Applied Physics 24PH101T Lecture Notes PDF

Applied Physics Course Code: 24PH101T Applied Physics July 28, 2024 1 / 76 Outline 1 Introduction to Vectors 6 Differential operator 2 Vector Algebra 7 Gradient, Divergence and Curl 3 Vectors in Component Form 8 Vector...

Applied Physics Course Code: 24PH101T Applied Physics July 28, 2024 1 / 76 Outline 1 Introduction to Vectors 6 Differential operator 2 Vector Algebra 7 Gradient, Divergence and Curl 3 Vectors in Component Form 8 Vector Integrals 4 Vector Triple Products 9 Fundamental Theorem of 5 Position, Displacement and Gradients, Divergences and Curl Separation Vectors 10 Curvilinear Coordinates Applied Physics July 28, 2024 2 / 76 Vectors – Introduction – (1) If you walk 4 km due north and then 3 km due east you will have gone a total of 7 km, but you’re not 7 km from where you set out – you’re only 5 km. The line which is 5 km is displacement which doesn’t equal to distance travelled. The straight line segments, like displacements require magnitude (length) as well as direction. Such objects are called Vectors. Applied Physics July 28, 2024 3 / 76 Vectors – Introduction – (2) The physical quantities velocity, acceleration, force and momentum are other examples of vectors. By contrast, quantities that have magnitude but no direction are called scalars: examples include mass, charge, density, and temperature. Notations (In books:) The vectors are denoted by bold face print. for example: A, B, etc. The magnitude of the vector is denoted by normal font or bold face within vertical lines. A or |A|. In diagram, vectors are denoted by arrows with the arrow head indicating direction. Applied Physics July 28, 2024 4 / 76 Vectors – Introduction – (3) Notations (to be used while writing): ⃗ B, he vectors are denoted by ‘→’ over the character. for example: A, ⃗ etc. The magnitude of the vector is denoted by normal font or character ⃗ with ‘→’ face within vertical lines. A or |A|. Minus A⃗ ie. −A, ⃗ is a vector with same magnitude as A ⃗ but of opposite direction. Applied Physics July 28, 2024 5 / 76 Vector Algebra – (1) The are four vector operations: addition and three kinds of multiplication. Addition (i) Vector Addition: Place the tail of B⃗ at the head of A; ⃗ the ⃗ ⃗ sum, A + B, is the vector from the tail of A ⃗ to the head of B.⃗ Addition is commutative, ie. A ⃗+B⃗ =B ⃗ + A. ⃗ Subtraction Addition   is also associative,  ie. ⃗ ⃗ ⃗ A+B +C =A+ B +C.⃗ ⃗ ⃗ For subtraction, it is basically addition  of ⃗ ⃗ ⃗ opposite vector, ie. A − B = A + −B. ⃗ Applied Physics July 28, 2024 6 / 76 Vector Algebra – (2) Multiplication by Scalar (ii) Multiplication by a scalar: Multiplication of a vector by a positive scalar a multiplies the magnitude but leaves the direction unchanged. (If a is negative, the direction is reversed.) Scalar  multiplication  is distributive, ie. ⃗ ⃗ ⃗ c A + B = c A + c B. ⃗ Dot product (iii) Dot product of two vectors: The dot product of two vectors is defined by, ⃗·B A ⃗ = AB cosθ, where θ is the angle they form when placed tail to tail. Applied Physics July 28, 2024 7 / 76 Vector Algebra – (3) Cross Product (iv) Cross product of two vectors: The cross product of two vectors is defined by, A⃗×B ⃗ = AB sinθ nb. nb is a unit vector (vector of magnitude 1) ⃗ and B. pointing perpendicular to the plane of A ⃗ Of course, there are two directions perpendicular to any plane: “in” and “out.” The ambiguity