vector calculus notes
DESCRIPTION
Brief notes on vector calculus, including uses of suffix notationsTRANSCRIPT
• Intro
• Recap : vector algebra, calculus
• Gradient, Divergence, Curl
Lec #2
Vector Calculus
PC 2131Electricity & Magnetism
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
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