1

PHY2009S“Fields and fluids”

Andy BufflerDepartment of Physics

University of Cape Town

Part A: Vector Calculus

for Physics

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The story so far….

Right handed coordinate system:Unit vectors kji ˆ ˆ ˆ

1 ˆ ˆ ˆ === kji

, ,

ˆ ˆ ˆ( ) ˆ ˆ ˆ( , , )

( , , )

( , , )

( ) ( , , )

( )

d

x y z x y z

x y z x y z

x x y y z z

x y z

x x y y z z

x y z

x x y y z z

A ,A ,A A A A

B B B B B B

A B A B A B

A A A

A B A B A B

c cA cA cA

A B A B A B

= = + +

= = + +

+ = + + +

− = − − −

− = + − = − − −

=

= + + =

=

A i j k

B i j k

A B

A

A B A B

A

A B

A B B A

i

i i 2 A ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ1 0

=

= = = = = =

A A

i i j j k k i j j k k i

i

i i i i i i

z

y

x

k j

i

Vector algebra

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( ) 0ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ 0 ; ;

× = − × × =

× = × = × = × = × = × =

A B B A A A

i i j j k k i j k j k i k i j

where and = ⊥ ⊥G G A G B

ˆ ˆ ˆ ( ) + ( ) + ( ) × = − − −y z z y z x x z x y y xA B A B A B A B A B A BA B i j k

easy to remember:always

ˆ ˆ ˆ

x y z

x y z

A A A

B B B

i j k

In polar form in 2D:

and

where is the angle between tails of and .

θcosAB=•BA kBA ˆsinθAB=×

θ BA

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5

The spherical polar coordinate system

2 2 2

ˆ ˆ ˆ

cos sin sin sin

cos

cos

tan

x y z

x

y

z

x y z

z

y

x

A A A

A AA A

A A

A A A A

AAAA

φ θφ θ

θ

θ

φ

= + +

=

=

=

= + +

=

=

A i j k

Spherical coordinates: A, θ, φ :

z

y

x

k ji

Ax

Az

Ay

Aθ

φ

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The cylindrical polar coordinate system

2 2

ˆ ˆ ˆ

cos sin

tan

x y z

x

y

z

x y

y

x

z

A A A

AA

A z

A A

AA

z A

ρ φρ φ

ρ

φ

= + +

=

=

=

= +

=

=

A i j k

Cylindrical coordinates: ρ, θ, z :

z

y

x

k ji

Ax

Az

Ay

Aθ

φρ

z

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ˆ ˆ ˆ( ) ( ) ( ) ( )

( )( ) ( )ˆ ˆ ˆ ( )

( ) ( )( ) ( )

( ) ( )( ) ( ) ( ) ( )

( ) ( )( ) ( ) ( )

x y z

yx z

t A t A t A t

dA tdA t dA td tdt dt dt dt

d d t d tt tdt dt dtd dc t d tc t t t c tdt dt dtd d t d tt t tdt dt

= + +

= + +

⎡ ⎤+ = +⎣ ⎦

⎡ ⎤ = +⎣ ⎦

⎡ ⎤ = +⎣ ⎦

A i j k

A i j k

A BA B

AA A

B AA B Ai i ( )

( ) ( )( ) ( ) ( ) ( )

tdt

d d t d tt t t tdt dt dt⎡ ⎤× = × + ×⎣ ⎦

B

B AA B A B

i

Differentiation of vector functions

Also:

then

If

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If position m

then the instantaneous velocity m s-1

and the instantaneous acceleration m s-2

( ) ( ) ( )ˆ ˆ ˆ( ) dx t dy t dz ttdt dt dt

= + +v i j k

( )( ) d ttdt

=va

( )( ) d ttdt

=rv

( )( ) ( )ˆ ˆ ˆ( ) yx zdv tdv t dv tt

dt dt dt= + +a i j k

ˆ ˆ ˆ( ) ( ) ( ) ( )t x t y t z t= + +r i j k

Example of the time derivatives of a vector function

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Consider a scalar function f(x,y) of two variables x

and y.

If the variable y

is kept fixed and x

is allowed to vary, the function f(x,yfixed

) will change, and the partial derivative of f(x,y) with respect to x

is defined in a similar way as the

ordinary derivative.

i.e.

for a small change in f(x,y) if x

is changed by ∂x

then

( , ) ( , ) ( , ) ( , )∂∂ = + ∂ − ≈ ∂

∂f x y f x x y f x y f x y x

x

xyxfyxxfyxf

x x ∂−∂+

=∂∂

→∂

),(),(lim),(0

Partial derivatives

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Example: If f(x,y) = x³cos y, then

( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , )

( , ) ( , )

∂ = + ∂ + ∂ −= + ∂ + ∂ − + ∂ + + ∂ −

∂ ∂≈ + ∂ ∂ + ∂

∂ ∂

f x y f x x y y f x yf x x y y f x y y f x y y f x y

f x y y x f x y yx y

( , , ) f f fdf x y z dx dy dzx y z∂ ∂ ∂

= + +∂ ∂ ∂

If there are small changes in both x

and y:

, 0lim ( , ) ( , )x y

df f f x y dx f x y dyx y∂ ∂ →

∂ ∂= ∂ = +

∂ ∂

In the limit as ∂x

→ 0 and ∂y

→0

where ∂f

, ∂x

and ∂y

are infinitesimally small.

2 33 cos sin f fx y x yx y∂ ∂

= = −∂ ∂

In 3D:

and

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Scalar and vector fields

A scalar or vector field is a distribution

of a scalar or vector quantity on a specified surface or throughout a specified region of space

such that there is a unique scalar or vector

associated with each position.

Fields may be time independent, e.g. or time dependent

( , , )T x y z( , , , )T x y z t

Examples of scalar fields:• Temperature, or• Potential, or

Examples of vector fields:• Electric field• Velocity

( )T r( , , )T x y z( , , )V x y z ( )V r

( )E r( )v r

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Scalar fields

A scalar field can be represented by specifying a finite number of scalar values

at strategic positions in the region of

interest.

It is also possible to draw contour curves

- continuous curves joining points where the scalar values are the same. In 3D space these contours are surfaces. Such representations are always incomplete, since an infinite number of contours or surfaces should really be drawn.

A third way of representing a scalar field is by a mathematical function.

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Scalar fields in 2D ....

... and 3D ...