is resolved by the right-hand rule: let your fingers point in the direction of the first vector and curl around (via the smaller angle) toward the second; then your thumb indicates the direction of nb. Applied Physics July 28, 2024 8 / 76 Vector Algebra – (4) Cross Product ⃗×B In the figure, A ⃗ points into the slide. ⃗ ×A On the other hand, B ⃗ points out of the slide. Note that A⃗×B ⃗ is itself a vector (hence the alternative name for this product is vector product). The  cross product   is distributive,   ie.  ⃗× B A ⃗ + C⃗ = A ⃗×B⃗ + A ⃗ × C⃗ Applied Physics July 28, 2024 9 / 76 Vector Algebra – (5) The cross product is not commutative, ie. ⃗ ×A B ⃗ ̸= A ⃗×B ⃗  butin fact, it is, ⃗ ×A B ⃗=− A ⃗×B ⃗ ⃗ × B| Geometrically, |A ⃗ is the area of the parallelogram generated by A ⃗ and ⃗ B. If two vectors are parallel, their cross product is zero. In particular, A⃗×A ⃗ = ⃗0 or 0. Here, ⃗0 or 0 is also a vector with magnitude 0. Applied Physics July 28, 2024 10 / 76 Vectors in Component Form – (1) For vector operations, it is easier to perform in its cartesian components x, y and z. Notations (to be used): The xb, yb and zb are unit vectors defining the directions parallel to x, y and z axes respectively. An arbitrary vector A ⃗ can be expanded and written as: ⃗ = Ax xb + Ay yb + Az zb. A The scalars/numbers Ax , Ay and Az are ⃗ along the three axes. components of A Applied Physics July 28, 2024 11 / 76 Vectors in Component Form – (2) (i) Vector Addition: Add like components ⃗+B A ⃗ = (Ax xb + Ay yb + Az zb) + (Bx xb + By yb + Bz zb) = (Ax + Bx ) xb + (Ay + By ) yb + (Az + Bz ) zb ⃗−B A ⃗ = (Ax xb + Ay yb + Az zb) − (Bx xb + By yb + Bz zb) = (Ax − Bx ) xb + (Ay − By ) yb + (Az − Bz ) zb (ii) Scalar Multiplication: Multiply each component with the scalar. ⃗ = cAx xb + cAy yb + cAz zb cA Applied Physics July 28, 2024 12 / 76 Vectors in Component Form – (3) Important Rule about xb, yb and zb xb · xb = yb · yb = zb · zb = 1 xb · yb = yb · zb = zb · xb = 0 xb × xb = yb × yb = zb × zb = 0 xb × yb = zb , yb × xb = −b z yb × zb = xb , zb × yb = −b x zb × xb = yb , xb × zb = −b y Applied Physics July 28, 2024 13 / 76 Vectors in Component Form – (4) (iii) Dot product: Multiply like components, and add ⃗·B A ⃗ = (Ax xb + Ay yb + Az zb) · (Bx xb + By yb + Bz zb) = (Ax Bx ) + (Ay By ) + (Az Bz ) ⃗ with itself is, A Dot product of A ⃗·A ⃗ = A2 = A2 + A2 + A2 x y z (iv) Cross Product or Vector Product ⃗×B A ⃗ = (Ax xb + Ay yb + Az zb) × (Bx xb + By yb + Bz zb) = (Ay Bz − Az By ) xb + (Az Bx − Ax Bz ) yb + (Ax By − Ay Bx ) zb Applied Physics July 28, 2024 14 / 76 Vectors in Component Form – (5) To calculate the cross product, it is easier to form the determinant whose ⃗ (in component form), and first row is xb, yb and zb, whose second row is A ⃗ whose third row is B (in component form). xb yb zb ⃗×B A ⃗ = Ax Ay Az Bx By Bz Applied Physics July 28, 2024 15 / 76 Example 1: Question: Find the angle between the face diagonals of a cube. Consider a cube of side length 1, The two face diagonals easiest to use, ⃗ = 1 xb + 1 zb and B A ⃗ = 1 yb + 1 zb In component form, ⃗·B A ⃗ =1·0+0·1+1·1=1 In “abstract” form, √ √ ⃗·B A ⃗ = |A|| ⃗ B|⃗ cosθ = 2 2 cosθ = 2 cosθ Comparing the two, one can calculate the angle. θ = 60◦. Applied Physics July 28, 2024 16 / 76 Vector Triple Products – (1) Scalar Triple Product ⃗ × C⃗ produces a vector. B ⃗ with this is possible. A dot product of A   ⃗· B A ⃗ × C⃗   ⃗· B Geometrically, |A ⃗ × C⃗ | is the volume of the parallelepiped which is ⃗ B generated by A, ⃗ and C⃗. Also, the product is cyclic in alphabetical order,       ⃗· B A ⃗ × C⃗ = B ⃗ · C⃗ × A⃗ = C⃗ · A⃗×B ⃗ Applied Physics July 28, 2024 17 / 76 Vector Triple Products – (2) The product is also cyclic in non-alphabetical order,       ⃗ · C⃗ × B A ⃗ =B ⃗· A ⃗ × C⃗ = C⃗ · B ⃗ ×A ⃗ In component form,   Ax Ay Az ⃗· B A ⃗ × C⃗ = Bx By Bz Cx Cy Cz Applied Physics July 28, 2024 18 / 76 Vector Triple Products – (3) Vector Triple Product ⃗ × C⃗ produces a vector. Since, B ⃗ with this is possible. A vector product of A   ⃗× B A ⃗ × C⃗ It can be simplified in the form of BAC–CAB rule:       A⃗× B ⃗ × C⃗ = B ⃗ A⃗ · C⃗ − C⃗ A⃗·B ⃗ Applied Physics July 28, 2024 19 / 76 Position, Displacement and Separation Vectors – (1) Position Vector The location of point in three dimensions can be described by listing its Cartesian coordinates. The vector to the point from the origin is called position vector. ⃗r = x xb + y yb + z zb Notations (to be used): ⃗r is reserved as position vector. p Its magnitude is, r = x 2 + y 2 + z 2. Unit vector pointing radially outward, ⃗r x xb + y yb + z zb rb = = p r x2 + y2 + z2 Applied Physics July 28, 2024 20 / 76 Position, Displacement and Separation Vectors – (2) Notations (to be used): The infinite displacement vector from point (x,y ,z) to (x + dx,y + dy ,z + dz) is, ⃗ = dx xb + dy yb + dz zb dl In electrodynamics, the typical problem involves source point where the charge is placed and a field point where the electric or magnetic field is measured. Here, the r⃗′ represents the source point and ⃗r represents the field point. The separation vector from source point to field point is represented by r⃗ r⃗ = ⃗r − r⃗′ Applied Physics July 28, 2024 21 / 76 Position, Displacement and Separation Vectors – (3) Notations (to be used): The magnitude of separation vector, r = |⃗r − r⃗′ | r⃗ ⃗r − r⃗′ r = = The unit vector in direction from r⃗′ to ⃗r , c r |⃗r − r⃗′ | In Cartesian form, r⃗ = x − x ′ xb + y − y ′ yb + z − z ′ zb


q r = (x − x ′ )2 + (y − y ′ )2 + (z − z ′ )2 (x − x ′ ) xb + (y − y ′ ) yb + (z − z ′ ) zb r =q c (x − x ′ )2 + (y − y ′ )2 + (z − z ′ )2 Applied Physics July 28, 2024 22 / 76 Example 2 – (1) Example 2: Find the separation vector r⃗ from the source point (2, 8, 7) to the field point (4, 6, 8). Determine its magnitude ( r ), and construct the unit r. vector c Separation Vector Source Point: r⃗′ = 2b x + 8b y + 7b z Field Point: ⃗r = 4b x + 6b y + 8b z Separation Vector, r⃗ = ⃗r − r⃗′ = 4b x + 6b z − 2b y + 8b x + 8b y + 7b z r⃗ x − 2b = 2b y + 1b z Applied Physics July 28, 2024 23 / 76 Example 2 – (2) Magnitude r⃗ x − 2b = 2b y + 1b z q Magnitude, r = 22 + (−2)2 + 12 √ r = 5 Unit Vector r⃗ r = r c x − 2b 2b y + 1b z r = c √ 5 2 2 1 r = √ xb − √ yb + √ zb c 5 5 5 Applied Physics July 28, 2024 24 / 76 “Ordinary” Derivatives Suppose we have a function of one variable: f (x). Question: df What does the derivative, dx , do for us? Answer: It tells us how rapidly the function f (x) varies when we change the argument x by a tiny amount, dx. If we increment x by an infinitesimal amount dx, then f changes by an amount df ; the derivative is the proportionality factor. Applied Physics July 28, 2024 25 / 76 Gradient – (1) Suppose, now, that we have a function of three variables - say, the temperature T (x, y , z) in this