14 lectures given in the autumn term to second year physicists as part of the Maths IV course

Lecturer: Professor Peter Main

About the course As the name implies, vector calculus is a combination of vector algebra and calculus. It is an elegant branch of mathematics that is extremely useful to physicists as it is the mathematics of fields. You will already have encountered electric, magnetic and gravitational fields, but you are soon to come across many more. Vector calculus gives you the mathematical tools to manipulate fields, to do calculations involving fields, to describe their properties and to characterise them precisely. It is an important branch of mathematical physics. As we live in a three-dimensional world, the variables we will deal with are x, y, z, i.e. three-dimensional space. Towards the end of the course, time will be introduced as an extra variable – not everything stands still and we need to be able to deal with things that move or change with time. What you need to know It is assumed that you are already familiar with both vector algebra and the calculus of functions of more than one variable. In particular, make sure you know about: Vector algebra: Cartesian components, dot and cross products, triple products. Calculus: partial differentiation, total differential, differential operators, multiple integrals. Coordinate systems: Cartesian, spherical polar, cylindrical polar.

Mathematics provides the framework to make difficult things easy.

http://www-users.york.ac.uk/~pm1/PMweb/teaching.htm

1

Revision Vector algebra A vector is a quantity that has both magnitude and direction. It is useful in physics because it can be used to represent velocity, acceleration, momentum, force, position, displacement and many other quantities. The following is a review of the essentials of vector algebra that will be used in this term’s mathematics course. Representation The algebraic symbol for a vector may be written in bold as a, or underlined as a. In a diagram it is represented by an arrow:

The magnitude of a can be written as a, which is a scalar quantity and its direction can be given by a unit vector n. The magnitude of n is unity, hence its name. It is a dimensionless quantity and is used only to define a direction.

We can therefore write a as an where the magnitude and direction are given separately. Cartesian components A vector is often described in terms of its Cartesian components, i.e. the components of the vector parallel to the x-, y- and z-axes.

The directions of the axes are given by the unit vectors i, j, and k so that a vector may be written in terms of its components as a = a1i + a2j + a3k

Pythagoras’s theorem gives the magnitude of the vector as 23

22

21 aaaa ++=

Dot product There are two ways of multiplying vectors, depending upon the context. The dot product of a and b is written as a.b and the result of the multiplication is a scalar. It is therefore also called a scalar product.

The result of the multiplication is a.b = a b cosθ where θ is the angle between the vectors. This definition shows that a.b = b.a, i.e. the vectors commute.

It can be seen in the diagram that b cosθ is the projection of b on a. This is often a useful way of thinking about scalar products, i.e. as a multiplied by the projection of b. It is also used to resolve one vector in the direction of another.

b

If the vectors are orthogonal to one another, i.e. at right angles, then a.b = 0 θ a

Also, the definition of the dot product leads to a.a = a2 b cosθ The dot products of the unit vectors i, j and k are therefore:

i.i = j.j = k.k = 1; i.j = j.i = 0; i.k = k.i = 0; j.k = k.j = 0

When the vectors are expressed in terms of their Cartesian components, so that a = a1i + a2j + a3k and b = b1i + b2j + b3k, the dot product becomes (a1i + a2j + a3k).(b1i + b2j + b3k). Multiplying out the brackets and using the dot products of the unit vectors given above results in

a.b = a1 b1 + a2 b2 + a3 b3

2

Cross product The other way of multiplying vectors is the cross product, written as a×b which results in a vector. It is therefore also called a vector product.

The vector product is defined by a×b = ab sinθ n where θ is the angle between the vectors and n is a unit vector perpendicular to both a and b such that a, b, n form a right-handed set.

This definition requires that a×b = – b×a since reversing the order of the vectors changes the sense of n, i.e. it points in exactly the opposite direction. You must always be careful of the order of the vectors when dealing with cross products.

a×b A geometric interpretation of the cross product is shown in the diagram. The two vectors a and b define the parallelogram and the magnitude of a×b is its area. The direction of a×b is normal to the plane of the parallelogram.

If the two vectors are parallel, the cross product gives the null vector. In particular, a×a = 0

The cross products of the unit vectors i, j and k are:

i×j = k; j×k = i; k×i = j; i×i = j×j = k×k = 0

Note that, in the first three relationships, the vectors are always in the same cyclic order.

With a = a1i + a2j + a3k and b = b1i + b2j + b3k, and using the above relationships, the cross product becomes

(a1i + a2j + a3k) × (b1i + b2j + b3k) = (a2b3-a3b2)i + (a3b1-a1b3)j + (a1b2-a2b1)k

Fortunately, there is an easier way of expressing this:

321

321

bbbaaakji

ba =×

Expansion of the determinant gives the same result as above. Scalar triple product Triple products of vectors frequently arise. The scalar triple product, written as a.b×c, results in a scalar quantity, hence its name. To make mathematical sense, the cross product must be evaluated first, giving a vector which is dotted with a

A geom

.

etrical interpretation of the scalar triple product is

b.c since the same three vectors are involved

c.a×b = -c.b×a = -b.a×c = -a.c×b

e a value of zero.

shown in the diagram. The three vectors define a parallelepiped. The magnitude of b×c gives the area of the base and its direction is normal to the base. The dot product with a therefore resolves a in the direction of b×c, giving the vertical height. Area of base multiplied by the height gives the volume of the parallelepiped.

We therefore have the result that a.b×c = a×and therefore give the same volume.

Similarly: a.b×c = b.c×a =

Clearly, if two of the vectors are parallel, the scalar triple product must hav

a θ

b

b×ca

c

b

3

Vector triple product t arises is the vector triple product, so called because it results in a

×(b×c) or (a×b)×c. Note that the brackets are necessary to indicate which

le products are awkward to deal with.

= (a.c) b – (a.b) c

traight line

The other triple product thavector quantity.

It is written as aproduct is performed first because (a×b)×c ≠ a×(b×c)

While providing a very compact expression, vector tripFor the purposes of algebraic manipulation, they are nearly always changed to an alternative expression using the standard identity:

a×(b×c)

S 3D space are most conveniently represented using vectors.

where r is the position vec on the line and the scalar

two points a and b, then the vector )

curve

Straight lines in

The position of the line is given by specifying a point on it, e.g. the point

a

b r

a, and its direction is given as parallel to the vector b: r(λ) = a + λ b tor of a point

variable λ moves the point along the line.

If the line is defined as going through the b-a is in the direction of the line and its equation is therefore: r(λ) = a + λ (b-a

Space A space curve is a curve in 3D space which may be used, for example, to describe the path of

e field.

vector of a point on the curve as a function of θ. As θ varies, the e.

arly to move the particle along the z-axis, making the

a particle in a forc

An example of a space curve is: r(θ) = a cosθ i + a sinθ j + b θ k where r(θ) is the position position vector traces out the curv

In this example, the i and j components trace out a circle of radius a as θ varies. In addition to this, the k component varies linespace curve into a helix lying along the z direction with a pitch of 2πb. Differentiation of vectors If the vector is defined as a function of one or more variables, the possibility arises of

n. This is a common operation in vector calculus. differentiating the vector functio

Taking the above space curve as an example, the derivative with respect to θ is obtained by differentiating each component separately:

kjir aadd

+−= θθθ

cossin b+

which can be rewritten as: dr = (-a sinθ i + a osθ j + b k) dθ

r is the d ve θ+

s it possible to mseries of infinitesimal steps dr.

c

The infinitesimal vector d isplacement required to mofrom r(θ) to r(θ+dθ), i.e. r( dθ) = r(θ) + dr

Since both r(θ) and r(θ+dθ) lie on the space curve, the vector dr must lie along the curve, so it is in the direction of the tangent. This make ove along a space curve in a

r(θ) r(θ+dθ)

dr

4

Vector calculus Scalar field Many scalar quantities have only a single value, e.g. mass of an electron, specific heat of copper, speed of light. However, the value of a physical quantity may depend upon position such as the air temperature in a large hall or the height above mean sea level of an area on the Earth’s surface.

The association of a particular value of a physical quantity with each point in a region of space is said to constitute a field. When the physical quantity is a scalar, the field is called a scalar field. Definition If to each point in a region of space there corresponds a scalar quantity φ, then φ(x,y,z) is known as a scalar field. Examples Temperature at each point within the Earth’s surface. Electric potential at every point in an electron optical system. φ(x,y,z) = x3y – z2 defines a scalar field. Representation Scalar fields are best represented as contour maps in 2 or 3 dimensions. Vector field Just as there are scalar fields, there are also vector fields. The velocity of a boat on a river can be represented by a single vector, but the velocity of the water in the river can not. The water velocity depends upon where it is measured. Definition If to each point in a region of space there corresponds a vector quantity V, then V(x,y,z) is known as a vector field. Examples Velocity at every point in a moving fluid. Magnetic field at every point in an electron microscope. V(x,y,z) = xy2i – 2yz3j + x2zk defines a vector field. Before deciding on a good representation for a vector field, let us graph the field V(x,y) = xi + yj. The magnitude of the vector is 22 yx + , which corresponds to distance from the origin.

