Chapter VIII Conservation Laws

Recommended problems: 8.1, 8.2, 8.3, 8.4, 8.5, 8.8, 8.9, 8.11, 8.12.

Charge and Energy

It is known that the charge in a volume is given by

dtrtQ

V

,

And the current flowing through the surface that bound the volume is

S

adJtI

If the current is flowing out of the surface (charge is decreasing in the volume),

then

dt

dQtI

SV

adJdt

Using the divergence theorem for the R.H.S we get

SV

dJdt

)1(Jt

Eq.(1) is the continuity equation which tells that if the total charge in some

volume changes, then exactly that amount of charge must have passed in or

out through the surface bounding the volume.

Previously, we found that the electric energy stored in a static charge

distribution is given by

)2(22

dEW oe

Likewise, the magnetic energy stored in a magnetic field is given by

)3(221

dBW

oe

So the total energy stored in the electromagnetic field is

)4(21202

1

dBEU

oem

Suppose we have some charge and current configuration which, at time t, produces E & B. The work done by electromagnetic forces on these charges in

an interval dt is

dtvEqdtvBvEqldFdW

vJdq

& dJEdW

The rate at which work is done on all the charges in a volume (the power) is

)5(

V

dJEdt

dW

Using Ampere’s law we have

t

ErJB ooo

)6(1

t

EEBEJE o

o

Using the rule BAABBA

BEEBBE

)7(BEtBBBE

Using Faraday’s law tBE

BEtBBBE

Substituting Eq.(7) into Eq.(6) we get

)8(1

t

EEBEtBBJE o

o

2

212

21 & B

tt

BBE

tt

EEBut

)9(12

221 BE

BE

tJE

ooo

Substituting Eq.(9) into Eq.(5) we get

)10(12

221

dBEd

BE

tdt

dW

VoV oo

Using the divergence theorem for the second integral we get

)11(12

221

SoV oo adBEd

BE

dt

d

dt

dW

From Eq.(4), it is clear that the first term represents the rate of energy stored in

the electromagnetic fields. The second term is then represents the rate at

which energy is carried out of the volume across its boundary surface.

Defining the Poynting vector S as the energy per unit time per unit area

transported by the em-fields, i.e.,

)12(1

BESo

It is sometimes called the energy flux density. Eq.(11) can now be written as

theoremsPoyntingadSdt

dU

dt

dW

S

em ')13(0

In words, the sum of the work done by all fields on charges in the volume, plus

the changes in the field energy within the volume, plus the energy that flows out

of the volume carried by the field must balance.

It is known that the total work done on the charges will increase their

mechanical energy (kinetic, potential, chemical, or whatever), i.e.,

)14(

Vmec

mec dudt

d

dt

dU

dt

dW

With umec is the mechanical energy density (energy per unit volume). Denoting

the electromagnetic energy density as uem with

)15(2

221

ooem

BEu

Using Eq.(14), Eq.(13) can be written as

0 SV

emmec adSduudt

d

Using the divergence theorem for the second integral we get

0

Vemmec dSuu

t

)16(0

Suu

temmec

Differential form of

Poynting’s theorem

Eq.(16) has the same form of the continuity equation. The charge density is

replaced by the energy density (mechanical + electromagnetic), and the current

density is replaced by the Pointing vector. S represents the flow of energy in

exactly the same way J represents the flow of charge.

Example 8.1 Calculate the rate of energy

passing in through the surface of a wire of

radius a and length l that carries a current I.

Solution: It is known that the work done on a

moving charge dq by the em-filed with potential

difference V between its ends is

a I

L

E

S

dqVLEdqLBvEdqLFdW

The rate of the work which will be dissipated as energy in the wire is

S

adSP

Let us prove this argument in terms of the Poynting’s theorem. From Eq.(13),

the rate of energy passing in through the surface is

VIdt

dqV

dt

dWP

Note that since E is along the axis of the wire and B point tangent to its surface,

then S must be radially inward.

IVaLaL

IVadSP

S

22

This energy shows up as Joule heating of the wire.

Now, if V is the potential difference between the ends of the wire then

a

IB

L

VE o

2&

aL

IVBES

o 2

1

Newton’s 3rd law in Electrodynamics

Consider a point charge q1 traveling along

the -ve x-axis at a constant speed v1 .