room. Here, We want to generalize the notion of “derivative” to functions like T , which depend not on one but on three variables. A derivative is supposed to tell us how fast the function varies, if we move a little distance. A theorem on partial derivatives states that:       ∂T ∂T ∂T dT = dx + dy + dz ∂x ∂y ∂z This tells us how T changes when we alter all three variables by the infinitesimal amounts dx, dy , dz. Applied Physics July 28, 2024 26 / 76 Gradient – (2) In fact, the equation of dT is a dot product:   ∂T ∂T ∂T dT = xb + yb + zb · (dx xb + dy yb + dz zb) ∂x ∂y ∂z   = ∇T⃗ ⃗ · dl ⃗ is the gradient of function T and it is a vector quantity. where, ∇T Geometrical Interpretation of the Gradient: ⃗ points in the direction of maximum increase of the The gradient ∇T function T. ⃗ | gives the slope (rate of increase) along this The magnitude |∇T maximal direction. Applied Physics July 28, 2024 27 / 76 Example 3: – (1) Example 3: p Find the gradient of r = x 2 + y 2 + z 2 (the magnitude of the position vector. Solution ⃗ = ∂r xb + ∂r yb + ∂r zb ∇r ∂x ∂y ∂z Applied Physics July 28, 2024 28 / 76 Example 3: – (2) ∂r ∂ p 2  = x + y 2z 2 ∂x ∂x 1 ∂ x2 + y2 + z2
= p 2 x 2 + y 2 + z 2 ∂x 2x x = p = 2 2 x +y +z 2 2 r Similarly ∂r y ∂r z = and = ∂y r ∂z r Applied Physics July 28, 2024 29 / 76 Example 3: – (3) Substituting, ⃗ = x xb + y yb + z zb ∇r r r r x xb + y yb + z zb = z ⃗r = = rb r Applied Physics July 28, 2024 30 / 76 “del” Operator – (1) ⃗ “multiplying” a The gradient has the formal appearance of a vector, ∇, scalar T :   ∂ ∂ ∂ xb + yb + zb T ∂x ∂y ∂z “del” operator The term in this equation is “del”: ⃗ = ∂ xb + ∂ yb + ∂ zb ∇ ∂x ∂y ∂z It is not a vector, in the usual sense. It doesn’t mean anything until we provide it with a function to act upon. Applied Physics July 28, 2024 31 / 76 “del” Operator – (2) “del” does not multiply T ; rather, it is an instruction to differentiate. ⃗ is a vector operator that acts upon T , To be precise, then, we say that ∇ not a vector that multiplies T. “del” operations Vector Multiplication 1 ⃗ – On a scalar function T, ∇T 1 ⃗ By a scalar a, Aa Gradient 2 By a vector B via dot ⃗ via dot ⃗ · B. ⃗ 2 On a vector function V product A ⃗ ·V⃗ – Divergence product, ∇ 3 By a vector B via cross ⃗ via cross ⃗ × B. ⃗ 3 On a vector function V product A ⃗ ⃗ product, ∇ × V – Curl Applied Physics July 28, 2024 32 / 76 The Divergence – (1) Divergence of Vector   ⃗ = ⃗ ·V ∂ ∂ ∂ ∇ xb + yb + zb · (Vx xb + Vy yb + Vz zb) ∂x ∂y ∂z ∂Vx ∂Vy ∂Vz = + + ∂x ∂y ∂z Geometrical Interpretation ⃗ is a measure of how much ⃗ ·V The name divergence is well chosen, for ∇ ⃗ the vector V spreads out or diverges from the point in question. Applied Physics July 28, 2024 33 / 76 The Divergence – (2) The vector function has a large positive The vector function The vector function divergence, (if the has zero divergence. has a positive arrows are pointed in, divergence. it would be a negative divergence). Applied Physics July 28, 2024 34 / 76 The Divergence – (3) Physical Examples of Divergence: Imagine standing at the edge of a pond. Sprinkle some sawdust or pine needles on the surface. If the material spreads out, then you dropped it at a point of positive divergence; if it collects together, you dropped it at a point of negative divergence. Water going down the drain Water coming out of sprinkler Negative divergence Positive divergence Applied Physics July 28, 