1 Vector alculus 1.1 ALAR A scalar is a quantity that has only magnitude, such as distance, mass, temperature or time. The value of a scalar is an ordinary number. Operations with scalars obey the rules of elementary algebra. The order of compounding is immaterial. 1. VETOR A vector is a quantity that has both magnitude and direction, such as displacement, force, velocity or momentum. A vector is represented either by a bold-faced letter such as P or by an arrow over the head of a letter such as P Æ. Graphically, a vector B is represented by the directed line segment AB as in ig. 1.1. The vector P Æ has a direction from A to B. The point A is called the initial P point (tail of the arrow) and the point B is called the terminal point ææ of P Æ (head of the arrow). The length AB of the line segment is the magnitude of P. The magnitude of P is denoted either by P or P P > 0 for any P π 0 P = 0 if and only if P = 0. t is called a Null Vector. Vectors are compounded geometrically. The order of compounding is immaterial. A ig. 1.1 1.3 EQUVALENE O VETOR P Q Two vectors P and Q are equal if they have the same magnitude and direction regardless of the position of their initial points. Thus, in ig. 1., P = Q. ommutative property: f A = B, then B = A Transitive property: f A = B and B =, then A =. ig. 1.
Mechanics of Particles, Waves & Oscillations 1.4 MULTPLATON O VETOR BY ALAR 1.4.1 calar Multiplication Let A be any vector and m any scalar. Then the vector ma as in ig. 1.3 is defined as follows: (a) The magnitude of ma is m A ; that is ma = m A (b) f m > 0, the direction of ma is that of A. (c) f m < 0, the direction of ma is opposite to that of A. ig. 1.3 (d) m (na) = (mn) A for any scalars m, n. Two non-ero vectors A and B are parallel if and only if there exists a scalar m such that B = ma Thus, the result of multiplying a given vector by a scalar is a vector parallel to the given vector. 1.4. Result of calar Multiplication (a) The ero vector if m = 0, we obtain the ero vector 0; i.e. (0) A = 0, where A is any vector. The ero vector 0 is a vector that has magnitude 0 and has any direction. (b) The negative of a vector f m = 1, we obtain the negative of vector A, indicated by A; that is, ( 1) A = A. Thus, a negative vector A is a vector with magnitude of A, but whose direction is opposite of that of A. (c) The unit vector f A π 0 and m = 1/ A = 1/A, the unit vector, e A = 1 A A. Thus, the vector e A is a vector whose magnitude is e A = 1, with the direction same as that of A. The vector A may be represented by the product of its magnitude and the unit vector e A ; that is; A = Ae A 1.5 ADDTON O VETOR (TRANGLE LAW) The sum or resultant of two vectors A and B is a vector which can be determined geometrically, ig. 1.4. f the tail of B is placed at the head of A, the resultant is the vector whose tail is at the tail of A and whose head is at the head of B. This geometric method of vector addition is known as the triangle law. A ig. 1.4 B
Vector alculus 3 1.6 UBTRATON O VETOR f A and B are two vectors, the difference (A B) is the sum, of A and ( B); that is: = A B = A + ( B) The vector subtraction is shown geometrically in ig. 1.5. 1.7 PROPERTE O VETOR ADDTON A + B = B + A (ommutative law) A + (B + ) = (A + B) + (Associative law) m(a + B) = ma + mb (Distributive law) (m + n) A = ma + na (calar distributive law) A + 0 = A A A = 0 1.8 RETANGULAR UNT VETOR i, j, k ig. 1.5 onsider a right-handed, rectangular co-ordinate system. The unit vectors i, j, k are shown in the directions of x, y and axes respectively, ig. 1.6. A vector A in three dimensions, can be represented by its initial point at the origin 0 and rectangular components (A 1, A, A 3 ) for its terminal point, ig. 1.7. ince the resultant of A î + A 1 ĵ + A 3 k, is the vector A, A = A î + A 1 ĵ + A 3 k, The magnitude of A is A = A = A1 + A + A3 Let A = a x i + a y j + a k and B = b x i + b y j + b k be two vectors where a x, a y, a are the x, y and components respectively, and b x, b y, b, are the corresponding components of B. Then the resultant of A and B is given by: x x y y R = ( a + b ) + ( a + b ) + ( a + ig. 1.6 ig. 1.7 The three unit vectors may be written as: i = (1, 0, 0); j = (0, 1, 0); k = (0, 0, 1). The Position Vector or Radius Vector r from O to the point (x, y, ) is written as r = xî + y ĵ + k and has magnitude r = r = x + y + x x ^ 1 A i ^ i ^ k ^ j 0 y 0 y ^ A j A ^ 3 A k
4 Mechanics of Particles, Waves & Oscillations Example 1 orces A and B expressed by the equation A = A 1 i + A j + A 3 k and B = B 1 i + B j + B 3 k, act on an object. ind the magnitude of the resultant of these forces. Resultant force R = A + B =(A 1 + B 1 )i + (A + B )j + (A 3 + B 3 )k R = ( A + B ) + ( A + B ) + ( A 1 1 3 The result can be extended to any number of forces. Example Determine the vector having initial points P (x 1, y 1, 1 ) and the terminal points Q(x, y, ) and find its magnitude. The position vector of P is r 1 = x 1 î + y 1 ĵ + 1 k The position vector of Q is r = x î + y ĵ + k n ig. 1.8, r 1 + R = r or R = r r 1 = (x î + y ĵ + k ) (x1 î + y 1 ĵ + 1 k ) = (x x 1 )î + (y y 1 ) ĵ + ( 1 ) k ig. 1.8 R = ( x - x ) + ( y - y ) + ( - 1 1 1.9 ALAR OR DOT PRODUT The calar or Dot Product, also called the nner Product, of two vectors A and B is denoted by A.B and is defined as a scalar given by, A. B = A B cos q = AB cos q (1) where q is the acute angle between A and B. We may regard the scalar product of two vectors as the product of the magnitude of one vector and the component of A the other vector in the direction of the first ig. 1.9. As the scalar product A.B is represented by a dot q between two vectors, it is called the dot product. rom B the definition of the scalar product (1), the angle A cos q between A and B is found from the formula cos q = A.B ig. 1.9 (1a) AB provided A π 0, and B π 0. f the angle q between two vectors is a right angle then the two vectors are said to be perpendicular or orthogonal and the condition for the vectors A and B to be orthogonal (A π 0, and B π 0) is seen from () to be A. B = 0; ondition for orthogonality ()