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For example, consider the hydrostatic pressure

p

in a lake. Assuming that the lake is flat and extensive, and introducing a coordinate system with the origin and x-y

plane on the surface

of the lake:

The pressure p

at depth z

is given by

p(x,y,z) = po

+ ρ g z

where po

= atmospheric pressure

A scalar field exists in a specified region only, called the domain

of the function. In the example above, the domain of

p(x,y,z) is restricted to on, or below, the surface of the lake.

xy

z

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Scalar fields are difficult to visualize in 3D, so we consider 2D representations first.Consider a map of terrain where the altitude z

is a function

of position . The use of contours

(lines of constant altitude) is a familiar

way of representing

Scalar fields and the gradient operator

( ) ( ) ( ) t x t y t= +r i j

( )z r

Now, how does z vary around (x,y)?If we walk along a path such as the

one shown, with a velocity ,

what would be ? ( )dz

dtr

( )( ) d ttdt

=rv

contours represented by= constant( )z r

x

yz

out

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The rate of change of the altitude z depends on the relative directions of

and , and is a maximum when .The vector “points uphill” and gives the magnitude and direction of the local slope of .

v z∇z∇v

( )z r

z∇

Then ( ) d d z z z dx z dyz dx dydt dt x y x dt y dt

⎛ ⎞∂ ∂ ∂ ∂= + = +⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

r

ˆ ˆ( ) ∂ ∂∇ = +

∂ ∂z zzx y

r i j

a vectorthe “gradient” of zcontour lines

direction of z∇

where

ˆ ˆ ˆ ˆ ( ) ( )z z x y z tx y t t

⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞= + + = ∇⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠i j i j r vi i

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E∇

Consider a topographic map of a hill showing the elevation E

as a

function of position.

= at onepoint whereE

= 400 m.

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The gradient vector field

of a scalar field specifies the magnitude and direction of the larger spatial rate of change of at any point (x,y,z) .

The gradient operator in 3D:

“gradient operator”can also be written as or

ˆ ˆ ˆ

ˆ ˆ ˆ

grad "del"

x y z

x y zφ φ φφ

φ φ

∂ ∂ ∂∇ = + +

∂ ∂ ∂∂ ∂ ∂

∇ = + +∂ ∂ ∂

∇ = =

i j k

i j k

φ φ∇φ

∇

:∇, , ∇ ∇ ∇

The gradient of scalar field

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Example:

Determine the gradient of the following scalar functions:

(a)

(b)

2 3 2 33x y z x yz+ +

cos yzx e x+

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This animation illustrates the gradient of a scalar function.

The animation shows: · the surface · a unit vector rotating about the point (1, 1, 0) · a rotating plane parallel to the unit vector · the traces of the planes in the surface · the tangent lines to the traces at (1, 1, f

(1, 1))

· the gradient vector (shown in green)

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( ) ( ) ( ) ,

ˆ ˆ ˆ ˆˆ ˆ( ) ( )

( ) or ( )

φ φ φφ

φ φ φφ

φ φ φ φ

∂ ∂ ∂= + +

∂ ∂ ∂∂ ∂ ∂

= + + + +∂ ∂ ∂

∴ = ∇ Δ = ∇ Δ

i

i i

d dx dy dzx y z

d dx dy dzx y z

d d

i j k i j k

r r

dφ∇ r

cosd dφ φ θ= ∇ r

Since

Then

where θ is the angle between and .

is maximum when or (i.e. when you move in the same direction as )

Therefore:• points in the direction of maximum increase

of the function.

• The magnitude gives the slope

(rate of increase) along this

maximal direction.

d cos 1 0 φ θ θ= =

φ∇

φ∇

φ∇

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In 2D think of standing on a hillside. The direction of steepest ascent (or descent) is the direction of the gradient at that position. the magnitude of the gradient is equal to the maximum of the slope.

In 3D think of the temperature T in a room as a function of position Then and the contours are surfaces defined by constant temperature.

The gradient is normal to these surfaces.A bee flying with velocity in a path would sense an instantaneous variation of temperature given by .

Positions at which are interesting and are called stationary points. These points include local maxima and minima and saddle points.

.ˆˆˆ kjir zyx ++=

vT∇

T∇vi

0φ∇ =

()t r

( )T T= r

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Example:

Determine the gradient of the following scalar function at the position (1, 2, 3) metres:

2 22 3x y yz zφ = + + ˚C

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1. The temperature gradient in 2 dimensionsLet T(x,y) be the temperature at the point .If the surfaces of constant temperature will be parabolas where c

is the constant

or .Shown are set of these surfacesfor c

= −1, 0, 1 and 2.

The temperature gradient at any point is given by .The vector is shown at the points (0,0) and (1,1) on the contour T = 0.

ˆ ˆx y= +r i j,),( 2 yxyxT +−=

2x y c− + =cxy += 2

ji ˆˆ2),( +−=∇ xyxTT∇

jji ˆˆˆ0 =+=∇T

jiji ˆˆ2ˆˆ)1(2 +−=+−=∇T

Examples of the gradient operator in physics

and at (1,1)

at (0,0)

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2. Spherically symmetric fields

It is common in physics to find fields which are only a function of a radial distance from some origin.

i.e. where .

)(rφ

2 2 2( ) ( ) r r x y zφ φ= = + +r

122 2 2( ) ( ) 1 ( ) ( ) ( ) 2

2r r d r d r x d rx y z x

x x dr dr r drφ φ φ φ−∂ ∂

= = + + =∂ ∂

Then

( )ˆ ˆ ˆ( ) ( )

( ) ˆ( )

x y z d rrr r r dr

d rrdr

φφ

φφ

∇ = + +

∇ =

i j k

r

Then

or

sinceˆ ˆ ˆ

ˆ .x y zr

+ += =

r i j krr

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3. Gravitational fields

The conservative gravitational and electrostatic force fields of nature can be written gradients of appropriate scalar potential fields.

For example the gravitational force field

due to mass m

at the origin can be written as

and in terms of the gradient of the gravitational potential,

The equipotential surfaces are spheres centred at the origin, and the gravitational field lines are radial.

g

Gmρr

ρ= −∇ =g

2ˆGm

r−

=g r

where .

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5. Electrical conductivityThe electrical current density in a medium of conductivity σ

is

given in terms of the electrical potential ϕ by

This is essentially Ohm’s law: for a rod of length l

, cross-sectional area A, carrying a current i, then j

= i

/A

and

where V

is the potential difference across the rod.Then .

4. Electrostatic fields… have the same form as the gravitational fields.Write for the electric field for a point charge q, at the origin

and where .