The direction of the vector is ⎟⎠⎞⎜

⎝⎛

xyarctan , which

always points directly away from the origin.

Placing a vector at each point in the diagram clearly gives a very clumsy picture of the field, although this is sometimes used. Representation Since the quantity that distinguishes a vector field from a scalar field is its direction, it is this which is plotted. This gives a clearer representation than when the magnitude is plotted as well. Thus, a vector field is most conveniently represented using field lines.

5

Field line A curve whfield at that poin

ose tan direction of the vector t is known as a field line. (You may also have used the terms line of force,

hich enable us to do many important calculations of field properties.

gent vector at each point is in the

stream line or flow line.) Having introduced fields, we are now going to look at three main measures of field characteristics w Gradient of a scalar field The first of the important field characteristics to be considered is the gradient of a scalar

eld. fi

For a function of a single variable we have gradientdy=

dxhich can be rearranged as dy = gradient × dx

φ for an infinitesimal

omparison with the above relationship suggests we can write

dφ = gradient . dl

his is the gradient of the scalar field φ and the standard way of writing it is

x

y w

This expression relates the amount by which the function changes, d a shift in the value of x, dx. y, for small Now consider a scalar field φ(x,y,z) and find the change in the value of vector displacement dl.

C

where it is clear that gradient must be a vector quantity and that a dot product with dl is required to produce the scalar quantity dφ.

T

dφ = gradφ . dl

To determi n expression for gradφ in terms of x, y and ne a z, we can express dl as

nd the total differential of φ is

dl = dx i + dy j + dz k

dzz

dyayx

dxd∂

∂ ∂+ +

∂ ∂∂

=φφφφ

s gradφ to be Putting these into the expression for dφ require

kjizyx ∂

+∂∂ ∂

+∂∂

=φφφφgrad

Note that φ is a scalar field and a vector can be expressed in terms of the gradient of a

that gradφ is field. However, not all vector fields scalar field in this way.

curs when gradφ and dl are parallel. when θ = 0, i.e. when the vectors are

um rate of increase of φ is therefore in the direction of gradφ.

e imp

Since dφ = gradφ . dl, the maximum value of dφ oc(Remember that a.b = ab cosθ, which has a maximumparallel.) The maxim

We now have th ortant results that:

Direction of gradφ = direction of maximum rate of increase of φ.

Magnitude of gradφ = maximum rate of increase of φ.

6

Vector d rential operator iffe ∇ ‘del’ It is convenient to express the gradient of a scalar field in terms of the vector differential

perator: ozyx ∂

∂+

∂∂

+∂∂

=∇ kji

Compare with the operator dxd which gives the gradient of f(x).

The gradient of the scalar field φ(x,y,z) can now be written as

kjizyx ∂

∂+

∂∂

+∂∂

=∇=φφφφφgrad

Mathematical example If φ(x,y,z) = 3x2y – y3z2, find gradφ at the point (1, -2, -1).

( )2323 zyyxSolution: zyx ⎟

⎠⎜⎝ ∂∂∂

−⎟⎞

⎜ ++=∇ kjiφ ⎛ ∂∂∂

( ) ( ) ( )2323 zyyxz

−∂

232232 33 zyyxy

zyyxx

∂+−

∂∂

+−∂

= kji ∂

( ) kj zyzy 322 23 −i xxy 236 −+=∇∴ φ

nt (1, -2, -1) we have

This is the gradient of φ(x,y,z), so at the poi kji 16912 −−−=∇φ

Manipulations of ∇ Here are some examples of algebraic and differential operations that may be carried out using the operator: ∇

( ) φθφθ ∇+∇=+∇ – normal rules of calculus apply

( ) ( ) φθφθφθ ∇+∇=∇ – rodthe p uct rule

Evaluate

An example which will be useful later is:

⎟⎠⎞

⎜⎝⎛∇

r1 where r = x i + y j + z k i.e. ( ) 2

1222 zyxr ++=

Since r is symmetric in x, y and z, it is sufficient only to determine ⎟⎞

⎜⎛

∂∂x

1 : ⎠⎝ r

( ) ( ) 32

322221222 2

211

rxxzyxzyx

xrx−=++−=++

∂∂

=⎟⎠⎞

⎜⎝⎛

∂∂ −−

We can now use this result to determine the other components of ⎟⎞

⎜⎛∇

1 : ⎠⎝ r

3333

1rr

zry

rx

rrkji −=−−−=⎟

⎠⎞

⎜⎝⎛∇∴

7

Another example which illustrates t mathemhe atics of the ∇ operator is to evaluate where r is defined above and a = a i + a j + a k is a constant:

( )a.r∇ 1 2 3

( ) ( ) akjia.r =++=++∇=∇ aaazayaa

321321 x

Directional derivative of a scalar field φ nd upon direction. The rate At any point in a scalar field , the rate of change of φ will depe

of change given by φ∇ corresponds to the direction and magnitude of the maximum gradient

The component of

at that point.

φ∇ in the direction of a unit vector n is given by , which is the of φ in the direction of n. It gives the rate of change of φ at (x, y, z) in

dφ = gradφ.dl

dl

φ.n∇directional derivativethe direction of n.

It is easily derived as follows:

Express dl as n then dφ = gradφ.n dl

.n.nSo that φφφ∇== grad

dld

Physical examples 1. Electric field Work done against an electric field E on moving a unit charge a vector distance dl is –E.dl. By definition, this is equal to dV, the potential difference between the

d is negative, work done by

rom the physical definition of potential: dV = – E.dl

a eady have: dφ = gradφ.dl

ends of the vector dl. (Remember that work done against a fielthe field is positive.)

F

For a scalar field φ, we lr

∴ from the mathematical definition of grad: dV = gradV.dl

Com arison of thp ese expressions gives:

Electric potential, V, i E, the gradient of V, is the corresponding vector f ld. T the above expression comes about because

to one of low potential, but gradV is in

lectri nt of the electric po ential is easily seen in a one-dimensional example.

E = – gradV

s a scalar field and the electric field ie he negative sign in

the direction of E is from a region of high potential the direction of increasing V.

That the e c field is the negative of the gradie t

E

+V -V

V

D

For the simple field in the diagram: metre/volts2VE = D

The slope (gradient) of the graph is DV2

−

So the magnitude and direction of the field is iEDV2

+=

ote that kjizV

yV

xVV

∂∂

+∂∂

+∂∂

=grad N

8

For V in volts and x, y, z in metres, the dimensions of gradV must be volts / metre. 2. Heat conduction

Fourier’s first law of heat conduction is:

lowing in the direction of Q.

eleg arises because heat flows from high temperature to low, but gradθ is in the

onduction you should already be familiar

where Q = quantity of heat/unit time/ unit area f k = thermal conductivity θ = temperature

Again, we have an ant relationship between a scalar field θ and a vector field Q. The negative signdirection of increasing temperature. A formula in heat cwith is:

Heat / unit time = D

ak × rea 21 θ−×

This can be seen as Fourier’s law in one dimension.

. Gravity

θ

3

avitation, the force exerted on mass From Newton’s law of gr

m by mass M is: 2rmMGF =

With vectors, this expression is: 3rmMG rF −=

where the negative sign is necessary since F and r are clearly in opposite directions.

Now gravitational potential is r

mMGV −=

From previous results, we obtain 3rr ⎠1grad mMGGV =⎟

⎞⎜⎛∇−= mM r⎝

ence, we have the relationship H

and the gravitational force field is minus the gradient of the gravitational potential. Line integral Here is a very important calculation on a scalar field.

Consider a particle constrained to move along the line AB in the

The line AB is represented by the space curve l(t).

g the article an infinitesimal distance dl along the line is dW = F.dl

oving the particle

from A to B is: A

W F.dl

line integral of F along the curve AB.

Q = – k gradθ

F = – gradV

M mF

r

k θ1 θ2

D

B

force field F.