Vector alculus 5 1.9.1 Properties of the calar Product A. B = B. A (ommutative law) A. (B + ) = A. B + A. (Distributive law) (ma). B = A. (mb) = m(a. B) A. A = A = A ; for every A. A. A = 0 ; if and only if A = 0, where m is an arbitrary scalar. î. î = ĵ. ĵ = k. k = 1 (3) î. ĵ = ĵ. k = k. î = 0 f A = A 1 î + A ĵ + A 3 k and B = B1 î + B ĵ + B 3 k then A. B = A 1 B 1 + A B + A 3 B 3 1 A. A = A = A + A + A 1 B. B = B = B + B + B 1.10 VETOR OR RO PRODUT 3 3 The vector product of the vector A and B, written as A B, is another vector: = A B (4) The magnitude of is given by = AB sinq (5) A B y q B x A where q is the acute angle between A and B. The direction of is that of the advance of a right-hand screw as A rotates towards B through the angle q ig. 1.10. B A ig. 1.10 1.10.1 Geometric nterpretation ig. 1.11 shows the parallelogram completed from the vectors A and B. Now, the height of the parallelogram B sinq = h. Thus, by (5), the magnitude of the vector product = A B, represents the area of the parallelogram whose B q h = B sin q sides are A and B. A 1.10. Properties of the Vector Product ig. 1.11 (i) A B = B A (Anticommutative law) (6) (ii) A (B + ) = A B + A (Distributive law) (ma) B = A (mb) = m A B (iii) A A = 0 U V (iv) A 0 = 0 for any A (7) W
6 Mechanics of Particles, Waves & Oscillations (v) î î = ĵ ĵ = k k = 0 î ĵ = k, ĵ k = î, k ĵ î = î ĵ, k ĵ = ĵ k, î k = k î î = ĵ, (8) (vi) f A = A 1 î + A ĵ + A 3 k and B = B1 î + B ĵ + B 3 k then, U V W A B = i j k A1 A A3 B B B 1 3 1.11 ALAR TRPLE PRODUT The scalar triple product of the three vectors A, B and is a scalar A. (B ) or simply, calar Triple Product = A. (B ) (9) Observe that the expression A (B. ) is meaningless since it implies the vector product of a scalar and a vector. f A = A î + A 1 ĵ + A k 3 and B = B î + B 1 ĵ + B k 3 then, = 1 î + ĵ + 3 k A. (B ) = A A A 1 3 B B B 1 3 1 3 The scalar triple product A. (B ) is also written as [AB]. 1.11.1 Geometric nterpretation of A. B ig. 1.1 shows a parallelepiped whose sides are A, B and. By (4), the vector product of B and is equal to the area for the parallelogram with adjacent sides B and. = B f h is the altitude of the parallelopiped then, h = A cos q where q is the angle between A and B. Therefore, the volume of the parallelopiped is V = h = A B cos q = A. B ig. 1.1 1.11. ondition for oplanarity Vector A, B and are coplanar if and only if A. B = 0; ondition for coplanarity (10) This is understandable since h = 0 in the event of coplanarity and so volume V = 0. The converse of (10) is also true.
Vector alculus 7 Example 3 f any two vectors in a scalar triple product are equal, show that the product is ero; that is, A. A = 0 ;. B = 0; A. B B = 0 The vector product A = P is perpendicular to A. Hence, A. P = 0 by (3), that is, A. A = 0. Also, since B is perpendicular to, their scalar product is ero, that is,. B = 0. Again, B B = 0 by (7). Hence A. 0 = 0. 1.11.3 undamental identify for the calar Triple Product A. B = A B. (11) This implies that in the scalar triple product the position of the dot and cross is immaterial. 1.1 VETOR TRPLE PRODUT The vector triple product of three vectors A, B and is the vector D given by: D = A (B ) (1) Here a parenthesis is essential since A B depends on the result of the product, whether we form A B first or B first. Thus, A (B ) π (A B) 1.1.1 undamental dentities for the Vector Triple Product A (B ) = (A. ) B (A. B) (13) (A B) = (A. ) B (B. ) A (14) Example 4 Evaluate (a) î. î ; (b) î. k (a) î. î = î î cos 0 = (1) (1) (1) = 1 (b) î. k = î k cos 90 = (1) (1) (0) = 0 Example 5 f A = A î + A 1 ĵ + A k 3 and B = B î + B 1 ĵ + B k 3 prove that A. B = A 1 B 1 + A B + A 3 B 3. A. B =(A î + A 1 ĵ + A k 3 ). (B î + B 1 ĵ + B k 3 ) = A 1 B î. î + A 1 1 B î. ĵ + A 1 B î. k 3 + A B 1 ĵ. î + A B ĵ. ĵ where we have used (3). + A B 3 ĵ. k + A 3 B 1 k. î + A 3 B k. ĵ + A 3 B 3 k. k = A 1 B 1 + A B + A 3 B 3
8 Mechanics of Particles, Waves & Oscillations Example 6 A man proceeds from his house 3 Km. due north. He then turns in the north-east direction and goes further far 4 km. How far is he from his house? What is the direction of the terminal point? a x =0 ; a y = 3 ; b x = 4 sin45 ; b y = 4 cos 45 x x y y r = ( a + b ) + ( a + b ) e j e j = 0 + 4 + 3+ 4 = 6.478 km. tan q = r y /r x = a y + a + b x b y = 3 + 4 =.06. 4 q = 64.1, north of east. x b y A a = y 3 km O 45 q 4 km b x B D Example 7 ig. 1.13 how that (a) î ĵ = k; (b) ĵ ĵ = 0 i j k (a) î ĵ = 1 0 0 0 1 0 i j k (b) ĵ ĵ = 0 1 0 0 1 0 = k = 0 Example 8 f  = x 1 î + y 1 ĵ + 1 k and B = x î + y ĵ + k, prove that : A B = i j k x1 y1 1 x y A B =(x 1 î + y 1 ĵ + k 1 ) (x î + y ĵ + k ) = x 1 x î î + x y î ĵ + x 1 î k + y 1 x ĵ î + y 1 y ĵ ĵ + y 1 ĵ k + 1 x k î + 1 y k ĵ + 1 k k = x 1 y k x 1 ĵ y 1 x k + y 1 î + 1 x ĵ 1 y î =(y 1 1 y ) î + ( 1 x x 1 ) ĵ + (x 1 y y 1 x ) k
Vector alculus 9 = i j k x1 y1 1 x y where we have used (8). Example 9 Prove that A (B ) = (A. ) B (A. B) A (B ) = A i j k B1 B B3 1 3 =(A î + A 1 ĵ + A k 3 ) [î (B 3 B 3 ) ĵ (B 1 3 B 3 1 ) + k (B 1 B 1 )] = k A 1 (B 1 3 B 3 1 ) ĵ A 1 (B 1 B 1 ) k A (B 3 B 3 ) + î A (B 1 B 1 ) + ĵ A 3 (B 3 B 3 ) + î A 3 (B 1 3 B 3 1 ) where we have used (8). \ A (B ) = î [B 1 (A + A 3 3 ) 1 (A B + A 3 B 3 )] + ĵ [B (A 1 1 + A 3 3 ) (A 1 B 1 + A 3 B 3 )] + k [B 3 (A 1 1 + A 3 ) 3 (A 1 B 1 + A B )] = î [B 1 (A 1 1 + A + A 3 3 ) 1 (A 1 B 1 + A B + A 3 B 3 )] + ĵ [B (A 1 1 + A + A 3 3 ) (A 1 B 1 + A B + A 3 B 3 )] + k [B 3 (A 1 1 + A + A 3 3 ) 3 (A 1 B 1 + A B + A 3 B 3 )] =(A 1 1 + A + A 3 3 ) (î B 1 + ĵ B + k B 3 ) (A 1 B 1 + A B + A 3 B 3 ) (î 1 + ĵ + k 3 ) (A. ) B (A. B) Example 10 Verify the Jacobi identity: A (B ) + B ( A) + (A B) = 0 Using the fundamental identity (13), A (B ) = (A. ) B (A. B) B ( A) = (B. A) (B. ) A (A B) = (. B) A (. A) B Add these identities to obtain the Jacobi identity. Example 11 Prove that (A B) ( D) = [A B D] [A B ] D