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1 ˆ ( )4

qrπε

=E r

ϕ= −∇E

E

)j(r( ) ( )r r ϕ= −∇j

V lϕ∇ = −

lAi Vσ=

ϕ is the electric potential (also spherically symmetric)

0

1 ( )4

qr

ϕπε

=

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ϕ(x,y)

xy

The electric potential in 2D ϕ(x,y) around a positive point charge

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7. DiffusionFor the diffusion of particles in a closed system

where is the density of particles at position is the current of the particlesis a constant (the diffusion coefficient)

6. Heat conductivityFor the flow of heat in the system (with no internal sources)

where: is the temperature at positionis the heat current at is a constant (the thermal conductivity)

( ) ( )K θ= − ∇j r r

)(rθ( )j rK

( ) ( )r D n= − ∇j r

)(rn)r(j

D

rr

r

(Fick’s law)

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A class of operations which leaves a function unchanged is referred to as the symmetry

of the function.

If these operation concern only the spatial coordinates of the function, then the symmetry is the spatial symmetry.

• Spherical symmetry: when the values of the field depend only on the distance from a fixed point, such as the origin).e.g. the light intensity field from a point source.

• Cylindrical symmetry: when the values of the field only depend on the perpendicular distance from a fixed straight line, such as an axis.e.g. the electric potential around a long thin charged conductor

222 zyxr ++=

22 yx +=ρ

Symmetry

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• Translational (Cartesian) symmetry: when the values of the field are independent of position with respect to a particular coordinate axis.e.g. the electric field between two parallel plates.

It useful to recognise spatial symmetries where they exist so that the appropriate coordinate system can be chosen so as to reduce the number of spatial variables to a minimum.

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A vector field is a vector function

of position.

Vector fields may be represented visually by field lines

which

are everywhere parallel to the local value of the vector function.These lines are sometimes called “lines of force” in mechanics and “stream lines” in fluid mechanics.

A vector field may also be represented by lines which are everywhere a tangent

to the vectors. Although we lose track of the

lengths of the vectors, we can keep track of the strength of the field by drawing lines far apart where the field is weak, and close where it is strong.

Vector fields

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Vector fields may also be represented mathematically, often using differential equations.

We adopt the convention that the number of lines per unit area at right angles to the lines is proportional to the field strength.

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36

The electric field around a point charge q

may be written as

where and

In Cartesian form

and

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ˆ( ) ( )4

qπε

=E r rr

rrrr => ˆ 0

2 2 2 2 2 2 3 22 2 20 0

ˆ ˆ ˆ ˆˆ ˆ( ) ( ) ( ) 4 ( ) 4 ( )

q x y z q x y zx y z x y zx y zπε πε

+ + + +∴ = =

+ + + ++ +

i j k i j kE r

kjir ˆˆˆ zyx ++=

Example of a vector field function of position: the electric field

2 2 2

ˆ ˆ ˆˆ x y z

x y z+ +

= =+ +

r i j krr

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This vector function is the sum of a 3 component vector

where and are the three scalar components of

2 2 2 3 20

2 2 2 3 20

2 2 2 3 20

( , , ) 4 ( )

( , , ) 4 ( )

( , , ) 4 ( )

x

y

z

qxE x y zx y z

qyE x y zx y z

qzE x y zx y z

πε

πε

πε

=+ +

=+ +

=+ +

).( , rEzyx EEE

The electric field continued

2 2 2 3 20

ˆ ˆ ˆ( )( ) 4 ( )

q x y zx y zπε+ +

=+ +

i j kE r

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Consider now the scalar product

of with a vector field :( )F r

div ( ) ( )

ˆ ˆ ˆ ˆˆ ˆ( ) = ( )

x y z

yx z

F F Fx y z

FF Fx y z

= ∇

⎛ ⎞∂ ∂ ∂∇ + + + +⎜ ⎟∂ ∂ ∂⎝ ⎠

∂∂ ∂= + +∂ ∂ ∂

F r F r

F r i j k i j k

i

i i

The divergence of a vector field

is a measure of how much spreads out (diverges) from the point in question.

Note that is a scalar field.

( )∇ F ri

∇

( )F r

( )∇ F ri

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Field lines have sources

, i.e. starting points in regions of positive divergence, and sinks

in regions of negative

divergence.

If a vector has zero divergence everywhere, then there are no sources or sinks anywhere, and the field lines form closed loops.

The divergence of a vector field continued

div > 0 atF

div = 0 atF

div < 0 atF

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Example:

Determine the divergence of the following vector functions:

(a) N

(b) N

2 2ˆ ˆ ˆ = 2 3 xy x y z+ +F i j k

2 3ˆ ˆ ˆ = y z x y xy+ +F i j k

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The animation shows a flow field using animated textures. The direction of the velocity field is indicated by the correlation in the textures. When animated, the texture patterns move in the direction of the velocity field. This flow field has a source at the origin, and the texture patterns diverge from that point.

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This animation shows a flow field with a source sitting in a constant downward flow.

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Line integrals occur whenever we need to integrate a scalar or vector function along a line or curve.

An example of a scalar line integralto determine the work done by the force acting along path of a moving particle :

An example of a vector line integral

to determine the force on a wire carrying an electric current in a magnetic field :

2

1

( )r

r

W d= ∫F r ri

( )C

I d= − ×∫F B r r

Line integrals

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Consider, for example, a comet moving along a path around the sun and let Cbe the segment of this path.

is the sun’s gravitational force acting on the comet at position .Both the magnitude and direction of varies with .

The work done by along the path segment C

is a scalar quantity.

For the small path segment shown:

and along the entire path C

:

Taking the limit of a large number of subpaths:

( )W ≈ ΔF r ri

1

( )n

i

W=

≈ Δ∑F r ri

1

( ) lim( ( ) )n

n iC

d→∞

=

= Δ∑∫F r r F r ri i

r( )F r

Scalar line integrals

( )F r

( )F r

iΔr

ir

C

i i+ Δr r

r

( )C

d∫F r ri is the scalar line integral

of along the path C.( )F r

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Also

1 1( ) ( ) ( )

n n

i i x i y i ii i i

dyF x F x xdx= =

⎧ ⎫⎛ ⎞∴ Δ = + Δ⎨ ⎬⎜ ⎟⎝ ⎠⎩ ⎭

∑ ∑F r ri

1( ) lim( ( ) )

n

i in iC

d→∞

=

= Δ∑∫F r r F r ri i

ˆ ˆ

i ii

i i ii

dyy xdx

dyx xdx

⎛ ⎞Δ ≈ Δ⎜ ⎟⎝ ⎠

⎛ ⎞∴Δ = Δ + Δ⎜ ⎟⎝ ⎠

r i j

ˆ ˆi i ix yΔ = Δ + Δr i j

ˆ ˆ ( ) ( ) ( )i x i y iF x F x= +F r i j

Evaluating line integralsA.