2024 35 / 76 Example 4 – (1) Example 4: Calculate the divergences of the functions, given in the figure. V⃗b = zb V⃗c = z zb V⃗a = ⃗r = x xb + y yb + z zb Applied Physics July 28, 2024 36 / 76 Example 4 – (2) V⃗a ⃗ · V⃗a = ∂x + ∂y + ∂z = 1 + 1 + 1 = 3 ∇ ∂x ∂y ∂z V⃗b ⃗ · V⃗b = ∂0 + ∂0 + ∂1 = 0 ∇ ∂x ∂y ∂z V⃗c ⃗ · V⃗c = ∂0 + ∂0 + ∂z = 1 ∇ ∂x ∂y ∂z Applied Physics July 28, 2024 37 / 76 The Curl – (1) Curl of Vector xb yb zb ∇ ⃗ = ⃗ ×V ∂ ∂ ∂ ∂x ∂y ∂z Vx Vy Vz       ∂Vx ∂Vy ∂Vx ∂Vz ∂Vy ∂Vx = − x+ b − y+ b − zb ∂y ∂z ∂z ∂x ∂x ∂y ⃗ is, like any cross product, a vector. The curl of a vector function V Curl is... ⃗ swirls around the point in question. a measure of how much the vector V Applied Physics July 28, 2024 38 / 76 The Curl – (2) The vectors here show a substantial curl, pointing in the z direction, as the natural right-hand rule would suggest. Applied Physics July 28, 2024 39 / 76 The Curl – (3) Magnetic field line around a magnet is the best example of curl. Similar behaviour is not observed in electric field for dipole. Applied Physics July 28, 2024 40 / 76 Example 5 – (1) Example 5: For the functions in given figure, calculate the Curls and comment on them. V⃗a = −y xb + x yb V⃗b = x yb Applied Physics July 28, 2024 41 / 76 Example 5 – (2) xb yb zb ⃗ × V⃗a = ∇ ∂ ∂ ∂ ∂x ∂y ∂z −y x 0   ⃗ × V⃗a = zb ∂x − ∂ − y ∇ = 2 zb ∂x ∂y V⃗a = −y xb + x yb xb yb zb ⃗ × V⃗b = ∇ ∂ ∂ ∂ ∂x ∂y ∂z 0 x 0 ⃗ × V⃗b = zb ∂x = zb ∇ ∂x V⃗b = x yb Applied Physics July 28, 2024 42 / 76 Product Rules – (1) For ordinary derivatives Product Rule Sum Rule: d dg df d df dg (f g ) = f +g (f + g ) = + dx dx dx dx dx dx Quotient Rule Multiplying with constant, α d df   g df − f dg d f dx (α f ) = α dx = dx 2 dx dx g g f and g are functions of variable x. ⃗ operator. Similar relations hold true for the ∇ Applied Physics July 28, 2024 43 / 76 Product Rules – (2) ⃗ and B If f and g are scalar functions and A ⃗ are vector functions of variable ⃗r. Sum rule Multiplying with constant, α ⃗ (f + g ) = ∇f ∇ ⃗ + ∇g⃗ ⃗ (αf ) = α∇f ∇ ⃗       ⃗ · A ∇ ⃗+B ⃗ =∇ ⃗ ·A⃗+∇ ⃗ ⃗ ·B ⃗ · αA ∇ ⃗ =α ∇ ⃗ ⃗ ·A       ⃗ × A ∇ ⃗+B ⃗ =∇ ⃗ ×A⃗+∇ ⃗ ⃗ ×B ⃗ × αA ∇ ⃗ =α ∇ ⃗ ×A ⃗ The product rules are not quite so simple. There are four types of product between scalars and vectors which provides different output. Applied Physics July 28, 2024 44 / 76 Product Rules – (3) Product output is scalar Product output is vector fg – product of two scalar ⃗ – scalar times vector functions fA ⃗·B ⃗ – dot product of two vector ⃗×B A ⃗ – cross product of two A functions vector functions Accordingly, there are six product rules Two for gradients ⃗ (fg ) = f ∇g ∇ ⃗ + g ∇f ⃗           ⃗ A ∇ ⃗·B ⃗ =A ⃗× ∇ ⃗ +B ⃗ ×B ⃗× ∇ ⃗ + A ⃗ ×A ⃗·∇ ⃗ B ⃗+ B⃗ ·∇ ⃗ A ⃗ Applied Physics July 28, 2024 45 / 76 Product Rules – (4) Two for divergences       ⃗ · fA ∇ ⃗ =f ∇ ⃗ +A ⃗ ·A ⃗ · ∇f ⃗       ⃗ · A ∇ ⃗×B⃗ =B ⃗· ∇ ⃗ −A ⃗ ×A ⃗· ∇ ⃗ ⃗ ×B Two for curl       ⃗ × fA ∇ ⃗ =f ∇⃗ ×A ⃗ −A ⃗ × ∇f ⃗           ⃗ × A ∇ ⃗×B⃗ = B ⃗ ·∇ ⃗ A ⃗− A ⃗·∇⃗ B⃗ +A ⃗ ∇ ⃗ −B ⃗ ·B ⃗ ∇ ⃗ ⃗ ·A Applied Physics July 28, 2024 46 / 76 Example 6 – (1) Example 6:       ⃗ =f ∇ ⃗ · fA Prove the product rule, ∇ ⃗ +A ⃗ ·A ⃗ · ∇f ⃗ Solution:   ⃗ ⃗ · fA LHS = ∇ =∇⃗ · (f Ax xb + f Ay yb + f Az zb) ∂ ∂ ∂ = (f Ax ) + (f Ay ) + (f Az ) ∂x ∂y ∂z ∂f ∂Ax ∂f ∂Ay ∂f ∂Az = Ax +f + Ay +f + Az +f ∂x ∂x ∂y ∂y ∂z ∂z Applied Physics July 28, 2024 47 / 76 Example 6 – (2)     ∂f ∂f ∂f ∂Ax ∂Ay ∂Az LHS = Ax + Ay + Az + f +f +f ∂x ∂y ∂z ∂x ∂y ∂z    ∂f ∂f ∂f = (Ax xb + Ay yb + Az zb) · xb + yb + zb ∂x ∂y ∂z    ∂ ∂ ∂ +f xb + yb + zb · (Ax xb + Ay yb + Az zb) ∂x ∂y ∂z     =f ∇ ⃗ ·A ⃗ +A ⃗ · ∇f ⃗ = RHS Applied Physics July 28, 2024 48 / 76 Line Integrals A line integral is an expression of the form, Z ⃗ b ⃗ · d ⃗l V ⃗ a V⃗ is a vector function, d ⃗l is the infinitesimal displacement vector, and the integral is to be carried out along a prescribed path P from point ⃗a to point ⃗b. To a physicist, the most familiar example of a line integral is the work done by a force F⃗ : W = F⃗ · d ⃗l. R Applied Physics July 28, 2024 49 / 76 Example 7: – (1) Calculate the line integral of the function ⃗v = y 2 xb + 2x(y + 1)b y from the point a = (1, 1, 0) to the point b = (2, 2, 0), along the paths (1) and (2). For Path 1: It can be split into two part. Path (i) and Path (ii). Path (i), Integral I1 Path (ii), Integral I2 d ⃗l = dx xb, y = 1, x : 1 → 2 d ⃗l = dy yb, x = 2, y : 1 → 2 ⃗v = xb + 4x yb ⃗v = y xb + 4 (y + 1) yb ⃗v · d ⃗lR= dx ⃗v · d ⃗l = 4 (y + 1) dy 2 R2 I1 = 1 dx = 1 I2 = 1 4 (y + 1) dy = 10 Applied Physics July 28, 2024 50 / 76 Example 7: – (2) For Path 1:, Total Integral IA , IA = I1 + I2 = 1 + 10 = 11 For Path 2 For Path 2 x = y , dx = dy , d ⃗l = dx xb + dx yb R2 IB = 1 x 2 dx + 2x (x + 1) dx x = y : 1 → 2, R2 IB = 1 3x 2 + 2x dx
⃗v = x 2 xb + 2x (x + 1) yb ⃗v · d ⃗l = x 2 dx + 2x (x + 1) dx IB = 10 Applied Physics July 28, 2024 51 / 76 Surface Integrals – (1) A surface integral is an expression of the form, Z ⃗ · d⃗a V S V⃗ is a vector function and the integral is to be carried out over specified surface S. Here d⃗a is an infinitesimal patch of area, with direction perpendicular to the surface. Applied Physics July 28, 2024 52 / 76 Surface Integrals – (2) Sign Ambiguity There are, of course, two directions perpendicular to any surface, so the sign of a surface integral is intrinsically ambiguous. Closed Surface If the surface is closed, (forming a “balloon”),then tradition dictates that “outward” is positive, but for open surfaces it’s arbitrary. ⃗ RIf V describes the flow of a fluid (mass per unit area per unit time), then ⃗ V · d⃗a represents the total mass per unit time passing through the surface, hence the alternative name, “flux”. Applied Physics July 28, 2024 53 / 76 Volume Integrals A volume integral is an expression of the form, Z T dτ V where T is a scalar function and dτ is an infinitesimal volume element. In Cartesian coordinates dτ = dx dy dz Volume integrals of vector function Z Z Z Z Z ⃗ dτ = V (Vx xb + Vy yb + Vz zb) dτ = xb Vx dτ+b y Vy dτ+b z Vz dτ Applied Physics July 28, 2024 54 / 76 Fundamental Theorem of Gradients Suppose T (x, y , z) is a scalar function of three variables. Starting at point ⃗a, a small distance d ⃗l change is made, the function T will change by an amount,   dT = ∇T ⃗ · d ⃗l The total change in T in going from ⃗a to ⃗b (along the path selected) is, Z ⃗ b   T (b) − T (a) = ⃗ ∇T · d ⃗l ⃗ a Applied Physics July 28, 2024 55 / 76 Fundamental Theorem of Divergences It states that: Z   I ⃗ ⃗ ∇ · V dτ = ⃗ · d⃗a V V S This theorem has at least three special names: Gauss’s theorem, Green’s theorem, or simply the divergence theorem. Geometrical interpretation ⃗ represents the flow of an incompressible fluid, then the flux of V If V ⃗ (RHS) is the total amount of fluid passing out through the surface, per unit time. Now, the divergence