vector calculus - minimal preparation course for …differentiation integration vector calculus...

DifferentiationIntegration

Vector CalculusMinimal preparation course for 1st year electromagnetism

Shinsuke Kawai

Department of Physics/University College, Sungkyunkwan University

Autumn semester 2010

Shinsuke Kawai Vector Calculus


Outline

1 DifferentiationDifferential operatorsGradient, divergence, rotation

2 IntegrationIntegrations in vector calculusIntegration formulaeMaxwell’s equations



Differential operatorsGradient, divergence, rotation

Differential operators

Physical observables we have studied are scalars (such asenergy, charge density, mass density, etc.) and vectors (velocity,electric field, magnetic field, etc.). There are also axial vectors,tensors, etc.

They are functions of positions (x , y , z), as well as of time t .

Let us forget about the time dependence now.

Differentiation by one variable (say, x) while treating others (y andz) as constants – partial differentiation ∂

∂x

Let us make a vector from the differential operators ∂∂x

, ∂∂y

, and∂

∂z

:

~— = (∂

∂x

,∂

∂y

,∂

∂z

). (1)




Products of vectors

For vectors~A = (Ax

,Ay

,Az

) and~B = (Bx

,By

,Bz

), we can define

the scalar product: ~A ·~B = A

x

B

x

+A

y

B

y

+A

z

B

z

.

the vector product:

~A⇥~

B =

��

~ı ~j

~k

A

x

A

y

A

z

B

x

B

y

B

z

��

= (Ay

B

z

�A

z

B

y

,Az

B

x

�A

x

B

z

,Ax

B

y

�A

y

B

x

).




Operation with ~—

On a scalar f (x ,y ,z),

Gradient:

grad f = ~—f = (∂ f

∂x

,∂ f

∂y

,∂ f

∂z

).

On a vector ~F(x ,y ,z),

Divergence: div

~F = ~— ·~F = ∂F

x

∂x

+ ∂F

y

∂y

+ ∂F

z

∂z

.

Rotation (or curl):

rot

~F = curl

~F = ~—⇥~

F = (∂F

z

∂y

� ∂F

y

∂z

,∂F

x

∂z

� ∂F

z

∂x

,∂F

y

∂x

� ∂F

x

∂y

).

Note: div is a scalar, whereas grad, rot are vectors.



Integrations in vector calculusIntegration formulaeMaxwell’s equations

Integration

In 3 dimensions we may consider:

Line integration: along d

~s

Surface integration: over an area vector d

~A

(|dA| is the area and the direction is perpendicular to the surface.Pointing outside if the surface is closed. Use the right hand rulewhen a closed line integral on the boundary curve is defined.)

Volume integration: over a volume element dv = dxdydz.




Integration formulae

There are formulae relating integrals over different dimentions.

The divergence theorem:volume integral over ⌃ $ surface integral over ∂⌃Z

⌃

~— ·~Fdv =I

∂⌃~F ·d~A.

Stokes’ theorem:surface integral over ⌃ $ line integral over ∂⌃Z

⌃(~—⇥~

F) ·d~A =I

∂⌃~F ·d~s.

~F =~

F(x ,y ,z) is a vector. ∂⌃ is the boundary of ⌃.




Proof of the divergence theorem

The divergence theorem isZ

⌃

~— ·~Fdv =I

∂⌃~F ·d~A.

Note that the theorem is additive, so it is enough to show it for a smallcube of size dx ⇥dy ⇥dz.

(x0,y0,z0)

(x0,y0+dy,z0)

(x0,y0,z0+dz)

(x0,y0+dy,z0+dz)(x0+dx,y0+dy,z0+dz)

(x0+dx,y0,z0)

(x0+dx,y0,z0+dz)

(x0+dx,y0+dy,z0)

dz

dydx

F(x0,y0,z0)F(x0+dx,y0,z0)→

→




Proof of the divergence theorem – cont.