Evaluating line integrals in the x-y plane using the equation of the curve:

Consider

In 2D:

If the path C

is described by y

= y(x) then

2

1

( ) ( ) ( )x

x yC x

dyd F x F x dxdx

⎧ ⎫⎛ ⎞∴ = +⎨ ⎬⎜ ⎟⎝ ⎠⎩ ⎭

∫ ∫F r ri

Each term is a function of x

only

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Example 1:

Calculate the work done by a force newtons

along the parabola C

specified by the equation from

the point A(1,0) metres to B(0,1) metres .

jiF ˆˆ 222 yxx +=21 xy −=

47

Example 2:

Calculate the work done by a force

newtons

along the path specified by the equation

from the point A(1,1) metres to B(2,8) metres .

2 ˆ ˆ(5 6 ) (2 4 )xy x y x= − + −F i j3y x=

48

It is often useful to specify a curve C

in terms of some non-Cartesian parameter t.

Each point on the curve then has coordinates (x(t), y(t), z(t)) where x(t), y(t) and z(t) are known functions of t.

For a small displacement on the i th subpath

where Δti

is the small change in t

moving from the beginning point to the end point of the subpath.

ˆ ˆ ˆ i i i ii i i

dx dy dzt t tdt dt dt

⎛ ⎞ ⎛ ⎞ ⎛ ⎞Δ ≈ Δ + Δ + Δ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

r i j k

iΔr

B.

Evaluating scalar line integrals using parametric equations of the curve.

49

2

1

( ) ( ) ( ) ( )t

x y zC t

dx dy dzd F t F t F t dtdt dt dt

⎧ ⎫⎛ ⎞ ⎛ ⎞ ⎛ ⎞∴ = + +⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭

∫ ∫F r ri

ˆ ˆ ˆ( ) ( ) ( ) ( ) x i y i z iF t F t F t= + +F r i j k

( ) ( ( ), ( ), ( ))x i x i i iF t F x t y t z t=

1 1

( ) ) ( ) ( ) ( )n n

i i x i y i z i ii i

dx dy dzF t F t F t tdt dt dt= =

⎧ ⎫⎛ ⎞ ⎛ ⎞ ⎛ ⎞Δ = + + Δ⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭

∑ ∑F r riThen

where

Also

ˆ ˆ ˆ i i i ii i i

dx dy dzt t tdt dt dt

⎛ ⎞ ⎛ ⎞ ⎛ ⎞Δ ≈ Δ + Δ + Δ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

r i j k

( ) ( ( ), ( ), ( ))z i z i i iF t F x t y t z t=( ) ( ( ), ( ), ( ))y i y i i iF t F x t y t z t=

Each term is a function of t only

50

Example 1:

Evaluate the scalar line integral of the vector field

along the curve C

defined by the

parametric equations where

a, b, and c

are constants and t goes from 0 to 1.

ˆ ˆ ˆ yz zx xy= + +F i j k32 ; ; ctzbtyatx ===

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Example 2:

What is the work done by a force

newtons

in moving a particle along the path metres

between t

= 0 and t

= 1 ?

ˆ ˆ ˆ(2 3) ( ) y xz yz x= + + + −F i j k2 3ˆ ˆ ˆ2t t t= + +r i j k

52

Line integrals can also be evaluated along closed curves or loops.

Consider the loop shown alongside. The line integral of a vector fieldfor one complete anticlockwise transversal of this loop starting at point A is alongthe path ABCDEFA.

This can be written:

C ABC CDE EFA

d d d d= + +∫ ∫ ∫ ∫F r F r F r F ri i i i

FB

A

CD

E

F

C∫ integral around a closed loop The order of these letters

indicate the direction

Line integrals around closed loops

53

Example:

Evaluate where

and C

is the closed loop APBQA shown alongside.

BQA: straight lineAPQ: quarter circle

ˆ ˆ(2 ) x y x= + −F i j

• C

d∫ F r

A(2,0)

B(0,2)

Q

P

x

y

54

Consider the scalar fieldMove from to along path shown:

• for the first line segment

• for the second line segment

for the whole curve C:

),,()( zyxφφ =rBA PP

( ) A APφ φ=r

( ) B BPφ φ=r

φ φΔ = ∇ Δi

1 1 1( )a Aφ φ φ φΔ = − = ∇ Δi

2 2 2( )b aφ φ φ φΔ = − = ∇ Δi

1 1 2 2( ) ( )b Aφ φ φ φ∴ − = ∇ Δ + ∇ Δi i

( )B A i ii

φ φ φ− = ∇ Δ∑ i

Line integrals cont….

at

Since

BP

AP

path C

1Δ 2Δ 3Δ

a bc

iΔiφ∇

at

A

B AB

dφ φ φ− = ∇∫ i Taking the limit → 0 as i

→∞Δ

55

• True for any curve C

joining points and

• If you define to mean “the line integral around a closed path”

then always

since

2

1

2 1( ) ( ) dφ φ φ φΔ = − = ∇∫r

r

r r i

1 2 r r

∫0dφ∇ =∫ i

0)()( 11 =− rr φφ

Generally:

1( )φ r

56

Later we will see that, from Stokes’ theorem for conservative field , everywhere(curl test

for a conservative field).

curl 0=G

Any vector field that can be expressed as the gradient of a scalar field is conservative.

i.e. if is a scalar field, then is a conservative field, i.e. f is a potential of .

( ) ( ) ( ) f f= −∇r G r rG

( )G r

Conservative fields

57

The line integral of a conservative field along the path between any two fixed points A and B depend only on the beginning point A and the end point B of the path.It is independent

of the path taken between the points.

Therefore if we asked to evaluate a line integral of a conservative field between two given points then we are free to choose a path so as to produce the simplest possible integral.

The line integral of a conservative field around any closed loop is always zero.

0 C

d =∫G ri G

The scalar potential of conservative field is defined by( ) ( ) ( )A B

AB

G d V V= −∫ r r r ri

i.e. for a conservative field

( )V r

Conservative fields continued

58

Example:

An electric field is described by:

N C-1

Is this field conservative?

What is the potential difference φb

−

φa between points

a

= (1, 2 , 3) and b

= (−4, 5, 6) metres in the field?