10 Mechanics of Particles, Waves & Oscillations Let A B =, then by (13), (A B) ( D) = ( D) = (. D) (. ) D = (A B. D) (A B. ) D = [A B D] [A B ] D Example 1 ind the angle between the vectors A = i + j k and B = 4 i + j +4 k A = ( ) + ( ) + (- 1) = 3 ; B = ( 4) + ( ) + ( 4) = 6 A. B = ( i + j k). (4 i + j + 4 k) = () (4) + () () + ( 1) (4) =8. cosq = A.B AB = 8 = 0.4444 ; q = 63.6. ( 3) ( 6) Example 13 ind the angles which the vector A = î ĵ + k makes with the co-ordinate axes. Let a, b, g be the angles which A makes with the positive x, y, axes respectively. A. î =(A) (1) cos a = ( ) + (- ) + ( 1) cos a = 3 cos a. A. î = ( î ĵ + k ). î = î. î ĵ. î + k. î = cos a = /3 = 0.6667 and a = 48.. imilarly, cos b = 0.6667 and b = 131.8, and cos g = 0.3333 and g = 70.5. The osines of a, b, g are called the direction cosines of A. Example 14 how that cos a + cos b + cos g = 1, where cos a, cos b and cos g are the direction cosines of a vector. Let A = A î + A 1 ĵ + A k 3. The direction cosines are given by cos a = A 1 /A, cos b = A /A, cos g = A 3 /A. cos a + cos b + cos g = A A A + + A A A = A A 1 + + A A 1 3 3 = A A = 1. 1.13 APPLATON O VETOR MULTPLATON 1.13.1 calar Product n general, when the force and displacement are not parallel, then the quantity of work
Vector alculus 11 is given by the component of the force parallel to the displacement, multiplied by the displacement: W =( cos q) d =. d. q 1.13. Vector Product d (a) Torque: n general, the torque (or moment of a force about O actually about an axis through O perpendicular to the paper) is defined as the magnitude of the force times its arm which is the perpendicular distance between the axis of rotation and the line of action of force; that is t = r sin q = r. Thus r represents the torque of about an axis through O and perpendicular to the plane of the paper. (b) Angular Velocity: Let a particle P move in a circular orbit about an axis through the point O with angular velocity w, ig. 1.16. ince the radius of the circle is r sin q, the magnitude of the linear velocity v is w (r sin q) = w r. Also, v must be perpendicular to both w and r. The quantities w, r and v form a right-handed coordinate system. We thus have the vector relation v = w r. (c) Angular Momentum: With reference to ig. 1.16 Angular momentum (J) is defined by the vector produced J = r p, where p the linear momentum is in the direction of v. The magnitude of J is given by J = rp sin q. r sin q 0 ig. 1.14 r q ig. 1.15 w r sin q P q r 0 ig. 1.16 p v 1.13.3 Triple calar Product onsider the vector r and the force lying in the xy plane, as in ig. 1.17. Then the torque of the force about -axis (which is perpendicular to the xy plane) is given by the vector product r. This special case was actually considered under 1.13.. However, if the axis L is inclined to the axis then the result must be modified. onsider the axis L which is inclined to the -axis. Take a unit vector n along the axis L. Then the torque about L is given by the Triple calar product, n.(r ), since it gives the component along the axis L. L n x 0 r y ig. 1.17 1.13.4 Triple Vector Product (a) Angular Momentum: Let a particle of mass m be at rest on a rotating rigid body such as the earth, ig. 1.18. Then the angular momentum of m about point 0
1 Mechanics of Particles, Waves & Oscillations is defined as, J = r (mv) = mr v. But since v = w r, we find J = mr (w v) (b) entripetal Acceleration: The centripetal acceleration of m in ig. 1.18 is a = w (w v). or the special case when r is perpendicular to w, the above result reduces to the familiar formula, a = w r so that the acceleration is towards the centre of the circle and of magnitude w r. w r m v 1.14 DERENTATON O VETOR Let R(u) be a vector which is a function of a single scalar variable u. Then, 0 ig. 1.18 D R D u = R( u + D u) - R( u) DR = R( u + Du ) R Du where D u denotes an increment in u, as in ig. 1.19. The derivation of the vector R (u) with respect to the scalar u is given by: O dr D R = Lim du Du Æ 0 D u = Lim R( u+ Du) - R( u) ig. 1.19 Du Æ 0 Du provided the limit exists. ince dr/du is itself a vector depending on u, if its derivative with respect to u exists, then it is denoted by d R/du. imilarly, higher order derivatives may be defined. f r = xi + yj + k, is the position vector of a moving particle P (x, y, ) in space, then, dr = dx î + dy ĵ + d k (15) and the velocity is v = d r dx î dy ĵ d k = + + (16) dt dt dt dt and the acceleration is a = d r d x i d y j d k = + + dt dt dt dt Here we have assumed that the unit vectors, i, j and k, remain fixed in space. Example 15 Using unit vectors show that the acceleration of a particle p moving on a circle or radius r with constant angular velocity dq/dt is given by a = w r. Referring to ig. 1.0, r = r cosq î + r sinq ĵ therefore, v = dr dr =. dt dq dq dt D R ( u + u ) R ( u)
Vector alculus 13 and therefore, =( r sinq î + r cosq ĵ) d q dt a = d v d r = dt dt K J = ( r cosq î r sinq ĵ) d q dt a = w r y v r q P x 1.14.1 Differentiation ormulae (i) d du (A + B) = d A db + du du ig. 1.0 (18) (ii) (iii) (iv) (v) d du (A. B) = A. d B da +. B (19) du du d du (A B) = A d B da + B (0) du du d du (fa) = f da df + A du du d du (A. B ) = A. B d db da + A. +. B () du du du d (vi) du {A (B )} = A B d du + A db du The order in the above products may be important. K J K J + d du (1) A (B ) (3) Example 16 A particle moves in a circle of radius r at constant speed v. Obtain the formula of centripetal accleration using the fact that r = r. r = onstant, and v = v. v = onstant. Differentiating the given equation with respect to time, r. ṙ =0 or r. v = 0 (i) v. v =0 or v. a = 0 (ii) Differentiating (i), r. a + v. v =0 or r. a = v (iii) By (i), r is perpendicular to v and by (ii), a is perpendicular to v. t follows that a and r are either parallel or antiparallel since the motion is in a plane. The angle q between a and r is either 0 or 180. Using the definition of scalar product in (iii), r. a = r a cosq = v (iv) ince cosq is negative, q = 180. By (iv), r a ( 1) = v or a = v /r.
