• Intro

• Recap : vector algebra, calculus

• Gradient, Divergence, Curl

Lec #2

Vector Calculus

PC 2131Electricity & Magnetism

2

Electrostatics in brief

We start with electrostatics; its theoretical ideas and mathematical techniques will be used again.

Electrostatics: the charge must be at rest or changes slowly in time “quasi-static”.

Force action at a distance

𝐅 =1

4𝜋𝜖0

𝑞𝑞′

𝓇2 𝓻

F

O

r

r′

𝓻

Coulomb’s law

F is the force that charge 𝑞′ feels due to charge 𝑞.

Force & Field

Action at a distance Field

F

E

Field is the modern perspective of force

And Vector Calculus is the mother-language of fields 4

Electric fields: why do we care?

The electric field is a vector field… and how to measure variation of field … needs vector calculus …

𝐄 =1

4𝜋𝜖0

𝑞

𝓇2 𝓻

𝐄 = lim𝑞′→0

𝐅

𝑞′

• It is very useful to find the force in two stages by introducing the concept of the electric field E.

• Although introduced in this way as a mathematical convenience, the electric field has important physical significance on its own, and is not merely a mathematical construct.

Recap: vector algebra

Notations

vector boldface E, r

unit vector hat 𝐢, 𝐣, 𝐤scalar plainface 𝑇

Normal vector

Basic vector operations

𝐀

𝐁

𝐀 + 𝐁𝐀

𝐁

𝐁 + 𝐀

𝐀

−𝐁

𝐀 − 𝐁

𝐀 + 𝐁 = 𝐁 + 𝐀(𝐀 + 𝐁) + 𝐂 = 𝐀 + (𝐁 + 𝐂)

𝐀–𝐁 = 𝐀 + (−𝐁)

Component form in Cartesian coordinates

• Unit vectors in Cartesian coordinates 𝐢 𝐣 𝐤

• A vector A in term of basis vectors 𝐀 = 𝐴𝑥 𝐢 + 𝐴𝑦 𝐣 +

𝐴𝑧 𝐤

• Adding vectors

𝐀 + 𝐁 = 𝐴𝑥 𝐢 + 𝐴𝑦 𝐣 + 𝐴𝑧 𝐤 + 𝐵𝑥 𝐢 + 𝐵𝑦 𝐣 + 𝐵𝑧 𝐤

= (𝐴𝑥+𝐵𝑥) 𝐢 + (𝐴𝑦+𝐵𝑦) 𝐣 + (𝐴𝑧+𝐵𝑧) 𝐤

• Multiplying with a scalar

𝑎𝐀 = (𝑎𝐴𝑥) 𝐢 + (𝑎𝐴𝑦) 𝐣 + (𝑎𝐴𝑧) 𝐤

7

𝐢

𝐣

𝐤

𝑥

𝑦

𝑧

Dot product

• 𝐀 ∙ 𝐁 ≡ 𝐴𝐵 cos 𝜃 parallel 𝐀 ∙ 𝐁 = 𝐴𝐵

perpendicular 𝐀 ∙ 𝐁 = 0

𝐀 ∙ 𝐁 = 𝐁 ∙ 𝐀 , 𝐀 ∙ 𝐀 = 𝐴2

𝐀 ∙ 𝐁 + 𝐂 = 𝐀 ∙ 𝐁 + 𝐀 ∙ 𝐂

8

𝐀

𝐁

𝐢

𝐣

𝐤

𝑥

𝑦

𝑧

Geometrical Interpretation

Vector component• 𝐢 ∙ 𝐢 = 𝐣 ∙ 𝐣 = 𝐤 ∙ 𝐤 = 1

• 𝐢 ∙ 𝐣 = 𝐢 ∙ 𝐤 = 𝐣 ∙ 𝐤 = 0

• 𝐀 ∙ 𝐁 = 𝐴𝑥 𝐢 + 𝐴𝑦 𝐣 + 𝐴𝑧 𝐤 ∙ 𝐵𝑥 𝐢 + 𝐵𝑦 𝐣 + 𝐵𝑧 𝐤

= 𝐴𝑥𝐵𝑥 + 𝐴𝑦𝐵𝑦 + 𝐴𝑧𝐵𝑧

Using suffix notation

• 𝐀 ∙ 𝐁 = 𝑖=13 𝐴𝑖𝐵𝑖 (Note: here i is not a unit vector; it

corresponds to x, y, z)