2ˆ ˆ ˆ(4 5 ) 2 (5 4) xy z x x= + + + −E i j k

59

Surface integration involves the summation of scalar or vector values over a surface.

Think of :• the hydrostatic forces on a wall• the flux of magnetic field lines• ... or rain striking your roof :

Surface integrals

normal to plane

raindirection

θ

60

Consider the surface S

and the scalar function of value

at a point on surface S.

Divide the surface into Nsurface elements of area and let be the position vector of a point on the surface element.

Then

( ) f r r

ir

1

( ) lim ( )N

i iN iS

f da f a→∞

=

= Δ∑∫ r r

; ( 1 to )ia i NΔ =

The scalar surface integral

surface integral

S

iriaΔ

61

Let and divide the surface S

into elements by a rectangular grid, then

and becomes where

( ) ( , )f f x y=rkjjk yxa ΔΔ=Δ

1 1 1

( ) ( , ) N m n

i i j k j ki k j

f a f x y x y n m N= = =

Δ Δ Δ ⋅ =∑ ∑∑r

Evaluating surface integrals the plane surfaces using Cartesian coordinates.

Consider the special case of a rectangular surface Swith the sides parallel to the coordinate axes.

This rectangular surface is bounded by the four straight lines:x = a; x = b; y = c

and y = d.

yy=d

y=c

x=a xx=bΔxj

Δyk

62

If we first do a double summation over the and then over the then the summation is

j kx yΔ Δ

1 1

( , )m n

j k j kk j

f x y x y= =

⎛ ⎞Δ Δ⎜ ⎟

⎝ ⎠∑ ∑

1

: ( , ) ( , )bn

k j kj a

n f x y x f x y dx=

→∞ Δ →∑ ∫As

here the coordinate yk

is fixed

( , )d b

c a

f x y dx dy⎛ ⎞

= ⎜ ⎟⎝ ⎠∫ ∫

( )1

lim ( , )m b

k kam kf x y dx y

→∞=

∴ Δ∑ ∫

( , )S

f x y da= ∫

63

Example

Evaluate the surface integralof the function over the rectangular surface Sshown alongside.

2( , )f x y xy=

y

2

1

0 x5

S

64

where the curves and describe the left hand and the right hand boundaries of the region.

If the surface is not rectangular then we can write:

( )( )

( )( ) ( , )

dh y

g yS c

f da f x y dx dy=∫ ∫ ∫r

)( )( yhxygx ==

Sometimes it more convenient to use cylindrical

or spherical

coordinate systems,

instead of a Cartesian coordinate system.

y

x

( )x g y=

( )x h y=S

d

c

65

Example

Evaluate the surface integralof the function on the triangular surface S

shown alongside.

2),( xyxf =

y

1

0 x½

S0

66

Consider a vector field

For a constant field and small area element

Flux ≡

“volume flow per unit time”

For a large surface S,flux through surface

)(rF

d= F ai

Fad

S

d= ∫F ai

Surface integrals and fluxad Fθ

(= F da

cosθ)

a surface integral

adF

S

67

Generally, “flux” is the surface integral of the normal component of the vector:

If then the source of is inside

S

then the source of is outside

S

0S

d >∫ F ai

For a closed surface of volume V:

Flux

where where is a unit vector pointing outwards at right angles to S.

ˆ S S

d da=∫ ∫F a F ni i

ˆ d da=a nn

< 0S

d∫ F ai

F

F

68

Example

Calculate the flux of the vector field

through the surface shown alongside.

z−3 m

x

−4 m

ˆ ˆ ˆ = (3 2 ) ( 2) yz zy xyz+ + − +F i j k

69

70

The volume of the cube

We wish to find the flux of a vector field through the surface of the cube.Net outward flux through shaded faces

x y z= Δ Δ ΔVΔ

Left face Right face

d d= +∫ ∫F a F ai i

F

ˆ ˆ ˆ ˆ( ) ( ) ( )

( ) ( )

( )

y y

y y

y yy y

F y x z F y y x z

F y x z F y y x z

F FF y x z F y y x z x y z

y y

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤Δ Δ − + + Δ Δ Δ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦= − Δ Δ + + Δ Δ Δ

∂ ∂⎛ ⎞= − Δ Δ + + Δ Δ Δ = Δ Δ Δ⎜ ⎟∂ ∂⎝ ⎠

j j j ji i

The divergence theorem

Consider a small cube whoseedges are aligned with the xyzaxes as shown.

z

xy

zΔ

yΔxΔ

y Δy

71

So the total outward flux from the surface of an infinitesimal cube is equal to the divergence

of the vector field multiplied by the

volume of the cube.

yx zFF F x y z

x y z∂⎛ ⎞∂ ∂

= + + Δ Δ Δ⎜ ⎟∂ ∂ ∂⎝ ⎠

surfacesof cube

d V= ∇ × Δ∫ F a Fi i

Divergence theorem …

2z

xy

zΔ

yΔxΔTotal outward flux from whole cube (six faces)

or

yx zFF Fx y z x y z x y z

x y z∂∂ ∂

= Δ Δ Δ + Δ Δ Δ + Δ Δ Δ∂ ∂ ∂

72

Then the total flux of out of V

=

The divergence of a vector at a point in space is the flux per unit volume in the neighbourhood of the point.

= outward flux of per unit volume at a point.div F

F

div dVF

div V

dV∫ F

• S V

d dV→ = ∇∫ ∫F a Fi

Divergence theorem …

3

F

For a finite volume V

the total flux is the sum of fluxes out of each part.

for any closed surface

Flux out of volume element dV

=

volume integral

VdV

surface S

Divergence theorem( or Green’s theorem or Gauss’ theorem)

73

The divergence tells us about whether is, on average, diverging

or converging

to the point:

0div lim S

V

d

V→= ∇ = ∫ F a

F Fi

i

F

Divergence theorem …

4

• S V

d dV= ∇∫ ∫F a Fi Divergence theorem

∇ Fi

From above:

div > 0 atFdiv = 0 atF

div < 0 atF

74

The mass M of material of uniform density ρ is simply ρV

where V

is the volume.

If the density is not uniform, then we can write .

It is possible to divide the block into small volume elements, each of which is small enough that the density does not vary significantly within it.

( )V

M dVρ= ∫ r

Volume integrals

i

N

ii VM Δ≈∑

=1)(rρ

( )ρ r

Then

or volume integral

75

Let be a scalar field in three dimensions and let B be the region of space (e.g. a cube) of total volume V

within the domain of .

