* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Vector Calculus - New Age International
Navier–Stokes equations wikipedia , lookup
Equations of motion wikipedia , lookup
Noether's theorem wikipedia , lookup
Electrostatics wikipedia , lookup
Probability amplitude wikipedia , lookup
Classical mechanics wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Path integral formulation wikipedia , lookup
Circular dichroism wikipedia , lookup
Minkowski space wikipedia , lookup
Photon polarization wikipedia , lookup
Field (physics) wikipedia , lookup
Derivation of the Navier–Stokes equations wikipedia , lookup
Lorentz force wikipedia , lookup
Metric tensor wikipedia , lookup
Work (physics) wikipedia , lookup
Vector space wikipedia , lookup
Euclidean vector wikipedia , lookup
Bra–ket notation wikipedia , lookup
1 Vector Calculus 1.1 SCALAR 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. 1.2 VECTORS 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 Æ by an arrow over the head of a letter such as P . Graphically, a vector B is represented by the directed line segment AB as in Fig. 1.1. The Æ vector P 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 ææÆ Æ of P (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. It is called a Null Vector. Vectors are compounded geometrically. The order of compounding is immaterial. 1.3 EQUIVALENCE OF VECTORS 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 Fig. 1.2, P = Q. Commutative property: If A = B, then B = A Transitive property: If A = B and B = C, then A = C. A Fig. 1.1 P Q Fig. 1.2 2 Mechanics of Particles, Waves & Oscillations 1.4 1.4.1 MULTIPLICATION OF VECTORS BY SCALARS Scalar Multiplication Let A be any vector and m any scalar. Then the vector mA as in Fig. 1.3 is defined as follows: (a) The magnitude of mA is |m||A|; that is |mA| = |m||A| (b) If m > 0, the direction of mA is that of A. (c) If m < 0, the direction of mA is opposite to that of A. Fig. 1.3 (d) m (nA) = (mn) A for any scalars m, n. Two non-zero 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. 1.4.2 Result of Scalar Multiplication (a) The zero vector if m = 0, we obtain the zero vector 0; i.e. (0) A = 0, where A is any vector. The zero vector 0 is a vector that has magnitude 0 and has any direction. (b) The negative of a vector If 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 If A π 0 and m = 1/|A| = 1/A, the unit vector, eA = 1 A. |A| Thus, the vector eA is a vector whose magnitude is |eA| = 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 eA; that is; A = AeA 1.5 ADDITION OF VECTORS (TRIANGLE LAW) The sum or resultant of two vectors A and B is a vector C which can be determined geometrically, Fig. 1.4. If the tail of B is placed at the head of A, the resultant C 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. C B A Fig. 1.4 Vector Calculus 1.6 3 SUBTRACTION OF VECTORS If A and B are two vectors, the difference (A – B) is the sum C, of A and (–B); that is: C = A – B = A + (–B) The vector subtraction is shown geometrically in Fig. 1.5. 1.7 Fig. 1.5 PROPERTIES OF VECTOR ADDITION A+ B= B+ A (Commutative law) A + (B + C) = (A + B) + C (Associative law) m(A + B) = mA + mB (Distributive law) (m + n) A = mA + nA (Scalar distributive law) A+ 0= A A– A= 0 z 1.8 RECTANGULAR UNIT VECTORS i, j, k ^ Consider a right-handed, rectangular co-ordinate system. j, k are shown in the directions of x, The unit vectors i, y and z axes respectively, Fig. 1.6. A vector A in three dimensions, can be represented by its initial point at the origin 0 and rectangular components (A1, A2, A3) for its terminal point, Fig. 1.7. Since the resultant of A1 i + A2 j + A3 k , is the vector A, A = A i + A j + A k , 1 2 k ^ j x Fig. 1.6 3 z The magnitude of A is A = |A| = A A12 + A22 + A32 Let A = ax i + ay j + az k and B = bx i + by j + bz k be two vectors where ax, ay, az are the x, y and z components respectively, and bx, by, bz, are the corresponding components of B. Then the resultant of A and B is given by: R = y 0 ^ i 2 2 ( a x + bx ) + (a y + by ) + ( a z + ^ A3 k ^ A1 i y 0 ^ A2 j x Fig. 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, z) is written as r = x i + y j + z k and has magnitude r = |r| = x 2 + y2 + z 2 4 Mechanics of Particles, Waves & Oscillations Example 1 Forces A and B expressed by the equation A = A1 i + A2 j + A3 k and B = B1 i + B2 j + B3 k, act on an object. Find the magnitude of the resultant of these forces. Resultant force R =A+ B = (A1 + B1)i + (A2 + B2)j + (A3 + B3)k R = ( A1 + B1 )2 + ( A2 + B2 )2 + ( A3 The result can be extended to any number of forces. Example 2 Determine the vector having initial points P (x1, y1, z1) and the terminal points Q(x2, y2, z2) and find its magnitude. The position vector of P is r = x i + y j + z k 1 1 1 1 The position vector of Q is r2 = x2 i + y2 j + z2 k In Fig. 1.8, r1 + R = r2 or R = r2 – r1 = (x2 i + y2 j + z2 k ) – (x1 i + y1 j + z1 k ) = (x – x ) i + (y – y ) j + (z – z ) k 2 R = 1.9 1 2 1 2 Fig. 1.8 1 ( x2 - x1 )2 + ( y2 - y1 )2 + ( z2 - SCALAR OR DOT PRODUCT The Scalar or Dot Product, also called the Inner 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 Fig. 1.9. As the scalar product A.B is represented by a dot q between two vectors, it is called the dot product. From B A cos q the definition of the scalar product (1), the angle between A and B is found from the formula Fig. 1.9 A.B (1a) cos q = AB provided A π 0, and B π 0. If 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 (2) to be A . B = 0; Condition for orthogonality (2) Vector Calculus 1.9.1 5 Properties of the Scalar Product A . B = B. A (Commutative law) A . (B + C) = A . B + A . C (Distributive law) (mA) . B = A . (mB) = m(A . B) A . A = |A|2 = A2 ; for every A. A . A = 0 ; if and only if A = 0, where m is an arbitrary scalar. i . i = j . j = k . k = 1 i . j = j . k = k . i = 0 If A = A1 i + A2 j + A3 k and B = B1 i + B2 j + B3 k then (3) A . B = A1 B1 + A2 B2 + A3 B3 A . A = A2 = A 21 + A 22 + A 23 B . B = B2 = B 21 + B 22 + B 23 1.10 VECTOR OR CROSS PRODUCT The vector product of the vector A and B, written as A × B, is another vector: C =A× B z A ¥B (4) y The magnitude of C is given by B q C = AB sinq (5) x where q is the acute angle between A and B. The direction of C is that of the advance of a right-hand screw as A rotates towards B through the angle q Fig. 1.10. A B¥A Fig. 1.10 1.10.1 Geometric Interpretation Fig. 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 C = |A × B|, represents the area of the parallelogram whose sides are A and B. 1.10.2 B q A Fig. 1.11 Properties of the Vector Product (i) A × B = –B × A h = B sin q (Anticommutative law) (ii) A × (B + C) = A × B + A × C (6) (Distributive law) (mA) × B = A × (mB) = m A × B (iii) A × A = 0 (iv) A × 0 = 0 for any A UV W (7) 6 Mechanics of Particles, Waves & Oscillations i = j × j = k × k = 0 j = k , j × k = i , k × i = j , j × i = – i × j , k × j = – j × k , i × k = – k × i = B i + B j + B k then, (vi) If A = A1 i + A2 j + A3 k and B 1 2 3 (v) i × i × U| V| W i A × B = A1 B1 1.11 j A2 B2 (8) k A3 B3 SCALAR TRIPLE PRODUCT The scalar triple product of the three vectors A, B and C is a scalar A . (B × C) or simply, Scalar Triple Product = A . (B × C) (9) Observe that the expression A × (B . C) is meaningless since it implies the vector product of a scalar and a vector. If A = A1 i + A2 j + A3 k and B = B1 i + B2 j + B3 k C = C i + C j + C k 1 then, A1 A . (B × C) = 2 3 A2 A3 B1 B2 C1 C2 B3 C3 The scalar triple product A . (B × C) is also written as [ABC]. 1.11.1 Geometric Interpretation of A . B × C Fig. 1.12 shows a parallelepiped whose sides are A, B and C. By (4), the vector product of B and C is equal to the area S for the parallelogram with adjacent sides B and C. S = |B × C| If h is the altitude of the parallelopiped then, h = |A| |cos q| where q is the angle between A and B × C. Therefore, the volume of the parallelopiped is Fig. 1.12 V = hS = |A| |B × C| |cos q| = |A . B × C| 1.11.2 Condition for Coplanarity Vector A, B and C are coplanar if and only if A . B × C = 0; Condition 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 Calculus 7 Example 3 If any two vectors in a scalar triple product are equal, show that the product is zero; that is, A . A × C = 0 ; C . B × C = 0; A . B × B = 0 The vector product A × C = P is perpendicular to A. Hence, A . P = 0 by (3), that is, A . A × C = 0. Also, since B × C is perpendicular to C, their scalar product is zero, that is, C . B × C = 0. Again, B × B = 0 by (7). Hence A . 0 = 0. 1.11.3 Fundamental identify for the Scalar Triple Product A . B× C = A× B . C (11) This implies that in the scalar triple product the position of the dot and cross is immaterial. 