+3; ap P pd+ Q az (3.21) By the same method you can compute the exp!ession for the gradient of a scalar in spherical coordinates . To ensure that you have understood the procedure, we would like you to do it 84 and check your answer with that given in Sec. 3.5. iou can similarly work out expressions for divergence, curl, and the . Coordinate Systems However, mathematical steps get involved. Normally we circumvent this inconvenience by working in terms of curvilinear coordinates. This is a general coordinate system and can be specialised to any system of interest. Let us, therefore, now proceed to understand the nature of curvilinear coordinate system.

3.4 CURVILIN COORDINATE SYSTEM

We know that in the Cartesian coordinate system in 3-0 space, the position of a point P is denoted by (x, y, z ). These can be related to other coordinates through a set of functional relations. You now know these relations for cylindrical and spherical coordinates (Eqs. (3.3)

and (34, respectively) . Let us now generalise these relations by introducing three new , quantities u,, u2, u, by writing the functional relationship as

You can invert these relations to express ul, u2, u, in terms of x, y, z:

You will note that the transformations given by Eqs. (3.22) and (3.23) give a one-to-one correspondence between (x, y, z ) and (u;, u,, u3) . That is ,you can label a given point either by the coo;dinates (x,y, z ) or (u,, u2, u3).The values (us u2,%) corresponding to a given point P (x, y, z) are called the curvilinear coordinates of P and Eq. (3.23) defines a curvilinear coordinate system. Geometrically, the transformation'frorn (4y, z ) to (u,, u2, u3) may be viewed as a mapping from the Cartesian space to curvilinear space. Let us choose u, = C, (a constant). Then, from Sec. 3.2 we would recall that

u, (3~.Y, z) = C1 (3.24a) represents a surface, say S, , in u-space, as shown in Fig. 3.14. Similarly, if you choose u2 = CZ and u, = C,,the equations

Fig. 3.14 : Schematic represenhtion of suhc& in cudinear ~oormate.system.

and represent surfaces S, and S3,say. These surfaces are also shown in Ag. 3.14. They intersect at point P whose Cartesian coordinates (+ y, z) can be obtained by solving Eqs. Vector Cak~llus From Sec. 3.2 you would recall that intersection of two surfaces at a point defines a coordinate curve through that point. By referring to Fig. 3.14, you will note that intersection of surfaces S2 and S, defines the u,-coordinate curve. Similarly, the intersection of S3and S, surfaces defines the u,-coordinate curve and so on. You will note that, unlike in the case of Cartesian coordinate system, the ui(i = 1, 2, 3) curves are not straight lines. For this reason, the directions of curvilinear coordinate axes are determined by drawing tangents to the coordinate curve at a point. Moreover, they vary from one point to another point in space, which means that you cannot define one coordinate system for every point in u-space. That is, for curvilinear coordinate system, you have to define unit vectors for each point separately. A vector F may be written in terms of its components F,, F2, F3 as

A A A where el, e,, and e, are respective unit vectors along tangents to (ul, u,, u,) coordinate lines at the given point. So far our discussion of curvilinear coordinate system has been somewhat abstract. To be precise, we should know the orthogonality condition for curvilinear coordinates. We should be able to specify the nature of unit vectors. This involves the calculation of partial derivatives. Let us first understand what a partial derivative is and how to calculate it. Partial Differentiation Let us re-examine Fig. 3.la. Here P (x, y) is a point in 2-0 Cartesian space. Starting from the origin 0, you can reach P along two different routes. You can move along x-axis, i.e. keep y fixed. Then keepingx fixed you can move parallel to y-axis. Alternatively, you can first go along y-axis and then keeping y fixed, move parallel to x-axis. We use a similar technique to compute changes in a which depends on two (or more) variables. From PHE-06 course on Thermodynamics and Statistical Mechanics you may recall many such situations. The temperature of a gas is a function of its volume and pressure. The internal energy of,a paramagnetic salt depends on temperature and magnetic field strength, and so on. In all such cases we express the change in the dependent variable as a sum of two terms. Each term is a product of the rate of change of dependent variable with respect to one variable, keeping the other constant, times the change in that variable. For instance, to calculate the change in temperature of the gas, we first keep volume constant and let pressure vary. Then - dp denotes the product of the rate of change of temperature with (::Iv (::Iv . . pressure at constant Vand the change dp in pressure. Next, we keepp fixed and calculate the change in temperature due to changes in volume and multiply the result with dV to obtain (=IpaT dV. Hence, the net change in temperature is given by

The subscripts Vand p denote that these variables have been.kept fixed. You may now consider a function of three variables X=Xbv, T) By induction, a change in X can be written as

Let us now proceed to discover the orthogonality condition for the curvilinear coordinate system.