( )f r

Then the volume integral of over the region V

is

1

( ) lim ( )N

i iV N i

f dV f V→∞

=

= Δ∑∫ r r

i 1 1 1 1

( ) ( )l

qN n m

i i j k l j kl k j

f V f x y z x y z= = = =

⎛ ⎞⎛ ⎞Δ = Δ Δ Δ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

∑ ∑ ∑ ∑r

( )( )( ) ( , , )f d b

iV e c af dV f x y z dx dy dz=∫ ∫ ∫ ∫r

( )f r

If Cartesian coordinates are used, then

If the region V is the rectangular block with faces parallel to the coordinate planes, then

z

x

yyΔ xΔ

z = f x = a

x = b

y = c y = d

z = e

zΔiVΔ

V

Evaluating volume integrals the plane surfaces using Cartesian coordinates.

76

Example

A cube of edge 2a is centred at the origin.What is the flux of the vector field through the surface of the cube?Check your answer by also applying the divergence theorem and working out the volume integral.

ˆ ˆ = (2 ) x y xy+ +F i j

77

... is the cross product

of with the vector field:

curl ( ) ( ) ( )= ∇× =F r F r G r

( )ˆ ˆ ˆ ˆˆ ˆ( ) x y zF F Fx y z

⎛ ⎞∂ ∂ ∂∇× = + + × + +⎜ ⎟∂ ∂ ∂⎝ ⎠

F r i j k i j k

ˆ ˆ ˆ y yx xz zF FF FF F

y z z x x y∂ ∂⎛ ⎞ ⎛ ⎞∂ ∂∂ ∂⎛ ⎞= − + − + −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

i j k

The curl of a vector field

a vector field

ˆ ˆ ˆ

x y z

x y zF F F

∂ ∂ ∂=

∂ ∂ ∂

i j k

∇

78

is a measure of how much the vector curls

around the point in question:

F×∇ F

curl = 0 atF curl = 0 atF curl > 0 atF

79

Example

Calculate the curl of the following vector functions:

(a)

(b)

2 2ˆ ˆ ˆ = 2 3 xy x y z+ +F i j k

3 2 4ˆ ˆ ˆ = 2 2 xz x yz yz− +F i j k

80

Interpretation of curl: an example from physics

The velocity of a point in a body rotating with angular velocity about an axis through the origin is given by

then

Look at the x

component:

v rω = ×v ω r

( )∇× = ∇× ×v ω r

( ) ( )( ) ( ) ( ) z yx x y z

∂ ∂∇× = ∇× × = × − ×

∂ ∂v ω r ω r ω r

( ) ( ) x y z xy z x zy zω ω ω ω∂ ∂

= − − −∂ ∂

2x x xω ω ω= + =... and similarly for the other components…

Therefore 2∇× =v ω (a constant)

∇× v is an example of a rotational field.

81

Consider a vector field and any closed loop C

in space.

( ) =C

d∫ F r i

The circulation of a vector field

)(rF

Circulation

= integral of the component of parallel to all the way around C.

the circulation

of around curve C

)(rFd

Then

Floop C

d

Fd

82

Circulation ... 2

Consider the velocity field in a liquid.

Imagine a tube of uniform cross section that follows an arbitrary closed curve.

If the liquid were suddenly frozen everywhere except inside the tube, then the liquid would circulate as shown.

solid

liquid

83

It can be shown that the circulation around C

is

the sum of the circulations around two partial loops.

contributions to these integrals due to path bc

cancel

If the original loop is aboundary of some surface, which is then divided into small number of areas, each approximately a square.then the circulation around C

is the sum

of the circulations around the little loops….

1 2

=loop loop curve C

d d d+∫ ∫ ∫F F Fi i i

C

C

1 2

b

c

84

How to find the circulation of each little square?

Choose the square small enoughsuch that does not vary much along any one side of the square.

1 2 3 4

( ) ( ) ( ) ( ) ( )d d d d d= + + +∫ ∫ ∫ ∫ ∫F r F r F r F r F ri i i i i

)(rF

y

y

x xx + dx

y + dy

1

24

3

Then

( , , ) ( , , )

( , , ) ( , , )

x dx y dy

x yx y

x y

x yx dx y dy

F x y z dx F x dx y z dy

F x y dy z dx F x y z dy

+ +

+ +

= + +

+ + +

∫ ∫

∫ ∫

85

[ ]( , , ) ( , , ) ( , , ) x x xF x y z F x y dy z dx F x y z dy dxy

⎛ ⎞∂− + = − ⎜ ⎟∂⎝ ⎠

( , , )( , , ) ( , , ) xx x

F x y zF x y dy z F x y z dyy

∂+ = +

∂

( , , ) ( , , ) ( , , ) ( , , )x dx y dy x y

x y x yx y x dx y dyF x y z dx F x dx y z dy F x y dy z dx F x y z dy

+ +

+ ++ + + + +∫ ∫ ∫ ∫

( ) d =∫ F r i

since

First and third terms give:

Second and fourth terms give:

[ ]( , , ) ( , , ) ( , , ) x x yF x dx y z F x y z dy F x y z dx dyx∂⎛ ⎞+ − = ⎜ ⎟∂⎝ ⎠

and the complete integral around the square is

( ) yx FFd dxdyy x

∂⎛ ⎞∂= − +⎜ ⎟∂ ∂⎝ ⎠

∫ F r i

86

where

0

1 1lim y x

dxdy

F Fd dxdydxdy dxdy x y→

∂⎛ ⎞∂= −⎜ ⎟∂ ∂⎝ ⎠

∫ Fi

( )y xz

F Fx y

∂ ∂= − = ∇×

∂ ∂F

curl or ( )∇×F F

( ) , , , ,x y zF F Fx y z

⎛ ⎞∂ ∂ ∂∇× = ×⎜ ⎟∂ ∂ ∂⎝ ⎠

F

ˆ ˆ ˆ y yx xz zF FF FF F

y z z x x y∂ ∂⎛ ⎞ ⎛ ⎞∂ ∂∂ ∂⎛ ⎞= − + − + −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

i j k

( ) yx FFd dxdyy x

∂⎛ ⎞∂= − +⎜ ⎟∂ ∂⎝ ⎠

∫ F r i

then

which is the z

component of

87

The circulation of around C

is the surface integral of normal component of .

( )S

( ) C

d d= ∇×∫ ∫F r F ai i

F×∇F

Of course, our little square need not be lying on the xy

plane, with sides neatly aligned parallel to the x

and y

axes.

We can extend the argument to any little square orientated in any way in 3D xyz space.