1.12 VECTOR TRIPLE PRODUCT The vector triple product of three vectors A, B and C is the vector D given by: D = A × (B × C) (12) Here a parenthesis is essential since A × B × C depends on the result of the product, whether we form A × B first or B × C first. Thus, A × (B × C) π (A × B) × C 1.12.1 Fundamental Identities for the Vector Triple Product A × (B × C) = (A . C) B – (A . B) C (13) (A × B) × C = (A . C) B – (B . C) A (14) Example 4 Evaluate (a) i . i ; (b) i . k (a) i . i = | i | | i | cos 0° = (1) (1) (1) = 1 (b) i . k = | i | | k | cos 90° = (1) (1) (0) = 0 Example 5 If A = A1 i + A2 j + A3 k and B = B1 i + B2 j + B2 + A3 B3. A . B = (A1 i + A2 j + A3 k ) . (B1 i + B2 = A B i . i + A B i . j + A B B3 k prove that A . B = A1 B1 + A2 j + B k ) 3 i . k + A B j . 1 1 1 2 1 3 2 1 + A2 B3 j . k + A3 B1 k . i + A3 B2 k . j + A3 B3 = A1 B1 + A2 B2 + A3 B3 where we have used (3). i + A2 B2 j . j k . k 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? ax = 0 ; ay = 3 ; bx = 4 sin45° ; by = 4 cos 45° ( a x + bx )2 + ( a y + by )2 r = 2 e0 + 4 2 j + e3 + 4 2 j = B 2 = 6.478 km. a y + by tan q = ry/rx = a x + bx = 3+4 m 4k by 4 5° A C ay = 3 km 2 = 2.06. 4 2 q = 64.1°, north of east. q O Example 7 bx Fig. 1.13 Show that (a) i × j = k; (b) j × j = 0 (a) i j k i × j = 1 0 0 = k 0 1 0 (b) i j k j × j = 0 1 0 = 0 0 1 0 Example 8 = x i + y j + z k and B = x i + y j + z k , prove that : If A 1 1 1 2 2 2 i A × B = x1 x2 j y1 y2 k z1 z2 A × B = (x1 i + y1 j + z1 k ) × (x2 i + y2 j + z2 k ) = x1 x2 i × i + x2 y2 i + y1 x2 j × i + y1 y2 + z x k × i + z y 1 2 1 2 × j + x1 z2 i × k j × j + y z j × k 1 2 k × j + z z k × k 1 2 = x1 y2 k – x1 z2 j – y1 x2 k + y1 z2 i + z1 x2 j – z1 y2 i = (y1 z2 – z1 y2) i + (z1 x2 – x1 z2) j + (x1 y2 – y1 x2) k D Vector Calculus i = x1 x2 j y1 y2 9 k z1 z2 where we have used (8). Example 9 Prove that A × (B × C) = (A . C) B – (A . B) C j i A × (B × C) = A × B1 B2 C1 C2 k B3 C3 = (A1 i + A2 j + A3 k ) × + k (B1 C2 – B2 C1)] = – k A (B C – B C ) – – k + j [ i (B2 C3 – B3 C2) – j (B1 C3 – B3 C1) j A (B C – B C ) 1 1 2 2 1 A2 (B2 C3 – B3 C2) + i A2 (B1 C2 – B2 C1) A (B C – B C ) + i A (B C – B C ) 1 1 3 3 2 3 3 1 3 2 3 1 3 3 1 where we have used (8). \ A × (B × C) = i [B1 (A2 C2 + A3 C3) – C1 (A2 B2 + A3 B3)] + j [B2 (A1 C1 + A3 C3) – C2 (A1 B1 + A3 B3)] + k [B (A C + A C ) – C (A B + A B )] 3 1 1 2 3 2 2 3 3 3 1 1 2 2 = i [B1 (A1 C1 + A2 C2 + A3 C3) – C1 (A1 B1 + A2 B2 + A3 B3)] + j [B2 (A1 C1 + A2 C2 + A3 C3) – C2 (A1 B1 + A2 B2 + A3 B3)] + k [B3 (A1 C1 + A2 C2 + A3 C3)– C3 (A1 B1 + A2 B2 + A3 B3)] = (A C + A C + A C ) ( i B + j B + k B ) 1 1 1 2 3 – (A1 B1 + A2 B2 + A3 B3) ( i C1 + j C2 + k C3) – (A . C) B – (A . B) C Example 10 Verify the Jacobi identity: A × (B × C) + B × (C × A) + C × (A × B) = 0 Using the fundamental identity (13), A × (B × C) = (A . C) B – (A . B) C B × (C × A) = (B . A) C – (B . C) A C × (A × B) = (C . B) A – (C . A) B Add these identities to obtain the Jacobi identity. Example 11 Prove that (A × B) × (C × D) = [A B D] C – [A B C] D 10 Mechanics of Particles, Waves & Oscillations Let A × B = F, then by (13), (A × B) × (C × D) = F × (C × D) = (F . D) C – (F . C) D = (A × B . D) C – (A × B . C) D = [A B D] C – [A B C] D Example 12 Find the angle between the vectors A = 2 i + 2 j – k and B = 4 i + 2 j +4 k A = (2)2 + (2)2 + ( -1)2 = 3 ; B = (4)2 + (2)2 + (4)2 = 6 A . B = (2 i + 2 j – k) . (4 i + 2 j + 4 k) = (2) (4) + (2) (2) + (–1) (4) = 8. cosq = 8 A.B = = 0.4444 ; q = 63.6°. (3) (6) AB Example 13 Find the angles which the vector A = 2 i – 2 j + k makes with the co-ordinate axes. Let a, b, g be the angles which A makes with the positive x, y, z axes respectively. A . i = (A) (1) cos a = (2)2 + ( -2)2 + (1)2 cos a = 3 cos a. A . i = (2 i – 2 j + k ) . i = 2 i . i – 2 j . i + k . i = 2 cos a = 2/3 = 0.6667 and a = 48.2°. Similarly, cos b = – 0.6667 and b = 131.8°, and cos g = 0.3333 and g = 70.5°. The Cosines of a, b, g are called the direction cosines of A. Example 14 Show that cos2a + cos2b + cos2g = 1, where cos a, cos b and cos g are the direction cosines of a vector. Let A = A1 i + A2 j + A3 k . The direction cosines are given by cos a = A1/A, cos b = A2 /A, cos g = A3 /A. cos2a + cos2b + cos2g = = A 21 A2 + A 22 A2 + A 23 A2 A 21 + A 22 + A 23 A2 = A2 A2 = 1. 1.13 APPLICATION OF VECTOR MULTIPLICATIONS 1.13.1 Scalar Product In general, when the force and displacement are not parallel, then the quantity of work Vector Calculus 11 is given by the component of the force parallel to the displacement, multiplied by the displacement: F W = (F cos q) d = F . d. 1.13.2 Fig. 1.14 F r 0 q nq r si Fig. 1.15 w rs in v q P p r q 0 Fig. 1.16 Triple Scalar Product Consider the vector r and the force F lying in the xy plane, as in Fig. 1.17. Then the torque of the force F about z-axis (which is perpendicular to the xy plane) is given by the vector product r × F. This special case was actually considered under 1.13.2. However, if the axis L is inclined to the z axis then the result must be modified. Consider the axis L which is inclined to the z-axis. Take a unit vector n along the axis L. Then the torque about L is given by the Triple Scalar product, n . (r × F), since it gives the component along the axis L. 1.13.4 d Vector Product (a) Torque: In 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 = Fr sin q = |r × F|. Thus r × F represents the torque of F 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, Fig. 1.16. Since the radius of the circle is r sin q, the magnitude of the linear velocity w × r|. v is w (r sin q) = |w 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 Fig. 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. 1.13.3 q Triple Vector Product L z n y 0 r x F Fig. 1.17 (a) Angular Momentum: Let a particle of mass m be at rest on a rotating rigid body such as the earth, Fig. 1.18. Then the angular momentum of m about point 0 12 Mechanics of Particles, Waves & Oscillations is defined as, J = r × (mv) = mr × v. But since v = w × r, we find J = mr × (w w × v) w v (b) Centripetal Acceleration: The centripetal accelw × v). For the eration of m in Fig. 1.18 is a = w × (w special case when r is perpendicular to w, the above w2 r so result reduces to the familiar formula, a = –w that the acceleration is towards the centre of the circle and of magnitude w2r. m r 0 1.14 DIFFERENTIATION OF VECTORS Fig. 1.18 Let R(u) be a vector which is a function of a single scalar variable u. Then, DR = R ( u + Du ) – R where D u denotes an increment in u, as in Fig. 1.19. The derivation of the vector R (u) with respect to the scalar u is given by: O R (u + Du ) DR R (u + D u) - R (u) = Du Du R (u ) dR = Lim Du Æ 0 du DR R (u + D u) - R (u) Fig. 1.19 = Lim Du Æ 0 Du Du provided the limit exists. Since dR/du is itself a vector depending on u, if its derivative with respect to u exists, then it is denoted by d2R/du2. Similarly, higher order derivatives may be defined. If r = xi + yj + zk, is the position vector of a moving particle P (x, y, z) in space, then, dr = dx i + dy j + dz k (15) and the velocity is v = dr dx i dy j dz k = + + dt dt dt dt (16) and the acceleration is a = d2r dt2 = d 2 x i + d 2 y j + d 2 z k dt 2 dt 2 dt 2 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 = –w2r. Referring to Fig. 1.20, r = r cosq i + r sinq j therefore, v = dr dr dq . = dt dq dt Vector Calculus 13 y dq = (– r sinq i + r cosq j ) dt dv d2 r a = = dt dt2 and = (– r cosq i – r sinq j ) therefore, a = – w2 r v P FG dq IJ H dt K r 2 q x 1.14.1 Differentiation Formulae Fig. 1.20 d dA dB (A + B) = + du du du (i) (18) (ii) d dB dA (A . B) = A . . B + du du du (19) (iii) d dB dA (A × B) = A × × B + du du du (20) d dA df A (fA) = f + du du du (iv) (v) (vi) (21) dC dB dA d +A . ¥C+ (A . B × C) = A . B × . B× C du du du du d {A × (B × C)} = A × du FG B ¥ dC IJ H du K + A× FG dB ¥ CIJ H du K + dA × (B × C) du (22) (23) The order in the above products may be important. 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 r2 = r . r = Constant, and v2 = v . v = Constant. Differentiating the given equation with respect to time, 2 r . r = 0 = 0 2 v . v or r . v =0 (i) or v . a =0 (ii) or r . a = – v2 Differentiating (i), r . a+ v . v = 0 (iii) By (i), r is perpendicular to v and by (ii), a is perpendicular to v. It 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 = –v2 Since cosq is negative, q = 180°. By (iv), |r| |a| (–1) = –v2 or a = v2/r. (iv) 14 Mechanics of Particles, Waves & Oscillations Example 17 If A and B are differentiable functions of a scalar u, prove that d dB dA (A . B) = A . . B + du du du Let A = A1 i + A2 j + A3 k and B = B1 i + B2 j + B3 k. d d (A . B) = (A1 B1 + A2 B2 + A3 B3) du du Then = FG A dB + A dB + A dB IJ H du du du K F dA B + dA B + dA B IJ + G H du K du du 1 1 1 =A . 2 2 2 1 3 3 2 3 3 dB dA . B + du du Example 18 If A and B are differentiable functions of a scalar u, prove that dB dA d + (A × B) = A × × B du du du d d (A × B) = du du i j k A1 B1 A2 B2 A3 B3 i = j k A1 A2 A3 dB1 du dB2 du dB3 du =A× + i dA1 du B1 j dA2 du B2 k dA3 du B3 dB dA + × B du du where we have used a theorem of differentiation of determinants 1.14.2 Rules for Partial Differentiation of Vector Functions If 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, ∂ ∂A ∂B + (A + B) = ∂u ∂u ∂u ∂ ∂A ∂f A (f A ) = f + ∂u ∂u ∂u (24) (25) Vector Calculus 15 ∂ ∂B ∂A + (A . B) = A . . B ∂u ∂u ∂u ∂ ∂B ∂A + (A × B) = A × × B ∂u ∂u ∂u Observe that the order of factor in (27) is important. (26) (27) 1.15 UNIT VECTORS IN PLANE POLAR CO-ORDINATES So far we have expressed vectors in terms of their rectangular components using the unit vectors i, j and k. In 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 Fig. 1.21. 