14 Mechanics of Particles, Waves & Oscillations Example 17 f A and B are differentiable functions of a scalar u, prove that d du (A. B) = A. d B da +. B du du Let A = A 1 i + A j + A 3 k and B = B 1 i + B j + B 3 k. d Then du (A. B) = d du (A 1 B 1 + A B + A 3 B 3 ) = A db 1 A db 1 + + A du du 3 db 3 du da1 + du B da du B da3 1 + + du B 3 = A. d B da +. B du du Example 18 f A and B are differentiable functions of a scalar u, prove that d du (A B) = A d B du da + B du K J K J d du (A B) = d du i j k A A A 1 3 B B B 1 3 = i j k A A A db du 1 3 db du db du 1 3 + i j k da du da du da du B B B 1 3 1 3 = A d B da + B du du where we have used a theorem of differentiation of determinants 1.14. Rules for Partial Differentiation of Vector unctions f A and B are differentiable vector functions of u, v and w, and f is a differentiable scalar function of u, v and w, then using u as an example, u (A + B) = A B + u u (4) f u ( A ) = f A f + A (5) u u
Vector alculus 15 u (A. B) = A. B A +. B (6) u u u (A B) = A B A + B (7) u u Observe that the order of factor in (7) is important. 1.15 UNT VETOR N PLANE POLAR O-ORDNATE o far we have expressed vectors in terms of their rectangular components using the unit vectors i, j and k. n many situations it is convenient to use other co-ordinate systems such as polar co-ordinates in two dimensions and spherical or cylindrical co-ordinates in three dimensions. Here we shall be concerned with the plane polar co-ordinates as in ig. 1.1. The point P has the co-ordinates r and q. The co-ordinate r is the radial distance of P from the origin O and the angle q is measured by the line OP with x-axis. The unit vector (that is, a vector of unit length) e r is along the line q = onstant, in the direction of increasing r (r-direction). The other unit vector e q is along the circle r = onstant, in the direction of increasing q(q direction) and tangential to the circle. These two unit vectors are mutually perpendicular to each other and they continuously change their orientation, although maintaining constant magnitude as the point P moves, unlike the rectangular unit vectors i and j which have fixed orientation in space. We can express the given vector in terms of its components in the direction e r and e q simply by finding its projection in these directions. But since their directions change from point to point, the derivatives of a vector in polar co-ordinates are obtained by differentiating the unit vectors as well as the components. This is in contrast with the rectangular unit vectors where we differentiate the components only. ig. 1.1 Example 19 Let (r, q) be the polar co-ordinates describing the position of a particle. f e r is a unit vector in the direction of the positive vector r and e q is a unit vector perpendicular to r and in the direction of increasing q, show that: e r = cosq i + sinq j, (i) e q = sinq i + cosq j. ince r r is a vector tangent to the curve q = onstant, a unit vector in the direction of r (increasing r) is given by, e r = r r r r ince r = xi + yj = r cosq i + r sinq j (ii)
16 Mechanics of Particles, Waves & Oscillations as in ig. 1.1, r r = cosq î + sinq ĵ, r = 1 r so that e r = cosq î + sinq ĵ. (iii) urther, r qis a vector tangent to the curve r = onstant. A unit vector in this direction is given by, e q = r r (iv) q q By (ii), so that (iv) yields: r = r sinq î + r cosq q ĵ, r q = r e q = sinq î + cosq ĵ (v) Example 0 how that (a) ė r = q. e q (b) ė q = q. e r (a) By (iii) of Ex. 19, ė r = d e r er dr er dq = + dt r dt q dt = (0) (ṙ) + ( sinq i + cosq j) ( ) q = q. e q where ṙ and q. mean dr dt and d q, respectively. dt (b) By (v) of Ex. 19, ė q = de q = eq dr eq dq + dt r dt q dt = (0) (ṙ) + ( cosq î sinq ĵ) (q. ) = q. e r Example 1 Prove that in polar coordinates (a) the velocity is given by v = ṙ e r + r q. e q and (b) the acceleration is given by a = ( r.. r. - q ) e + ( r q.. + r.. q. ) e q (a) r = re r v = d r dr r d er = er + dt dt dt r.... r r r = re + re = re + rqe q where use has been made of Example (0).
(b) a = d v d.. = ( r e r + r q ė q ) dt dt =.. r r. r r r.. e ė r.. e e r. + + q + q + q e q q =.. r r.. e r r.. e r.. + ( q + e + e q ) q q q q + r....... = ( r - rq ) e + ( rq+ rq) eq r Vector alculus 17 where we have used the results of Ex. 0 and Ex. 1 (a). The above expressions for velocity and acceleration will be used in hapter 3 & 4. 1.16 ELD n general, physical quantities have different values at different points in space. Thus, for example, the temperature in a room varies from one place to another, being higher near a fire place and lower near an open window. imilarly, the electric field near a point charge is larger than at points farther from it. The expression field is used to imply both the region and the value of the physical quantity in the region (electric field, gravitational field etc.). f the physical quantity is a scalar (for example temperature) then we are concerned with a scalar field. f the quantity is a vector (for example electric field, velocity etc.) then we speak of a vector field. 1.17 GRADENT, DVERGENE AND URL 1.17.1 The Del Operator The vector differential operation Del, symbol, is defined by = i + j+ k = i + j + k (8) x y x y Del is also called Nabla. t enjoys the properties of both a vector as well as a differential operator. The operator del is not a vector in the geometrical sense since it has no scalar magnitude but it does transform properly so that it may be treated formally as a vector. t is found useful in defining Gradient, Divergence, url and Laplacian. 1.17.1.1 Properties of the Del Operator f we multiply a scalar quantity u with this vector operator or, we obtain u = i j k + + u = x y KJ i u j u k u + + = grad u. x y (i) f we form the scalar product of the del operator with a vector v, we obtain, according to the definition of scalar product, the sum of the products of corresponding components:. v = i j k + + x y. v + v + v KJ e = v v x y v + + = div v. x y i j k x y j