Cross product

• 𝐀 × 𝐁 ≡ 𝐴𝐵 sin 𝜃 𝐧 𝐀 × 𝐀 = 0

𝐁 × 𝐀 = −(𝐀 × 𝐁)

𝐀 × 𝐁 + 𝐂 = (𝐀 × 𝐁) + (𝐀 × 𝐂)

9

𝐢

𝐣

𝐤

𝑥

𝑦

𝑧

Geometrical Interpretation

Vector component • 𝐢 × 𝐢 = 𝐣 × 𝐣 = 𝐤 × 𝐤 = 0

• 𝐢 × 𝐣 = − 𝐣 × 𝐢 = 𝐤 , 𝐣 × 𝐤 = − 𝐤 × 𝐣 = 𝐢 , 𝐤 × 𝐢 = − 𝐢 × 𝐤 = 𝐣

• 𝐀 × 𝐁 = 𝐢

𝐴𝑥

𝐵𝑥

𝐣𝐴𝑦

𝐵𝑦

𝐤𝐴𝑧

𝐵𝑧

= … try to complete this !

Using suffix notation

q

A

B

n̂right-hand rule

• (𝐀 × 𝐁)𝑖= 𝐴𝑗𝐵𝑘 − 𝐴𝑘𝐵𝑗 , where (ijk) is a cyclic permutation of (123)

Vector Identities

𝐀 ∙ 𝐁 × 𝐂 = 𝐀 × 𝐁 ∙ 𝐂

𝐀 × 𝐁 × 𝐂 = 𝐁 𝐀 ∙ 𝐂 − 𝐂 𝐀 ∙ 𝐁

𝐀 × 𝐁 ∙ 𝐂 × 𝐃 = 𝐀 ∙ 𝐂 𝐁 ∙ 𝐃 − (𝐀 ∙ 𝐃)(𝐁 ∙ 𝐂)

Levi-Civita Tensor

(𝐀 × 𝐁)𝑖=

𝑗=1

3

𝑘=1

3

𝜖𝑖𝑗𝑘𝐴𝑗𝐵𝑘

Levi-Civita Antisymmetric Tensor

Try and compare with previous results…

𝜖𝑖𝑗𝑘 = 0, unless i, j and k are all different

𝜖123 = 𝜖231 = 𝜖312 = 1𝜖213 = 𝜖132 = 𝜖321 = −1

Using Einstein summation convention(Any repeated suffix is summed from 1 to 3)

(𝐀 × 𝐁)𝑖= 𝜖𝑖𝑗𝑘𝐴𝑗𝐵𝑘

𝐀 × 𝐁 = …

𝐀 × 𝐁 = …

𝐀 ∙ 𝐁 = 𝐴𝑖𝐵𝑖

𝐀 ∙ 𝐁 = …

(𝐀 × 𝐁)𝑖=

𝑗=1

3

𝑘=1

3

𝜖𝑖𝑗𝑘𝐴𝑗𝐵𝑘

𝐀 ∙ 𝐁 =

𝑖=1

3

𝐴𝑖𝐵𝑖

I admire the elegance of your method of computation; it must be nice to ride through these fields upon the horse of true mathematics while the like of us have to make our way laboriously on foot. —Albert Einstein