It encounter a second identical point

charge q2 traveling along the -ve y-axis

with the same constant speed v2 .

It is clear that the e. force between them

is repulsive and obeying Newton’s 3rd law.

What about the magnetic force? q1 will create a m.field into the page at the

position of q2. Therefore, the m.force on q2 is

ivBqkjvBqBvqF ˆˆˆ12121221

Similarly, q2 will create a m.field out of the page at the position of q1. Therefore,

the m.force on q1 is

jvBqkivBqBvqF ˆˆˆ121212112

It is clear now that the net electromagnetic force on q2 due to q1 is equal in

magnitude but not opposite to the force acting on q1 due to q2 , in apparent

violation of Newton’s 3rd law. In electrostatics and magetostatics the 3rd law

holds but in electrodynamics it doesn’t.

Since Newton’s 3rd Law is intimately related to conservation of linear

momentum, which depends on the cancelation of the internal forces due to the

3rd law. Therefore, it would seem that electrodynamical phenomena would then

also seem to violate conservation of linear momentum.

The conservation of linear momentum in electrodynamics is rescued by

proposing that the electromagnetic fields carry momentum as well as energy, as

stated in the previous chapters. Whatever momentum is lost to the particles is

gained by the fields.

Thus, in electrodynamics, the electric charges plus the electromagnetic fields

accompanying the electric charges together conserve total linear momentum p. Thus, Newton’s 3rd Law is not violated after all, when this broader perspective

on the nature of electrodynamics is fully understood.

Maxwell’s Stress Tensor

The electromagnetic force acting on a charge within a volume V is

)17(

VV

dBJEdBvEF

The force per unit volume is then

)18(BJEf

It should be noted that in the case that the two point charges are moving

parallel to each other Newton’s 3rd law hold and so the linear momentum

carried by the electromagnetic fields is zero for this special case.

Note that this special-case situation is related to the case of parallel electric

currents attracting each other, e.g. two parallel conducting wires carrying steady

currents I1 and I2 .

Eo

t

EBrJ o

o

&

Using Maxweel’s equations (i) & (iv) we get

)19(Bt

EBEEf o

oo

t

BEB

t

E

t

BEBut

But from Faraday’s law we have

t

BE

EEBt

E

t

BE

Eq.(19) now becomes

)20(BEt

BBEEEEf o

oo

To let Eq.(20) looks more symmetric and since 0 B

We can add to Eq.(20) the term BB

)21(1

BEt

BBBBEEEEf oo

o

Using the rule ABBAABBABA

EEEEEEE

222

EEEEE

221 BBBBB

2

21&

Eq.(21) Now reads

)22(

1 2212

21

BEt

BBBBBEEEEE

f

o

oo

We now introduce the Maxwell’s Stress Tensor (a 3×3 matrix), the nine

elements of which are defined as:

)23(1 2

212

21 BBBEEET ijji

oijjioij

With i,j= 1,2,3 refer to the coordinates x, y, and z. The Kronecker delta is

defined as

ji

jiij 0

1

It is clear from Eq.(23) that Maxwell’s tensor is symmetric under the interchange

of the indices i & j , i.e., Tij = Tji . Now the 9-elements are:

22222221

112

1zyx

ozyxoxx BBBEEETT

22222221

222

1xzy

oxzyoyy BBBEEETT

22222221

332

1yxz

oyxzozz BBBEEETT

yxo

yxoyxxy BBEETTTT

1

2112

zxo

zxozxxz BBEETTTT

1

3113

zyo

zyozyyz BBEETTTT

1

3223

Because it carries 2-indices Tij is called double vectors and denoted by T

The dot product between and a vector a is another vector b such that T

)24(3

1

i

ijijj Taab T

Replacing the vector a by and noting the definition of Eq.(23) we get

3

1

221

3

1

221 1

iijjii

oiijjiioj BBBEEE

T

)25(1 2

21

221

BBBBB

EEEEE

jjjo

jjjo

j

T

Comparing Eq.(22) and Eq.(25) we get for the force per unit volume

)26(t

STf oo

The total force on the charges in the volume V is then

Voo

VV

dSdt

ddTdfF

Using the divergence theorem for the 1st integral we get

)27(

Voo

S

dSdt

dadTF

Note the following important points about the physical nature of this result:

(I) In the static case the second term on the RHS in the above equation

vanishes, then the total EM force acting on the charge configuration contained

within the (source) volume v can be expressed entirely in terms of Maxwell’s

Stress Tensor at the boundary of the volume v , i.e. on the enclosing surface S :

(II) Physically, is the force per unit aria (or stress) acting on the surface.