Then in 3 dimensions:

where S

is any surface bounded by C.

This is Stokes’

Theorem

88

This animation shows a flow field using animated textures. The direction of the velocity field is indicated by the correlation in the textures. When animated, the texture patterns move in the direction of the velocity field. This flow field is a pure circulation. There is no source here, the fluid simply moves in circles with no destruction or creation of fluid particles.

89

This animation shows a flow field which has a circulation sitting in a constant downward flow.

90

Example

Verify Stokes theorem forN

where S

is the surface of the cube delimited byx

= 0 , y

= 0 , z

= 0 , x

= 2 , y

= 2 , z

= 2 above the xy-plane

(all dimensions in metres)

ˆ ˆ ˆ = ( 2) ( 4) y z yz xz− + + + −F i j k

91

From Stokes’ theorem, if everywhere, then the circulation

around any loop is zero.

Therefore the line integral of from 1 to 2 along path a

must

be the same as the line integral along path b.

i.e. depends only on the location of points 1 and 2.

It follows that any vector field whose curl is zero

is equal to the gradient of some scalar function.

i.e. If then for some .

Then is conservative.

Curl free and divergence free fields

0∇× =F

2

1

d∫Fi

0∇× =F ϕ= ∇F ϕ

a

b

1

2

dFi

F

92

If then for some

There is a similar theorem if :

Suppose now we have any scalar field

then

and from Stokes’ theorem

Curl free and divergence free fields ... 2

If then for some0∇× =F ϕ= ∇F ϕ

div 0=F

div 0=F = ∇×F D D

ϕ

always

(over any surface)

0dϕ∇ =∫ i

( )S

0dl dϕ ϕ∇ = ∇× ∇ =∫ ∫ ai i

( ) 0ϕ∇ × ∇ =

93

Rotational and irrotational fields

If everywhere, then the field is said to be “irrotational”.

If somewhere, then the field is said to be “rotational”

Consider a vector field .

0∇× =F

0∇× ≠F

F

Examples of vector fields

18 examples of vector fields follow ...Although the fields are presented in 2D, try to imagine what thefield would look like in 3D.

94

Constant

ˆ = aF i

0∇× =F

0∇ =iF

This is a very simple field where the field vectors are pointing in the same direction everywhere with the same magnitude. The potential decreases linearly in the direction the particles are moving. The divergence and curl are both zero.

95

Linear to y-axis

ˆ = − k xF i

0∇× =F

∇ = −i kF

In this field, the particle is being pulled towards the y-axis with a force proportional to its distance from it. An object attached to a spring has a force on it which is linear to its displacement. If you consider many objects attached with springs to the y-axis, and you ignore friction, you would see a force field similar to this one. The divergence of this field is negative everywhere. So as the particles move through the field they get more dense.

96

Inverse to y-axis

1 ˆ = −x

F i

0∇× =F

2

1 ∇ =ix

F

In this field, the particle is being pulled towards the y-axis with a force inversely proportional to its distance from it. This field may be associated with electrostatic attraction. If we calculate the electric field around an infinitely long conductor in a straight line with a uniform charge along its length, we find that the field strength is inversely related to the distance from the line. So if we take a cross section of the field with the line running down the centre, we would see something like this inverse radial field.

97

ˆ = yF i

ˆ ∇× = −F k

0 ∇ =iF

In this field, particles far from the x-axis are moving faster than those close to it. Even though the particles are not moving in a circle, this field has a negative curl everywhere, which means clockwise rotation. For each particle, the force on the upper side is stronger than the force on the lower side, causing the particle to rotate clockwise. The particles themselves do not move in a circular path because the force has no y-component. The divergence of this field is zero everywhere, as we might expect, since the density of particles is the same everywhere.

98

2 ˆ = yF i

ˆ 2y∇× = −F k

0 ∇ =iF

In this field, particles far from the x-axis are moving faster than those close to it. Even though the particles are not moving in a circle, this field has curl everywhere except on the x-axis. For each particle above the y-axis, the force on the upper side is stronger than the force on the lower side, causing the particle to rotate clockwise; the effect is the opposite below the y-axis. The particles themselves do not move in a circular path because the force has no y-component. The divergence of this field is zero everywhere, as we might expect, since the density of particles is the same everywhere.

99

( ) ( )3 32 2

2 3

2 2 2 2

ˆ

ˆ ˆ

= − = −

= − −+ +

r rx y

x y x y

r rF

i j

0∇× =F

3

1 ∇ =ir

F

This field is associated with gravitational and electrostatic attraction, for example. The gravitational field around a single planet and the electric field around a single point charge are similar to this field. In three dimensions, the divergence of this field is zero; but in two dimensions the divergence is positive everywhere, and is especially large near the centre.

1/r2

single

100

Linear radial

0∇× =F

2 ∇ = −i k rF

( ) ( )1 12 22 2 2 2

ˆ

ˆ ˆ

= − = −

= − −+ +

kr kkx ky

x y x y

F r r

i j

101

1/r rotational

2 2 2 2

ˆ ˆ ˆ + = − = −+ +y x

r x y x yθF i j

0∇× =F

0 ∇ =iF

In this field, the particle is being pulled in a circle around the centre. The speed of its motion is inversely proportional to its distance from the centre. If we calculate the magnetic field around an infinitely long conductor in a straight line with a uniform current along its length, we find that the field strength is inversely related to the distance from the line. So if we take a cross section of the field along a plane with the charged line intersecting the plane in the centre, we would see something like this inverse radial field. Also, if we look at the velocity of particles in a tub which is being drained, we would see something similar to this field.

102

The divergence of this field is zero everywhere as we would expect for a magnetic field. Surprisingly, the curl is also zero everywhere except at the origin, even though the particles are clearly moving in a circle. This is because the force on the far side of the particles (the side away from the centre) is slightly weaker than the force on the near side; this is just enough to cause the particles to turn slightly in the opposite direction of the force. This counterbalances the rotation caused by the circular motion of the particles. So, the particles remain pointed in the same direction as they go around in a circle.

... 1/r rotational continued ...

103

1/r2

rotational

( ) ( )3 32 2

2 2 2 2 2

ˆ ˆ ˆ + = − = −+ +

y xr x y x y

θF i j

3

1 ∇× = −r

F 0 ∇ =iF

So if we take a cross section of the field along a plane with the wire intersecting the plane in the centre, we would see something like the inverse square radial field.

The divergence of this field is zero everywhere as we would expect for a magnetic field. However, note that the curl is negative

everywhere; this means that the curl

vector is in the negative z-direction. Does this make sense?