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) er is along the line q = Constant, in the direction of increasing r (r-direction). The other unit vector eq is along the circle r = Constant, 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 er and eq 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 differFig. 1.21 entiate the components only. Example 19 Let (r, q) be the polar co-ordinates describing the position of a particle. If er is a unit vector in the direction of the positive vector r and eq is a unit vector perpendicular to r and in the direction of increasing q, show that: er = cosq i + sinq j, (i) eq = – sinq i + cosq j. Since ∂r ∂r is a vector tangent to the curve q = Constant, a unit vector in the direction of r (increasing r) is given by, er = ∂r ∂r ∂r ∂r Since r = xi + yj = r cosq i + r sinq j (ii) 16 Mechanics of Particles, Waves & Oscillations as in Fig. 1.21, so that ∂r = cosq i + sinq j , ∂r er = cosq i + sinq j . ∂r ∂r = 1 (iii) Further, ∂r ∂q is a vector tangent to the curve r = Constant. A unit vector in this direction is given by, eq = By (ii), ∂r ∂q ∂r ∂q (iv) ∂r = –r sinq i + r cosq j , ∂q ∂r ∂q = r so that (iv) yields: eq = – sinq i + cosq j (v) Example 20 . . . . Show that (a) er = q eq (b) eq = – q er (a) By (iii) of Ex. 19, der ∂er dr ∂er dq . + = er = dt ∂r dt ∂q dt . = (0) (r) + (– sinq i + cosq j) (q ) . = q eq . dr dq . where r and q mean and , respectively. dt dt (b) By (v) of Ex. 19, deq ∂e q dr ∂e q dq . = eq = + dt ∂r dt ∂q dt . . . = (0) (r) + (– cosq i – sinq j ) (q) = – q er Example 21 . . Prove that in polar coordinates (a) the velocity is given by v = r er + r q eq and (b) the acceleration is given by .. .2 .. .. . a = (r - r q )er + (r q + 2 r q) e q (a) r = rer v = dr dr de er + r r = dt dt dt . . . . = r er + r er = r er + r q e q where use has been made of Example (20). Vector Calculus 17 .. dv d . (r er + r q e q ) = dt dt .. . . . .. . . = r er + r er + r q e q + r q e q + r q e .. .. . . . . = r e r + r (q e q ) + r q e q + r q e q + r . .. .. .. = (r - r q2 ) er + (r q + 2 r q) e q (b) a = where we have used the results of Ex. 20 and Ex. 21 (a). The above expressions for velocity and acceleration will be used in Chapter 3 & 4. 1.16 FIELDS In 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. Similarly, 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.). If the physical quantity is a scalar (for example temperature) then we are concerned with a scalar field. If the quantity is a vector (for example electric field, velocity etc.) then we speak of a vector field. 1.17 GRADIENT, DIVERGENCE AND CURL 1.17.1 The Del Operator The vector differential operation Del, symbol —, is defined by ∂ ∂ ∂ ∂ ∂ ∂ i+ j+ k = i (28) +j + k ∂x ∂y ∂z ∂x ∂y ∂z Del is also called Nabla. It 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. It is found useful in defining Gradient, Divergence, Curl and Laplacian. — = 1.17.1.1 Properties of the Del Operator If we multiply a scalar quantity u with this vector operator or —, we obtain FG H IJ K ∂ ∂ ∂ ∂u ∂u ∂u +j + k u = i = grad u. —u = i +j +k ∂x ∂y ∂z ∂x ∂y ∂z (i) If 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: FG H ∂ ∂ ∂ — . v = i +j + k ∂x ∂y ∂z = IJ . ev i + v K x ∂ v x ∂v y ∂v z + + = div v. ∂x ∂y ∂z + v k z yj j 18 Mechanics of Particles, Waves & Oscillations (ii) If we form the vector product of the del operator with v then we obtain FG H IJ × (v i + v j + v k ) K ∂v I F ∂v ∂v I +jG H ∂z - ∂x JK + ∂z JK ∂ ∂ ∂ +j + k — × v = i ∂x ∂y ∂z F GH ∂v z = i ∂y y x y z z x = Curl v. (iii) If we form the scalar product of the del operator with itself then we obtain, FG i ∂ + j ∂ + k ∂ IJ ◊ FG i ∂ + j ∂ + k ∂ IJ = H ∂x ∂y ∂z K H ∂x ∂y ∂z K ∂2 ∂x 2 + ∂2 ∂y 2 + ∂2 ∂z 2 = —2 = Laplacian (iv) The operations with del operator are distributive with respect to addition. That is, —(v1 + v2) = —v1 + —v2 (v) The definition of del operator is independent of the co-ordinate system. 1.17.2 The Gradient Let f (x, y, z) be a differentiable scalar field in a certain region of space x, y, z. Then the gradient of f, symbol —f or grad f, is defined by, —f = FG ∂ i + ∂ j + ∂ k IJ f = H ∂x ∂y ∂z K ∂f ∂f ∂f i+ j+ k ∂x ∂y ∂z (29) 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, z) in the direction n. From the calculus, ∂f ∂f ∂f (30) dx + dy + dz. ∂x ∂y ∂z Let r be the position vector to the point P (x, y, z). r = x i + y j + z k df = r+ If we move to the point Q (x + dx, y + dy, z + dz), as in Fig. 1.22, (31) dr = dx i + dy j + dz k Using (29) and (31), we can take the dot product of dr and —f to yield df as in (30), d f = dr . —f (32) Q z Dr Dr P 0 r y x Fig. 1.22 1.17.2.1 Geometrical Interpretation of —f Consider a surface f (x, y, z) = c (33) Vector Calculus 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. Consequently from (32), dr . —f = 0 (34) Now —f is a vector which is completely determined once f has been differentiated, and since neither dr is zero 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 = Constant. 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, z) = Constant, as in Fig. 1.23. In 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 ds is the directional derivative along n. Since |n| = 1, (35) df = |—f| cosq ds Thus, df/ds (namely |—f|) if we go in the that is df/ds = (36) Fig. 1.23 is the projection of —f on the direction n. The largest value of df/ds occurs if we go in the direction of —f (that is q = 0). On the other hand opposite direction (that is q = 180°) f has the largest rate of decrease, –|—f|. Example 22 If f = x2 y – xz3, find —f. FG H ∂ ∂ ∂ +j + k —f = i ∂x ∂y ∂z IJ ex y - xz j K 2 3 = (2xy – z3) i + x2 j – 3xz2 k . Example 23 If f = 1 , where r = r FG H x 2 + y 2 + z 2 , show that —f = – ∂ ∂ ∂ +j + k —f = i ∂x ∂y ∂z = - IJ ex K 2 r r3 . + y2 + z 2 1 1 1 . 2 x i - . 2 y j - . 2 z k r - ( xi + yj + zk ) 2 2 2 =- 3. = 2 2 2 32 2 2 32 2 (x + y + z ) r (x + y + z ) 20 Mechanics of Particles, Waves & Oscillations Example 24 Find a unit vector normal to the surface, x2y + xz = 2 at the point (1, –1, 1). FG H IJ K ∂ ∂ ∂ (x2y + xz) —( x 2 y + xz ) = i +j + k ∂x ∂y ∂z = (2xy + z) i + x2 j + x k = – i + j + 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 j - i + j + k k - i + + = 3 3 3 ( -1)2 + (1)2 + (1)2 Example 25 Find the directional derivative of f = x2yz + 3xz2 at (1, –1, 2) in the direction 2i + j – 2k. FG H IJ e K ∂ ∂ ∂ —f = i x 2 yz + 3xz +j + k ∂x ∂y ∂z = (2xyz + 3z2) i + x2z j + (x2 y + 6xz) k = 8 i + 2 j + 11 k at (1, –1, 2) The unit vector in the direction of 2 i + j – 2 k is n = 2 i + j - 2k (2)2 + (1)2 + ( -2)2 = 2 1 2 i + j - k. 3 3 3 The required directional derivative is —f . n = (8 i + 2 j + 11 k ) . FG 2 i + 1 j - 2 k IJ H3 3 3 K 4 16 2 22 + = - . 3 3 3 3 Since this is negative, f decreases in this direction. = 1.17.2.2 Properties of the Gradient — (C f) = C —f — (f + Y) = —f + —Y — ( fY) = f—Y + Y—f (37) (38) (39) where f and Y are differentiable scalar functions in some region in space and C is a constant. 1.17.2.3 An Example of a Gradient Consider 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. Vector Calculus 21 Example 26 Prove that —( fY ) = f—Y + Y—f —( fY ) = ∂ ∂ ∂ ( fY)i + ( fY)j + ( fY)k ∂y ∂z ∂x FG H = f = f IJ FG K H ∂Y ∂f ∂Y ∂f i+ f +Y +Y ∂x ∂x ∂y ∂y IJ K FG ∂Y i + ∂Y j + ∂Y k IJ + Y FG ∂f i + ∂f j + ∂f k IJ H ∂x ∂y ∂z K H ∂x ∂y ∂z K = f—Y + Y—f . Example 27 Find the angle between the surfaces x2 + y2 + z2 = 4 and z = x2 + y2 – 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 x2 + y2 + z2 = 4 at (1, –1, 1) is —f1 = — (x2 + y2 + z2) = 2xi + 2yj + 2zk = 2i – 2j + 2k A normal to z = x2 + y2 – 1 or x2 + y2 – z = 1 at (1, –1, 1) is —f 2 = — (x2 + y2 – z) = 2xi + 2yj – zk = 2i – 2j – k. b—f g . b—f g = 1 2 —f1 —f2 cosq, where q is the required angle. Then (2i – 2j + 2k) . (2i – 2j – k) = |2i – 2j + 2k| |2i – 2j – k| cosq 4 + 4 + –2 = or cosq = (2)2 + ( -2)2 + (2)2 6 12 9 = 1 3 (2)2 + ( -2)2 + ( -1)2 cosq = 0.5773. Thus the acute angle is q = 54.7°. 1.17.3 The Divergence Let V(x, y, z) = V1 i + V2 j + V3 k be a differentiable vector field at each point (x, y, z) in a certain region of space. Then the divergence of V, symbol — . V or div V is defined by FG ∂ i + ∂ j + ∂ k IJ . (V i + V H ∂x ∂y ∂z K F ∂ V + ∂ V + ∂ V IJ = G H ∂x ∂y ∂z K —.V = 1 1 Observe that — . V π V . —. 2 3 2 j + V k ) 3 (40) 22 Mechanics of Particles, Waves & Oscillations 1.17.3.1 Physical Significance of Divergence Consider the flow of a fluid of density r(x, y, z) with velocity v(x, y, z), through a small parallelpiped ABCDEFGH (Fig. 1.24) of dimensions dx, dy, dz. We will first calculate the amount of fluid passing along the x-direction through the face EFGH per unit time. The y and z components of the velocity v contribute nothing to the flow through EFGH. The mass of fluid entering EFGH per unit time is given by rvx dy dz. The mass of the fluid leaving the face ABCD per unit time is LMrv N x + Fig. 1.24 OP Q ∂ (rv x )dx dy dz. ∂x The net rate of flow out for these two faces is simply the difference between these two flows, or ∂ (rv x ) dxdydz. ∂x If we also take into consideration the other two faces we find that the total loss of mass per unit time is Net rate of flow out = LM ∂ (rv N ∂x x) + ∂ ∂ ( rv y ) + ( rv z ) ∂y ∂z O Q dxdydz. 