3.4.1 Orthogonality Condition Consider a point P (x, y, t ) in Cartesian space. If its position vectoris r, you would recall from Unit 1 that dr= r&+j^dy+~& and square of the distance between two neighbouring points along a curve is given bv Coordinate Systems 1 (ds), = dr . dr = (clx), + (d~)~+ (dz), (3.27) Using Eq. (3.26)you can express clx, dy, dz in terms of u,, u,, u, through the relations

ak- ()- du, + (:tJ- du, + (ii)- du,

and

For converiience, we have dropped the subscripts u,, u,, u,. On substituting for (d~)~,(dy), and (dz), in Eq. (3.27) and collecting the coefficients of dui duj (i, j - 1,2, 3) you can rewrite Eq. (3.27) as

where

are referred to as the metric coefficients of the curvilinear coordinate system. In the following example we have illustrated the calculation of metric coefficients for cylindrical coordinate system.

Example 6 In cylindrical coordinates, u, = p, u, - $, u3 = z and

Compute gii's and the arc length. Solution From Eq. (3.29), we have 1 Vector Calculus i i

Similarly, you will find that gZ3= g,, = 0. A coordinate system is said to be orthogonal if gd= p, Vi # j. In cylindrical coordinates, the expression for square of the arc element is given by

(d~)~= (d~)~ + p2 (d~$)~ + (d~)~ In plane polar coordinates, the third term in this expression would drop out.

SAQ 7

Spend 10 min In spherical polar coordinates, u, = r, u2= 0, u3 = (I . Using Eq. (3.28), show that the expression for square of the arc element is given by I (d~)~= (dr)2 + r2 + r2 sin28 (d@)2 I

In Example 6 you have seen that the condition for a curvilinear coordinate system to be ' orthogonal is

So for an orthogonal curvilinear coordinate system, Eq. (3.28) for square of arc length takes a very compact form:

= (h, d~,)~+ (h2 d~~)~+ (h3d~3)2 (3.31) 2 where we have p~tgij = hi (i = 1, 2,3). This means that for cylindrical polar coordinate system

and h3=h,=l Since you have worked out SAQ 7, you can show that for spherical polar coordinate system

and

h3 a; h$ = r sin0 (3 .Y 3) ktus pause for a minute and ask: What physical meaning can we assign to the coefficients h,, h,, b?To answer this question, we note that when an element of arc ak is directed along the u, coordinate line, du, = dug = 6. Then Eq. (3.31 ) gives

(u!.~)~= (d~,)~= h12 (dU1)' 8 8

d so that koorctinate Systems

That is, the length of the arc ds along the ul- coordinate line is the product of hl and the differential of u,. Similarly, you can show that the components of length of arc ck along the u, and u3 coordinate lines are

ds2 = h, du, (3.34b)

Since a's, and dui are real , you can say that hi (i = 1,2,3) are positive quantities. Sometimes, they are referred to as scale factors. You may now ask: Do the scale factors need to be numbers? From Eqs. (3.32) and (3.33) you will note that these can have any dimensions but the product hidui must have the dimensions of length. As pointed out earlier, to be able to express a vector in curvilinear coordinates we must know the'nature of unit vectors. Do they form an orthogonal triad? Let us now learn to compute these.

3.4.2 Unit Vectors Having discovered the condition of orthogonality for curvilinear coordinate system, you can lo~ically,ask:How to specify the unit vectors associated with an orthogonal curvilinear coordinate system? To discover the answer to this question, let us express the position vector r in terms of u,, u,, u3:

fou will recognise that the symbol -ar denotes the derivative of r with respect to a aui particular variable ui = (i = 1,2,3) . This implies that if we fix u, and u3,r becomes a function of u, alone. That is, the terminus of r will move along the ulcoordinate line in the u-cobrdinate system (Unit 1). So the vector Y1 Ar

Fig. 3.15 U2, Us= MtUL will be tangent to the coordinate line u, at point P. Similarly, we can say that vector ar ar P 1 -and -are, respectively, tangent to u, and u3-coordinate lines, as shown in Fig. 3.15. If au, au3 we denote these vtctors by ai, i.e.