A

l(t)

O

F dl

The work done by the field in movin p

Therefore, the total work done by the field in mB

∫=

This is the

9

If the integral is performed round a closed path, returning to the starting point, it is written as

∫CF.dl where C labels the path.

ly pe tions of A and B and also pon the path between them. We will see shortly how line integrals are evaluated.

lar potential

The value of a line integral will normal de nd upon the posiu Conservative field and sca

tween a scalar and a vector field is that the vector field F is obtained

tive field, F, is that the line integral of F between any This is proved as follows:

Work done against the field A

d φφφ ===−= ∫∫ Agrad .dlF.dl

s only upon the tential difference between the ends of the path.

sed path in a conservative field must be

A special relationship beas the gradient of the scalar field φ, i.e. F = – gradφ. In this case, F is known as a conservative field and φ is its scalar potential.

n important property of a conservaAtwo points is independent of the path between the points.

BB

AB

B

A∫ φ−

i.e. the work depend po

It is clear from this that the line integral round a clo

∫ =C

0F.dl for any closed path C, if and only if F = – gradφ.

ons rvat e fields

zero:

Examples of c e iv A gravitational field is conservative since the gravitational potential (or potential energy) of

istance from the centre of the Earth, i.e. work done depends is independent of the path.

done is a points and not upon the path.

Non-conservative fields

an object depends only upon its donly upon difference in height and

Since an electrostatic field is given by E = – gradV, E must be conservative. Work function only of potential difference between start and end

rce field is always opposite to the direction of motion, i.e. Friction – the direction of the fo

∫ ≠ 0F.dl

magnetic field is not conservative since Ampère’s law gives: A

∫ ≠= I 0H.dl C

rection will gain energy and

otor.

H

C

I

work done in taking a unit magnetic pole round a closed

current through = the loop

loop in a magnetic field

A particle taken round a closed loop in the appropriate ditherefore can do work. This occurs in an electric m Evaluation of line integrals B

A

l(t)

O

F dl Line integrals have already been defined. We will see

how they are evaluated. B

now

F.dl represents the work done by the force field F in ∫A

moving a particle along the line from A to B.

10

The recipe for evaluating line integrals is as follows. An actual example will be given in the ction. next se

1. Obtain an expression for the space curve AB, for example: l(t) = f1(t) i + f2(t) j + f3(t) k

. Differentiate to obtain dl: 2 dtdfdfdf

⎟⎞

⎜⎛ ++= kjidl 321

dtdtdt ⎠⎝

f t

3. Obtain an expression for the vector field: F(x,y,z) = F1(x,y,z)i + F2(x,y,z)j + F3(x,y,z)k

4. Substitute for x, y, z from 1. to obtain F(t), i.e. x = f1(t), y = f2(t) and z = f3(t). This makes F a function of t only, which is the value o he force field along the path – it doesn’t matter what the force field is elsewhere: F(t) = F1(t)i + F2(t)j + F3(t)k

5. Now form F.dl from 2. and 4. dtdf

FdfFdfF ⎟⎞

⎜⎛ ++ 321

dtdtdt ⎠321

6. Determine the values of t at the ends of the path: tA, tB

⎝

7. Set up the integral: ∫∫ ⎟⎠

⎜⎝

++= B

AtAdt

dtF

dtF

dtF 3

32

21

1F.dl ⎞⎛tB dfdfdf

8. Evaluate the integral. It should now be in the form of an ordinary integral in terms of tthe single variable, .

Example If kjiF 22 2014)63(),,( zxzyyxzyx +−+= , evaluate ∫C

F.dl from (0, 0, 0) to (1, 1, 1): 2 3a) along the path x = t, y = t , z = t b) along the straight line joining the points.

olution: a) If x = t, y = t2, z = t3, then l(t) = t i + t2 j + t3 k S

and dl = (i + 2t j + 3t2 k) dt

then

t = 1, l(t) = (1, 1, 1)

Set up the integral:

If kjiF 22 2014)63(),,( zxzyyxzyx +−+=

kjiF 7522 2014)63()( ttttt +−+=

The end points of the path: when t = 0, l(t) = (0, 0, 0) and when

( ) [ ] 564360289 10

10731

0

962 =+−=+−= ∫∫ tttdttttCF.dl

b 0) (1 t, z = t

ji

, 1, 1) is x = t, y =

dt)k

) The straight line joining (0, 0,

tttt thatso)( dlkjil

to

( ++=++=∴ also 14)63()( tttt +−+= kjiF 322 20t

( )3

135113201161

0

143232 =⎤

⎢⎣⎡ +−=+−=

3 0⎥⎦

∴ ∫∫ tttdttttCF.dl

The value of the integral clearly dep

ends upon the path, so the field F cannot be conservative.

Level surfaces Consider a scalar field same va ce.

φ without discontinuities. Connect all points in the region having the lue of φ. This will form a surface known as a level surfa

11

Let dl0 lie within a level surface, then the difference in the value of φ, dφ, betw gradφ een the ends of the vector dl0 is zero.

= gradφ . dl0

In general, neither gradφ nor dl0 will be the null vector, so the dot

dl0But we have already dφ = gradφ . dl φ1so that 0

product requires gradφ and dl0 to be perpendicular, i.e. gradφ is normal to the level surface. Mathematical consequence

, z) = constant defines a surface in three-dimensional space – a level surface. φ(x, yφ∇ is always normal to this surface.

Physical consequence

lectr E. Since E = – gradV, the

field vector must be normal to the equipotential surface.

id is a surface of equal gravitational potential. The direction of the gravitational field is therefore normal to this surface.

a heated body, level surfaces of temperature are called isotherms. The direction of heat s normal to the isotherms.

ctor area

Consider an equipotential surface V(x, y, z) = V0 in an e ic field

The surface of a static liqu

The steepest way uphill is perpendicular to the contours.

Inflow is alway Ve

ctor rea. It is not immediately obvious that areas can be represented m thematically as vectors,

e of the vector gives the size of the area.

di ction e vector is normal to the plane of the

The vector contains no information about the shape of the area.

o demonstrate that an area can behave as a vector quantity, consider a hollow cube filled A. The force on each face

F = P A

However, force is a vector quantity and the direction of the force will be normal to the face. The quantity PA must therefore also be a vector normal to the face of the cube. Since

F, so we can write:

F = P A

A concept which is essential to the understanding of the properties of fields is that of vea abut it is easy to demonstrate. We start with the definition.

S If the vector S represents a plane area then:

The magnitud

The re of tharea.

Twith a gas at a pressure P. Let the area of each face of the cube be due to the internal pressure will therefore be given by:

F

pressure is a scalar, we must conclude that A is a vector parallel to

12

If the area is on a curved surface, we can take part of the area smalenough to be regarded as a plane and represent the s

l mall area by the dS

infinitesimal vector dS. Flux An important measure of a vector field is its flux. Consider a fluid moving with a velocity given by the vector field F(x, y, z). Its rate of flow (volume / time) may be obtained by observing its

by the vector S:

at F and S are orthogonal, no fluid

there

|S| cosθ

= F . S

d of calling it flow rate (because is simply the Latin for flow.

n therefore refer to the flux of an electric or gravitational field and measure it as F.S.

s done using an infinitesimal rea dS, i.e. flux at a point = F . dS.

S F velocity through an area represented flow rate = speed × area = |F| × |S|

If the area is reoriented so th S will pass through the area, so the measured flow rate will be zero:

F

flow rate = 0 A more general case is afforded by orienting the area so

Fθ

is an angle θ between F and S. The area presented to the flow is now |S| cosθ:

S flow rate = |F| × All of these results can be obtained as the dot product of the two vectors involved:

i.e. flow rate

Not all vector fields correspond to a flowing substance (very fewmeasurement can be made on any vector field. Instea

do), yet the same

nothing may be flowing) a generic term is used, i.e. flux. Flux We ca

It is more useful to measure flux at a point in a field and this ia Calculation of dS for any surface Calculations of flux require a value of dS, the element of vector area. The easiest way of achieving this is to expres ion of the surface in parametric form such that r(λ, µ) s the equatgives the position vector of a point on the surface for any λ, µ.