18 Mechanics of Particles, Waves & Oscillations (ii) f we form the vector product of the del operator with v then we obtain i v = i j k + + x y (v KJ x î + v y ĵ + v k ) + KJ = v v y v x v - j y KJ + - x = url v. (iii) f we form the scalar product of the del operator with itself then we obtain, i j k + + x y KJ i j + + k x y = KJ + + x y = = Laplacian (iv) The operations with del operator are distributive with respect to addition. That is, (v 1 + v ) = v 1 + v (v) The definition of del operator is independent of the co-ordinate system. 1.17. The Gradient Let f (x, y, ) be a differentiable scalar field in a certain region of space x, y,. Then the gradient of f, symbol f or grad f, is defined by, f = i + j+ k f = x y KJ f f f i + j+ k (9) x y Observe that f defines a vector field. The component f in the direction of a unit vector n is given by f. n and is called the direction derivative of f in the direction n. This is the rate of change of f at (x, y, ) in the direction n. rom the calculus, df = f f f x dx + y dy + d. (30) Let r be the position vector to the point P (x, y, ). Q r = xî + y ĵ + k f we move to the point Q (x + dx, y + dy, + d), as in ig. 1., dr = dx î + dy ĵ + d k (31) Using (9) and (31), we can take the dot product of dr and f to yield df as in (30), d f = dr. f (3) x 0 r + Dr r P D r ig. 1. y 1.17..1 Geometrical nterpretation of f onsider a surface f (x, y, ) = c (33)
Vector alculus 19 where c is a particular constant. Let us select an infinitesimal displacement dr of r, and consider only those displacements which are tangential to the surface described by (33). As long as we move along this surface, f has the constant value and df = 0. onsequently from (3), dr. f = 0 (34) Now f is a vector which is completely determined once f has been differentiated, and since neither dr is ero nor in general f, according to (34), f is perpendicular to dr where dr denotes a change from P to Q with Q remaining on the surface f = onstant. We therefore conclude that f is normal to all possible tangents to the surface at P so that f must necessarily be normal to the surface f (x, y, ) = onstant, as in ig. 1.3. n general, dr. f = dr f cosq where q is the angle between the unit vector n and the vector f. Let dr = ds so that df = n. f (35) ds is the directional derivative along n. ince n = 1, df = f cosq (36) ig. 1.3 ds Thus, df/ds is the projection of f on the direction n. The largest value of df/ds (namely f ) occurs if we go in the direction of f (that is q = 0). On the other hand if we go in the opposite direction (that is q = 180 ) f has the largest rate of decrease, that is df/ds = f. Example f f = x y x 3, find f. Example 3 e f = i j k + + x y - x x y KJ 3 =(xy 3 ) î + x ĵ 3x k. f f = 1 r, where r = x + y +, show that f = r r. 3 e f = i j k + + x + y + x y KJ j = 1 - - - 1 1. x i. y j. k = - ( xi + y j + k ) r =- 3 ( x + y + ) ( x + y + ) r 3 3.
0 Mechanics of Particles, Waves & Oscillations Example 4 ind a unit vector normal to the surface, x y + x = at the point (1, 1, 1). ( x y+ x) = i j k + + x y KJ (x y + x) =(xy + ) î + x ĵ + x k = î + ĵ + k at the point (1, 1, 1). A unit vector normal to the surface is obtained by dividing the above vector by its magnitude. Hence the unit vector is - i + j+ k = - i j k + + (- 1) + () 1 + () 1 3 3 3 Example 5 ind the directional derivative of f = x y + 3x at (1, 1, ) in the direction i + j k. e f = i j k + + x y + 3x x y KJ =(xy + 3 ) î + x ĵ + (x y + 6x) k =8î + ĵ + 11 k at (1, 1, ) The unit vector in the direction of î + ĵ k is i + j-k n = = 1 i + j - k. ( ) + ( 1) + (-) 3 3 3 The required directional derivative is f. n =(8î + ĵ + 11 k 1 ). i + j - k 3 3 3 = 16 + - = - 4 3 3 3 3. ince this is negative, f decreases in this direction. 1.17.. Properties of the Gradient ( f) = f (f + Y) = f + Y ( fy ) = f Y + Y f (39) where f and Y are differentiable scalar functions in some region in space and is a constant. 1.17..3 An Example of a Gradient onsider the lines of equal pressure (isobars) marked on a weather map. The direction of the wind is then given, apart from the earth s rotation, by the direction of greatest pressure drop, which is perpendicular to the isobars, and the strength of the wind is given by the magnitude of the pressure drop. K J (37) (38)
Vector alculus 1 Example 6 Prove that ( fy ) = f Y + Y f ( fy ) = ( fy) i+ ( fy) j+ ( fy) k x y K J + + Y Y i + j+ k x y KJ = f Y f f Y f + Y î Y x x y y = f Y = f Y + Y f. + Y KJ f f f i + j + k x y Example 7 ind the angle between the surfaces x + y + = 4 and = x + y 1 at the point (1, 1, 1). The angle between the surfaces at the point is the angle between the normal to the surfaces at the point. A normal to x + y + = 4 at (1, 1, 1) is f 1 = (x + y + ) = xi + yj + k = i j + k A normal to = x + y 1 or x + y = 1 at (1, 1, 1) is f = (x + y ) = xi + yj k = i j k. b f 1 g. b f g = f 1 f cosq, where q is the required angle. Then (i j + k). (i j k) = i j + k i j k cosq 4 + 4 + = ( ) + (- ) + ( ) ( ) + (- ) + (- 1) cosq 6 or cosq = 1 9 = 1 3 = 0.5773. Thus the acute angle is q = 54.7. 1.17.3 The Divergence Let V(x, y, ) = V 1 î + V ĵ + V 3 k be a differentiable vector field at each point (x, y, ) in a certain region of space. Then the divergence of V, symbol. V or div V is defined by. V = i + j+ k x y = x V + y V + V Observe that. V π V.. 1 3. (V KJ î + V 1 ĵ + V k 3 ) KJ (40) KJ
Mechanics of Particles, Waves & Oscillations 1.17.3.1 Physical ignificance of Divergence onsider the flow of a fluid of density r(x, y, ) with velocity v(x, y, ), through a small parallelpiped ABDEGH (ig. 1.4) of dimensions dx, dy, d. We will first calculate the amount of fluid passing along the x-direction through the face EGH per unit time. The y and components of the velocity v contribute nothing to the flow through EGH. The mass of fluid entering EGH per unit time is given by rv x dy d. The mass of the fluid leaving the face ABD per unit time is L NM rv x O QP + ( rvx) dx dy d. x The net rate of flow out for these two faces is simply the difference between these two flows, or Net rate of flow out = ( rv x ) dxdyd. x f we also take into consideration the other two faces we find that the total loss of mass per unit time is L NM ( rvx) + ( rvy) + ( rv) x y Q dxdyd. so that the quantity within square brackets represents the loss of mass per unit time per unit volume and is called the Divergence. (rv). This is the physical meaning of divergence. div (rv) may not be ero either because of the time variation of density or the existence of sources and sinks. Let Y = source density minus sink density = net mass created per unit time per unit volume. r = time rate of increase of mass per unit volume. t Then rate of increase of mass per unit volume = rate of creation minus rate of outward flow. n symbols, the balance equation becomes r = Y. (rv) t alling V = rv,. V = Y- r (41) t n the absence of sources and sinks, Y = 0, and (41) reduces to O ig. 1.4