Exercise #1

𝐀 ∙ 𝐁 × 𝐂 = 𝐀 × 𝐁 ∙ 𝐂Prove this identity

(𝐀 × 𝐁)𝑖= 𝜖𝑖𝑗𝑘𝐴𝑗𝐵𝑘𝐀 ∙ 𝐁 = 𝐴𝑖𝐵𝑖

𝐀 ∙ 𝐁 × 𝐂 = 𝜖𝑖𝑗𝑘𝐴𝑖𝐵𝑗𝐶𝑘

Left side:

Right side:

And because …

(𝐀 × 𝐁) ∙ 𝐂 = 𝜖𝑘𝑖𝑗𝐴𝑖𝐵𝑗𝐶𝑘

𝜖𝑘𝑖𝑗 = −𝜖𝑖𝑘𝑗= +𝜖𝑖𝑗𝑘 LHS=RHS

Exercise #2

𝐀 × 𝐁 × 𝐂 = 𝐁 𝐀 ∙ 𝐂 − 𝐂 𝐀 ∙ 𝐁

𝐀 × 𝐁 × 𝐂 = 𝜖𝑖𝑗𝑘𝐴𝑗(𝐁 × 𝐂)𝑘= 𝜖𝑖𝑗𝑘𝜖𝑘𝑙𝑚𝐴𝑗𝐵𝑙𝐶𝑚

𝐁 𝐀 ∙ 𝐂 − 𝐂 𝐀 ∙ 𝐁 = 𝐵𝑖𝐴𝑗𝐶𝑗 − 𝐶𝑖𝐴𝑗𝐵𝑗

Prove this identity

Left side:

Right side:

(𝐀 × 𝐁)𝑖= 𝜖𝑖𝑗𝑘𝐴𝑗𝐵𝑘𝐀 ∙ 𝐁 = 𝐴𝑖𝐵𝑖

Exercise #2

𝐀 × 𝐁 × 𝐂 = 𝐁 𝐀 ∙ 𝐂 − 𝐂 𝐀 ∙ 𝐁

𝐀 × 𝐁 × 𝐂 = 𝜖𝑖𝑗𝑘𝐴𝑗(𝐁 × 𝐂)𝑘= 𝜖𝑖𝑗𝑘𝜖𝑘𝑙𝑚𝐴𝑗𝐵𝑙𝐶𝑚

Identity of the Levi-Civita tensor:𝜖𝑖𝑗𝑘𝜖𝑘𝑙𝑚 = 𝛿𝑖𝑙𝛿𝑗𝑚 − 𝛿𝑖𝑚𝛿𝑗𝑙

𝛿𝑖𝑗 = 1 if 𝑖 = 𝑗0 if 𝑖 ≠ 𝑗

Kronecker delta tensor

𝐁 𝐀 ∙ 𝐂 − 𝐂 𝐀 ∙ 𝐁 = 𝐵𝑖𝐴𝑗𝐶𝑗 − 𝐶𝑖𝐴𝑗𝐵𝑗

LHS= 𝜖𝑖𝑗𝑘𝜖𝑘𝑙𝑚𝐴𝑗𝐵𝑙𝐶𝑚 = 𝐴𝑗𝛿𝑖𝑙𝐵𝑙𝛿𝑗𝑚𝐶𝑚 − 𝐴𝑗𝛿𝑖𝑚𝐶𝑚𝛿𝑗𝑙𝐵𝑙 = 𝐴𝑗𝐵𝑖𝐶𝑗 − 𝐴𝑗𝐶𝑖𝐵𝑗 =RHS

𝛿𝑖𝑗𝑢𝑗 = 𝑢𝑖

Prove this identity

Left side:

Right side:

So …

(𝐀 × 𝐁)𝑖= 𝜖𝑖𝑗𝑘𝐴𝑗𝐵𝑘𝐀 ∙ 𝐁 = 𝐴𝑖𝐵𝑖

Recap: vector calculus

Variation of a scalar field:

• Gradient

Variation of a vector field:

(A vector field can have two different types of variation)

• Divergence

• Curl

15

Gradient

• A function with one variable 𝑓(𝑥)

• The derivative 𝑑𝑓

𝑑𝑥indicates how fast f varies by x

• 𝑑𝑓 =𝑑𝑓

𝑑𝑥𝑑𝑥

Proportionality factor/ gradient

How a scalar field varies … (we will use later e.g. for potential)

Gradient

• A function with three variable 𝑇(𝑥, 𝑦, 𝑧)

• The derivative indicates how fast the function varieswhich depends on the direction of movement

• 𝑑𝑇 =𝜕𝑇

𝜕𝑥𝑑𝑥 +

𝜕𝑇

𝜕𝑦𝑑𝑦 +

𝜕𝑇

𝜕𝑧𝑑𝑧

• In form of a dot product 𝑑𝑇 =𝜕𝑇

𝜕𝑥 𝐢 +

𝜕𝑇

𝜕𝑦 𝐣 +

𝜕𝑇

𝜕𝑧 𝐤 ∙ 𝑑𝑥 𝐢 + 𝑑𝑦 𝐣 + 𝑑𝑧 𝐤

= 𝛁𝑇 ∙ 𝑑𝐥

𝛁𝑇 ≡𝜕𝑇

𝜕𝑥 𝐢 +

𝜕𝑇

𝜕𝑦 𝐣 +

𝜕𝑇

𝜕𝑧 𝐤

(𝛁 or ‘del’)

The differential operator(vector)

𝛁𝑇 ≡𝜕

𝜕𝑥 𝐢 +

𝜕

𝜕𝑦 𝐣 +

𝜕

𝜕𝑧 𝐤 𝑇

Vector Operator

Three ways the operator ‘del’ can act:

On a scalar function T: 𝛁𝑇 (The gradient)

On a vector v via the dot product: 𝛁 ∙ 𝐯 (The divergence)

On a vector v via the cross product: 𝛁 × 𝐯 (The Curl)

Multiplying normal vectors:

𝛁 ≡𝜕

𝜕𝑥 𝐢 +

𝜕

𝜕𝑦 𝐣 +

𝜕

𝜕𝑧 𝐤

𝑎𝐀

𝐀 ∙ 𝐁

𝐀 × 𝐁

The Divergence

How much the vector v diverges from the point in question

𝛁 ∙ 𝐯 = 𝐢𝜕

𝜕𝑥+ 𝐣

𝜕

𝜕𝑦+ 𝐤

𝜕

𝜕𝑧∙ 𝑣𝑥 𝐢 + 𝑣𝑦 𝐣 + 𝑣𝑧 𝐤

=𝜕𝑣𝑥𝜕𝑥

+𝜕𝑣𝑦

𝜕𝑦+𝜕𝑣𝑧𝜕𝑧

Examples:

In Levi-Civita tensor: 𝛁 ∙ 𝐯 =𝜕𝑣𝑖

𝜕𝑥𝑖

• Variation of a vector field along its direction

𝐯

𝐯

𝑣𝑜𝑙𝑢𝑚𝑒

In class problem #1

Consider the vector function 𝐅 𝐫 = 𝐶𝑥 𝐢. Calculate the divergence 𝛁 ∙ 𝐅 and interpret the result geometrically.

( )F r

x

y

• The net flux of F at r• The flow further away from origin is larger• In the region 𝑥 < 0 the divergence is also positive

(flux away from r)

• Assuming 𝐶 > 0• The divergence at arbitrary point r :

Answers:

𝛁 ∙ 𝐅 =𝑑𝐹𝑥𝑑𝑥

= 𝐶

The Curl

How much the vector v curls around the point in question

𝛁 × 𝐯 =

𝐢𝜕

𝜕𝑥𝑣𝑥

𝐣𝜕

𝜕𝑦𝑣𝑦

𝐤𝜕

𝜕𝑧𝑣𝑧

= 𝐢𝜕𝑣𝑧𝜕𝑦

−𝜕𝑣𝑦

𝜕𝑧+ 𝐣

𝜕𝑣𝑥𝜕𝑧

−𝜕𝑣𝑧𝜕𝑥

+ 𝐤𝜕𝑣𝑦

𝜕𝑥−𝜕𝑣𝑥𝜕𝑦

Examples:

In Levi-Civita tensor: (𝛁 × 𝐯)𝒊= 𝜖𝑖𝑗𝑘𝛻𝑗𝑣𝑘 = 𝜖𝑖𝑗𝑘𝜕𝑣𝑘

𝜕𝑥𝑗

• Variation of a vector field across its direction

In Class Problem #2

Consider the vector function 𝐆 𝐫 = 𝐶𝑥 𝐣. Calculate the curl 𝛁 × 𝐆 and interpret the result geometrically.

• Circulation of G around r

• Assuming 𝐶 > 0

Answers:

( )G r

x

y

• The curl at arbitrary point r :

𝛁 × 𝐆 = 𝐤𝑑𝐺𝑦

𝑑𝑥= 𝐶 𝐤

Additional exercises

• 𝛁 × 𝛁𝑓 = 0

• 𝛁 ∙ 𝛁 × 𝐅 = 0

Using Levi-Civita tensor, prove !

(𝐀 × 𝐁)𝑖= 𝜖𝑖𝑗𝑘𝐴𝑗𝐵𝑘𝐀 ∙ 𝐁 = 𝐴𝑖𝐵𝑖

• (𝛁 × 𝛁𝑓)𝒊= 𝜖𝑖𝑗𝑘𝛁𝑗𝛁𝑘𝑓 = 𝜖𝑖𝑗𝑘𝜕2𝑓

𝜕𝑥𝑗𝜕𝑥𝑘

• 𝛁 ∙ 𝛁 × 𝐅 = 𝜖𝑖𝑗𝑘𝛁𝑖𝛁𝑗𝐅𝑘 = 𝜖𝑖𝑗𝑘𝜕2𝐹𝑘

𝜕𝑥𝑖𝜕𝑥𝑗

Recap: Curvilinear Coordinates

Cylindrical Coordinates

Infinitesimal displacement:

Volume element :

𝑥 = 𝑠 cos𝜙

𝑦 = 𝑠 sin𝜙

𝑧 = 𝑧

Unit vectors: 𝐬, 𝛟, 𝐳

𝑑𝑙𝑠 = 𝑑𝑠 𝑑𝑙𝜙= 𝑠𝑑𝜙 𝑑𝑙𝑧 = 𝑑𝑧

𝑑𝐥 = 𝑑𝑠 𝐬 + 𝑠𝑑𝜙 𝛟 + 𝑑𝑧 𝐳

𝑑𝜏 = 𝑠 𝑑𝑠 𝑑𝜙 𝑑𝑧

Spherical Polar Coordinates

radial

polar

azhimutal

𝑥 = 𝑟 sin 𝜃 cos𝜙

𝑦 = 𝑟 sin 𝜃 sin𝜙

𝑧 = 𝑟 cos 𝜃

Infinitesimal displacement:

Volume element :

Unit vectors: 𝐫, 𝛉, 𝛟

𝑑𝑙𝑟 = 𝑑𝑟 𝑑𝑙𝜃= 𝑟𝑑𝜃 𝑑𝑙𝜙 = 𝑟 sin 𝜃 𝑑𝜙

𝑑𝜏 = 𝑑𝑙𝑟 𝑑𝑙𝜃 𝑑𝑙𝜙= 𝑟2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙

Summary & Outlook

What we have learnt:

• Basic vector operations

• Dot and Cross products

• Vector identities and suffix notations

• Gradient, Divergence & Curl

What next:

• We will apply to electrostatics (Electric fields)