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 zero 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. In symbols, the balance equation becomes ∂r = Y – — . (rv) ∂t Calling V = rv, ∂r ∂t In the absence of sources and sinks, Y = 0, and (41) reduces to —. V = Y- (41) Vector Calculus 23 ∂r = 0; Equation of continuity (42) ∂t If 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 zero is called solenoidal. —. V + ∂r = 0, then (41) reduces to ∂t If —. 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. In 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. For 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 (42) also has an application in the interpretation of the wave function in Quantum Mechanics. Example 28 Calculate — . (ff), where f = u(x, y, z) i + v(x, y, z) j + w(x, y, z) k . — . (ff) = ∂ ∂ ∂ (fu) + (fv) + (fw) ∂x ∂y ∂z =f FG ∂u + ∂v + ∂w IJ + FG u ∂f + v ∂f + w ∂f IJ H ∂ x ∂ y ∂ z K H ∂x ∂y ∂z K = f— . f + f . —f Example 29 Calculate — . f if f = r/r3. (inverse-square force) — . (r–3 r) = r–3 — . r + r . —r–3 But FG H ∂ ∂ ∂ +j + k — . r = i ∂x ∂y ∂z = ∂x ∂y ∂z = 3 + + ∂x ∂y ∂z and by problem (16), —rn = nrn–2 r IJ . ( i x + j y + k z) K 24 Mechanics of Particles, Waves & Oscillations — r–3 = –3 r–5 r so that \ — . (r–3 r) = 3r–3 – 3r–5 r . r = 3r–3 – 3r–3 = 0 Thus the divergence of an inverse square force is zero. 1.17.4 The Laplacian The Laplacian, symbol —2 is the divergence of a gradient. FG H ∂ ∂ ∂ +j + k — . —f = i ∂x ∂y ∂z = —2 = where ∂2 f ∂x ∂ + 2 ∂2 f ∂y 2 + ∂x 2 ∂ 2 + ∂2 f ∂z 2 + ∂y 2 2 ∂2 IJ . FG i ∂f + j ∂f + k ∂f IJ K H ∂x ∂y ∂z K = — 2f = Laplacin. ∂z 2 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 2 3 FG H ∂ ∂ ∂ +j + k = i ∂x ∂y ∂z = IJ . efA i + fA j + fA k j K 1 2 3 ∂ ∂ ∂ (fA1 ) + (fA2 ) + (fA3 ) ∂x ∂y ∂z = f = j ∂A1 ∂A ∂A ∂f + f 2 + f 3 + A1 + ∂x ∂y ∂z ∂x FG ∂f i + ∂f j + ∂f k IJ . e A i + A j + A k j H ∂x ∂y ∂z K F ∂ + j ∂ + k ∂ IJ . e A i + A j + A k j + f G i H ∂x ∂y ∂z K 1 2 1 3 2 3 = (—f) . A + f (— . A) Example 31 f = x2 y – 2xz3, find —2 f. If F∂ GH ∂x 2 —2 f = 2 + ∂2 ∂y 2 + ∂2 ∂z 2 I x y - 2xz j JK e 2 3 = 2y – 12xz. 1.17.5 The Curl If V(x, y, z) is a differentiable vector field then the Curl or rotation of V, symbol — × V, Curl V or rot V is defined by Vector Calculus 25 — × V= FG ∂ i + ∂ j + ∂ k IJ × (V H ∂x ∂y ∂z K i ∂ = ∂x Vx j ∂ ∂y Vy i + V2 j + V3 k ) k ∂ ∂x Vz ∂ ∂ ∂ = ∂y ∂z i - ∂x Vx V y Vz F ∂V GH ∂y 1 ∂ ∂ j + ∂x ∂z Vz Vx I FG JK H ∂ ∂y V ∂V y ∂V x ∂V z i+ j+ ∂z ∂z ∂x Curl V represents the rotation or vorticity in the fluid. If the flow is irrotational, = z - IJ K — × V = 0; Condition 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. Curl v = — × (w w × R). By identity (13), A × (B × C) = B (A . C) – C(A . B). Therefore, Curl v = w(— . R) – R(— . w). But, — . R = 3, by Example (29). Also, R (— . w) = (w w . —)R since w is a constant vector. RS T UV W ∂ ∂ ∂ Therefore, (w w . —)R = w x ∂x + w y ∂y + w z ∂z ( i x + j y + k z) = i wx + j wy + k wz = w Therefore, Curl v = 3w – w = 2w. 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 Significance of the Curl The physical significance of the curl is brought about by considering the circulation of fluid around a differential loop in the xy-plane, Fig 1.25. Circulation1234 = z z z v x ( x, y) dx + v y ( x, y) dy - v 1 2 3 Use the Taylor expansion about the point (x0, y0), taking into account the displacement of line segment 3 from 1 and 2 from 4. v y ( x0 + dx, y0 ) = v y ( x0, y0 ) + F ∂v I dx + . . . GH ∂x JK y x0 , y0 26 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 vy with y is cancelled by the corresponding term in the fourth integral. Circulation1234 = vx (x0, y0)dx y ∂v L O dx P dy + Mv ( x , y ) + ∂x N Q L O ∂v dy P ( - dx ) + v + Mv ( x , y ) + y ∂ N Q F ∂v - ∂v I dxdy = G H ∂x ∂y JK y x 0 0 y 0 0 x 0 y 0 + dy (x 0 + 2 4 x 0, y0 y x 3 (– x 0 1 Fig. 1.25 y x Dividing by dxdy, we have Circulation per unit area = — × v z The circulation about the differential area in the xy-plane is given by the z-component of — × v. In fluid dynamics — × v is called the “vorticity”. In principal, the curl — × v at (x0, y0) can be determined by inserting a (differential) paddle wheel into the moving fluid at the point (x0, y0). 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. 1.17.5.2 Examples of the Curl of Vector Field (i) The curl of a vector field implies circulation or vortex motion (rotation). If the fluid velocity v has a ‘curl’ at some point then that signifies the existence of vorticity at that point. Since the line integral of a conservative field A around any closed path z is zero, that is A . dr = 0, the conservative fields have zero curl at all points of space. (ii) An example of the conservative vector field is the electrostatic field E. Therefore, curl E = 0. (iii) For waves in an elastic medium, if the displacement U is irrotational, — × U = 0, plane waves (or spherical waves at large distances) become longitudinal. If 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 S (secondary waves) (See 13.1.2) (iv) If 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. It can be shown that curl v = 2w (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 Calculus 27 Example 32 Prove that Curl of a gradient is zero. — × (—f) = = i i ∂ ∂x j ∂ ∂y k ∂ ∂z ∂f ∂x ∂f ∂y ∂f ∂z F ∂ f - ∂ f I - jF ∂ f - ∂ GH ∂y ∂z ∂z ∂y JK GH ∂x ∂z ∂z 2 2 2 2 = 0 since terms in brackets cancel in pairs. Example 33 Prove that Curl Curl V = grad div V – —2 V. By identity (13), A × (B × C) = B (A . C) – (A . B) C. Putting A = —, B = — , C = V, — . —) V = grad div V – —2 V. — × (— — × V) = — (— — . V) – (— Example 34 — × V) = 0 Show that the divergence of a curl is zero, that is — . (— FG i ∂ + j ∂ + k ∂ IJ H ∂ x ∂ y ∂z K . = FG i ∂ + j ∂ + k ∂ IJ H ∂ x ∂ y ∂z K . = ∂ ∂ ∂ ∂ ∂ ∂y ∂z ∂x ∂x V V ∂y V x y z = ∂ ∂ ∂ ∂x ∂y ∂z ∂ ∂ ∂ ∂x ∂y ∂z Vx V y Vz i ∂ ∂x Vx LM MMi N ∂ ∂y Vy j ∂ ∂y Vy k ∂ ∂z Vz ∂ ∂ ∂z - j ∂x Vx Vz ∂ ∂ ∂z + ∂z Vz = 0 since two rows of the determinant are identical. ∂ ∂ k + x ∂ ∂z Vz Vx 28 Mechanics of Particles, Waves & Oscillations Example 35 If A and B are irrotational, prove that A × B is solenoidal. By problem — × A = 0 and — × B = 0 It follows that B . (— — × A) = 0 A . (— — × B) = 0 — × B) = 0 Subtracting B . (— — × A) – A . (— By problem (28), 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 F(r). (a) Show that the field is irrotational. (b) What should be the function F(r) so that the field is solenoidal? (a) That Curl A = 0 for the central field (criterion for the conservative forces) is proved in Chapter 4. (b) If the field is solenoidal, then, — . r F(r) = 0 ∂ ∂ ∂ [ xF (r )] + [ yF (r )] + [ zF ∂x ∂y ∂z =0 ∂F ∂F ∂F +F+y +F+z =0 ∂x ∂y ∂z ∂F x ∂F y +y +z 3F (r ) + x =0 ∂r r ∂r r F+x FG IJ H K 3F(r) + FG IJ H K FG ∂F IJ FG x H ∂r K H 2 + y2 + z 2 r I JK =0 But x2 + y2 + z2 = r2 ∂F ∂r = -3 F r Integrating, ln F = – 3 ln r + In C, where ln C = constant. C ln F = ln C – ln r3 = ln 3 r Therefore F = C/r3. The field is A = r/r3 (inverse square law). therefore 1.17.6 Table of Vector Identities Involving — Here A and B are differentiable vector functions, and f and Y are differentiable scalar functions of position (x, y, z) and r is the position vector. 1. —(f + Y) = —f + —Y 2. —(fY) = f—Y + Y—f 3. — . (A + B) = — . A + — . B Vector Calculus 29 —f) . A + f(— — . A) 4. — . (fA) = (— 5. — × (A + B) = — × A + — × B —f) × A + f(— — × A) 6. — × (fA) = (— — × A) – A . (— — × B) 7. — . (A × B) = B . (— — . A) – (A . —)B + A(— — . B) 8. — × (A × B) = (B . —)A – B(— — × A) + A × (— — × B) 9. — (A . B) = (B . —)A + (A . —) B + B × (— —f) = div grad f = Laplacian f 10. — . (— = ∂2 f 2 + ∂2 f 2 ∂2 f + ∂x ∂y ∂z 2 —f) = 0. The curl of gradient of f is zero. 11. — × (— — × A) = 0 The divergence of curl of A is zero. 12. — . (— — × A) = —(— — . A) – —2A 13. — × (— —f × —Y) = 0 14. — . (— 15. — . r = 3 16. — × r = 0 17. (A . —) r = A 1.18 VECTOR INTEGRATION 1.18.1 Line integral This is an extension of the line integral described in Scalar 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) z C f dr (b) z C A . dr (c) z C A × dr where f is a scalar, A is a vector, and r is the position vector r = xi + yj + zk. Each of these is a line integral along the curve C. The result of integration is a vector for (a), a scalar for (b) and a vector for (c). When the space curve C 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 z f dr. We can similarly write for (b) and (c). The movement along the closed written as C curve C 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. It must be emphasised that in a line integral there is only one independent variable because we are constrained to remain on a curve. In two dimensions, the equation of a curve becomes y = F(x), where x is the independent variable. In three dimensions, we could take x as the independent variable and find y and z as functions of x. Alternatively, the parameter t may be taken as independent variable so that x, y, z 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 Calculation of Work Done by a Varying Force on a Body Using the Line Integral (b) dr Work done by a force on an object which undergoes an infinitesimal vector displacement dr can be written as dW = F . dr. In general, the force F acting on the object varies from point to point. For example, the force on a charged z particle in an electric field would be a function of x, y, z. However, along a curve x, y, z are related by the B equation of the curve. Since along a curve there is only one independent variable, we can write F and dr = i dx F + j dy + k dz as functions of a single variable. The A integral of dW = F . 