Eq. (3.35) can be rewritten as 3 dr-2 a,dui (3.37) i-1 The vectors ai are called base vectors in the curvilinear coordinate system. Can you draw an L-4 A analogy between the base vectors a1,hz,as and the unit vectors 1, j, k ? The two sets of vectors are analogous in that you can resolve any vector F into its components Fl, Fz,F, , as shown in Fig. 3.15. However a,, a,, a, are not unique; these are defined at a point and vary ,from point to point. Are they unit vectors? To discover the answer to this question, you should compute the dot product dr . dr. Since we are considering an orthogonal coordinate system, I i, I 1 Vector Calculus ; ai.aj=O i#j I i Then, on comparing the coefficients of (d~,)~,(d~,)~ and (duJ2 with Eq. (3.31) , you will get Jail= hi( i = l,2,3) $ 1 so that the base vector ai can be written as

where :i is a unit vector along ui-coordinate line. Using this result in Eq. (3.37) we find that 3

A For a given i, ei can be oxpressed as

Before proceeding further, you must study the following example carefully where we have illbstrakcl the merhdd to evaluate the unit vectors for cylindrid,+d spherical polar coordinat~s.

------Example 7 Starting from Eq. (3.40) evaluate 2;s for cylindrical and spherical polar coordinate systems. Solution The cylindrical polar coordinate system is defined by ul = p, u, - 4, u, = z, h, - 1, h, = p, and h, =1. So from Eq. (3.40) we can write

A A = - i sin+ t j cos+ and

A -k (iii) You will note that (i), (ii), (iii) are identical with Eqs. (3.9a,.b, c). ------The spberica! polar coordinate system is defined by ul- r, u2 = 8, u3 -. + and h, = 1, hZ= r, h3 = r sine. Again referring to Eq. (3.40), we cnn write

A A el mer==--ar == (kinems+t frsine sh+ rcos0) hl dr ar + 2 - ?sine co* t ;sine sin+ t i; case

A A lar la A (i nine cos$ +irsin0 sin$ t k rcose) % - e, = i; = ; = i'cose cos+ + j'co~e sin$ - k sine and

A A 1ar me =--=-- I a (frsinB CO~+ Jtsin0 sin$ + C r cos8) '3 + rsineam

A A = - i sin+ + j cos+ 90 3.43 Vector Differential Operators Coordinate Systems Now that we have introduced the curvilinear coordinate system, we will develop expressions for vector differential operators - the divergence, the curl and the Laplacian. Let us begin with t6e computation of divergence of a vector field.

Divergence I Consider a vector field F. We can express it in terms of its components F,,F2, F3 along A A A el, e;! and e3 as

In summation notation, it can be written in a compact form :

We have preferred to use this notation as it makes expressions look more elegant. The divergence of F is then given by

From Unit 2, you would recall that divergence of the product of a scalar and a vector is the sum of the scalar thes the divergence of the vector and the dot product of the vector with gradient of the scalar. So you can write 3

Thus, to evaluate V . F7you must know V . 2i and V Fi . From Unit 2 you would recall that V Fi is a vectorhaving the sagnitude and direction of the maximum space rate of change of F,. So the component of VF,(u,, u,, u,J in the direction normal to the surface defined by ui = constant is given by

where dsi is a differential length in the direction of increasing ui.This direction is indicated .by the unit vector :p Using Eq. (3,Jqa) we can rewrite it as

1 where h, is scale factor. By repeating Eq.(3.42) for u2 and u3 and adding the result vectorially, we find that

This shows that in curvilinear coordinates, the del operator can be represented as

A Let us now proceed to evaluate V . e,. Since ii'sare orthogonal, we can write

and A A A e, =.el x e, Y i bi , A f Vector Calculus This means that evaluation of V . e, requires calculation of

To evaluate the terms on the left hand side of Eq. (3.45) you would recall from Unit 2 that if s v . (AXB) =B .(vXA) Y is a scalar, then -A.(pXB) Vx (VY) =o Let us take Y = ui.Then, we have vx(vui)=o Using Eq. (3.44), we can write

so that A

Using the iderltity V x (aA) = a(V x A) -A x V a, where a is a scalar, you can write

so that

Hence

On combining this result with Eq. (3.49, you will get

AAA Using the orthogonality condition of unit vectors el,e2, e,, you can simplify this expression

You may note that though g2 is a unit vector, its curl does not necessarily vanish.