(III) More precisely, Tij is the force per unit aria in the ith direction acting on an

element of surface oriented in the jth direction. Thus the diagonal element (Txx ,

Tyy , Tzz ) represents pressure, while the off-diagonal elements (Txy , Txz , Tyz )

represent shears..

T

Example 8.2 Determine the

net EM force acting on the

“northern” hemisphere of a

uniformly charged solid non-

conducting sphere of radius R

and charge Q.

Solution: The boundary surface of the upper hemisphere consists of 2-parts: a

hemispherical bowl at radius R, and a circular disk lying in the x-y plane (at

=/2). Since we have static case the 2nd term of Eq.27) vanish. For the

hemispherical bowl we have

zyxddRrddRad ˆcosˆsinsinˆcossinsinˆsin 22

The electric field on the surface of the charged sphere is:.

zyxR

Qr

R

QE

oo

ˆcosˆsinsinˆcossin4

1ˆ

4

122

Due to the symmetry of the problem, the net force acting on the hemispherical

bowl is along the z-direction, thus we only need to calculate, from Eq(24)

zzzyzyxzxi

iizz TdaTdaTdaTdaaddf

3

13

T

Now from Eq.(23) we have

coscossin4

112

2

R

QBBEET

oozx

ozxoxz

cossinsin4

112

2

R

QBBEET

oozy

ozyoyz

222

221

22222221

sincos4

1

2

1

R

Q

BBBEEET

oo

yxzo

yxzozz

ddRdax cossin22

ddRday sinsin22

ddRdaz cossin2

ddRR

Q

ddRR

Q

ddRR

Qdf

oo

oo

ooz

cossinsincos4

1

sinsincossinsin4

1

cossincoscossin4

1

2222

221

222

2

222

2

ddR

Qdf

ooz sincossincossincoscossin

4

1 2221222

2

ddR

Qdf

ooz sincossincoscossin

4

1 22212

2

ddR

Qdf

ooz sincoscossin

4

1 2221

2

ddR

Qdf

ooz cossin

4

12

21

The force acting on the hemispherical bowl is

2

0

2

0

2

21 cossin

4

1

ddR

QF

ooz

2

1/2

2

2

84

1

R

QF

oz

For the circular disk we have

zdrrdzdrdrad ˆˆsin2

Since now we are inside the sphere we have for the electric filed

23

23

ˆcosˆsinsinˆcossin4

1ˆ

4

1

zyxR

Qrr

R

QrE

oo

yxR

QrE

o

ˆsinˆcos4

13

zzzzzzyzyxzxi

iizz TdaTdaTdaTdaTdaaddf

3

13

T

222

2

321222

21 sincos

4

1r

R

QEEET

ooyxzozz

drdrR

QTdadf

oozzzz

32

321

4

1

R

ooz ddrr

R

QF

0

2

0

32

321

4

1

2

R4/4

2

2

164

1

R

QF

oz

The force acting on the disk is now

The total force acting on the northern hemisphere is then

zR

Qz

R

Qz

R

QFFF

ooodiskbowl ˆ

16

3

4

1ˆ

164

1ˆ

84

12

2

2

2

2

2

Conservation of Momentum

According to Newton’s 2nd law we have

dt

PdF mech

From Eq.(27) we can write now

)28(

Voo

S

mech dSdt

dadT

dt

Pd

Where Pmech is the total mechanical momentum of the particles contained in the

volume V.

From Eq.(28) we can say that the 2nd integral represents the momentum stored

in the EM fields, i.e.,

)29(

VooEM dSP

While the 1st integral represents the momentum per unit time flowing inward

through the surface.

Using Eq.(29) Eq.(28) can be written as.

dt

PdadT

dt

Pd EM

S

mech

adTdt

Pd

dt

Pd

S

EMmech

)30(adTdt

Pd

S

total

Conservation of momentum in

Electrodynamics

Any increase in the total momentum is equal to the momentum brought in by

the em-fields.