The vector (λ, µ) → (λ+dλ, µ) lies in the surface and is λλ

d∂∂r

The vector (λ, µ) → (λ, µ+dµ) lies in the surface and is µµ

d∂∂r

hese vectors define an element of area given by T (λ, µ) (λ+dλ, µ)

(λ, µ+dµ) µ

µd

∂∂r

λλ

∂∂∂r

13

µλµλ

dd∂∂

×∂∂

=rrdS

Remember that the magnitude of a cross product gives the area of the parallelogram defined two vectors and that the direction of the cross product is normal to the plane of the by the

vectors. Example – cylindrical surface Cylindrical polar coordinates are defined by: x = r cosθ; y = r sinθ; z .

point on a cylinder of radius a is therefore: x = a cosθ; y = a sinθ; z = z.

oint:

= zA

kjir zaaz ++= θθθ sincos),( This makes the position vector of the p

dzdaadzdz

θθθθθ

1000cossin−=

∂∂

×∂∂

=∴kji

rrdS dz

a dθ a

dθ z

( ) ( ) dzdadzdaa θθθθθθ jiji sincossincos +=+

his is a unit vector normal to the surface multiplied by the e

xample – spherical surface

=

Tscalar a dθ dz. This is seen to be the magnitude of dS from thdiagram. E

are defined by: x = r cosφ sinθ; y = r sinφ sinθ; z = r cosθ point on a sphere of radius a is therefore: x = a cosφ sinθ; y = a sinφ sinθ; z = a cosθ

Spherical polar coordinatesA

This makes the position vector of the point: jir kθθθφθφθ cosasinsincossin),( aa ++=

φθφθφθ

θφθφθφθφθ

ddaa

aaadd0cossinsinsinsinsincoscoscos

−−=

∂∂

×∂∂

=∴kji

pherical surface multiplied by a2 e of dS from the diagram.

rrdS

( ) φθθφθφθθ dda kji cossinsincossinsin2 ++=

his is a unit vector normal to the s

adθ

θ

Tsinθ dθ dφ, which is seen to be the magnitud Surface integral The calculation of flux normally involves the use of a surface integral.

Consider a surface S in a vector field F(x, y, z). Let dSelement of area on S that dS is always normal to the

be an such

The flux of F through dS is F . dS

surface. By convention, dS points outwards if S is a closed surface.

Therefore, total flux through S ∫∫= dS.F S

S

dS F

φ asinθdφ

a

z

y

x

14

where the integral is taken over the whole surface S. Th is a surface integral. Note that it is a double integral.

the integral of the normal component of F over the surface.

ritten as but it is still a double integral.

is

Since dS = n dS where n is a unit vector, the surface integral can be written as ∫∫S

dSn.F

which is

∫S

dS.FA surface integral may sometimes be w

If the integral is over a closed surface, it may be written as ∫∫∫SS

FdS.F or dS.

Evaluation of a surface integral The evaluation of a surface integral is carried out in the same way as a line integral, except there are now two variables instead of one.

Example: Evaluate the surface integral where a S is

rved surface of the cylinder x2 + y2 = 16 in the first octant between z = 0 and

∫∫S

dS.F kjiF zyxzzyx 23),,( −+= nd

z = 2. the cu

( )zz ,sin4,cos4)Solution: The equation of the curved surface of a cylinder is ,(θ =r

From prev

θθ

ious results, dzdzdz

θθθθθ

)0,sin,(cos4=∂∂∂

×∂

=rr

∂dS

osθ and y = 4 sinθ. Substitute these into the expression On the surface of the cylinder, x = 4 cfor F(x, y, z) to obtain the value of F on the surface:

kjiF zzz 2)sin4(3cos4),( θθθ −+=∴

Now form F . dS: ( ) dzdz θθθθ sincos4cos4 +=dS.F

The surface integral is therefore ( )∫∫∫∫==

+=2

0

2

0

cossin4cos4zS

dzdz θθθθ

π

θ

dS.F

∫∫∫∫ +=2

0

2

0

2

0

2

0

2sin8cos4 dzddzzdππ

θθθθ

[ ] [ ] 2.1.42cos218

21sin4 2

0

2

0

2

0

220 =⎥⎦

⎤⎢⎣⎡−+⎥⎦

⎤⎢⎣⎡= zz

ππ

θθ 242.21

21

=⎟⎠⎞

⎝++ .8 ⎜

⎛

Volume integral After line and surface integrals come volume integrals. Howtriple inte

ever, these are just ordinary grals and don’t need any special tricks for their evaluation. An exam

olume.

volume dV. The integral will measure the total charge in the volume V.

ple of a volume integral is ∫∫∫

V

dVφ where the field of integration is a v

If φ is charge density, then φdV is the charge contained in the therefore

15

The concept of divergence The second important characteristic after the gradient of a scalar field is the divergence of a

Enclose a volume V in the vector field F(x, y, z) by an arbitrary closed surface S. The total

vector field.

outward flux of F through S is given by ∫∫F.dS . Now consider the fS

ollowing:

or sinks. What goes in also comes out, so the inward flux is equal to the The field inside S is continuous, i.e. there are no sources

outward flux making 0=∫∫ dS.F S

A source of the field inside S gives a net outward flux through the surface, making 0>∫∫

S

dS.F

at There is a sink of the field (a sink is the opposite of a source) inside S. This gives a net inward flux through the surface so th 0<∫∫

S

dS.F

It is clear that the surface integral can be used as a detector of sources or sinks of the field. It th of sources inside S.

For distributed sources, a more useful measure is the source strength per unit volume given

by

gives a measure of the total streng

∫∫S

VdS.F1 where V is the volume enclosed by S.

It is even more useful to pin-point the sources and sinks. This is done by shrinking the volume V to the point where the surface S encloses an infinitesimal volume dV. This will give a measure of flux / unit volume diverging from a point. We therefore define the divergence of the field as

∫∫∆→∆S

VV 0=F

limdiv dS.F1 where ∆V is the volume enclosed by the surface S.

The limit is finite and independent of the shape of the volume element ∆V. Note that divF defines a . scalar field An expression for the divergence of a vector field Consider the elementary rectangular volume ABCDEFGH whose edges are parallel to the oordinate axes and of lengths ∆x, ∆y, ∆z. he volume is centred on the point (x, y, z) in the vector field F(x, y, z) = F1 i + F2 j + F3 k

cT

16

Assume that ∆x, ∆y and ∆z are small enough to approximate the value of the field F over

the surface integral

each face of the box by the value at the mid point of the face.

Now evaluate ∫∫S

the box by calculating the flux through each face separately.

dS.F over the surface of

G

B

HE z

A D

C

F ∆y ∆x

∆z

B ( ) ⎟

⎠⎞

⎜⎝⎛ ∆

∆∆∂∂

+∆∆=211xzyF

xzyFdS.F For the face EFGH:

c m

F.dS at the mid plane + This comes from:

∆−= 1 yFdS.FFor the face ABCD:

Adding these gives the total outward f

l outward f x in theSimilarly, the tota lu

nd in the z-direction it is a

The value of the surface integral over

∴ ∫∫ dS.FS

The volume of the box = ∆V

=∴lim

divF→∆V

Divergence can be expressed in terms

kjiF. ⎟⎟⎠

⎜⎜⎝ ∂

+∂

+∂

=∇zyx

Mathematical examples

⎞⎛ ∂∂∂

Refer back to the vector field V(x, Imagine that it represents the velobut the vectors further away fromfrom the origin increases. Where do

city the

field detects the sources and sinks oanswer this question:

div =∇= V.V

There is positive divergence everywhewater of the same strength over the wh

hange of F.dS uponoving to the front face

x y

( ) ⎟⎠⎞

⎜⎝⎛ ∆−∆∆−

∂∂

+∆21xzyF

xz

lux in the x-direction as zyxxF

∆∆∆∂∂ 1

y-direction is zyxy

F∆∆∆

∂∂ 2

zyxz

F∆∆∆

∂∂ 3

the whole surface is the sum of these:

zyxz

FFF⎜⎜⎛ ∂

+∂

+∂

= 321

yx∆∆∆⎟⎟

⎠

⎞

⎝ ∂∂∂

= ∆x ∆y ∆z

∫∫∂∂∂

=FFF1 dS.F∂∂∂∆

SzyxV0

++ 321

of the vector differential operator∇ . Consider:

( ) Fkji. div321 =∂

+∂

+∂

=++zyx

FFF 321 ∂∂∂ FFF

y) = x i + y j that was sketched in the first lecture. small, stance

es the extra water come from? The divergence of the ivergence should

of flowing water. The flow rate near the origin isorigin show that the flow rate must increase as di

f the field, so a calculation of the d

( ) 1+=+⎟⎟⎠

⎜⎜⎝ ∂

+∂

ji.ji yxyx

21 =⎞⎛ ∂∂

ield to produce this particular flow pattern. re (independent of x and y) so there must be a source of ole f

17

A more adventurous example of the calculation of divergence is:

( ) ( )zyyxzyxzyyxzyx 22div 2222 =++∇=++ kji.kji zyyxzyyxzy 5242 +=++

hen m s involving

um of derivatives.

)

∇ : The normal rules of calculus apply w anipulating expression

( ) B.A.BA. ∇+∇=+∇ - the derivative of a sum is the s

( ) ( ) A.A. ( A.∇+∇=∇ φφφ - the product rule. Since divergence is a scalar quantity, the ust also be scalars and this dictates where the dot erators

efore we consider a of we must first look at a very important theorem.