SAQ 8 Spend 5 min Starting from Eq. (3.46) prove that

a (3.47b) vi;3-~(l--k~)h3h3 hZ au2 hl aul

On inserting these results in Eq. (3.49, you would obtain A va$' -&,(%L~L)~ h2 h, au, h, au, $(k~-+d)~,hg h2 a3 I au,

A A A A A A A A Sincee,. e,=O=e,. eland e2.&,=l=e,. e,,tbexp&onfor?. e,takesavery 92 compact form :

1 i i. b6%3" i Cqordinate Systems

We can generalise this result as

T~Bresult shows that like curl :,, div tiis also positive definite . On substituting for V Fifrom Eq. ( 3.43) and V . cifrom Eq. (3.48) in Eq. (3.41), we get

SAQ 9 Show that in Cartesian coordinates, cylindrical coordinates and spherical polar coordinates, Spend I0 min divergence of a vector can be expressed as

a and V.F-- [sine p (rZF.)l r - (sin0 F,)+ r r2 sine ae "a4I

Example 7 A pariicle is moving in space. Show that divergence of its position vector is invariant under coordinate transformation. Solution To show this, let us compute!. r:or Gartesian cylindrical and spherical polar coordinates. In Cartesian coordinates,'r - i x + j y + kr so that

In cylindrical coordinates A A r=pe +ze, P and

In this case r = p, r+- 0 and r, - z. Hence, 1 Vector Calculus

Sirnilarl4, in spherical polar coordinates A r=re,

and

Here, r, = r, r, = r+ = 0. Hence

1 = -[3r 2 sine] = 3 9 sme Since the vdue of V . r comes out to be the same in all coordinate systems, we say that it is 'invariant.

Following the procedure outlined in arriving at Eq. (3.49), we would like you to show that

I

3 A A Hint : Write V x F = V x (V x 4) - ei (vF,)]and substitute

A for V x e,md VFi. In cylindrical and spherical polar coordinates, Eq. (3.50) takes the form

and

I Example 8 The magnetic potential of a single current loop in thexy-plane is given by 1 v = v x [V x {A+ (r, 8)) Express it in spherical polar coordinates. - -

Solution Coordinate Spiems

A A er re0 nine :+ v-VX- 1 -aa -a 9sme dr d8 a4 0 0 rsin0A4(r,8)

Taking the curl again, we obtain

n A er reg A -a -a rsinee+ v = ----- dr -a 9sine ae 1 a 84 --a (?-&€kt,) (rsin8AJ 0 r%ine a8 mine ar By expanding the determinant, we have la 1 1 a 1 a rA ) +-- --a - G ( + ] 9 80 [sine (sine~~]} I I; L J L JJ Laplace Operator Like the del operator, the Lap1,ace operator also finds applications in fluid mechanics, electromagnetism, elasticity, propagation of waves and quantum mechanics. It is therefore important to give expressions for V* in curvilinear, spherical and cylindrical coordinates. You will use these in the next unit as well as in Block 2 of FHE-05 course. We know that

where f is a scalar. Using Eq. (3.43), you can write

Inserting the result contained in Eq. (3.49), this expression becomes

I I

In the expanded form, you can write (3.52)

SAQ 10 Express Eq. (3.52) in cylindrical and spherical polar coordinates. Spend 10 rnin

Let us now summarise the contents of this unit.

3.5 SUMMARY

@ The cylindrical coordinates (p, $, z), of a point P in space are elated to its Cartesian' coordinates (x, y, z )by x-pcos$, y=psin$, Z=Z The plane polar coordinates of the projection of P on to thexy plane are p and $. @ The spherical polar coordinates (r, 0, $) of a point P are related to its Cartesian coordinates (x, y, z)by x =rsin€Icos$ y - r sin0 sin$ 9 z = r cos0 Vector Calculus O The generalised curvilinear coordinates are related to Cartesian coordinates through the relation Ui = u,'(x, y, z) O The square of length of an arc and differential element of vector area in generalized curvilinear coordinates are given by (03)~= h! (d~,)~+ h: (duJ2 + h: (duJ2 A A ~A=IT?h3du2du3 e^, + h,h, du, du, e2 + /~,h~du,du~e, @ In curvilinear coordinates, the grad, div, curl and the Laplacian operator can be expressed as

I I and 3 1

i#j#k A AA. 0 @ For Cartesian coordinates, u,= x, u2 = y and u3 = Z, hl h2 = h, = I, e, = i, e2 =J, h A and e, = k. @ For cylindrical coordinates, ul = p, u2 = $, u:, =z, hi = hp = I, h2 = h$ = p, h, =.h, = 1, AAAA A A el= e, , e,= eb and e3 = e, @ For spherical polar coordinates, u, = r, u2 = 0, u, = $, h, = h, = 1, h2 = h, = r, h, = h$ rsin0 AAAA A A el=e,, e,=qand e3=e,+. @ In cylindrical coordinates