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 F in moving an x object say from A to B, can be determined (Fig. 1.26). Fig. 1.26 1.18.1.2 Examples of Line Integral (i) The line integral, z F . dr represents the work done by the force along the curve C. (ii) The quantitative relationship between current i and the magnetic field B is given by Ampere’s law z B . dl = m0i (iii) The potential difference between two points A and B is related to the electric field by the line integral B z - E . dl = VB – VA A B (iv) If A = grad f, then the integral z A . dr depends only on initial and final values of A f and is independent of the path. Also z A . dr = 0. z Conversely, 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 (See Example 43). (v) Magnetostatics also provides a physical example of the third type of line integral of a loop of wire C carrying a current I placed in a magnetic field B, then the force dF on a small length dr of the wire is given by dF = I dr × B, so that the total vector force on the loop is F = I z dr ¥ B. C Example 37 Evaluate z C f dr if f = f(x, y, z). Vector Calculus 31 Using the differential displacement vector dr = dx i + dy j + dz k , we find. z f dr = c z z z z f(dx i + dy j + dz k ) = i fdx + j fdy + k fdz c c c c where we have taken i , j and k outside the integral as they have constant magnitude and direction. Example 38 Evaluate vector. z C z C A . dr if A = A1(x, y, z) i + A2(x, y, z) j + A3(x, y, z) k and r is the radius A . dr = = z z C C ( A1 i + A2 j + A3 k ) . (dxi + dyj + ( A1dx + A2 dy + A3 dz). If A is the force F on a particle moving along C, this line integral represents the work done by the force. Example 39 z Evaluate C A . dr from the point P (0, 0, 0) to Q (1, 1, 1) along the curve r = i t + j t2 + k t3 with x = t, y = t2, z = t3, where A = xy i + z j + xyz k . Since x = t, y = t2, z = t3, y = x2, z = x3, dy = 2x dx, dz = 3x2 dx. z C A . dr = z z 1 = (xy i + z j + xyz k ) . ( i dx + j dy + k dz) = 1 1 z z x 3 dx + 2 x 4 dx + 3 x 8 dx = 0 0 0 59 60 z (xydx + zdy + xyzdz) y (1 , 1 ) 2 Example 40 Evaluate z C y =x A . dr around the closed curve C defined by y= x y = x2 and y2 = x, with A = (x – y)i + (x + y)j. z C A . dr = = z z z [(x – y) i + (x + y) j ] . [ i dx + j dy] C 1 = Fig. 1.27 1 (x – x2)dx + 0 z (x + x2) . 2xdx (along y = x2) 0 0 + x 0 (x – y)dx + (x + y)dy C 2 z 1 0 (y2 – y) . 2ydy + z (y2 + y)dy (along y2 = x) 1 = 2/3 If the movement was opposite to the indicated direction then the result would have 2 been - . 32 Mechanics of Particles, Waves & Oscillations Example 41 Evaluate z C A × dr if A = A1 i + A2 j + A3 k i A × dr = A1 dx j A2 dy k A3 dz = i(A2dz – A3dy) + j(A3dx – A1dz) + k(A1dy – A2dx). So the integral is: z C z A × dr = i ( A2 dz - A3 dy) + j z C ( A3 dx - A Example 42 Solve 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. Consider a planet of mass m around a heavy Sun of mass M at S Fig. 1.28. 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 Sun, and the polar angle q, measured with respect to the polar axis. The force acting on the planet due to the Sun is F = – (GmM/r3)r P m r q M k S Fig. 1.28 By Newton’s second law, GmM r so that Now 3 r = m d2 r dt 2 = m dv dt dv GM = - 3 r dt r d dv dr + (r × v) = r × × v dt dt dt (46) (47) dv d (r × v) = r × (48) dt dt since the second term on the right hand side of (47) vanishes, being the cross product of parallel vectors. Using (46) in (48) Therefore d (r × v) = r × dt FG - GM rIJ H r K 3 = 0 r × v = h = const . vector Vector Calculus 33 or r× dr dt =h (49) dr = twice the area of the sector, we get 2(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. Further, using the fact that if two vectors in a triple scalar product are identical, the product is zero, we find r . [r × dr/dt] = r . h = 0. Thus r remains perpendicular to the fixed vector h, and the motion is planar. Since r ¥ Taking the cross product of (46) with h, GMr dv ¥ h = - 3 ¥ h = - GMr × (r × v) dt r r3 where we have used (49). Also, (50) d dv dh dv ¥h + ¥h (v × h) = × v= dt dt dt dt (51) since h is constant. Combining (50) and (51), d GMr (v × h) = - 3 × (r × v) dt r Now, r = r er, where er is the unit vector. Hence der dr dr er + v= = r dt dt dt (52) so that (52) is reduced to FG H de r GM d (v × h) = - 3 r ¥ r ¥ r dt r dt = - GM LMFG e NH r . IJ K = – GM er × FG e H r ¥ de r dt IJ K IJ K der de er - (er . er ) dt d where we have used the identity (13), and the relations er . er = 1 and er . der d (v × h) = GM dt dt der dr = 0. (53) Integrating (53), we find v × h = GM er + k whence (54) r . (v × h) = r . (GM er + k) = GM r . er + r . k = GMr + rk cosq where k is an arbitrary constant vector with magnitude k, and q is the angle between k and er. Using the result of problem 9. 34 Mechanics of Particles, Waves & Oscillations r . (v × h) = (r × v) . h = h . h = h2 h2 GM + K cosq This is the polar equation of conic section. For 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 Sun as one of the foci. Kepler’s third law can be deduced without further use of vectors as indicated in Chapter 4. r = Example 43 (a) If F = —f, where f is single-valued and has continuous partial derivatives, show that the work done is moving a particle from point A [x1, y1, z1] to B[x2, y2, z2] in this field is independent of the path joining the two points. z z z FGH z z (b) Conversely, if C F . dr is independent of the path C joining any two points show that there exists a function f such that F = —f. B (a) Work done = B z F . dr = A B = A B = A —f . dr A ∂f ∂f ∂f i+ j+ k ∂x ∂y ∂z IJ K ∑ (dx i + dy j + dz k ) ∂f ∂f ∂f dx + dy + dz ∂x ∂y ∂z B = A df = f(B) – f(A) = f(x , y , z ) – f(x , y , z ). 2 2 2 1 1 1 B II Thus the integral depends only on the points A and B, not on the path joining them. It is assumed that f(x, y, z) is single valued at A and B. In Fig. I A 1.29 two paths I and II are shown. The above statement would then imply that the result of integration along these two paths would be Fig. 1.29 identical. (b) Suppose the line integral is independent of the path C, that is, F is a conservative field, then ( x , y, z ) f(x, y, z) = z ( x1 , y1 , z1 ) ( x , y, z ) F . dr = z ( x1 , y1 , z1 ) F . dr ds ds df dr =F . . ds ds dr df dr = 0. But = —f . ; so that (—f - F) . ds ds ds Since this result must be valid irrespective of (dr/ds), we deduce F = —f. Differentiating, Vector Calculus 35 Example 44 (a) If F is a conservative field, prove that Curl F = — × F = 0, that is F is irrotational. (b) Conversely, if — × F = 0, prove that F is conservative. (a) If F is a conservative field then by Example 43, F = —f. Then curl F = — × —f = 0 where we have used the fact that the curl of a gradient of f is zero. i ∂ (b) If — × F = 0, then ∂x F1 so that j ∂ ∂y F2 k ∂ = 0, ∂z F3 ∂F3 ∂F2 = ∂y ∂z (55) ∂F1 ∂F3 = ∂z ∂x (56) ∂F1 ∂F2 = ∂y ∂x (57) Required to prove that F = —f Work done to move the particle along a path C joining (x1, y1, z) and (x, y, z) in the force field F is z C F1 (x, y, z)dx + F2 (x, y, z)dy + F3 (x, y, z)dz With the choice of a path consisting of straight line segments from (x1, y1, z1) to (x, y1, z1) to (x, y, z1) to (x, y, z) and calling f(x, y, z) the work done along this particular path, we have y x f(x, y, z) = z x1 F1 ( x, y1 , z1 )dx + z F2 ( x, y, z1 ) y1 ∂f = F3 (x, y, z), since differentiation of the first two terms on the right hand ∂z side give zero. whence ∂f = F2(x, y, z1) + ∂y z z z z1 z = F2(x, y, z1) + z1 ∂F3 ( x, y, z ) dz ∂y ∂F2 ( x, y, z ) dz ∂z z = F2(x, y, z1) + F2(x, y, z) |z1 36 Mechanics of Particles, Waves & Oscillations = F2(x, y, z1) + F2(x, y, z) – F2(x, y, z1) = F2(x, y, z) where we have used (55). ∂f = F1(x, y1, z1) + ∂x y z z y1 y = F1(x, y1, z1) + y1 ∂F2 ( x, y, z1 )dy + ∂x ∂F1 ( x, y, z1 ) dy + ∂y z z z1 ∂F3 ( x, y, z ) dz ∂x z z z1 ∂F1 ( x, y, z ) dz ∂z y z = F1(x, y1, z1) + F1(x, y, z1) |y1 + F1(x, y, z) |z1 = F1(x, y1, z1) + F1(x, y, z1) – F1(x, y1, z1) + F1(x, y, z) – F1(x, y, z1) = F1(x, y, z) where we have used (56) and (57). ∂f ∂f ∂f i+ j+ k = —f. If follows that F = F1 i + F2 j + F3 k = ∂x ∂y ∂z We conclude that a necessary and sufficient condition that a field F be conservative is that Curl F = — × F = 0. Example 45 (a) Show that F = (3x2y + z3) i + x3 j + 3xz2 k is a conservative force field, (b) Find the scalar potential, (c) Find the Work done in moving an object in this field from (1, –1, 2) to (2, 1, 1). (a) It is sufficient to show that Curl F = 0. —× F = i ∂ ∂x 3x 2 y + z 3 j ∂ ∂y x3 k ∂ ∂z 3xz 2 = 0 Thus F is a conservative force field. (b) df = F. dr = (3x2y + z3)dx + x3dy + 3xz2dz = (3x2ydx + x3dy) + (z3dx + 3xz2dz) = d(x3y) + d(z3x) = d(x3y + z3x) f = x3y + z3x + constant. (c) Work done = f(2, 1, 1) – f(1, –1, 2) = 3. 