Semester II, 2017-18 Department of Physics, IIT Kanpur PHY103A: Lecture # 3 (Text Book: Intro to Electrodynamics by Griffiths, 3rd Ed.) Anand Kumar Jha 08-Jan-2018 Notes • The first tutorial is tomorrow (Tuesday). • Updated lecture notes will be uploaded right after the class. • Office Hour – Friday 2:30-3:30 pm • Phone: 7014(Off); 962-142-3993(Mobile) [email protected]; [email protected] • Tutorial Sections have been finalized and put up on the webpage. • Course Webpage: http://home.iitk.ac.in/~akjha/PHY103.htm 2 3 Summary of Lecture # 2: 2 Gradient of a scalar 1 0 1 + + 2 ≡ � � � 3 3 2 1 0 1 2 3 Divergence of a vector 2 1 ⋅ = + + 0 ⋅ 1 Curl of a vector × 2 2 2 1 0 1 2 1 × = � � � 0 1 = + + 2 2 1 0 1 2 3 − � − � − � Integral Calculus: The ordinary integral that we know of is of the form: In vector calculus we encounter many other types of integrals. ∫ Line Integral: � ⋅ Vector field Infinitesimal Displacement vector Or Line element = + + If the path is a closed loop � then the� line �integral is written as • When do� we ⋅ need line integrals? Work done by a force along a given path involves line integral. 4 Example (G: Ex. 1.6) Q: = + ( + 1) ? What is the line integral from A to2 B along path (1) and (2)? � 2 � Along path (1) We have = + . (i) = ; =1; �= � = 1 (ii) = ; =2; = 4( 2 + 1) =10 � ∫ ⋅ ∫ 2 � ∫ ⋅ ∫1 Along path (2): = + ; = ; = = ( + 2 + ) =10 � � 2 2 2 ∫ ⋅ ∫1 2 =11-10=1 This means that if represented the force vector, ⋅ ∮ it would be a non-conservative force 5 Surface Integral: or � ⋅ � ⋅ Vector field Infinitesimal area vector Closed loop • For a closed surface, the area vector points outwards. • For open surfaces, the direction of the area vector is decided based on a given problem. • When do we need area integrals? Flux through a given area involves surface integral. 6 Example (Griffiths: Ex. 1.7) Q: = + + 2 + 3 ? Calculate the Surface integral. 2 “upward 2 and� outward” is� the positive − direction� (i) x=2, = ; = = 4 = 4 =16 22 � ⋅ 2 ∫ ⋅ ∫0 ∫0 (ii) =0 (iii) =12 ∫ ⋅ (iv) =-12 ∫ ⋅ (v) =4 ∫ ⋅ (vi) =-12 ∫ ⋅ ∫ ⋅ = 16 + 0 + 12 12 + 4 12 = 8 7 � ⋅ − − Volume Integral: ( , , ) • In Cartesian� coordinate system the volume element is given by = . • One can have the volume integral of a vector function which V as = + + Example∫ (Griffiths:∫ � Ex.∫ 1.8) � ∫ � Q: Calculate the volume integral of = over the volume of the prism 2 We find that integral runs from 0 to 3. The integral runs from 0 to 1, but the integral runs from 0 to 1 only. Therefore, the volume integral is given by − 3 = 1− 1 3 8 8 2 �0 �0 �0 The fundamental Theorem of Calculus: The integral of a derivative = = ( ) over a region is given by the value of the function at the � � − boundaries Example 22 = = = 22 2 2 2 − � � • In vector calculus, we have three different types of derivative (gradient, divergence, and curl) and correspondingly three different types of regions and end points. 9 The fundamental Theorem of Calculus: The integral of a derivative over a = ( ) region is given by the value of the function at the boundaries � − The fundamental Theorem for Gradient: The integral of a derivative (gradient) = ( ) over a region (path) is given by the value of the function at the boundaries � ⋅ − (end-points) 푃 The fundamental Theorem for Divergence (Gauss’s theorem): The integral of a derivative (divergence) = over a region (volume) is given by the value of the function at the boundaries � ⋅ � ⋅ (bounding surface) The fundamental Theorem for Curl (Stokes’ theorem): The integral of a derivative (curl) over a × = region (surface) is given by the value of the function at the boundaries (closed- � ⋅ �푃 ⋅ 10 path) Spherical Polar Coordinates: 0 ; 0 ; = sin cos , = sin sin , = cos 0 ≤ ≤ ∞ ≤ ≤ A vector in the spherical polar coordinate is given by = A + A + A ≤ ≤ 2 = sin� cos� + sin� sin + cos � = cos cos � + cos sin � sin� = sin + cos � � � −Griffiths:� Prob 1.37 � − � � The infinitesimal displacement vector in the spherical polar coordinate is given by dl= + + (In Cartesian system we have = + + ) � � � � � � 11 Spherical Polar Coordinates: = sin cos , = sin sin , = cos A vector in the spherical polar coordinate is given by = A + A + A = sin� cos� + sin� sin + cos � = cos cos � + cos sin � sin� = sin + cos � � � −Griffiths:� Prob 1.37 � − � � The infinitesimal displacement vector in the spherical polar coordinate is given by dl= + + (In Cartesian system we have = + + ) dl= �+ � + sin� � � � � The infinitesimal� �volume element: = = sin 2 The infinitesimal area element (it depends): = = sin (over the surface of2 a sphere) � � 12 Cylindrical Coordinates: 0 ; = s cos , = sin , = 0 ; ≤ ≤ ∞ ≤ ≤ 2 A vector in the cylindrical coordinates is given by = A + A + A −∞ ≤ ≤ ∞ = cos� +�sin � = sin + cos � � � = � Griffiths: Prob 1.41 − � � � � The infinitesimal displacement vector in the cylindrical coordinates is given by dl= + + (In Cartesian system we have = + + ) � � � � � � 13 Cylindrical Coordinates: = s cos , = sin , = A vector in the cylindrical coordinates is given by = A + A + A � = cos� +sin � � = sin � + cos � = � − � � Griffiths: Prob 1.41 � � The infinitesimal displacement vector in the cylindrical coordinates is given by dl= + + (In Cartesian system we have = + + ) dl= �+ � +� � � � The infinitesimal� �volume� element: = = The infinitesimal area element (it depends): = = (over the surface of a cylinder) � � 14 Gradient, Divergence and Curl in Cartesian, Spherical-polar and Cylindrical Coordinate systems: • See the formulas listed inside the front cover of Griffiths 15 .