Vector alculus 3. V + r = 0; Equation of continuity (4) t f there is no gain of fluid anywhere then. V = 0. This is called the continuity equation for an incompressible fluid. The fluid is said to have no sources or sinks since it is neither created nor destroyed at any point. A vector such as V whose divergence is ero is called solenoidal. r f = 0, then (41) reduces to t. V = Y (43) The above treatment is equally applicable to electric and magnetic fields where v is replaced by E or B and the quantity corresponding to outflow of a metal substance is called flux. n the case of an electric field, the so-called sources and sinks are the electric charges and the equation analogous to (43) is div D = Y (44) where Y is the charge density and D is the electric displacement. or the magnetic field the sources are assumed to be magnetic poles. However, free magnetic poles do not exists so that div B = 0 (45) Equation (44) and (45) constitute two of the celebrated equations in Electromagnetism, originally due to Maxwell. The continuity equation (4) also has an application in the interpretation of the wave function in Quantum Mechanics. Example 8 alculate. (ff), where f = u(x, y, ) î + v(x, y, ) ĵ + w(x, y, ) k.. (ff) = ( fu) + ( fv) + ( fw) x y = f u v w + + x y = f. f + f. f Example 9 alculate. f if f = r/r 3. (inverse-square force). (r 3 r) = r 3. r + r. r 3 But. r = i j k + + x y = x y + + = 3 x y and by problem (16), r n = nr n r f f f + u + v + w KJ x y KJ. KJ (îx + ĵy + k)
4 Mechanics of Particles, Waves & Oscillations so that r 3 = 3 r 5 r \. (r 3 r) = 3r 3 3r 5 r. r = 3r 3 3r 3 = 0 Thus the divergence of an inverse square force is ero. 1.17.4 The Laplacian The Laplacian, symbol is the divergence of a gradient.. f = i j k + + x y. KJ = f f f x + y + = f i f f j k f + + x y where = x + y + = Laplacin. Example 30 Prove that. (fa) = ( f). A + f(. A), where f is a scalar and A is a vector.. (fa) =. fa i + fa j+ fa k e 1 3 = i j k + + x y j. f KJ e + f + f = ( fa1) + ( fa) + ( fa3) x y = f A x f A f A f + + + A1 + y x 1 3 = f f f i + j+ k x y + f i + j + k x y =( f). A + f (. A) Example 31 f f = x y x 3, find f. 1.17.5 The url f = + + x y = y 1x. KJ A i A j A k 1 3. A + A + A KJ e KJ e e x y - KJ i j k 1 3. A i + A j + A k 3 x j 1 3 f V(x, y, ) is a differentiable vector field then the url or rotation of V, symbol V, url V or rot V is defined by j j j
V = = = i + j+ k x y i j k x y x Vx Vy V y V V i - x j+ V V KJ (V 1 î + V ĵ + V 3 k ) x V y x x V Vector alculus 5 = V V y Vx V - i + - j+ y KJ x KJ url V represents the rotation or vorticity in the fluid. f the flow is irrotational, V = 0; ondition for irrotationality. The curl may be used to describe the motion of a rigid body rotating about an axis with uniform angular velocity w. v = w R, is the linear velocity of any point in the body with radius vector R. url v = (w R). By identity (13), A (B ) = B (A. ) (A. B). Therefore, url v = w(. R) R(. w). But,. R = 3, by Example (9). Also, R (. w) = (w. )R since w is a constant vector. R Therefore, (w. )R = U w w w x + y + T V x y W (î x + ĵy + k ) = î w x + ĵ w y + k w = w Therefore, url v = 3w w = w. Thus the curl of the linear velocity of any point of a rigid body is equal to twice the angular velocity. 1.17.5.1 The Physical ignificance of the url The physical significance of the curl is brought about by considering the circulation of fluid around a differential loop in the xy-plane, ig 1.5. x y 1 3 - + irculation 134 = v ( x, y ) dx v ( x, y ) dy v Use the Taylor expansion about the point (x 0, y 0 ), taking into account the displacement of line segment 3 from 1 and from 4. v y ( x + dx, y ) = v 0 0 y v y ( x0, y0) + dx... H G x K J + x0, y0
6 Mechanics of Particles, Waves & Oscillations The higher-order terms will drop off in the limit dx Æ 0. A correction term for the variation of v y with y is cancelled by the corresponding term in the fourth integral. irculation 134 =v x (x 0, y 0 )dx L N M L NM + v y ( x, y ) 0 0 + v x y O Q P O dx dy v + v x ( x x0, y0 ) + v y dy QP (- dx) + = v y v x - dxdy x y Dividing by dxdy, we have KJ y y x y 0 0 + dy 4 x y 0, 0 3 1 ig. 1.5 ( x 0 + ( x 0 irculation per unit area = v The circulation about the differential area in the xy-plane is given by the -component of v. n fluid dynamics v is called the vorticity. n principal, the curl v at (x 0, y 0 ) can be determined by inserting a (differential) paddle wheel into the moving fluid at the point (x 0, y 0 ). The rotation of the small paddle wheel would be a measure of the curl, and its axis along the direction of v which is perpendicular to the plane of circulation. Whenever the curl of a vector v vanishes, v = 0 (47) and v is called irrotational. 1.17.5. Examples of the url of Vector ield (i) The curl of a vector field implies circulation or vortex motion (rotation). f the fluid velocity v has a curl at some point then that signifies the existence of vorticity at that point. ince the line integral of a conservative field A around any closed path is ero, that is A. dr = 0, the conservative fields have ero curl at all points of space. (ii) An example of the conservative vector field is the electrostatic field E. Therefore, curl E = 0. (iii) or waves in an elastic medium, if the displacement U is irrotational, U = 0, plane waves (or spherical waves at large distances) become longitudinal. f V is solenoidal,. U = 0, then the waves becomes transverse. The displacement of a seismic wave may be resolved into a solenoidal part and an irrotational part. The irrotational part corresponds to the longitudinal P (primary) earthquake waves. The solenoidal part yields the slower (secondary waves) (ee 13.1.) (iv) f v is the linear velocity then curl v may be used to describe the motion of a rigid body rotating about an axis with uniform angular velocity w. t can be shown that curl v = w (see 1.17.5). Thus the curl of the linear velocity of any point of a rigid body is equal to twice the angular velocity.
Vector alculus 7 Example 3 Prove that url of a gradient is ero. ( f) = i j k x y f x f y f f f f = i - j y ykj - x - = 0 since terms in brackets cancel in pairs. Example 33 Prove that url url V = grad div V V. By identity (13), A (B ) = B (A. ) (A. B). Putting A =, B =, = V, ( V) = (. V) (. ) V = grad div V V. Example 34 how that the divergence of a curl is ero, that is. ( V) = 0 i + j + k x y = i + j + k x y = x y V V KJ. i j k x y Vx Vy V L. i y - j k KJ x + V V V V NM - x + y V V y x x V y x x = x y x y Vx Vy V = 0 since two rows of the determinant are identical.