8 In spherical ~01a;coordinates

A h A e, reH r sin 13 eg

V F = ------a r2 sin9 ar a0 d+ Coora~nateSystems 1 3.6 TEMINAL QUESTIONS

1. Show that

and Fig. j.I 6

d e^$, + A #.A -= - sin6 4 e, - cose 4 ee dt 2 4 p = constant 2. A particle is moving through space. Express its acceleration in spherical polar coordinates. 3. Verify Eq. (3.50) 1 A rigid body is rotating about a fixed axis with a qonstant w Take w I 4. . I to be along the z-axis. Using spherical polar ~oot'din~tes,calculate i) v = w x r * )------and ii) V x v. Y --.------5. A central force fiela is given by

A cos8 2 ro A rO F = e, + eg - sine r3 r3 Fig. I3.17 Calculate V x F.

3.7 SOLUTIONS AND ANSWERS P = constant -

Self Assessment Questions Z= maetaat (plane) 1. Refer to Fig. 3.16. You will note that x - 1 sin4 and 1 y=lcos$--at2 2 tJSy4 = copstrgt 2. The &-surfaces for the cylindrical coordinate system are coaxial right circular (halgplane) cylinders having the z-axis for their common axis. This is shown in the figure 3.17 F~Q.3. 18

4. The solid is a conical section of a sphere. It is illustrated in Fig. 3.19,

A 5. r-peptzi Insert the value of lp. This gives

r = p (fcos$ + j" sin$) + & z A small change in r gives rise to changes in p; $, and z. So we can write fl dr - dp (;cos$ +$sin@) + p (- ;sin$ t ices$) d$ + 2 & A =epdp+Pd$$+$& s Y 6. In Cartesian coordinates, an element of volume is defined as a Fig. 3.19 We use the same definition in non-Cartesian coordinates. 1 Vector Calculus In cylindrical coordinates

= (dp) (pd4) dz = P dpd$ dz In spherical polar coordinates

= (dr) (rd0) (!sin0 d$)

7. From Eq. (3.29) we know that metric coefficients are given by

ax ax ay ay at. a +--+--at- .a GK aul au2 a~,a~,

ax ax a~ at.

--+- ** +-- ar a0 + ar ae ar ae = (sin0 cost$)(r cos0 cost$)+ (sin0 sin$) (r cos0 sin$) + cos0 (- r sine) = r sin0 cos0 - r sin0 cos0

Similarly, you can show that g13 = 823 '0. Hence, I d r = g1, + g22 (du2l2 + g33 (du3I2

I n gll (dd2.+g22 (a2+ 933 (d$I2 i 2 = (dr)2t r2 (dQ2 + r sin20 (d$)2 I 8. We know that i t , Vxe3--- 1~ e3xVh3 P I 98 h3 ! I i J Substituting for V h,, we find that Coordinate Systems

9.' In curvilinear coordinates

In Cartesian coordinates, h, a h, = h3 = 1, ul = x, u2 = y, u3 a z, F1= F,, F2 = Fy and F3 - F,. Hence,

In cylindrical coordinates, hl = hp= 1, h2 = hg = p, h3 = h, = 1

u1 = p, u2 = $, U, = Z, F1 F,,, F2 = F4 and F3 Fr Ilence

In spherical coordinates, hl - hr = 1, h2 - ho = r, h3 - 1% = rsin0,

u1 = r, u2 0, u3 = @, F1 = Fr , F2 = F,, and F3 = F4 Hence,

For cylindrical coordinates hl = 1, h2 = p, 15 P. l,ul = p, u2 - Q, and u3 z. Hence

Similarly, for spherical coordinates ul= r, u2 = 0, ug = $, hl - 1, h2 - r and h3 - r sine. Hence, xta12 v2f=&1.2 sine [& dr sine 3+ $ ae) 8) (sin8 a+)] I

Terminal Questions 1. We know that 4- mei + sin+] 6) and Ine4=-sin+ftc~f (ii) Vector Calculus Then A

- and,,

de^ A=[-ws+i'-sin4$6=-;+&dt (iii)