measures the “spreading out” of the vectors from a point – a place of high divergence is like a “faucet”, pouring out liquid. Applied Physics July 28, 2024 56 / 76 Example 8 – (1) Example 8: Test the divergence theorem for the function, ⃗v = (xy ) xb + (2yz) yb + (3zx) zb for the the volume shown in figure with sides of length 2. Divergence Theorem Z   I ⃗ ⃗ ∇ · V dτ = ⃗ · d⃗a V V S Applied Physics July 28, 2024 57 / 76 Example 8 – (2) Solution The surface integral for all 6 individual faces have to be calculated and then summed up for the RHS. The volume integral has to be calculated for LHS. For XY plane at z = 2 d⃗a =Rdx dy zb, z R= 2, ⃗v · d⃗ Ra2 = 6x dx dy , 2 I1 = ⃗v · d⃗a = 0 6x dx 0 dy = 24. For XY plane at z = 0 R d⃗a = −dx dy zb, z = 0, ⃗v · d⃗a = 0, I2 = ⃗v · d⃗a = 0. Applied Physics July 28, 2024 58 / 76 Example 8 – (3) For YZ plane at x = 2 d⃗a =Rdy dz xb, x R= 2, ⃗v · d⃗ Ra2 = 2y dy dz, 2 I3 = ⃗v · d⃗a = 0 2y dy 0 dz = 8. For YZ plane at x = 0 R d⃗a = −dy dz xb, x = 0, ⃗v · d⃗a = 0, I4 = ⃗v · d⃗a = 0. For XZ plane at y = 2 d⃗a =Rdz dx yb, y R= 2, ⃗v · d⃗ R a2 = 4z dz dx, 2 I5 = ⃗v · d⃗a = 0 4z dz 0 dx = 16. For XZ plane at y = 0 R d⃗a = −dz dx yb, y = 0, ⃗v · d⃗a = 0, I6 = ⃗v · d⃗a = 0. Applied Physics July 28, 2024 59 / 76 Example 8 – (4) Divergence: ⃗ · ⃗v = y + 2z + 3x ∇ Volume Integral Z   Z ⃗ ∇ · ⃗v dτ = (y + 2z + 3x) dx dy dz V Z Z Z 2 = (y + 2z + 3x) dzdy dx x y 0 Z Z 2 Z 2 = (2y + 4 + 6x) dy dx = (12 + 12x) dx x 0 0 = 48 Applied Physics July 28, 2024 60 / 76 Fundamental Theorem for Curls Special Name: Stoke’s Theorem Z   I ⃗ · d⃗a = ⃗ ×V ∇ ⃗ · d ⃗l V S P Geometrical Interpretation The curl measures the “twist” of the vectors V ⃗ ; a region of high curl is a whirlpool. Now, the integral of the curl over some surface (or, more precisely, the flux of the curl through that surface) represents the “total amount of swirl”. ⃗ · d ⃗l is sometimes called the circulation of V ⃗. H Indeed, V Applied Physics July 28, 2024 61 / 76 Spherical Polar Coordinates – (1) A point P in Cartesian coordinates is given (x, y , z), but sometimes it is more convenient to use spherical coordinates (r , θ, ϕ). r is the distance from the origin (the magnitude of the position vector ⃗r ), θ (the angle down from the z–axis) is called the polar angle, and ϕ (the angle around from the x–axis) is the azimuthal angle. Applied Physics July 28, 2024 62 / 76 Spherical Polar Coordinates – (2) Applied Physics July 28, 2024 63 / 76 Spherical Polar Coordinates – (3) Relationship to Cartesian 1 x = r sinθ cosϕ Line element 2 y = r sinθ sinϕ Along rb, dlr = dr 3 z = r cosθ Along θb dlθ = r dθ Along ϕ, b dlϕ = r sinθ dϕ ⃗ Vector A d ⃗l = dr rb + r dθθb + r sinθ dϕϕb ⃗ = Ar rb + Aθ θb + Aϕ ϕb A Volume Element rb, θb and ϕb are unit vectors in the dτ = r 2 dr sinθ dθ dϕ direction of increasing coordinates. Applied Physics July 28, 2024 64 / 76 Spherical Polar Coordinates – (4) Area element, r is constant d a⃗r = dlθ dlϕ rb d a⃗r = r 2 sin dθ dϕ rb Area element, θ is constant d a⃗θ = dlr dlϕ θb d a⃗θ = r dr sinθ dϕ θb Area element, ϕ is constant d a⃗ϕ = dlr dlθ ϕb d a⃗ϕ = r dr dθ ϕb Applied Physics July 28, 2024 65 / 76 Spherical Polar Coordinates – (5) Gradient: ⃗ = ∂T rb + 1 ∂T θb + 1 ∂T ϕb ∇T ∂r r ∂θ r sinθ ∂ϕ Divergence: ⃗ = 1 ∂ r 2 Vr + 1 ∂ (sinθ Vθ ) + 1 ∂Vϕ ⃗ ·V