8 Mechanics of Particles, Waves & Oscillations Example 35 f A and B are irrotational, prove that A B is solenoidal. By problem A = 0 and B = 0 t follows that B. ( A) = 0 A. ( B) = 0 ubtracting B. ( A) A. ( B) = 0 By problem (8), left hand side of the above equation is equal to. (A B). Therefore,. (A B) = 0, so that (A B) is solenodial. Example 36 A central field A in space is given by A = r (r). (a) how that the field is irrotational. (b) What should be the function (r) so that the field is solenoidal? (a) That url A = 0 for the central field (criterion for the conservative forces) is proved in hapter 4. (b) f the field is solenoidal, then,. r (r) = 0 But x + y + = r [ x x ( r )] + [ y y ( r )] + [ =0 x y + + + + + x y x y 3 ()+ r x y H G r K J + r H G r K J + r 3(r) + x + y + r K J r r therefore = -3 r ntegrating, ln = 3 ln r + n, where ln = constant. ln = ln ln r 3 = ln r 3 Therefore = /r 3. The field is A = r/r 3 (inverse square law). 1.17.6 Table of Vector dentities nvolving Here A and B are differentiable vector functions, and f and Y are differentiable scalar functions of position (x, y, ) and r is the position vector. 1. (f + Y) = f + Y. (fy) = f Y + Y f 3.. (A + B) =. A +. B KJ =0 =0 =0
4.. (fa) = ( f). A + f(. A) 5. (A + B) = A + B 6. (fa) = ( f) A + f( A) 7.. (A B) = B. ( A) A. ( B) 8. (A B) = (B. )A B(. A) (A. )B + A(. B) 9. (A. B) = (B. )A + (A. ) B + B ( A) + A ( B) 10.. ( f) = div grad f = Laplacian f = f f f + + x y 11. ( f) = 0. The curl of gradient of f is ero. 1.. ( A) = 0 The divergence of curl of A is ero. 13. ( A) = (. A) A 14.. ( f Y) = 0 15.. r = 3 16. r = 0 17. (A. ) r = A 1.18 VETOR NTEGRATON Vector alculus 9 1.18.1 Line integral This is an extension of the line integral described in calar calculus. Any integral which is to be evaluated along a curve is called a line integral. Examples of Line integral in vector calculus are: (a) f dr (b) A. dr (c) A dr where f is a scalar, A is a vector, and r is the position vector r = xi + yj + k. Each of these is a line integral along the curve. The result of integration is a vector for (a), a scalar for (b) and a vector for (c). When the space curve forms a closed path which is assumed to be a simple closed curve, that is, a curve which does not intersect itself anywhere, the line integral (a) is written as f dr. We can similarly write for (b) and (c). The movement along the closed curve is said to be positive or counterclockwise if the enclosed region always lies to the left, and negative or clockwise if the enclosed region lies always to the right. A line integral means an integral along a curve or a line, that is a single integral in contrast to a double integral over a surface or area, or a triple integral over a volume. t must be emphasised that in a line integral there is only one independent variable because we are constrained to remain on a curve. n two dimensions, the equation of a curve becomes y = (x), where x is the independent variable. n three dimensions, we could take x as the independent variable and find y and as functions of x. Alternatively, the parameter t may be taken as independent variable so that x, y, are all functions of t. Thus, the line integral is evaluated by writing it as a single integral using one independent variable.
30 Mechanics of Particles, Waves & Oscillations 1.18.1.1 alculation of Work Done by a Varying orce on a Body Using the Line ntegral (b) Work done by a force on an object which undergoes an infinitesimal vector displacement dr can be written as dw =. dr. n general, the force acting on the object varies from point to point. or example, the force on a charged particle in an electric field would be a function of x, y,. However, along a curve x, y, are related by the B equation of the curve. ince along a curve there is only one independent variable, we can write and dr = i dx + j dy + k d as functions of a single variable. The A integral of dw =. dr along the given curve is then y reduced to an ordinary integral of a function of one variable, and the total work done by in moving an x object say from A to B, can be determined (ig. 1.6). 1.18.1. Examples of Line ntegral (i) The line integral,. dr represents the work done by the force along the curve. (ii) The quantitative relationship between current i and the magnetic field B is given by Ampere s law B. dl = m 0 i (iii) The potential difference between two points A and B is related to the electric field by the line integral B - E. dl = V B V A A B (iv) f A = grad f, then the integral A. dr depends only on initial and final values of A f and is independent of the path. Also A. dr = 0. onversely, if A. dr = 0, then there must exist a scalar point function f such that A = grad f. The vector field is said to be conservative. Examples of conservative fields are gravitational field and electric field (ee Example 43). (v) Magnetostatics also provides a physical example of the third type of line integral of a loop of wire carrying a current placed in a magnetic field B, then the force d on a small length dr of the wire is given by d = dr B, so that the total vector force on the loop is = dr B. Example 37 dr ig. 1.6 Evaluate f dr if f = f(x, y, ).
Using the differential displacement vector dr = dxî + dy ĵ + d k, we find. f dr = f( dx i + dy j+ d k ) c = i fdx + j fdy + k fd c c c c Vector alculus 31 where we have taken î, ĵ and k outside the integral as they have constant magnitude and direction. Example 38 Evaluate A. dr if A = A 1 (x, y, ) î + A (x, y, ) ĵ + A 3 (x, y, ) k and r is the radius vector. A. d r = ( A A A ) ( dx dy 1i+ j+ 3k. i+ j+ = ( Adx+ Ady+ Ad). 1 3 f A is the force on a particle moving along, this line integral represents the work done by the force. Example 39 Evaluate A. dr from the point P (0, 0, 0) to Q (1, 1, 1) along the curve r = î t + ĵt + k t 3 with x = t, y = t, = t 3, where A = xyî + ĵ + xy k. ince x = t, y = t, = t 3, y = x, = x 3, dy = x dx, d = 3x dx. A. dr Example 40 = (xyî + ĵ + xy k ). (î dx + ĵdy + k d) = 1 1 1 3 4 8 = xdx+ xdx+ 3 xdx= 59 60 0 0 0 (xydx + dy + xyd) y y = x (1, 1) Evaluate A. dr around the closed curve defined by y = x and y = x, with A = (x y)i + (x + y)j. y = x A. d r 0 x = [(x y)î + (x + y) ĵ]. [î dx + ĵdy] ig. 1.7 = (x y)dx + (x + y)dy 1 1 = (x x )dx + (x + x ). xdx (along y = x ) 0 0 0 0 + (y y). ydy + (y + y)dy (along y = x) 1 1 = /3 f the movement was opposite to the indicated direction then the result would have been -.