If we calculate the magnetic field around a current carrying wire in the z

direction (out of the page),

we find that the field strength is inversely related to the square of the distance from the line.

104

linear rotational

ˆ ˆ ˆ + = = −r y xF θ i j

ˆ 2∇× =F k 0 ∇ =iF

The speed of its motion is proportional to its distance from the centre. Since angular velocity is defined as the product of a particle's velocity with its distance from the centre of rotation, we can say that the angular velocity is constant everywhere.

The divergence of this field is zero everywhere, as we might expect, since the density of particles is the same everywhere. The curl is positive and constant everywhere as we might expect for a field that causes particles to rotate.

If we have a rotating disk and we look at the velocities of the particles in the disk, and treated those velocities as a vector field, we would see something similar to the field.

105

Constant rotational

2 2 2 2

ˆ ˆ ˆ + = = −+ +

y k x kkx y x y

F θ i j

∇× =kr

F

0 ∇ =iF

In this field, the particle is being pulled in a circle around the centre. The speed of its motion is constant everywhere. But, its angular

velocity is not constant everywhere;

it is greater near the origin, since angular velocity is defined as the product of a particle's velocity with its distance from the centre of rotation. If we have a rotating disk and we look at the velocities of the particles in the disk, we would find that particles near the centre are moving slower than those near the outer edge of the disk. So, particles in a constant rotational field do not act like particles in a rotating disk.

106

Consider the force on the far side of the particles (the side away from the centre) as opposed to the near side. The near end of each particle is closer to the centre, so it has less ground to cover in order to make a single rotation around the centre. (The near end has to move in a circular path which has a smaller radius, which means it is shorter.) The far end has more ground to cover because it is farther away. But if the force on both near and far ends is the same, this will cause the near end to rotate more than the far end. This effect will be greater than closer the particle is to the centre. So, this causes particles to rotate faster near the centre.

Constant rotational cont

The divergence of this field is zero everywhere. The curl is inversely proportional to the distance from the centre. The particles near the centre are rotating faster than those near the outer edges, even though they are moving at the same speed.

107

Rotation & expansion (spiral)

ˆ ˆ ˆˆ ( ) + ( ) = + = − +r x y x yF r θ i j

ˆ 2∇× =F k

2 ∇ =iF

In this field, the particle is being pushed in a circular path around the centre. The speed of its motion is proportional to its distance from the centre. Since angular velocity is defined as the product of a particle’s velocity with its distance from the centre of rotation, we can say that the angular velocity is constant everywhere. So, we have a constant positive curl everywhere. In addition to the circular movement, the field has a radial component which pushes particles away from the origin. So the field is everywhere expanding; this gives us a positive divergence everywhere.

108

Saddle

ˆ ˆ + = − x yF i j

0∇× =F

0∇ =iF

109

2 2ˆ ˆ ( ) + ( ) = − +x y x yF i j

ˆ 2∇× =F k

2 2∇ = +i x yF

In this field, the particle is being pushed in a circular path around the centre. The speed of its motion is proportional to its distance from the centre. So, we have a constant positive curl everywhere. In addition to the circular movement, the field has a radial component which varies depending on where the particle is. In the lower left corner, the field has a negative divergence which causes particles to contract; in the upper right corner, the field has a positive divergence which causes particles to expand.

110

2 2ˆ ˆ ( ) + ( ) x y x y= + −F i j

( ) ˆ 2 2x y∇× = −F k

0∇ =Fi

2ˆ ˆ + x x=F i j

ˆ 2x∇× =F k

1∇ =Fi

111

Some important double vector operators

2.

3.

4.

5.

1. curl (grad ) ( ) 0V V= ∇× ∇ =

div (curl ) ( ) 0= ∇ ∇× =F Fi2 2 2

22 2 2div (grad ) ( ) + V V VV V V

x y z∂ ∂ ∂

= ∇ ∇ = ∇ = +∂ ∂ ∂

i

For a scalar field V

and a vector field :F

where the Laplacian operator = 2 2 2

22 2 2 +

x y z∂ ∂ ∂

∇ = +∂ ∂ ∂

2 2 22 2 2 2

2 2 2ˆ ˆ ˆ + = x y zF F F

x y z∂ ∂ ∂

∇ = + ∇ +∇ +∇∂ ∂ ∂

F F FF i j k

2curl (curl ) ( ) ( )= ∇× ∇× = ∇ ∇ −∇F F F Fi

... and there are many others ...

112

General approach to proving a vector identity

Expand the expression into components

( )∇ ∇×Fi

For example, div (curl ) ( ) ??= ∇ ∇× =F Fi

ˆ ˆ ˆy yx xz zF FF FF F

y z z x x y⎡ ⎤∂ ∂⎛ ⎞ ⎛ ⎞∂ ∂∂ ∂⎛ ⎞= ∇ − + − + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦

i j ki

y yx xz zF FF FF F

x y z y z x z x y∂ ∂⎛ ⎞ ⎛ ⎞∂ ∂∂ ∂∂ ∂ ∂⎛ ⎞= − + − + −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

2 22 22 2y yx xz z

F FF FF Fx y x z y z y x z x z y

∂ ∂∂ ∂∂ ∂= − + − + −∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

0=

113

Summary

ˆ ˆ ˆgrad = x y zφ φ φφ φ ∂ ∂ ∂

∇ = + +∂ ∂ ∂

i j k

ˆ ˆ ˆx y z∂ ∂ ∂

∇ = + +∂ ∂ ∂

i j kdel operator

div yx zFF F

x y z∂∂ ∂

= ∇ = + +∂ ∂ ∂

F Fi

ˆ ˆ ˆcurl y yx xz zF FF FF F

y z z x x y∂ ∂⎛ ⎞ ⎛ ⎞∂ ∂∂ ∂⎛ ⎞= ∇× = − + − + −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

F F i j k

Laplacian operator2 2 2

22 2 2 +

x y z∂ ∂ ∂

∇ = +∂ ∂ ∂

114

Summary ... 2

2

1

2 1 ( ) ( ) dφ φ φ φΔ = − = ∇∫r

r

r r i

S

d∫F ai

• S V

d dV= ∇∫ ∫F a Fi

C

d∫ Fi

( )S

( ) C

d d= ∇×∫ ∫F r F ai iStokes’

theorem:

Circulation

=

Flux

= = (average normal comp. of )•(surface area)

= (average tangential comp. of )•(distance around C)

for any closed surfaceDivergence theorem:

F

F

Potential difference