The divergence theorem

terms on the right hand side m opshould be. B pplications divergence to physics,

S in a vector field F. Total flux through the surface

Consider a finite volume V bounded by a simple closed surface

∫∫=S

dS.F

Now, from the definition of divergence: ∫∫='

1divSdV

dS'.FF where S’ bounds dV

However, flux diverging from V = flux through S

the flux diverging from the volume element dV = divF dV

Therefore, the total flux diverging from V = ∫∫∫ dVFdiv V

∫∫∫ ∫∫=∴V S

dVdiv dS.FF where the surface S is the boundary of the volume V

This is the divergence theorem, which can be expressed in words as:

The surface integral of the normal component of a vector field F taken over a simple closed e olume

Definition: A simple closed surface can be deformed continuously into a sphere without intersecting itself. Therefore, a torus (the shape of a ring) is not a simple closed surface.

ip between a volume and a surface integral can verting one to the other. However, its importance in physics

surface S is equal to the volume integral of the divergence of F tak n over the vbounded by S.

Comment: Mathematically, this relationshbe used to evaluate either by conis immense as we shall soon see. Electric field Gauss’s law of electrostatics states that:

Total electric flux through a closed surface = Total electric charge

enclosed by the surface

∫∫∫∫∫ =∴VS 0

where ρ = charge density (coulombs / metre

dVρε1dS.E

and ε0 = permittivity of free space = 8.85 × 10 farad / metre

3) -12

18

From the divergence theorem: ∫∫∫∫∫ = dVEdS.E div VS

Equate the two volume integrals: ∫∫∫∫∫∫ = dVdV ρε1divE

VV 0

Since the volume is arbitrary, the integrands must be equal, so we have:

0div ε

ρ=E

and this is the differential form of Gauss’s law.

tects and locates the sources and sinks of field, we see from Gauss’s law that the sources of an

These are the places where

charges and ne here, where the field lines are continuous, the divergence is zero.

It is instructive to examine the dimensions of the three integrals in the above analysis:

Since divergence deaelectric field are electric charges.the field lines begin and end, as seen in the sketch of a very simple electric e at positive field. Divergence is positiv

gative at negative charges. Elsew

∫∫∫∫∫∫VV

dV dSEdiv0ε ∫∫==

S

dV .Eρ

Their dimensions must all be the same to make physical sense. Let [x] mean the dimensions of x, then we h

1st integral:

1

ave:

[ ][ ] metresvolt=×

=metrecoulombmetremetreoulomb 3 =⎥

⎦

⎤⎢⎣

⎡faradfaradmetre

cdV 30

1ε

ρ

Note that volts

farads = , which comes from the formula for capacitance coulombsV

C =

2

Q

nd integral: [ ][ ] metresvoltE == 32div metrevolts

metre

Note that [E] = V / m, but since

dV

zyx ∂EEE ∂

+∂

∂+

∂∂

= 321divE , [divE] = V / m2.

3rd integral: [ ][ ] metresvoltdSE == 2metresmvolts etres

Magnetic field

es do not exist (which is everywh etH, must be zero. i =the field lines are continuous and do not begin or end anywhere. This can be seen in the magnetic field in the diagram.

where is

In regions where single magnetic polere) the divergence of a ma

We can therefore write dgnvH

ic field, 0, i.e. N S

Bar magnet

Any vector field for which divF = 0 everycalled solenoidal.

19

[Electric fields are conservative whereas magnetic fields are solenoidal.] The Laplace operator

, E, is given by E = -gradV We have seen that an electric field

From Gauss’s law we also have 0

divερ

=E

0graddiv

ερ

−=V Eliminate E to give

his result requires that the scalar field V be differT entiated twice. The differential operator is:

2

2

2

2

2

22graddiv

yxzyx ⎜⎜⎝

⎛+

∂∂

+∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=∇=∇∇= kji.kji.

and this is the Laplace operator, usually written as 2∇ . The

rewritten as

zyxz ∂∂

+∂∂

+∂∂

=⎟⎟⎠

⎞∂∂

above equation may now be

0

2

ερ

−=∇ V which is known as Poisson’s equation.

regions where there are no free charges, the equation is reduced to

Laplace’s equation is one of the most fundamental equations in mathematical physics. It is a second order partial differential equation, having applications in electrostatics, magnetostatics,

o solve it in a subsequent maths course.

pt of curl of a vector field

In

which is Laplace’s equation. 02 =∇ V

hydrodynamics, heat flow and many other fields. Its solution is the scalar field V(x, y, z). You will learn how t

The conce The third of the important field characteristics (after grad and div) that we need to know is

.

Recall that the value of the line integral round any closed loop in a conservative field is

called curl

φgradif0 −==∫ Fdl.F always zero, i.e.

A zero value for this integral is rather special, indicating that the field is special, i.e. it is tegral will not vanish sures an important

a vector field, F, as shown. To keep ction of F be everywhere the same, but

llow the magnitude to vary as in the diagram. We will now tegral of F.dl around the loop. With the

top of

its now parallel to F. The value of F.dl on all four

kes vector n, the normal to the loop, so it has

conservative. In general, this in and its value meaproperty of the vector field. Consider the following.

Take a plane loop inthings easy, let the dire

F

F

aevaluate the line innormal to the loop at right angles to the field, the value of F.dlwill be zero on the vertical sides, large and negative on the side and small and positive on the bottom side. The sumthese is certainly not zero.

In the second diagram, the loop has been reoriented sonormal is sides of the loop is now zero. Clearly the value of the line integral depends upon the orientation of the loop. This mait a function of the

20

the property of a vector.

If F is a force field, the quaHowever, the same measure c

ntity measured by the line integral is work done by the field. an be made on any vector field, whether it is a force field or not.

more meaningful to refer p and circulation per unit area is the curl of the field. We shr king t loop to a point nd define the curl of the field

by the formula:

The generic name for the quantity measured is circulation. It is circulation to the area of the loomeasure circulation at a point by in he a

∫∆=

→∆ CSSdl.FF.n 1lim

curl where the curve C is the boundary of the area ∆S 0

This expression gives the component of curlF in the direction of the unit vector n which is normal to the plane of the curve C. Clearly, curlF is a vector field.

Sign convention: the direction of circulation round the loop and the direction of the normal

-ze wher We

Curl of a vector field in tesian coord a

to the loop form a right-handed screw.

The curl of a vector field will be non ro ever the field possesses shear (as in the diagrams above) and certain kinds of rotation. will examine this more closely when we consider applications in physics.

Car tesin We will derive an expression for the curl of a vector field

by evaluating the line integral ∫∆∆ Czydl.F1 round the

closed loop ABCD in the vector field F = F1i + F2j + F3k.

CD

∆z

The sides of the loop are parallel to the coordinate axes and

hown, the integral will evaluate the positive x-

elementary loop ABCD is at (x, y z).

are of lengths ∆y and ∆z. The sign convention means that for the direction of circulation scomponent of curlF.

The centre of the ,

Assume ∆y and ∆z are both small enough so the value of the vector field along each line can be approximated by its value at the mid point of the line.

Now ∫∫∫∫ ∫ +++=ADCB

dl.Fdl.Fdl.Fdl.Fdl.F DCBA

( ) ( ) ⎟⎟⎠

⎞⎟⎠⎞

⎜⎝⎛ ∆

2y ⎜⎜

⎝

⎛∆

∂∂

+∆+⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ ∆−∆

∂∂

+∆=2 3322 zF

yzFzyF

zyF

( ) ( ) ⎟⎟⎠

⎞⎟⎠⎞

⎜⎝⎛ ∆−∆

23yzF ⎜⎜

⎝

⎛−

∂∂

+∆−+⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ ∆

∆−∂∂

+∆−+2 322 y

zFzyFz

yF

( ) ( ) zyF ∆∆∂

− zy ∂∂ 23 yzF ∆∆

∂= zy

zF

y⎜⎝ ∂∂

F∆∆⎟⎟

⎠

⎞⎜⎛ ∂

−∂

= 23

∴ the x-component of curlF = ∫ ∂∂

−∂∂

=∆∆ z

FyF

zy231 dl.F

∆y

z

y

x

B A

21

This expression can be used as a pattern to obtain the other components of curlF giving:

kjiF ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂

∂+⎟

⎠⎞

⎜⎝⎛

∂∂

−∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂

∂−

∂∂

=yF

xF

xF

zF

zF

yF 123123curl

Just as grad and div can be expressed using the ∇ operator, curl can too. Consider

Fkji curl123123 =⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂

∂+⎟

⎠⎞

⎜⎝⎛

∂∂

−∂∂

+⎟⎟⎠

⎞∂

∂−

∂ yF

xF

xF

zF

zF

yF

kjF ⎜⎜

⎝

⎛ ∂=∂

∂∂

∂∂

∂=×∇

FFFzyx

i

321

We now have a complete set of field characteristics:

∇=grad φφ )fieldvector(

FFF.F

×∇=∇=

curldiv

field)(vectorfield)(scalar

As before, the normal rules of calculus apply:

( ) BABA ×∇+×∇=+×∇ - the derivative of a sum is the sum of derivatives.