3 Mechanics of Particles, Waves & Oscillations Example 41 Evaluate A A dr = dr if A = A 1 î + A ĵ + A 3 k i j k A1 A A3 dx dy d = i(a d A 3 dy) + j(a 3 dx A 1 d) + k(a 1 dy A dx). o the integral is: A dr = i ( Ad Ady) j ( Adx A - 3 + 3 - Example 4 olve Kepler s problem by the Vector method. The solution to Kepler s problem can be obtained by solving a differential equation for the motion of a planet in a gravitational field. Here we demonstrate the power of the Vector method in solving the same problem. onsider a planet of mass m around a heavy un of mass M at ig. 1.8. Let k be a fixed vector along an arbitrary reference line which is taken as the polar axis. The co-ordinates of the planet at any instant are defined by r, the radial distance from the un, and the polar angle q, measured with respect to the polar axis. The force acting on the planet due to the un is = (GmM/r 3 )r By Newton s second law, M r q m P ig. 1.8 k so that Now Therefore GmM r 3 r = m d r = m d v dt dt dv GM = - dt r r 3 (46) d dt (r v) = r d v dr + v dt dt (47) d dt (r v) = r d v dt since the second term on the right hand side of (47) vanishes, being the cross product of parallel vectors. Using (46) in (48) d dt (r v) = r -GM r r 3 = 0 r v = h = const. vector K J (48)
Vector alculus 33 or r d r dt = h (49) r ince r d = twice the area of the sector, we get (da/dt) = h, that is equal areas dt are swept out in equal intervals of time. This is Kepler s second law of planetary motion. urther, using the fact that if two vectors in a triple scalar product are identical, the product is ero, we find r. [r dr/dt] = r. h = 0. Thus r remains perpendicular to the fixed vector h, and the motion is planar. Taking the cross product of (46) with h, dv GMr h = - h 3 = - GM r dt r 3 r (r v) (50) where we have used (49). Also, d dt (v h) = d v dh dv h + v = h dt dt dt (51) since h is constant. ombining (50) and (51), d dt (v h) = - GM r (r v) (5) 3 r Now, r = r e r, where e r is the unit vector. Hence v = d r = r d er dr + er dt dt dt so that (5) is reduced to d GM (v h) = - r dt r d e r r 3 dt r K J L = - d r H G GM K J - N M e r. dt = GM e r e de r dt e de e r ( e r.e r) d where we have used the identity (13), and the relations e r. e r = 1 and e r. d e drr d dt (v h) = GM d e r dt ntegrating (53), we find r K J = 0. (53) v h = GM e r + k (54) whence r. (v h) = r. (GM e r + k) = GM r. e r + r. k = GMr + rk cosq where k is an arbitrary constant vector with magnitude k, and q is the angle between k and e r. Using the result of problem 9.
34 Mechanics of Particles, Waves & Oscillations r. (v h) = (r v). h = h. h = h r = h GM + K cosq This is the polar equation of conic section. or planetary motion these conic sections are closed curves so that we deduce Kepler s first law which states that the orbits of the planets are ellipses with the un as one of the foci. Kepler s third law can be deduced without further use of vectors as indicated in hapter 4. Example 43 (a) f = f, where f is single-valued and has continuous partial derivatives, show that the work done is moving a particle from point A [x 1, y 1, 1 ] to B[x, y, ] in this field is independent of the path joining the two points. (b) onversely, if. dr is independent of the path joining any two points show that there exists a function f such that = f. B B (a) Work done =. dr = f. dr A A B f f f = i + j+ k (dxî x y + dy ĵ + d k ) A B KJ f f f = x dx + y dy + d A B = df = f(b) f(a) = f(x, y, ) f(x 1, y 1, 1 ). A Thus the integral depends only on the points A and B, not on the path joining them. t is assumed that f(x, y, ) is single valued at A and B. n ig. 1.9 two paths and are shown. The above A statement would then imply that the result of integration along these two paths would be identical. ig. 1.9 (b) uppose the line integral is independent of the path, that is, is a conservative field, then ( x, y, ) ( x, y, ) f(x, y, ) =. dr =. d r ds ds ( x1, y1, 1) ( x1, y1, 1) Differentiating, df dr =.. ds ds But df dr = f. ; so that ( f - = ds ds dr ). 0. ds ince this result must be valid irrespective of (dr/ds), we deduce = f. B
Vector alculus 35 Example 44 (a) f is a conservative field, prove that url = = 0, that is is irrotational. (b) onversely, if = 0, prove that is conservative. (a) f is a conservative field then by Example 43, = f. Then curl = f = 0 where we have used the fact that the curl of a gradient of f is ero. (b) f = 0, then i j k x y 1 3 = 0, so that 3 y = (55) 1 = x 3 (56) = 1 (57) x y Required to prove that = f Work done to move the particle along a path joining (x 1, y 1, ) and (x, y, ) in the force field is 1 (x, y, )dx + (x, y, )dy + 3 (x, y, )d With the choice of a path consisting of straight line segments from (x 1, y 1, 1 ) to (x, y 1, 1 ) to (x, y, 1 ) to (x, y, ) and calling f(x, y, ) the work done along this particular path, we have x f(x, y, ) = ( x, y, ) dx ( x, y, ) x1 y + 1 1 1 1 y1 whence f = 3 (x, y, ), since differentiation of the first two terms on the right hand side give ero. f y = (x, y, 1 ) + = (x, y, 1 ) + 1 1 3 y = (x, y, 1 ) + (x, y, ) ( x, y, ) d ( x, y, ) d 1
36 Mechanics of Particles, Waves & Oscillations = (x, y, 1 ) + (x, y, ) (x, y, 1 ) = (x, y, ) where we have used (55). y f x = 1 (x, y 1, 1 ) + ( x, y, 1) dy + x x = 1 (x, y 1, 1 ) + y1 1 y 1 ( x, y, 1) dy + y y1 = 1 (x, y 1, 1 ) + 1 (x, y, 1 ) y y 1 3 ( x, y, ) d 1 1 + 1 (x, y, ) 1 ( x, y, ) d = 1 (x, y 1, 1 ) + 1 (x, y, 1 ) 1 (x, y 1, 1 ) + 1 (x, y, ) 1 (x, y, 1 ) = 1 (x, y, ) where we have used (56) and (57). f f f f follows that = 1 î + ĵ + 3 k = i + j+ k = f. x y We conclude that a necessary and sufficient condition that a field be conservative is that url = = 0. Example 45 (a) how that = (3x y + 3 ) î + x3 ĵ + 3x k is a conservative force field, (b) ind the scalar potential, (c) ind the Work done in moving an object in this field from (1, 1, ) to (, 1, 1). (a) t is sufficient to show that url = 0. = i j k x y 3x y + x 3x 3 3 Thus is a conservative force field. = 0 (b) df =. dr = (3x y + 3 )dx + x 3 dy + 3x d =(3x ydx + x 3 dy) + ( 3 dx + 3x d) = d(x 3 y) + d( 3 x) = d(x 3 y + 3 x) f = x 3 y + 3 x + constant. (c) Work done = f(, 1, 1) f(1, 1, ) = 3. 1.18. urface ntegrals Let a surface be divided into infinitesimal elements each of which may be considered as a vector d. ntegrals that involve the differential element d of the surface area are called urface ntegrals. There are three types of surface integrals.