Using f (x0 +dx) = f (x0)+∂ f

∂x

��x=x0

dx +O(dx

2),

Z

⌃

~— ·~Fdv =Z

⌃

✓∂F

x

∂x

+∂F

y

∂y

+∂F

z

∂z

◆dxdydz

=Z

dydz (Fx

(x0 +dx)�F

x

(x0))

+Z

dzdx (Fy

(y0 +dy)�F

y

(y0))

+Z

dxdy (Fz

(z0 +dz)�F

z

(z0))

=I

∂⌃~F ·d~A.




Proof of Stokes’ theorem

Stokes’ theorem isZ

⌃(~—⇥~

F) ·d~A =I

∂⌃~F ·d~s.

Again, it is enough to show it for a small area of size dx ⇥dy .

dx

dy(x0,y0)

F(x0,y0)→

→

→

→

F(x0+dx,y0+dy)

F(x0+dx,y0)

F(x0,y0+dy)




Proof of Stokes’ theorem – cont.

Z

⌃(~—⇥~

F) ·d~A =Z

⌃

✓∂F

y

∂x

� ∂F

x

∂y

◆dxdy

=Z

dx

∂F

y

∂x

dy �Z

dy

∂F

x

∂y

dx

=Z

(Fy

(x0 +dx ,y ,z)�F

y

(x0,y ,z))dy

�Z

(Fx

(x ,y0 +dy ,z)�F

x

(x ,y0,z))dx

=I

∂⌃~F ·d~s.




Maxwell’s equations – Gauss (electric)

In the differential representation,

~— ·~E =re0

Performing the volume integration and applying the divergencetheorem, Z

⌃

~— ·~Edv =I

∂⌃~E ·d~A =

1e0

Z

⌃rdv .

q =R⌃ rdv is the electric charge contained in the region ⌃. Hence

I

∂⌃~E ·d~A =

q

e0.




Maxwell’s equations – Gauss (magnetic)


~— ·~B = 0

Performing the volume integration and applying the divergencetheorem, Z

⌃

~— ·~Bdv =I

∂⌃~B ·d~A = 0.




Maxwell’s equations – Faraday


~—⇥~E =�∂~B

∂ t

Performing the surface integration and applying Stokes’ theorem,

Z

⌃(~—⇥~

E) ·d~A =I

∂⌃~E ·d~s =� ∂

∂ t

Z

⌃

~B ·d~A.

�B

=R⌃~B ·d~A is the magnetic flux. Hence,

I

∂⌃~E ·d~s =�∂�

B

∂ t

.




Maxwell’s equations – Ampère-Maxwell


~—⇥~B = µ0~j + e0µ0

∂~E∂ t

Performing the surface integration and applying Stokes’ theorem,

Z

⌃(~—⇥~

B) ·d~A =I

∂⌃~B ·d~s = µ0

Z

⌃

~j ·d~A+ e0µ0

∂∂ t

Z

⌃

~E ·d~A.

I =R⌃~j ·d~A is the current and �

E

=R⌃~E ·d~A is the electric flux.

Hence, I

∂⌃~B ·d~s = µ0I + e0µ0

∂�E

∂ t

.


overview of Electromagnetism

Maxwell’s equations1. Gauss (electric)

2. Gauss (magnetic)

3. Faraday

4. Ampère-Maxwell

Coulomb

Biot-Savart

Integral expressions

I~E · d ~A = q/✏0

I~B · d ~A = 0

I~E · d~s = �@�B

@t

I~B · d~s = µ0I + ✏0µ0

@�E

@t

Differential expressions

d ~B =µ0

4⇡

Id~s⇥ ~n

r2

r · ~E = ⇢/✏0

r · ~B = 0

r⇥ ~E = �@ ~B

@t

r⇥ ~B = µ0~j + ✏0µ0

@ ~E

@t

~E =1

4⇡✏0

q

r2~n

~F = q( ~E + ~v ⇥ ~B)Lorentz force

(special case)

vector calculus - minimal preparation course for …differentiation integration vector calculus...

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