( ) AAA ×∇+×∇=×∇ φφφ - the product rule. Note that in the right hand side terms being v

Conservative field revisited: Since curlF is measured

the cross operator always results ectors.

by ∫ dl. ∫ = 0dl.FF and for ust have curlF = 0

siest test for a conservative field. all closed paths in a conservative field, then a conservative field meverywhere. This is the ea

Note that we now have:

Mathematical example

2If A = x y i – 2xz j + 2yz k, find curl curlA.

on: Because curlA is a vector field, its curl can also be determined, so the field is differentiated twice:

curl curlA =

Soluti

( ) ( ) jkjikji

A x

xzxzzyx

yzxzyxzyx 22

202222 22

+=

−−+∂

∂∂

∂∂

∂=

−∂

∂∂

∂∂

∂×∇=×∇×∇

conservative field has curlF = 0

ivF = 0 solenoidal field has d

22

Rigid body rotation The first of our physical examples examines a rotating field to see what its value of curl is. The system considered is a rotating disc. The linear velocity of each point on the disc constitutes a vector field and the linear velocity is given by

where ω is the angular velocity, which points along the axis of rotation, and r is the position vector of a point on the disc

Let us now obtain an expression for the curl of v. Define ω as (ω1, ω2, ω3) and r as (x, y, z):

rωv ×=

xyzxyzzyx

zyx211332

321)(curl

ωωωωωω

ωωω

−−−∂

∂∂

∂∂

∂=×∇=××∇=kji

rωv kji

( ) ( ) ( ) ωkji 2332211 =+++++= ωωωωωω

This shows that the circulation is constant over the whole field v (its value is independent of

Comment: Perhaps it is culation and therefore a non-zero value of curl. However, before jumping to conclusions, let us examine nother rotating system in the

re concentric circles. A field

.

ater down a plug-hole

position).

no surprise that a rotating system has lots of cir

a same way.

The field lines of v for the rotating disc awith virtually the same lines is the velocity of water flowing down a plug-hole. The field lines cannot be exactly concentric circles, but they can be very close, so we will assume they are to simplify the mathematics. It will be instructive to compare the two fields W As before, we will determin ield. First of all we need to etermine how the water mov dy like the rotating disc.

er of mass m at radius r with angular velocity ω.

ld = moment of inertia × angular velocity = Iω

e the curl of the linear velocity fes - it doesn’t behave as a rigid bod

Consider a small parcel of wat

Angular momentum about the centre of the fieIgnoring viscosity and friction, angular momentum is constant.

22 constantgivingconstant

rmrmI ===∴ ωωω

rmr

rmrv constantconstant

2 ==== ω that is, r

v 1∝ This makes the linear velocity

A vector field whose field lines are concentric circles and whose magnitude is 1/r is:

22 yxxy

++−

=jiv

We can now calculate curlv as

ω

r

23

kkji

⎢⎣

⎡+

++−+

=

−∂

∂∂

∂∂

∂=×∇ −

yxxyxx

yxxy

zyxv 12)(1

0

2222

22 ⎥⎦

⎤+−

++

− yyxy

yxyx

2)( 2222

222

2

0k =⎥⎦

⎤⎢⎣

⎡++

−+

= 222

22

22 )(222yx

yxyx

Comment: This result of obtaining zero curl for a whirlpool may be surprising if we had tried to predict the result. What is it telling us about the flow of water? What is the essential difference between this and the rotating disc?

If we place a marker on the rotating disc, it will be carried around a field line, i.e. it will interact with the f changing orientation as it goes. It will complete one rotation for

e float a cork on the whirlpool, it will also interact

ction of curl.

Stokes’s theorem

Rigid disc: rv ∝

ield,

each revolution of the disc. This is obvious.

If wwith the field, but how will its orientation change? This is not so obvious and we really need to do the experiment. The OU video of this (shown as part of the course) shows that the orientation of the cork does not change as it is carried around by the field. This lack of rotation is what is predicted by the zero curl.

Curl is seen to be a measure of that property of a vector field which will change the orientation of an object interacting

ith the field. w

The axis of rotation is the dire

The speed of rotation is measured by the magnitude of curl. Just as the concept of divergence gave rise to the very important divergence theorem, there is an equally important theorem involving curl – Stokes’s theorem.

Consider a surface S bounded by a closed curve C in a vector field F.

ea ABCD on S and its adjacent .

Consider also an element of arelement ADEF, enlarged in the lower diagram

For ABCD: dl.Fdl.F ⎟⎞

⎜⎛ +++= ∫∫∫∫∫

ADCB

⎠⎝ DCBAABCD

and for ADEF: dl.Fdl.F ⎟⎠⎞

⎜⎝⎛ +++= ∫∫∫∫∫

D

A

A

F

F

E

E

DADEF

Whirlpool: rv 1∝

dS

S C

C D E

B A F

24

∫∫ −=D

A

A

Ddl.Fdl.F But

Therefore ∫∫∫∫∫∫∫∫∫ =⎟⎠⎞

⎜⎝⎛ +++++=⎟

⎟⎠

⎞⎜⎜⎝

⎛+

ABCDEF

A

F

F

E

E

D

D

C

C

B

B

AADEFABCD

dl.Fdl.Fdl.F

which is around the boundary of the combined area.

From the definition of curl: ∫= dl.FdS.Fcurl where the line integral is round the

boundary of dS.

tegration of this expression over the whole surface S is achieved by adding up the o

taken round a simple closed of the curl of F taken over

duced to a point without

e and a line integral can be it is an extremely important

Incontributions f adjacent elements of area. As each new element is added, the line integral on the right is always round the boundary of the combined area as seen in the above analysis. The final result is therefore:

∫∫∫ =C

dl.FdS.FS

curl where C is the boundary of S.

This is Stokes’s theorem which can be expressed in words as:

The line integral of the tangential component of a vector field F curve C is equal to the surface integral of the normal componentany surface having C as its boundary.

Definition: A simple closed curve can be continuously reintersecting itself, i.e. it doesn’t form a knot.

Comment: Mathematically, this relationship between a surfacused to evaluate either by converting one to the other. However, theorem of immense use in physics as we shall soon see. Summary of integral relationships We now have a set of three quite remarkable integral relationships and it is worth looking at

em all together: th

Divergence theorem: ∫∫∫∫∫ =dV dS.FFV

div where surface S is theS

boundary of volume V.

∫S

∫Stokes’s theorem: ∫ =C

dl.FdS.Fcurl where curve C is the boundary of surface S.

ative field: φφφ −=∫ dl.grad where A and B are the ends of a defined curve.

ics – grad, div and curl – that we have been considering. In each case, the expression on the right involves only the b

n the left. The left-hand-side integrals can therefore be evaluated without knowing the detailed behaviour of the field inside the boundary.

ConservB

ABA

They involve the three major field characteristoundary of the field

of integration o

25

Ampère’s law Ampère’s law for a steady current relates the magnetic field around a current-carrying conductor to the electric current flowing along the

H

I

conductor:

IC

=∫ H.dl

Be careful about the physical dimensions in this expression. Note that:

B = µ0 H in vacuo H = magnetising force ampere / metre 2B = magnetic induction weber / metre µ0 = permeability of free space = 4π×10-7 henry / metre

rom Stokes’s theorem: F ∫∫∫ =SC

H.dScurl where S is any surfH.dl ace having C as its

2

where the surface integral is over the same surface

r I

s must be equal, so that:

boundary.

Now express the current in terms of current density J, measured in amps / m , i.e. current flowing through unit area. The vector direction gives the direction of flow of the current.