1.18.2 Surface Integrals Let a surface S be divided into infinitesimal elements each of which may be considered as a vector dS. Integrals that involve the differential element dS of the surface area are called Surface Integrals. There are three types of surface integrals. Vector Calculus 37 (a) zz fdS zz (b) S A . dS (c) S zz A ¥ dS S The result of (a) is a vector, that of (b) is a scalar and that of (c) is a vector. If we are dealing with a closed surface, the surface integrals are written as zz zz or f dS S A . dS or S zz A ¥ dS S For closed surfaces, it is usual to assume that the positive direction of the normal extends outward from the surface. (A closed surface is one that has no boundary and completely encloses a bounded region in space). The surface integral zz V . dS is called S the flux of V through the surface, for if V is the product of density and velocity of fluid, the integral represents the amount of fluid flowing through a surface in unit time. Alternatively, the vector V may refer to electric, magnetic, or gravitational force. 1.18.2.1 Examples of Surface integral (1) The total electric charge on the surface of a shell is given by z S r(r) ds. (2) The electric flux fE is given by the surface integral, fE = z E . ds Also gauss’ law is given by e0 z E . ds = q where E is the electric field, e0 the permittivity and q the electric charge. (3) The electromagnetic flux of energy out of a given volume V bounded by a surface S is z S ( E ¥ H ) . ds . 1.18.3 Volume Integrals Let dV = dx dy dz be the volume element. Since this is a scalar we can form only two volume integrals. (a) zzz V (b) f dV zzz V A dV (a) being a scalar and (b) a vector. 1.18.3.1 Examples of Volume Integral (1) Total mass of a fluid contained in a volume V is give by z z V r(r ) dV . (2) Total linear momentum of a fluid is given by r( r ) v(r) dV, where v(r) is the V velocity field in the volume element dV. (3) Consider a small volume element dV situated at position r; its linear momentum rdV . r , where r = r(r) is the density distribution and its angular momentum J is . (r ¥ r) r dV J = z V 38 Mechanics of Particles, Waves & Oscillations . putting r = w × r yields J = z V [r ¥ (w ¥ r )] rdV - z V ( r . w )r r y 1.19 GREEN’S THEOREM IN THE PLANE d zz S d ∂v ( x, y) dxdy = ∂x b ∂v ( x, y) dxdy = x ∂ a zz c z S d z C c a Fig. 1.30 v (b, y) - v (a, y) dy b x Let u(x, y) and v(x, y) be functions with continuous first partial derivatives. Consider the double integral of ∂ v(x, y) over the rectangle S as in Fig. 1.30. We shall ∂x show that the double integral is equal to a line integral around the boundary of the rectangle. We first perform the x integration to find. (58) c We now evaluate v(x, y) dy in the counterclockwise direction so that S is always towards left as we move around the closed curve. We note that along the horizontal sides of S, integrals are zero since dy = 0. Along the right side, x = b, and y ranges from c to d. Along the left side, x = a, and y ranges from d to c. Hence, d z v( x, y) dy = c z c z v (b, y) dy + v (a, y) dy = c d Combining (58) and (59), zz S ∂v dx dy = ∂x d z v (b, y) - v (a, y) dy (59) c z v dy. (60) z u dx. (60a) c Similarly, - zz S ∂u dx dy = ∂y c Adding (60) and (61), z c (u dx + v dy) = zz FGH S IJ K ∂v ∂ u dx dy ∂x ∂y Formula (62) is called Green’s theorem in the plane. This result which was proved for a rectangle can be generalized to any arbitrary figure in the xy-plane bounded by a closed curve as in Fig. 1.31. Let the figure S be divided into a set of small rectangles. For each rectangle the result of (62) is valid. The double integral over the whole area S is equal to the sum of all the double integrals in (62). However, it is seen that for two typical small rectangles P and Q, the line integrals are in the opposite (61) Fig. 1.31 Vector Calculus 39 sense over the common boundary LM and therefore get cancelled. What is left over is simply the line integral around C. Hence the result (62) is equally valid for an arbitrary figure in a plane. Thus, with the use of Green’s theorem, the line integral around a closed path can be evaluated or a double integral over the area enclosed can be found out. Example 46 z Verify Green’s theorem in the plane for ( x - y) dx + ( x + y) dy where C is the closed curve C of the region bounded by y = x and y = x2. Let u =x– y ∂u = –1 ∂y ; v =x+ y ∂v =1 ∂x The curves y = x and y = x2 intersect at (0, 0) and (1, 1) z z FGH ; IJ K ∂v ∂u dx dy = ∂x ∂y zz 1 - ( -1) dxdy S 1 x z z = 2 x=0y=x 1 = 2 ze 0 dxdy = 2 2 x - x 2 dx = j LM OP MN dyPQ dx 1 O L2 2M x - x P 3 Q N3 1 x 0 x2 zz 32 3 1 0 = 2 3 This is in agreement with the result obtained for the line integral in Ex. 40. 1.20 STROKES’ THEOREM This theorem states that if S is an open, two-sided surface bounded by a simple closed curve C, then, if A has continuous derivatives, z C A . dr = zz S ( — ¥ A ) . n dS = zz ( — ¥ A ) . dS S where C is traversed in the positive direction. In words, “The line integral of the tangential component of a vector A taken around a simple closed curve C is equal to the surface integral of the normal component of the curl of A taken over any surface S having C as its boundary”. This theorem relates an integral over an open surface to the line integral around the curve bounding the surface. Proof Let S be a surface bounded by closed contour C. Let C¢ be its projection on the xy-plane as in Fig. 1.32. Let a vector at each point on the surface be defined as A = A1i + A2j + A3 k. We associate a point P(x, y) on the plane with every point on the surface. Since on the surface, z is a function of x and y, that is z = f(x, y), a function A1(x, y, z) becomes A1(x, y, z) = f(x, y), since the value of f must be equal to the value of A1 on C. We may similarly consider projections on yz and xz-planes. 40 Mechanics of Particles, Waves & Oscillations We are required to prove that zz (— ¥ A ) . n dS = zz z z b dS S g — ¥ A1i + A2 j + A3 k . n dS S S = C A . dr y 0 R x dx dy — ¥ ( A1 i) . n dS is transformed. S Now, i ∂ — × (A1 i) = ∂x A1 P ¢( x, y, z ) C where C is boundary of S and n is a unit vector perpendicular to the surface at any point. We can find out how a typical term zz n j ∂ ∂y 0 FG ∂A H ∂z 1 P ( x, y ) Fig. 1.32 k ∂A ∂A1 ∂ = j 1 -k ∂z ∂y ∂z 0 = [— × (A1i) . n dS] = C¢ n. j- IJ K ∂A1 n . k dS ∂y (62) Since z = f(x, y) is taken as the equation of S, the position vector to any point of S is r = xi + yj + zk = xi + yj + f(x, y)k. ∂z ∂f ∂r k = j+ k = j+ ∂y ∂y ∂y But ∂r is a vector tangent to S and thus perpendicular to n, so that ∂y ∂r ∂z n. n.k = 0 = n . j + ∂y ∂y or n . j = - ∂z n.k ∂y (63) Substitute (63) in (62) FG ∂A H ∂z 1 n.j - FG H IJ K [— × (A1 i] . ndS = - or Now, on S, IJ K ∂A1 ∂z ∂A1 ∂A1 n . k dS = - ∂z ∂y n . k - ∂y n . k dS ∂y FG ∂A H ∂y 1 + ∂A1 ∂z ∂ z ∂y IJ K n . k dS (64) A1(x, y, z) = A1[x, y, f(x, y)] = F(x, y) Therefore, Using (65) in (64) ∂F ∂A1 ∂A1 ∂z = + ∂y ∂y ∂z ∂y [— × (A1 i)] . n dS = - ∂F ∂F n . k dS = dx dy ∂y ∂y (65) Vector Calculus 41 Therefore, zz z S [— × (A1 i] . n dS = zz R ∂F dx dy ∂y – where R is the projection of S on the xy-plane. By Green’s theorem for the plane, the last integral equals C¢ F dx where C' is the boundary of R. Since at each point (x, y) of C' the value of F is the same as the value of A1 at each point (x, y, z) of C, and since dx is the same for both C and C', we must have z C¢ zz or S F dx = [— ¥ ( A1 i)] . n dS = z z C¢ C A1dx A1dx (66) Similar equations are obtained by considering the projections of S on the xz and yzplanes. zz zz S S [— ¥ ( A2 j)] . n dS = [— ¥ ( A3 k )] . n dS = z z C C A2 dy (67) A3 dz (68) Adding (66), (67) and (68), we get the desired result zz (— ¥ A ) . n dS = z C A1dx + A2 dy + A3 dz = z C A . dr (69) Green’s theorem in the plane is a special case of Stoke’s theorem. 