(iv)

and A , ;+ = - sin+ i + cos$ j Hence

n A A 9dt = - sine cos+ ;8 - cos8 sin 4 f$- sin0 sin+ j 6 + case corm j 6 - cose C Q

.A .

and

Multiply (iii) by. sin0 and (iv) by cos0. You will get $ sine - sin2@cob+ ;+ sin28 sinti+ cod sin0 C

On adding theseA we find Athat A sin0 ;r + as8ee - case i + sin4 j Hence

'2. We know that A -A . A v=rer+r9e0+rsh9+e+ Differentiate with respect to time. The result is

. , d A .# A i=a=re,+r--e,+r0e~+r8;~dt

+ r as04 9 f + r sintli$ &, h A In terminal question 1 you have evaluated c,, ee, and e+.Inserting these results, you will get a=i:;+i(8i0+sin0($+ib&+r8& +isin~/i+r6(-Q;+6cos0i4) Coordinate Systems

A A A On collecting the coefficients of e,, eo and e@you will get

Hence a, = j: - r (6j2 - r sin28 (4)2

---- I (r2 e)- r sine mse (412 r dt and a4=tsin0$+2r8d,cose+2i.+sin0

=Vx(Fl il)+vx(F2 {)+vx (F3g3) Now A A VX(F~~~)-F~(V~~~)-~~xVF1

Similarly

A A A

Hence

We can write analogous expressions for other two terms:

A a

and A A A Vector Calculus On adding these results and collecting the coefficients of, el,e2 and eg we get

A

This can be written as

Ih,fl hZG h3231 .

4. i) In spherical polar coordinates

and

Hence

A AA = iio r (cosO e, - sin0 e,) x e,

AA =-ro sine (e, xe,)

= w r sine %

Hence

VxF-- 1 .- a- -a -a r2 sine ,dr &I a4 2 ro cos8 r, sine 0 . ? -P

3ru ; 2r0 -&r sin IrSw $(-pSln~+-sin~ P I . -eq7(2k3:)sine

102 UNIT 4 INTEG ION OF SCA AND VECTOR FIELDS

Structure 4.1 htroduction Objectives 4.2 Integration of a Vector with Respect to a Scalar Integrals Involving Scalar and Vector Products of Vectors 4.3 Multiple Integrals Double Integral Triple Integrals 4.4 Line Integral of a Field Path of Integration Types of Line Integrals Evaluation of Line Integrals 4.5 Surface Integral of a Field Surface of Integration Types of Surface Integrals Evaluation of Surface Integrals 4.6 Volume Integral of a Field 4.7 Vector Integral Theorems Gauss' Divergence Theorem Stokes' Theorem Green's Theorem 4.8 Summary 4.9 Terminal Questions 4.10 Solutions and Answers

4.31 INTRODUCTION

In the previous unit you have studied about different coordinate systems. In the process you have learnt to represent the gradient of a scalar field and the divergence and curl of a vector field in several coordinate systems. In Unit 2 you have studied about differentiation of vectors. In this unit you will learn about integration of scalar and vector fields. These integrals find many applications in physics. For example, the work done by a force or the magnetic field due to a current-carrying conductor can be expressed as a line integral. The flux of a magnetic field can be represented as a surface integral. Line, surface and volume integrals find application in determining the potential due to a continuous distribution of matter or charges. We shall discuss some of these examples here. These integrals form the foundations of many important equations in physics like the Maxwell's equation of Electromagnetism, the equation of continuity and so on. In this unit, first we shall discuss the integration of a vector with respect to a scalar. You will find that this will basically be an extension of the idea'of ordinary integral. This finds application in determining the trajectory of a particle if its equation of motion is known. We shall also discuss the integration of scalar and vector products of vectors with respect to a scalar. Next we shall discuss the integration of scalar and vector fields with respect to coordinates. As you, know a scalar or a vector field may be a function of one or more coordinates. Their integrals which we shall come across will necessarily not be in terms of single variable, So we shall learn about double and triple integrals. You will see that the three kinds of field integrals, i.e. the line, surface and volume integrals are respectivclylhe exfensions of ordinary, double and triple integrals. Finally, we shall discuss how one kind of a field integral can be transformed into another type. In the process you will learn to apply the vector integral theorems - namely the Gauss' divergence theorem, Stokes' theorem and Green's theorem. Here, we shall only state these theorems without proof. However, if you are interested in knowing the proofs you may go through the Appendix. These theorems provide us with very elegant methods for arriving at several fundamental equations of physics. In the next block, we shall take up Statistics and Probability and their applications in physics.