We find that: ∫∫=I .JS

S as in Stokes’s theorem, i.e. flux of J = current.

dS

Substitute fo from Ampère’s law and make use of Stokes’s theorem to give

∫∫∫∫ =S

dS.JdS.HS

curl

Since the surface S is arbitrary, the integrand

curlH = J

and this is the differential form of Ampère’s law. Grad in spherical polar coordinates So far, we have always used Cartesian coordinates to develop expressions for curl, div, grad

rical polar coordinates as well.

ld f(r, θ, φ). We already know that df = gradf . dl

and 2∇ . However, it is useful to have the equivalent expressions in cylindrical and sphe

Consider a scalar fie

The total differential of f is: φφ

θθ

dfdfdrrfdf

∂∂

+∂∂

+∂∂

=

From the diagram: φθrdl ˆsinˆˆ φθθ drdrdr ++=

where are dimensionless unit vectors in the directions f increasing r, θ, φ respectively, so that

magnetic sed loop

= current through loop Circulation of field round a clo

φθr ˆ,ˆ,ˆ o 0ˆ.ˆˆˆˆˆ,1ˆ.ˆˆ.ˆˆˆ ====== φθφ.rθ.rφφθθr.r

r sinθ dφ

dr z

φ

θ r dθ

y

x

26

φθr ˆˆˆ φθ AA ++ Let grad Af r=

Then ( ) ( )φθθφθ drAdrdrdrAAAf rr ˆsinˆˆˆˆˆ.grad +=++++= φθr.φθrdl φθθ φθ rAdrA sin+

Comparing terms in df and gradf . dl gives:

d

φθθ φθ ∂∂

=∂∂

=∂∂

=f

rAf

rA

rfAr sin

1;1;

φθr ˆsin1ˆ1ˆgrad

φθθ ∂∂

+∂∂

+∂∂

=∴f

rfff

rr Div in spherical polar coordinates Consider an elementary e ABCDEFGH centred around the point ( , ) in the vector field

volum

all enough so

e at the mid point.

r θ, φ ˆ φθrF ˆˆ),,( φθφθ FFFr r ++=

Assume that ∆r, ∆θ, ∆φ are all sm the field F on each face of ABCDEFGH can be approximated by its valu

To determine divF, use ∫∫∆→∆=

SVVdS.FF 1

0lim

div

where S is the surface o

Flux through face ABCD (= F.dS) =

f ∆V.

( ) ⎟⎠⎞

⎜⎝⎛ ∆

−∆∆−∂∂

+∆∆− sin2rFr θ2

sin2 rrFr r φθθφθ

Flux through face EFGH = ( ) ⎟⎠⎞

⎜⎝⎛ ∆

∆∆∂∂

+∆∆2

sinsin 22 rrFr

rF rr φθθφθθ

∴ total outward flux in r̂ -direction = ( )( )rrFr r ∆∆∆

∂∂ φθθsin2

Similarly, total flux in θ̂ -direction = ( θθ θ∂

)( )θφ ∆∆∆∂ r rF sin

Total flux in the φ̂ -direction = ( )( )φθ ∆∆∆φ φ∂∂ rr F

( ) ( ) ( )⎤⎡ ∂∂∂= φθ

φθ

θθ φθ ∆∆∆⎥

⎦⎢⎣ ∂

+∂

+∂

∴ ∫∫ 2dS.F rFrFrFrr r sinsin

= r

S

2Elementary volume ∆V sinθ ∆r ∆θ ∆φ

( ) ( ) ( )φF θ φθθ

θ rF

rFr

rr r ∂∂

+∂∂

+∂∂

=sin

sinsin111lim 2

2dS.FθVV S∆→∆

= ∫∫0divF 1

r̂ E

θ̂ C

φ̂

D F

G

H

∆r

r∆θ

A

rsinθ∆φ B

27

The Laplacian operator in spherical polar coordinates The Laplacian operator, , is given by div grad and we have just developed expressions

h grad and div above. 2∇

for bot

⎟⎟⎠

⎞⎜⎜⎛ ∂

==∇ divgraddiv2

⎝ ∂∂

+∂∂

+∂

φθr ˆsin1ˆ1ˆ

φθθ rrr

2

2

2222

22

sin1sin

sin11

φθθθ

θθ ∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=∇∴rrr

rrr

Comment: You are not expected to remember complicated expressions like thes You are rl and in Carte ns, spheric d

r, you are expected to be able to use

e. 2∇ given the complete set of expressions for grad, div, cu

cylindrical polars in the formula book in exams. Howevethem. You will come across them again in physics courses.

sia al an

28

2∇ IN DIFFGRAD, DIV, CURL AND ERENT C

Cartesian coordinates

OORDINATE SYSTEMS

Cylindrical polar coordinates

kjizyx ∂

∂+

∂∂

+∂∂

=φφφφgrad zθr ˆˆ1ˆ)(

zff

rrffgrad

∂∂

+∂∂

+∂∂

=θ

z

Fy

FxFdiv

∂∂

+∂

∂+

∂∂

= 321F ( ) ( ) ( )⎥⎦⎤

⎢⎣⎡

∂∂

+∂∂

∂∂

= zr rFz

FFrrr

div θθ1)(F

+

321 FFFzyxcurl ∂

∂∂

∂∂

∂=kji

F

zr FFrFr

r

rcurl

θ

φθ ∂∂

∂∂

∂∂=

zθrF

ˆˆˆ1)(

2

2

2

2

2

22

zyx ∂∂

+∂∂

+∂∂

=∇ 2

2

2

2

22 11

zrrr

rr ∂∂

+∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=∇θ

Spherical polar coordinates

φr ˆsin11ˆ)(

φθθ ∂∂

+∂∂

+∂∂

=f

rf

rrffgrad

( ) ( ) ( )φθ φθθ

θθF

rF

rFr

rrdiv r ∂

∂+

∂∂

+∂∂

=sin1sin

sin11)( 2

2F

φθ θφθ

θ

θFrFrF

r

rr

rcurl

r sin

ˆsinˆˆ

sin1)( 2 ∂

∂∂

∂∂

∂=φθr

F

2

2

2222

22

sin1sin

sin11

φθθθ

θθ ∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=∇rrr

rrr

29

The continuity equation w bring in time as an extra variable. Most things move or change with

l with that. We will no time and we have to be able to dea

Consider a substance of density ρ(x, y, z, t) with a flow rate (flux) given by J(x, y, z, t) as amount of substance / unit area / unit time.

Total amount of substance in V =

Consider also an arbitrary volume V bounded by the closed surface S.

∫∫∫V

dVρ

∫∫∫∂∂

V

dVt

ρ rate of increase of substance in V = ∴

Assuming ρ to be continuous in V with respect to space and time, we can reverse the order of

∫∫∫∫∫∫ ∂∂

=∂∂

VV tt operations so that

∴ rate of increase of substance in V = ∫∫∫ ∂∂

V

dVtρ

Now measure the same quantity in a different way and equate the two results.

ount of substance flowing through element of area on S in unit time = J . dS

ecause dS points outwards, this measures amount of substance lost from V.

Am

B

∫∫ rate of decrease of substance from V = ∴S

dS.J

But, from the divergence theorem ∫∫∫∫∫ =VS

dVJdS.J div

rate of decrease of substance in V = ∫∫∫V

dVJdiv ∴

If t e samhere are no sources or sinks in V, these two results must be th e. We therefore have

∫∫∫∫∫∫ −=∂∂

VV

dVdVt

Jdivρ

Since V is arbitrary, the integrands must be equal. This gives the continuity equation as:

0div =∂∂

+tρJ

Examples 1. Fluid (gas or liquid)

J = ρv with dimensions mass /unit area / unit time [divJ] = mass / vol / time where v = velocity and ρ = density [ ]t∂∂ρ = mass / vol / time

30

If ρ increases at a point making 0>∂∂ tρ , this can only be achieved by a net flow of fluid towards the point making divJ < 0 such that 0div =∂∂+ tρJ .

If the fluid is incompressible, then 0=∂∂ tρ so that divJ = 0. This means that divv = 0 and v must therefore be a solenoidal field. 2. Electricity J = current density (coulombs / area / time) [divJ] = coulombs / vol / time

ρ = charge density (coulombs / vol) [ ]t∂∂ρ = coulombs / vol / time

A build-up of charge at a point, making 0>∂∂ tρ , is only possible by a net flow of charge towards the point, making divJ < 0 such that 0div =∂∂+ tρJ .

In a good electrical conductor, there can be no build-up of charge. This means that 0=∂∂ tρ so that divJ = 0 and .

J is solenoidal

3. Heat J me) [divJ] = heat / vol / time

ρ = heat density = quantity of heat / vol

= heat flow (quantity of heat / area / ti

[ ]t∂∂ρ = heat / vol / time

= etemperaturheatspecificmass ×× vol

= density of heat conductor × specific heat × temperature

Th

Q = heat flow; ρ = density of conductor; σ = specific heat; θ = temperature

ith time, the continuity equation for heat becomes:

e more usual symbols used for heat are:

Assuming ρ and σ to be constant w

0div =∂∂

+tθσρQ

( )0<∂∂ tθ The temperature can only decrease by a net flow of heat away, i.e. divQ > 0 such that 0div =∂∂+ tθσρQ .