∇ r 2 ∂r r sinθ ∂θ r sinθ ∂ϕ Curl:     ⃗ = ⃗ ×V 1 ∂ ∂Vθ 1 1 ∂Vr ∂ ∇ (sinθ Vϕ ) − rb + − (rVϕ ) θb r sinθ ∂θ ∂ϕ r sinθ ∂ϕ ∂r   1 ∂ ∂Vr b + (rVθ ) − ϕ r ∂r ∂θ Applied Physics July 28, 2024 66 / 76 Cylindrical Polar Coordinates – (1) A point P in Cartesian coordinates is given (x, y , z), but sometimes it is more convenient to use cylindrical coordinates (s, ϕ, z). s is the distance to P from the z–axis, ϕ (the angle around from the x–axis) is the azimuthal angle and z is the same as Cartesian coordinates. Applied Physics July 28, 2024 67 / 76 Cylindrical Polar Coordinates – (2) Relationship to Cartesian 1 x = s cosϕ Line element 2 y = s sinϕ Along sb, dls = ds 3 z =z Along ϕb dlϕ = s dϕ Along zb, dlz = dz ⃗ Vector A d ⃗l = dsb s + s dϕϕb + dz zb ⃗ = As sb + Aϕ ϕb + Az zb A Volume Element sb, ϕb and zb are unit vectors in the dτ = ds s dϕ dz direction of increasing coordinates. Applied Physics July 28, 2024 68 / 76 Cylindrical Polar Coordinates – (3) Area element, s is constant d a⃗s = dlϕ dlz sb d a⃗s = s sin dϕ dz sb Area element, ϕ is constant d a⃗ϕ = dls dlz ϕb d a⃗ϕ = ds dz ϕb Area element, z is constant d a⃗z = dls dlϕ zb d a⃗z = ds s dϕ zb Applied Physics July 28, 2024 69 / 76 Cylindrical Polar Coordinates – (4) Gradient: ⃗ = ∂T sb + 1 ∂T ϕb + ∂T zb ∇T ∂s s ∂ϕ ∂z Divergence: ⃗ = 1 ∂ (sVs ) + 1 ∂Vϕ + ∂Vz ⃗ ·V ∇ s ∂s s ∂ϕ ∂z Curl:     ⃗ = ⃗ ×V 1 ∂Vz ∂Vϕ ∂Vs ∂Vz b ∇ − sb + − ϕ s ∂ϕ ∂z ∂z ∂s   1 ∂ ∂Vs + (sVϕ ) − zb s ∂s ∂ϕ Applied Physics July 28, 2024 70 / 76 Problems for homework – (1) Problem 1: Find the angle between the body diagonals of a cube. Problem 2: Use the cross product to find the components of the unit vector nb perpendicular to the shaded plane. Problem 3: Prove the BAC-CAB rule by writing out both sides in component form. Applied Physics July 28, 2024 71 / 76 Problems for homework – (2) Problem 4: h  i h  i h  i Prove that, A⃗× B⃗ × C⃗ + B ⃗ + C⃗ × A ⃗ × C⃗ × A ⃗×B ⃗ =0 Problem 5: The height of a certain hill (in meters) is given by, h (x, y ) = 10 2xy − 3x 2 − 4y 2 − 18x + 28y + 12
where, y is the distance (in km) north and x is the distance east of a city. 1 Find the location of the top of the hill. 2 Compute the maximum height of the hill. 3 Estimate the steepness of the slope and the direction of maximum steepness, at point 1 km north and 1 km east of the city. Applied Physics July 28, 2024 72 / 76 Problems for homework – (3) Problem 6: Calculate the divergences of the following functions. 1 A⃗ = x 2 xb + 3xz 2 yb − 2xz zb 2 B⃗ = xy xb + 2yz yb + 3zx zb C⃗ = y 2 xb + 2xy + z 2 yb + 2yz zb
3 Problem 7: Calculate the curl of the functions in Problem 6. Problem 8: Prove all the six product rules for the “del” operator. Applied Physics July 28, 2024 73 / 76 Problems for homework – (4) Problem 9: Calculate the line integral of the function, ⃗v = x 2 xb + 2yz yb + y 2 zb for the following paths. 1 (0,0,0) → (1,0,0) → (1,1,0) → (1,1,1) — Path 1 2 (0,0,0) → (0,0,1) → (0,1,1) → (1,1,1) — Path 2 3 The direct straight line — Path 3 4 (0,0,0) → (1,1,1) by Path (1) and then return (1,1,1) → (0,0,0) by path (3) Applied Physics July 28, 2024 74 / 76 Problems for homework – (5) Problem 10: Test the divergence theorem for the function, ⃗v = r 2 cosθ rb + r 2 cosϕ θb − r 2 cosθ sinϕ ϕb


using the octant of a sphere shown in Figure. Applied Physics July 28, 2024 75 / 76 Problems for homework – (6) Problem 11: Compute the line integral of, ⃗v = r cos2 (θ) rb − [r cos (θ) sin (θ)] θb + 3r ϕb   along the path shown in Figure. Applied Physics July 28, 2024 76 / 76