PHY103A: Lecture # 3 (Text Book: Intro to Electrodynamics by Griffiths, 3Rd Ed.)

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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

𝛁𝛁𝑇𝑇 1 + + 2 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝛁𝛁𝑇𝑇 ≡ 𝒙𝒙� 𝒚𝒚� 𝒛𝒛� 3 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 3 2 1 0 1 2 3 Divergence of a vector 2

1 𝛁𝛁 ⋅ 𝐕𝐕 = + + 0 𝜕𝜕𝑣𝑣𝑥𝑥 𝜕𝜕𝑣𝑣𝑦𝑦 𝜕𝜕𝑣𝑣𝑧𝑧

𝛁𝛁 ⋅ 𝐕𝐕 1 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 Curl of a vector × 2 2

2 1 0 1 2 1 𝛁𝛁 𝐕𝐕 × = 𝒙𝒙� 𝒚𝒚� 𝒛𝒛� 0 𝜕𝜕 𝜕𝜕 𝜕𝜕 𝛁𝛁 𝐕𝐕 𝜕𝜕𝜕𝜕 𝜕𝜕𝑦𝑦 𝜕𝜕𝑧𝑧 1 𝑣𝑣𝑥𝑥 𝑣𝑣𝑦𝑦 𝑣𝑣𝑧𝑧 = + + 2

𝑧𝑧 𝑦𝑦 𝑥𝑥 𝑧𝑧 𝑦𝑦 𝑥𝑥 2 1 0 1 2 𝜕𝜕𝑣𝑣 𝜕𝜕𝑣𝑣 𝜕𝜕𝑣𝑣 𝜕𝜕𝑣𝑣 𝜕𝜕𝑣𝑣 𝜕𝜕𝑣𝑣 3 − 𝒙𝒙� − 𝒚𝒚� − 𝒛𝒛� 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 Integral Calculus: The ordinary integral that we know of is of the form: In vector calculus we encounter many other types of integrals.𝑏𝑏 ∫𝑎𝑎 𝑓𝑓 𝑥𝑥 𝑑𝑑𝑑𝑑 Line Integral:

𝑏𝑏 � 𝐕𝐕 ⋅ 𝑑𝑑𝐥𝐥 Vector field𝑎𝑎 Infinitesimal Displacement vector Or Line element = + +

If the path is a closed𝑑𝑑𝐥𝐥 𝑑𝑑 loop𝑑𝑑 𝒙𝒙� then𝑑𝑑𝑑𝑑 the𝒚𝒚� line𝑑𝑑𝑑𝑑 𝒛𝒛�integral is written as

• When do� we𝐕𝐕 ⋅ 𝑑𝑑need𝐥𝐥 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)? 𝐕𝐕 𝑦𝑦 𝒙𝒙� 2𝑥𝑥 𝑦𝑦 𝒚𝒚�

Along path (1) We have = + .

(i) = ; =1;𝒅𝒅𝒅𝒅 𝑑𝑑𝑑𝑑 𝒙𝒙�= 𝑑𝑑𝑑𝑑𝒚𝒚� = 1 (ii) = ; =2; = 4( 2 + 1) =10 𝒅𝒅𝒅𝒅 𝑑𝑑𝑑𝑑 𝒙𝒙� 𝑦𝑦 ∫ 𝐕𝐕 ⋅ 𝒅𝒅𝒅𝒅 ∫ 𝑦𝑦 𝑑𝑑𝑑𝑑 2 𝒅𝒅𝒅𝒅 𝑑𝑑𝑑𝑑 𝒚𝒚� 𝑥𝑥 ∫ 𝐕𝐕 ⋅ 𝒅𝒅𝒅𝒅 ∫1 𝑦𝑦 𝑑𝑑𝑑𝑑 Along path (2): = + ; = ; = = ( + 2 + ) =10 𝒅𝒅𝒅𝒅 𝑑𝑑𝑑𝑑 𝒙𝒙� 𝑑𝑑𝑑𝑑𝒚𝒚� 𝑥𝑥 𝑦𝑦 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 2 2 2 ∫ 𝐕𝐕 ⋅ 𝒅𝒅𝒅𝒅 ∫1 𝑥𝑥 𝑥𝑥 2𝑥𝑥 𝑑𝑑𝑑𝑑 =11-10=1 This means that if represented the force vector, 𝐕𝐕 ⋅ 𝒅𝒅𝒅𝒅 ∮ it would be a non-conservative force 𝐕𝐕 5 Surface Integral:

� 𝐕𝐕 ⋅ 𝑑𝑑𝐚𝐚 � 𝐕𝐕 ⋅ 𝑑𝑑𝐚𝐚 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 “upward𝐕𝐕 2 𝑥𝑥𝑥𝑥and𝒙𝒙� outward”𝑥𝑥 is𝒚𝒚� the 𝑦𝑦positive𝑧𝑧 − direction𝒛𝒛�

(i) x=2, = ; = = 4 = 4 =16 𝑑𝑑𝒂𝒂 𝑑𝑑2𝑑𝑑𝑑𝑑2𝑑𝑑 𝒙𝒙� 𝑉𝑉 ⋅ 𝑑𝑑𝒂𝒂 2𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧 ∫ 𝑽𝑽 ⋅ 𝑑𝑑𝒂𝒂 ∫0 ∫0 𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧 (ii) =0 (iii) =12 ∫ 𝐕𝐕 ⋅ 𝑑𝑑𝐚𝐚 (iv) =-12 ∫ 𝐕𝐕 ⋅ 𝑑𝑑𝐚𝐚 (v) =4 ∫ 𝐕𝐕 ⋅ 𝑑𝑑𝐚𝐚 (vi) =-12 ∫ 𝐕𝐕 ⋅ 𝑑𝑑𝐚𝐚 ∫ 𝐕𝐕 ⋅ 𝑑𝑑𝐚𝐚 = 16 + 0 + 12 12 + 4 12 = 8

7 � 𝐕𝐕 ⋅ 𝑑𝑑𝐚𝐚 − − Volume Integral:

( , , )

• In Cartesian� 𝑇𝑇 𝑥𝑥 𝑦𝑦 𝑧𝑧coordinate𝑑𝑑𝑑𝑑 system the volume element is given by = .

• One can have the volume𝑑𝑑𝑑𝑑 integral𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 of a vector function which V as = + +

𝑥𝑥 𝑦𝑦 𝑧𝑧 Example∫ 𝐕𝐕𝑑𝑑𝑑𝑑 (Griffiths:∫ 𝐕𝐕 𝑑𝑑𝑑𝑑 𝒙𝒙� Ex.∫ 1.8)𝐕𝐕 𝑑𝑑 𝜏𝜏𝒚𝒚� ∫ 𝐕𝐕 𝑑𝑑𝑑𝑑 𝒛𝒛� Q: Calculate the volume integral of = over the volume of the prism 2 𝑇𝑇 𝑥𝑥𝑥𝑥𝑧𝑧 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, the𝑧𝑧 volume integral is given by 𝑦𝑦 𝑥𝑥 − 𝑦𝑦 3 = 1−𝑦𝑦 1 3 8 8 2 �0 �0 �0 𝑥𝑥𝑥𝑥𝑧𝑧 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 The fundamental Theorem of Calculus:

The integral of a derivative 𝑏𝑏 = 𝑏𝑏 = ( ) over a region is given by the 𝑑𝑑𝑓𝑓 value of the function at the � 𝐹𝐹 𝑥𝑥 𝑑𝑑𝑑𝑑 � 𝑑𝑑𝑑𝑑 𝑓𝑓 𝑏𝑏 − 𝑓𝑓 𝑎𝑎 boundaries 𝑎𝑎 𝑎𝑎 𝑑𝑑𝑑𝑑