In steady state heat conduction, the temperature will be constant with time so that 0=∂∂ tθ giving divQ = 0 and Q must be solenoidal.

Heat conduction equation

We have already seen that heat conduction is described by Fourier’s first law of heat onduction: Q = - k gradθ where k is thermal conductivity

divQ = - k div gradθ = c

θ2∇− k Therefore

But from the continuity equation t∂

∂−=

θσρQdiv

This is also known as Fourier’s second law of heat conduction.

Eliminating divQ gives the heat conduction equation tk ∂

∂θ=∇

σρθ2

31

The quantity σρk is known as the diffusivity. thermal

02 =∇ θ0=∂∂For steady state heat conduction, tθ t

iffusion equation

hen D

ction equation will also describe the more general process of diffusion, as heat .

gradc

fusion coefficient, D, = area / time

divergence of both sides of the above equation to get:

The heat conduconduction is a particular example of diffusion

The diffusion coefficient of a substance is defined by:

flow rate = - diffusion coefficient × concentration gradient

This is described by the equation c v = - D

where c = concentration (mass / vol) and v = velocity (m / s)

This makes the dimensions of the dif

Take the

cDcDc 2graddiv)(div ∇−=−=v

But the continuity equation gives 0)(div =∂∂

+cc v t

) gives the diffusion equation Eliminating div(cv

tDcc

∂∂

=∇12

Some partial differential equations of physics We can now write down and compare three very important partial differential equations in mathematical physics: Laplace’s equation

ectricity, m

tational potential, temperature etc., depending upon the application.

This equation has applications in el agnetism, gravitation and steady-state heat flow among many others. The scalar field φ may be electric potential, gravi

02 =∇ φ

Laplace (1749-1827) used it to study the gravitational attraction of extended masses. Diffusion equation The obvious applications are in heat conduction and diffusion.

artial p ssure, depending upon the

ped Fourier series as a means of solving it. We will come across Fourier series later this term.

ave equation

tD ∂∂

=∇φφ 12

The diffusion coefficient, D, becomes k / ρσ for heat flow. The scalar field φ may be concentration, temperature, p reapplication. Joseph Fourier (1768-1830) develo

W

e in the first year. It You will have encountered the wave equation in your waves cours

32

describes the behaviour of all waves that propagate at coconstant prof i.e. they do not change shape as they tr

nstant speed and ile, avel.

The field φ may be displacement, pressure, electric field, magnetic field the wave agnetic wave etc.).

is the speed of propagation.

2

2

22 1

tv ∂∂

=∇φφ

etc. depending upon the nature of (sound wave, electromv Useful relationships We have already seen a number of vector relationships that give the grad, div or curl of sums

ble in exams if you need to use them,

presented here w e second order

s the scalar potential of

idal field is zero. It is clear, therefore, that curlF produces a solenoidal field and F is known as the vector potential of that field.

complished by expanding the field into its Cartesian components, i.e. F = F1i + F2j + F3k so that

and products of fields. A more complete list is availabut they will also be given in exam questions as appropriate. Four more relationships are

hich are both useful and interesting. They all involvderivatives. 1. curl gradφ = 0 gradφ produces a conservative field, where φ ithat field, and the curl of a conservative field is everywhere zero. 2. div curlF = 0 We know that the divergence of a soleno

Proof of the relationship is ac

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂

∂∂∂

+⎟⎠⎞

⎜⎝⎛ ∂∂∂⎞⎛ ∂∂∂ FFF 123

∂−

∂∂+⎟⎟

⎠⎜⎜⎝ ∂

−∂∂

=×∇∇=yF

xF

zxF

zyzyx123curliv F.F d

012

23123 −∂∂

∂+

∂∂∂

−∂∂

∂+

∂∂∂

−∂∂

∂=

xzF

xyF

zyF

zxF

yxF 22222

=∂ F

e Laplace operator.

∂∂ yz

. div gradφ = φ2∇ This produces th3

4. ( ) ( ) FdivFgradF.FFcurlFcurl 22 ∇−=∇−∇∇=×∇×∇= This is one ofamous vector relationships as we shall soon see. It will be used in one oremarkable pieces of mathematical physics ever and will form a suitable c

f the more f the most

limax to this course.

r triple

= F. However, since

To prove this relationship, use the standard relationship for dealing with vectoroducts, i.e. a×(b×c) = (a.c) b � (a.b) c p

Let a = b = ∇ and c ∇ is an operator, the order of terms on the right-s a×(b×c) = b (a.c) – (a.b) c

is is exactly the relationship we ish to prove.

law revisited

hand-side becomes important. If we rewrite the relationship aso the operators come first in each term, we can see that thw Ampère’s

ith time because div curlH = divJ

for any differentiable vector field A, making divJ = 0

However, the continuity equation gives

We have already derived the differential form of Ampère’s law for steady currents, curlH = J

his does not allow variations of J wT

But div curlA = 0

0div =∂∂

+tρJ so that

t∂∂

−=ρJdiv

33

There is clearly a time-dependent term missing from Ampère’s law, but making it consistent with the continuity equation.

we can find it by

E

es

Substitute for ρ in the continuity equation from Gauss’s law divE = ρ / ε0, i.e. ρ = ε0 div

( ) 0divdiv 0 =∂∂

+ EJ εt

The continuity equation becom

Changing the order of the differential operators gives

0divdivdiv 00 =⎟⎠⎞

⎜⎝⎛

∂∂

+=⎟⎠⎞

⎜⎝⎛

∂∂

+ttEJEJ εε

It would appear, then, that J in Ampère’s law should be replaced by t∂

∂E+J 0ε to allow for

time variations.

Therefore, Ampère’s law becomes t∂∂

+=EJH 0curl ε

Faraday’s law of electromagnetic induction Faraday’s l

S

N

aw states that:

Expressing this mathematically gives:

voltage round loop = - rate of change of magnetic flux through loop

∫∫∫ ∂ SC tdiagram and S is a surface with

∂−= H.dSE.dl 0µ where C is the loop in the

C as its boundary.

remBy Stokes’s theo : ∫∫∫ =SC

E.dSE.dl curl

∫∫∫∫∫∫ ∂∂

−=∂∂

−=∴SSS tt

dSH .

Since the surface S is arbitrary, the integrands must be equal. This gives the differential form

H.dSE.dScurl 00 µµ

of Faraday’s law as:

t∂∂

−=HµEcurl 0

Just to make sure that the physical dimensions are correct, we see that [E] = V / m so that [curlE] = V / m2.

lso, µ0 = permeability of free space = 4π × 10-7 henrys / metre and henry = volt sec / amp A

20 / mV

smmAt==⎥⎦⎢⎣ ∂

µ H = magnetising force measured in amp / metre so that AsV⎤⎡ ∂H

Note that B = µ0 H = magnetic induction measured in weber / metre2.

34

Maxwell’s equations Let us now assemble the various laws of electricity and magnetism that we have encountered

divE = ρ /

Gauss’s law of magnetostatics divH = 0

in their differential forms:

Gauss’s law of electrostatics ε0

t∂∂

−=HE 0curl µ Faraday’s law

t∂∂

+=EJH 0curl ε Ampère’s law

These are known as Maxwell’s equations.

We will use them in their simplest possible form which is in free space where ρ = 0 and J = 0.

divE = 0 _____ (1)

The equations are now expressed as:

t∂∂

−=HE 0curl µ _____ (3)

_ (2) t∂

∂=

EH 0curl εdivH = 0 ____ _____ (4)

e 2divgrad ∇−

curlcurldivgrad −

Using equations (1) and (3):

Now use the vector id ntity: Ecurlcurl = EE

and rearrange it as: E2 =∇ EE

( )HE curl2

t∂

=⎟⎞

⎜⎝⎛

∂∂

=∇ µµ Hcurl00 t∂⎠

2

2

00002

ttt ∂∂

=⎟⎠⎞

⎜⎝⎛

∂∂and from (4):

∂∂

=∇EEE εµεµ

Similarly, we can obtain: 2

2

002

t∂∂

=∇HH εµ

These are of the same mathematical form as the wave equation 2

22 1 ∂

=∇φφ 2 tv ∂

Amazingly, we have discovered waves in electric and magnetic fields (actually, Maxwell got there first), which propagate at a speed of 001 εµ m / s.

henrys / metre and 0 = 8.85419 10 farads / metre, we find that the 8 / s, a num

I hope you have enjoyed the course.

With µ0 = 4π × 10-7 ε × -12

speed of the waves is 2.99792 × 10 m ber which you ought to recognise as thespeed of light. Mathematical physics doesn’t come any better than this!

35