1.21 DIVERGENCE THEOREM A method of reducing triple integrals to double integrals is provided by the Divergence theorem of Gauss. z To prove: zzz V — . A dV = zz zzz dS 1 S If A is the flux density of an incompressible fluid then we have shown that — . AdV gives the total amount of fluid flowing out of the volume dV per second. The total flow from a large volume is — . AdV which must be equal to the rate of flow zz A . dS S1 q1 n^ 1 A . n dS q2 dS 2 S2 n^ 2 0 R across all the surfaces of the given volume x This is the proof of Divergence (Gauss) theorem. An Fig. 1.33 analytical proof follows. Let S be a closed surface which is such that any line parallel to the co-ordinate axes cuts S in utmost two points. Let the equation of the upper portion S1 be z = f1(x, y) and that of the lower portion S2, z = f2(x, y). Let the vector field be given by A = A1 i + A2 j + A3 k. Consider the projection of the surface on the xy-plane, called R. 42 Mechanics of Particles, Waves & Oscillations zzz zzz FGH — . A dV = IJ K ∂A1 ∂A2 ∂A3 + + dV ∂x ∂y ∂z Now, zzz V ∂A3 dV = ∂z zzz zz zz V = ∂A3 dzdxdy = ∂z LM MN f1 ( x , y ) R f2 OP PQ ∂A3 dz dx dy z ∂ ( x, y) zz z f1 A3 ( x, y, z ) | dx dy = z = f2 [ A3 ( x, y, f1 ) - A3 ( x, y, f2 )]dy dx R For the upper portion S1, dy dx = cosq1 dS1 = k . n1 dS1 since the normal n1 to S1 makes an acute angle q1 with k. For the lower portion S2, dy dx = –cosq2 dS2 = –k . n2 dS2 since the normal n2 to S2 makes an obtuse angle q2 with k. Then, zz zz A3 ( x, y, f1 )dy dx = R zz zz A3 k . n1 dS1 S1 A3 ( x, y, f2 )dy dx = - A3 k . n 2 dS2 S2 R Therefore, zz zz zz A3 ( x, y, f1 )dy dx – = A3 k . n 1 dS1 + S1 = zz A3 ( x, y, f2 )dy dx R R zz A3 k . n 2 dS2 S2 A3 k . n dS S So that zzz V ∂A3 dV = ∂z zz A3 k . n dS (70) S Similarly from considerations of projections of S on the yz and xz-planes we have zzz zzz V V ∂A1 dV = ∂x ∂A2 dV = ∂y zz zz dS A1 i . n (71) A2 j . n dS (72) S S Adding (70), (71) and (72) zzz FGH V IJ K ∂A1 ∂A2 ∂A3 dV = + + ∂x ∂y ∂z zz S ( A1 i + A2 j + A3 k ) . n dS Vector Calculus 43 zzz or — . A dV = V zz dS A.n (73) S The Divergence theorem gives a passage from volume integral to surface integral. It is a generalization of Green’s theorem in the plane and is called Green’s theorem in space. 1.22 GREEN’S THEOREMS 1.22.1 Green’s First Identity or Theorem zzz [ f— 2 Y + (—f) . (—Y ] dV = zz (74) ( f—Y ) . dS S V This is easily obtained by substituting A = f—Y in (73). 1.22.2 Green’s Second Identity or Theorem zzz ( f — 2 Y - Y— 2 f) dV = V zz (75) ( f—Y - Y—f). dS S This is obtained by interchanging f and Y in (74) and subtracting the result from (74). Green’s identities are most frequently encountered as transformation formulae in mathematical physics. Example 47 Verify Stokes’ theorem for A = (x – y) i + yz2 j – y2z2 k where S is the upper half surface of the sphere x2 + y2 + z2 = 1 and C its boundary. The boundary C of S is a circle in xy-plane of radius unity and centre at the origin. Let x = cos q, y = sin q, z = 0, 0 £ q £ 2 p be the parametric equations of C. Then, z C A . dr = z C ( x - y) dx + yz 2 dy - y 2 z 2 dz 2p = z (cos q - sin q) ( - sin q dq) = p n 0 Also, —× A = S i j ∂ ∂ ∂x ∂y x - y yz 2 k ∂ ∂z - y2 z 2 R C Fig. 1.34 = i(–2yz2 – 2yz2) + j(0) + k(1) = k since z = 0. Then zz S (— ¥ A ) . n dS = zz S k . n dS = zz dx dy R since n . k dS = dx dy and R is the projection of S on the xy-plane. But the area of the circle of unit radius is p. Thus Stokes’ theorem is verified. 44 Mechanics of Particles, Waves & Oscillations Example 48 Given A = 3y i + x j + 2z k , find z ≥ 0. z (— × A) . n dS over the hemisphere x2 + y2 + z2 = a2, First method i ∂ ∂x 3y —× A = j ∂ ∂y x k ∂ ∂z 2z = – 2 k . Since this plane area is in the xy-plane, we have n = k ; (— × A) . n = –2 k . k = –2 so the integral is –2 z ds = – 2 . pa2 = – 2pa2 Second method Using Stoke’s theorem we evaluate z C A . dr = = z z z C A . dr around the circle x2 + y2 = a2 in the xy-plane. (3 y i + x j + 2 z k ) . (dx i + dy j + dz k ) 3 y dx + x dy + 2 z dz. Put x = a cosq, dx = – a sinq dq ; y = a sinq, dy = a cosq dq; z = 0, with 0 £ q £ 2p. z A . dr = – 3a2 2p 2p 0 0 z sin 2 q dq + a 2 z cos2 q dq = – 2pa2. Example 49 Use the divergence theorem to show that radius R and A = i x3 + j y3 + k z3. zz But A . dS = — . A = zzz z S A . dS = 12 pR5 where S is the sphere of 5 — . AdV ∂ 3 ∂ 3 ∂ 3 x + y + z ∂x ∂y ∂z = 3x2 + 3y2 + 3z2 = 3(x2 + y2 + z2) = 3R2, since x2 + y2 + z2 = R2. z A . dS = zzz 3R 2 dV = z = 12 p R 4 dR = z (3R 2 ) (4 p R 2 dR) 12 pR5 . 5 Example 50 Evaluate z A . n dS over the closed surface of an open cylinder, of height h and radius R. Vector Calculus 45 z By the divergence theorem, this is equal to —. A = z A . n dS = Surface of cylinder — . AdV over the volume of the cylinder. ∂x ∂y ∂z + + = 3. ∂x ∂y ∂z z z — . A dV = 3 dV = 3V Volume of cylinder = 3 times volume of cylinder = 3pR2h. Example 51 Show that the radius vector r = xi + yj + zk is irrotational. Solution To show —× r =0 —× r = i ∂ ∂x x j ∂ ∂y y k ∂ ∂z z FG ∂z - ∂y IJ H ∂y ∂ z K F ∂z - ∂x IJ + k FG ∂y - ∂x IJ – j G H ∂x ∂z K H ∂x ∂y K = i =0 + 0 + 0 = 0 Example 52 If f = 4x3 + 3yz2 – z3, find —2f at (1, –1, –1). Solution — 2f = ∂2 f + ∂2 f + ∂2 f ∂x 2 ∂y 2 ∂z 2 = 24x + 0 + (6y – 6z) = [(24) (1) + (6) (–1) – (6) (–1)] = 24 Example 53 Show that A . (B ¥ C) = ( A ¥ B) . C Solution A . (B ¥ C) = (A1 i + A2 j + A3 k) ∑ i B1 C1 j B2 C2 k B3 C3 46 Mechanics of Particles, Waves & Oscillations A1 = = A2 A3 B1 B2 B3 C1 C2 C3 C1 C2 C3 A1 A2 A3 B1 B2 B3 =– C1 C2 C3 B1 B2 B3 A1 A2 A3 = C . ( A ¥ B) = ( A ¥ B ) . C Example 54 Find — FG 1 IJ , where r is position vector. Hr K 2 Solution — FG 1 IJ = FG i ∂ + j ∂ + k ∂ IJ (x H r K H ∂x ∂y ∂z K 2 2 + y2 + z 2 = - 2 ( xi + yj + zk) - (2 xi + 2 yj + 2 zk) = 2 2 2 2 r4 (x + y + z ) = -2r r4 Example 55 Prove that — . r = 3 (r is a position vector). [OU March’97] [OU March’99] Solution —. r = = FG i ∂ + j ∂ + k ∂ IJ . (xi + yj + zk H ∂x ∂ y ∂z K ∂x ∂y ∂z + + = 1 + 1 + 1 = 3 ∂x ∂y ∂z [Refer Example 29] Example 56 For scalar function f(x, y, z) and a vector A = Ax i + Ay j + Az k show that — × (fA) = f (— × A) + —f × A. Solution — × (f A) = — × (fA1 i + fA2 j + fA3 k) = i ∂ ∂x fA1 j ∂ ∂y fA2 k ∂ ∂z fA3 Vector Calculus 47 LM ∂ (fA ) - ∂ (fA )OP i + L ∂ (fA ∂z N ∂y Q MN ∂z L ∂A + ∂f A - f ∂A - ∂f A OP = Mf ∂z ∂z N ∂y ∂y Q L ∂A + ∂f A - f ∂A - ∂fA OP j + LMf ∂A + Mf ∂x ∂x Q N ∂z ∂z N ∂x LF ∂A - ∂A IJ i + FG ∂A - ∂A I = f MG NH ∂y ∂z K H ∂z ∂x K LF ∂f A - ∂f A IJ i + FG ∂f A - ∂f + MG NH ∂y ∂z K H ∂z ∂x = 3 3 2 3 1 2 3 3 1 3 2 = f(— × A) + i ∂f ∂x A1 1 j ∂f ∂y A2 k ∂f ∂z A3 = f(— × A) + (—f) × A Example 57 If f is a scalar quantity show that curl (grad f) = 0. Solution i ∂ ∂x = ∂f ∂x FG ∂f i + ∂f j + ∂f kIJ . H ∂x ∂y ∂ z K j ∂ ∂y ∂f ∂y k ∂ ∂z ∂f ∂z LM ∂ FG ∂f IJ - ∂ FG ∂f IJ OP i - L ∂ FG ∂f N ∂y H ∂z K ∂z H ∂y K Q MN ∂x H ∂z L ∂ F ∂f I ∂ FG ∂f IJ OP k + M G JN ∂x H ∂y K ∂y H ∂x K Q L ∂ f - ∂ f OP i - LM ∂ f - ∂ f OP = M N ∂y∂z ∂z∂y Q N ∂x∂z ∂z∂x Q = 2 =0 2 2 LM ∂ (fA ) - ∂ (fA )OP k ∂y N ∂x Q 2 1 2 3 1 3 — × (—f) = — × + 2 2 2 + OP Q ∂fA2 ∂A ∂f -f 1 A1 k ∂x ∂y ∂y 48 Mechanics of Particles, Waves & Oscillations Example 58 Find —f where f = log|r| Solution |r| = xi + yj + zk. Then |r| = —f = = and f = log |r| = 1 — log (x2 + y2 – z2) 2 RS T 1 R = Si 2 T x = x 2 + y2 + z 2 1 ∂ ∂ i l log ( x 2 + y 2 + z 2 ) + j 2 ∂x ∂y 2 2x 2y + +j 2 + y2 + z 2 x + y2 + z 2 r x i + y j + zk = 2 2 r r Example 59 Prove that — . ( A + B) = — . A + — . B . Solution Let A = A1i + A2 j + A3 k, B = B1i + B2 j + B3 k. Then, — . ( A + B) = FG ∂ i + ∂ j + ∂ k IJ . H ∂x ∂y ∂z K ( A1 + B1 ) i + ( A2 + B2 ) j + ( A3 + = ∂ ∂ ∂ ( A1 + B1 ) + ( A2 + B2 ) + ∂x ∂y ∂z = ∂A1 ∂B1 ∂A2 ∂B2 ∂A3 + + + + + ∂x ∂x ∂y ∂y ∂z = FG ∂ i + ∂ j + ∂ kIJ . (A i + A j + A k) H ∂x ∂y ∂z K F ∂ ∂ j + ∂ kIJ . (B i + B j + B k) + G i+ H ∂x ∂y ∂z K 1 2 1 = —. A + —.B Example 60 Show that div ( fA) = f div A + A . grad f . 3 2 3 1 log (x2 + y2 +z2) 2 Vector Calculus 49 Solution div ( fA ) = — ( fA ) = — . ( fA1i + fA2 j + fA3 k ) = ∂ ∂ ∂ (fA1 ) + ( fA2 ) + ( fA3 ) ∂x ∂y ∂z = ∂A ∂A ∂f ∂f A1 + f 1 + A2 + f 2 + ∂x ∂x ∂y ∂y = FG ∂f i + ∂f j + ∂f k IJ . (A i + A j + A k ) H ∂x ∂y ∂z K F ∂ ∂ j + ∂ k IJ . (A i + A j + A + f G i + H ∂x ∂y ∂z K 1 2 1 3 2 3 k ) = (—f) . A + f (— . A ) Example 61 If A = iy + j(x2 + y2) + k(yz + zx), then find the value of curl A at (1, –1, 1) Solution Curl A = — ¥ A = i ∂ ∂x y j ∂ ∂y x 2 + y2 k ∂ ∂ ∂ 2 ∂ ∂y ( yz + zx ) ( x + y 2 ) – j ( yz + zx ) = i ∂z ∂y ∂z ∂x ∂z yz + zx LM N + k OP Q LM ∂ (x N ∂x 2 + y2 ) - ∂y ∂y LM N OP Q OP Q = (z – 0) i – z j + (2x – 1) k = z i – z j + (2x – 1) k = (1) i – (1) j + [(2)(1) – 1] k = i – j + k Example 62 Find the value of the constant ‘a’ for which the vector A = i(x + 3y) + j(y – 2z) + k(x + az) is a solenoidal vector. Solution A vector A is solenoidal if its divergence is zero. div A = — . A = = FG i ∂ + j ∂ + k ∂ IJ . H ∂x ∂y ∂z K i ( x + y) + j ( y - 2 z ) + k ( x + az ) ∂ ∂ ∂ ( x + y) + ( y - 2z) + (x + ∂x ∂y ∂z 50 Mechanics of Particles, Waves & Oscillations =1 + 1 + a = 0 a = –2 Example 63 Evaluate div F where F = 2x3z i – xy2z j + 3y2 x . Solution FG H ∂ ∂ ∂ +j + k div F = — . F = i ∂x ∂y ∂z = IJ K . (2x3z i – xy2z j + 3y2x k ) ∂ ∂ ∂ (2 x 3 z ) + ( - xy 2 z ) + (3 y 2 ∂x ∂y ∂z = 6x2z – 2xyz + 0 = 6x2z – 2xyz Example 64 If f = xy2 i + 2x2yz j – 3yz2 k find curl f at (1, –1, 1) Solution i ∂ Curl f = — × f = ∂x xy 2 = j k ∂ ∂ ∂y ∂z 2 x 2 yz - 3 yz 2 LM ∂ (- 3yz ) - ∂ (2x yz)OP i – L ∂ (- 3yz ) - ∂ (xy )O j PQ ∂z ∂z N ∂y Q MN ∂x L∂ O ∂ ( xy )P k + M (2 x yz ) x y ∂ ∂ N Q 2 2 2 2 2 = [– 3z2 – 2x2y] i – [0 + 0] j + [4xyz – 2xy] k = (– 3z2 – 2x2y) i + (0) j + (4xyz – 2xy) k = – i – 2 k where we have put x = 1, y = –1 and z = 1. Example 65 Show that the force F = (y2 – x2) i + 2xy j is conservative. Solution To show — ¥ F = 0 —¥F = i ∂ ∂x y2 - x 2 j ∂ ∂y 2 xy k ∂ ∂z 0 LM N OP LM Q N ∂ ∂ (2 xy) - j 0 ( y2 = i 0 ∂z ∂z 2 Vector Calculus 51 = + k LM ∂ (2xy) - ∂ ( y N ∂x ∂y 2 - x2 ) OP Q = 0 + 0 + k (2y – 2y) =0 Example 66 zz Find F . n ds , where F = 4xzi – y2j + yzk and S is the surface of a cube bounded by x S = 0, x = 1, y = 0, y = 1, z = 0 and z = 1 Solution By the divergence theorem, the required integral is equal to zzz — . FdV = zzz LMN zzz z z V V = ∂ ∂ ∂ (4 xz ) + (- y2 ) + (y ∂x ∂y ∂z V 1 = 1 (4z - y) dV = 1 1 1 z z z (4 z - y) dz dy dx x=0 y=0 z=0 2 z 2 - yz x=0 y=0 z =1 z =0 1 dy dx = 1 z z (2 - y) dy dx = x=0 y=0 3 2 QUESTIONS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. If A . B = A . C, must we conclude that B = C? If |A . B| = |A| |B|, what can you say about A and B? If |A + B|2 = |A|2 + |B|2, what can you say about A and B? If A . B = |A × B|, what can you say about the angle between A and B? If A × B = A × C, must we conclude that B = C? Give the geometric interpretation of the identity (A – B) × (A + B) = 2A × B. Give the geomertric interpretation of the identity (A + B) . (A – B) = A2 – B2 The Work-Energy theorem states that the work done on a particle in moving it from A to B is equal to the change in Kinetic energies at these points rather than change in potential energies of these points. Why? In the vector notation the inverse square law of gravitation is written as F = GMmr . How do you accept the use of r3 term in the denominator rather than r2 r3 to describe the inverse square law? Is the order of multiplication important in forming Scalar products? Vector products? Give a physical example of (a) Triple scalar product, (b) Triple vector product. Would you regard Del as a vector, an operator or both? 52 Mechanics of Particles, Waves & Oscillations 13. What is the criterion for a force to be conservative? 14. We can form three types of line integrals and three types of surface integrals but only two types of volume integrals. Why? 15. Does Stoke’s theorem apply to the Fig. 1.35 shown? 16. For what kind of surface is Green’s theorem applicable? 17. For what kind of surface is Stokes’ theorem applicable? 18. For what kind of surface is the Divergence theorem of Fig. 1.35 Gauss applicable? 19. Stokes’ theorem gives a passage from integral to integral. 20. The Divergence theorem gives a passage from integral to integral. 21. (a) Derive an expression for the divergence of a vector field in Cartesian coordinates. (b) Give the physical significance of divergence of vector field. 22. Explain the terms (a) divergence (b) curl of a vector field. Derive expressions in terms of Cartesian components for div A and curl A. Give examples from Physics where these ideas are used. 23. What is meant by curl of a vector function? Derive an expression for it in Cartesian coordinates. 24. (a) Explain the terms (i) Div (ii) Curl (iii) Gradient, with examples. (b) Prove the following: 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. (i) — (u A ) = A . —u + u— . A What are line, surface and volume integrals? Derive Gauss’ divergence theorem. Explain the terms gradient, divergence and curl bringing out their use in physics. Explain curl of a vector with two examples. State and prove Stokes’ theorem. Define curl and div. of a vector. State and prove Green’s theorem. Prove Curl V = 2 w. Explain what is meant by a scalar field and a vector field. Define curl of a vector and explain its physical significance. Explain the physical meaning of divergence of vector and define the curl of a vector. Prove that ‘the gradient of a scalar field is vector’ and explain the physical significance of gradient. Derive an expression for the divergence of a vector field. Explain how a del operator obeys all the rules of differential calculus. What are the properties of vector product? What are their physical applications? Explain the physical significance of divergence with two examples. Explain the physical significance of line integral. Explain the physical significance of gradient of a scalar field with two examples. Prove that if A, B, C are non-zero vectors and (A × B) . C = 0 then A, B, C are coplanar. What is a scalar triple product and a vector triple product? Vector Calculus 53 PROBLEMS 1. Find the work done in moving an object along a vector, r = 2i + 3j – k if the applied force is F = i – 2j – 6k. (Hint: Work done = F . r) [Ans. 2] 2. Prove that |A × B|2 + |A . B|2 = |A|2 |B|2. 3. Evaluate A × B, where A = –3 i – 4 j + 2 k and B = 2 i – j – k . [Ans. 6 i + j + 11 k ] 4. Prove that A × B = –B × A. 5. If A × B = 0 and if A and B are not zero then show that A is parallel to B. 6. Find the angle between A = i – 2j + 2k and B = 2i – 3j + 6k. [Ans. q = 17°45¢] 7. Prove that (A × B) × C = (A . C)B – (B . C)A. 8. Prove that (A × B) × C = A × (B × C) if and only if (C × A) × B = 0. 9. Prove that A . (B × C) = B . (C × A) = C . (A × B). (Hint: Write the triple products in the form of determinants and use the theorem of determinant which states that the interchange of two rows of a determinant changes its sign). 10. Prove that (A × B) × (C × D) = [CDA] B – [CDB] A. 11. Prove that if A, B and C are noncoplanar and A × (B × C) = (A × B) × C = 0, then A, B and C are mutually perpendicular. 12. If a and b are unit vectors and q is the angle between them, show that 1 1 |a - b| = sin q . 2 2 13. Prove that (A – B) × (A + B) = 2A × B. 14. If r = sin t i + cos t j – t k, show that |d2r/dt2| = 1. 15. Show that 16. 17. 18. 19. 20. 21. 22. d dC dB dA (A . B × C) = A . B × + A . × C+ . B × C where A, B, C are du du du du differentiable functions of a scalar u. Show that —r n = nrn – 2r. A particle moves such that its position vector at any time t is given by r = cos wt i + sin wt j, where w is a constant. Show that (a) the velocity v of the particle is perpendicular to r. (b) r × v = a constant vector w k. If f(x, y, z) = 2x2y – y2z3, find grad f at the point (2, 1, –1). [Ans. 8i + 10j – 3k] Prove that —(f + Y) = —f + —Y. If f = ln|r|, prove that —f = er/r2. Find the directional derivative of f = x2yz + xz2 at (1, –1, –1) in the direction i – 2j – 2k. [Ans. 11/3] 2 Find a unit normal to the surface x y + 2xz = 4 at the point (2, –2, –3). LM Ans. 1 i - 2 j - 2 kOP N 3 3 3 Q 54 Mechanics of Particles, Waves & Oscillations 23. Prove that — × (A × B) = (B . —)A – B(— — . A) – (A . —)B + A(— — . B). 24. Prove that —(A . B) = (B . — — × A) + A × (— — × B). —)A + (A . —)B + B × (— 25. Show that A = (y2 cos x + z3)i + (2y sin x – 4)j + (3xz2 + 2)k is a conservative field. 26. If r is the position vector, show that curl r = 0. 27. If f = 3x3 – y2z, find —2f at (1, 1, 8). —f) . A + f(— — . A). 28. Prove that — . (fA) = (— — — — × B). 29. Prove that . (A × B) = B . (— × A) – A . (— 30. Determine the constant C so that the vector V = (x + 2y)i + (y – z)j + (x – Cz)k is solenoidal. 3 2 31. If f = 1/r, prove that —f = – r/r , where r = 32. If f = 1/r, prove that —2f = 0. 33. Prove that — ¥ LM r OP Nr Q 2 2 [Ans. 2] [Ans. 2] 2 x +y +z . = 0. 34. Prove that Curl (fgrad f) = 0. 35. If a = a xi + b yj + g zk, show that —(a . r) = 2a, where r is the position vector. 36. If a is a constant vector and r is a position vector, prove that —(a . r) = a. 1 — v2 – v × (— — × v). 37. Prove that (v . —) —)v = 2 (Hint: In the table of vector identities, entry No, 9, set A = B = V). 38. If f = 2xyz2, F = xyi – zj + x2k and C is the curve x = t2, y = 2t, z = t3 from t = 0 to t = 1, evaluate the line integrals (a) z C f dr z (b) F × dr [Ans. (a) 39. If F = 4xyi – y2j, evaluate z C 8 4 i+ j+ k 11 5 (b) 2 7 -9 i- j+ k ] 10 3 5 F . dr where C is the curve in the xy-plane, y = 2x2, from (0, 0) to (1, 2). 40. If A = (2x2 + 4y) i – 7yz j + 10xz2 k , evaluate the path C with x = t, y = t2, z = t3. z C [Ans. - 2 3 ] A . dr from (0, 0, 0) to (1, 1, 1) along [Ans. 3] z 41. Verify Green’s theorem in the plane C (3xy + 2y2)dx + x2dy where C is the closed curve of the region bounded by y = x and y = x2. 42. Verify Stokes’ theorem for A = (x – y)i – yz2j – y2zk, where S is the upper half surface of the sphere x2 + y2 + z2 = 1 and C is its boundary. [Ans. p] 43. Evaluate zzz — ¥ A dV , where A = yi – xj and R is any three-dimensional region R with volume V. [Ans. –2Vk] Vector Calculus 55 44. Find zz r . dS over the surface S of the sphere x2 + y2 + z2 = a2. [Ans. 4pa3] S 45. Show that curl (A × B) = A div B – B div A. 46. Determine the angle between the vectors. A = i + 2 j + 3 k and B = 3 i – 2 j + 4 k . [Ans. 54°54¢] 47. Show that the triangle with sides A = 3 i + 2 j + k, B = 3i + j - 5k and C = 2i + j - 4k is right angled. [Hint. Show that A . C = 0] 48. If A = 2xi + 2yj + 3zk, find curl A. [Ans. 0] 2 3 2 49. If f (x, y, z) = 3x y – y z , find the value of grad f at the point (1, –2, –1). [Ans. –8i – 12j – 16k] 50. If A = 3i – 4j + 5k and B = 2i + 3j – 4k find the magnitudes of ( A + B) and ( A - B). [Ans. 3 3 and 131 ]