Example

22 𝑏𝑏 = 𝑏𝑏 𝑥𝑥 = = 𝑑𝑑 22 𝑏𝑏 2 2 2 𝑥𝑥 𝑏𝑏 − 𝑎𝑎 � 𝑥𝑥𝑑𝑑𝑑𝑑 � 𝑑𝑑𝑑𝑑 𝑎𝑎 𝑎𝑎 𝑑𝑑𝑑𝑑 𝑎𝑎

• 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 𝑏𝑏 = ( ) region is given by the value of the 𝑑𝑑𝑑𝑑 function at the boundaries � 𝑑𝑑𝑑𝑑 𝑓𝑓 𝑏𝑏 − 𝑓𝑓 𝑎𝑎 𝑎𝑎 𝑑𝑑𝑑𝑑 The fundamental Theorem for Gradient: The integral of a derivative (gradient) 𝑏𝑏 = ( ) over a region (path) is given by the value of the function at the boundaries � 𝛻𝛻𝑇𝑇 ⋅ 𝑑𝑑𝐥𝐥 𝑇𝑇 𝑏𝑏 − 𝑇𝑇 𝑎𝑎 (end-points) 𝑎𝑎 𝑃𝑃𝑃𝑃𝑃𝑃푃 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 � 𝛻𝛻 ⋅ 𝐕𝐕 𝑑𝑑𝑑𝑑 � 𝐕𝐕 ⋅ 𝑑𝑑𝐚𝐚 𝑉𝑉𝑜𝑜𝑜𝑜 𝑆𝑆𝑢𝑢𝑢𝑢𝑢𝑢 (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- �𝑆𝑆𝑢𝑢𝑢𝑢𝑢𝑢 𝛻𝛻 𝐕𝐕 ⋅ 𝑑𝑑𝐚𝐚 �𝑃𝑃𝑎𝑎𝑎𝑎�𝐕𝐕 ⋅ 𝑑𝑑𝐥𝐥 10 path) Spherical Polar Coordinates: 0 ; 0 ; = sin cos , = sin sin , = cos 0 ≤ 𝑟𝑟 ≤ ∞ ≤ 𝜃𝜃 ≤ 𝜋𝜋 𝑥𝑥A vector𝑟𝑟 𝜃𝜃 in the𝜙𝜙 spherical𝑦𝑦 polar𝑟𝑟 coordinate𝜃𝜃 𝜙𝜙 is given𝑧𝑧 𝑟𝑟 by 𝜃𝜃 = A + A + A ≤ 𝜙𝜙 ≤ 2𝜋𝜋 𝐀𝐀 = sin𝒓𝒓𝒓𝒓� cos𝜽𝜽𝜽𝜽� + 𝝓𝝓sin𝝓𝝓� sin + cos

𝒓𝒓� = cos𝜃𝜃 cos 𝜙𝜙 𝐱𝐱� + cos𝜃𝜃 sin𝜙𝜙 𝐲𝐲� sin𝜃𝜃𝐳𝐳� = sin + cos � 𝜽𝜽 𝜃𝜃 𝜙𝜙 𝐱𝐱� 𝜃𝜃 𝜙𝜙 𝐲𝐲� −Griffiths:𝜃𝜃𝐳𝐳� Prob 1.37 𝝓𝝓� − 𝜙𝜙𝐱𝐱� 𝜙𝜙𝐲𝐲� The infinitesimal displacement vector in the spherical polar coordinate is given by dl= + + (In Cartesian system we have = + + )

𝑑𝑑𝑑𝑑𝒓𝒓𝒓𝒓� 𝑑𝑑𝑑𝑑𝜽𝜽𝜽𝜽� 𝑑𝑑𝑑𝑑𝝓𝝓𝝓𝝓� 𝑑𝑑𝐥𝐥 𝑑𝑑𝑑𝑑 𝒙𝒙� 𝑑𝑑𝑑𝑑𝒚𝒚� 𝑑𝑑𝑑𝑑𝒛𝒛�

11 Spherical Polar Coordinates:

= sin cos , = sin sin , = cos

𝑥𝑥A vector𝑟𝑟 𝜃𝜃 in the𝜙𝜙 spherical𝑦𝑦 polar𝑟𝑟 coordinate𝜃𝜃 𝜙𝜙 is given𝑧𝑧 𝑟𝑟 by 𝜃𝜃 = A + A + A

𝐀𝐀 = sin𝒓𝒓𝒓𝒓� cos𝜽𝜽𝜽𝜽� + 𝝓𝝓sin𝝓𝝓� sin + cos

dl= 𝑑𝑑𝑑𝑑𝒓𝒓𝒓𝒓�+ 𝑑𝑑𝑑𝑑𝜽𝜽𝜽𝜽� + 𝑑𝑑𝑑𝑑sin𝝓𝝓𝝓𝝓� 𝑑𝑑𝐥𝐥 𝑑𝑑𝑑𝑑 𝒙𝒙� 𝑑𝑑𝑑𝑑𝒚𝒚� 𝑑𝑑𝑑𝑑𝒛𝒛�

The infinitesimal𝑑𝑑𝑑𝑑𝒓𝒓� 𝑟𝑟𝑟𝑟𝑟𝑟 𝜽𝜽�volume𝑟𝑟 𝜃𝜃element:𝜃𝜃𝜃𝜃𝝓𝝓� = = sin 2 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝒓𝒓𝑑𝑑𝑑𝑑𝜽𝜽𝑑𝑑𝑑𝑑𝝓𝝓 𝑟𝑟 𝜃𝜃𝑑𝑑𝑑𝑑𝑑𝑑𝜃𝜃𝑑𝑑𝑑𝑑 The infinitesimal area element (it depends): = = sin (over the surface of2 a sphere) 𝑑𝑑𝐚𝐚 𝑑𝑑𝑑𝑑𝜽𝜽𝑑𝑑𝑑𝑑𝝓𝝓𝒓𝒓� 𝑟𝑟 𝜃𝜃𝑑𝑑𝜃𝜃𝑑𝑑𝑑𝑑𝒓𝒓� 12 Cylindrical Coordinates:

0 ; = s cos , = sin , = 0 ; ≤ 𝑠𝑠 ≤ ∞ 𝑥𝑥 𝜙𝜙 𝑦𝑦 𝑠𝑠 𝜙𝜙 𝑧𝑧 𝑧𝑧 ≤ 𝜙𝜙 ≤ 2𝜋𝜋 A vector in the cylindrical coordinates is given by = A + A + A −∞ ≤ 𝑧𝑧 ≤ ∞ 𝐀𝐀 = cos𝒔𝒔𝒔𝒔� +𝝓𝝓𝝓𝝓�sin 𝒛𝒛𝒛𝒛� 𝐀𝐀 = sin + cos 𝐬𝐬� 𝜙𝜙 𝐱𝐱� 𝜙𝜙 𝐲𝐲� = � Griffiths: Prob 1.41 𝝓𝝓 − 𝜙𝜙𝐱𝐱� 𝜙𝜙𝐲𝐲� 𝐳𝐳� 𝐳𝐳� The infinitesimal displacement vector in the cylindrical coordinates is given by dl= + + (In Cartesian system we have = + + )

𝑑𝑑𝑑𝑑𝒔𝒔𝒔𝒔� 𝑑𝑑𝑑𝑑𝝓𝝓𝝓𝝓� 𝑑𝑑𝑑𝑑𝒛𝒛𝒛𝒛� 𝑑𝑑𝐥𝐥 𝑑𝑑𝑑𝑑 𝒙𝒙� 𝑑𝑑𝑑𝑑𝒚𝒚� 𝑑𝑑𝑑𝑑𝒛𝒛�

13 Cylindrical Coordinates:

= s cos , = sin , =

𝑥𝑥 𝜙𝜙 𝑦𝑦 𝑠𝑠 𝜙𝜙 𝑧𝑧 𝑧𝑧 A vector in the cylindrical coordinates is given by = A + A + A

𝒔𝒔 𝝓𝝓 � 𝒛𝒛 𝐀𝐀 = cos𝒔𝒔� +𝝓𝝓sin 𝒛𝒛� 𝐀𝐀 𝐬𝐬� = sin𝜙𝜙 𝐱𝐱� + cos𝜙𝜙 𝐲𝐲� = 𝝓𝝓� − 𝜙𝜙𝐱𝐱� 𝜙𝜙𝐲𝐲� Griffiths: Prob 1.41

𝐳𝐳� 𝐳𝐳� The infinitesimal displacement vector in the cylindrical coordinates is given by dl= + + (In Cartesian system we have = + + )

dl= 𝑑𝑑𝑑𝑑𝒔𝒔𝒔𝒔�+ 𝑑𝑑𝑑𝑑𝝓𝝓𝝓𝝓� +𝑑𝑑𝑑𝑑𝒛𝒛𝒛𝒛� 𝑑𝑑𝐥𝐥 𝑑𝑑𝑑𝑑 𝒙𝒙� 𝑑𝑑𝑑𝑑𝒚𝒚� 𝑑𝑑𝑑𝑑𝒛𝒛�

The infinitesimal𝑑𝑑𝑑𝑑𝒔𝒔� 𝑠𝑠𝑑𝑑𝑑𝑑 𝝓𝝓�volume𝑑𝑑𝑑𝑑𝒛𝒛� element: = =

The𝑑𝑑𝑑𝑑 infinitesimal𝑑𝑑𝑑𝑑𝒔𝒔𝑑𝑑𝑑𝑑𝝓𝝓𝑑𝑑𝑑𝑑𝒛𝒛 area𝑠𝑠 𝑑𝑑element𝑑𝑑𝑑𝑑𝜙𝜙𝑑𝑑𝑑𝑑 (it depends): = = (over the surface of a cylinder) 𝑑𝑑𝐚𝐚 𝑑𝑑𝑑𝑑𝝓𝝓𝑑𝑑𝑑𝑑𝒛𝒛𝒔𝒔� 𝑠𝑠 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝒔𝒔� 14 Gradient, Divergence and Curl in Cartesian, Spherical-polar and Cylindrical Coordinate systems:

• See the formulas listed inside the front cover of Griffiths