If the volume v is all space, then no linear momentum can flow into or out of v through the bounding surface S. Thus, in this situation, from Eq(3)

0dt

Pd total

constant EMmechtotal PPP

Defining the density of linear momentum (momentum per unit volume) as

)31(&

VEMEM

Vmechmech dPdP

Comparing Eq.(29) with the 2nd definition of Eq.(31) we get

)32(SooEM

adTddt

dd

dt

d

SVEM

Vmech

In terms of the densities of momentum, the relation of conservation of

momentum can be written as

Using the divergence theorem for the R.H.S. and rearange we get

0

V

EMmech dTtt

)33(0

T

t

EMmech

Differential form of conservation of

momentum in Electrodynamics

Comparing Eq.(33) with the continuity equation we conclude that physically

represents linear momentum flux density, playing the same role of J (current

density) in the continuity equation.

T

-

Example 8.3 A long coaxial cable of length L consists of an inner conductor

(radius a) and outer conductor (radius b). The coax cable is connected to a

battery at one end and a resistor at the other end, as shown in figure. The inner

conductor carries uniform charge λ and a steady current I = Izˆ (i.e. flowing to

the right). The outer conductor has the opposite charge and current. Calculate

the EM momentum carried by the EM fields associated with this system.

Solution: The fields inside the cable are: ss

Eo

ˆ2

1

ˆ2 s

IB o

zs

Is

s

IS

oo

ˆ4

ˆˆ4 2222

EM energy stored in the EM fields is flowing down the cable in the +zˆ -

direction, from battery to resistor. The power transported is

b

ao

dsss

IadSP

2

1

4 22

a

bI

s

dsIP

o

b

ao

ln22

bas

dsldEVBut

o

b

ao

ln22

IVP

b

a

o

VooEM Lsds

sz

IdSP

2

1ˆ

4 22

From Eq.(29) the EM momentum is

zabLI

P oEM ˆln

2

Macroscopically, we have a static problem: we have a cable (at rest in the lab

frame), a static electric charge distribution producing a static E-field and a

battery producing a DC current that produces a static B -field. How can there be

any net macroscopic linear momentum?

The answer is: there is a hidden mechanical momentum associated with the

flow of current, and this hidden momentum cancels the EM-momentum so that

the total momentum is still zero as it should be.

Now suppose that the resistor R increases with time, so the current decreases.

The changing m-field will introduces an induced E-filed according to

mdt

dldE

Note that since the m-field is counterclockwise and the current is decreasing we

expect an induced e-field along the +ve z-axis. To find this induced e-field at a

point inside the cable we select a rectangular loop as shown

oo

s

s

om

ssIw

s

LdsIadBBut

o

lnln2

2

s so

I

E

w

modt

dwsEwsEldE )()(

zKsdt

dIsE o ˆln

2)(

This field will exert a force on the +ve and the –ve charges according to

za

b

dt

dILzKb

dt

dILzKa

dt

dILELF ooo ˆln

2ˆln

2ˆln

2

The total momentum imparted to the cable as the current drops to zero is

za

bLIdtFP o

mech ˆln2

The total momentum imparted to the cable as the current drops to zero is

Angular Momentum

In the preceding sections we learn that the electromagnetic fields carry energy

and momentum according to:

dBEUo

em212

021

Vooem dSP

The energy per unit volume and the momentum per unit volume are:

21202

1 BEuo

em

BES oooem

Now the angular momentum per unit volume is:

)34(BErrl oemem

Example 8.4 Imagine a very long solenoid with

radius R, n turns per unit length, and current I. Coaxial with the solenoid are two long cylindrical

shells of length l, one inside the solenoid at

radius a, carries a charge +Q uniformly

distributed over its surface; the other, outside the

solenoid at radius b, carries charge –Q. When

the current in the solenoid is gradually reduced,

the cylinders begin to rotate. Where does the

angular momentum come from?

Solution: The e-field between the cylinders is

bsass

Q

lE

o

ˆ2

1

And the m-field between inside the solenoid is

RzznIB o ˆ

Rsas

Q

l

nIBE o

oem

ˆ

2

The linear momentum density is then

bsazl

nIQ

s

Q

l

nIssrl oo

emem

ˆ

2ˆ

2ˆ

The angular momentum density is then

The total angular momentum now is

zaRnIQ

zlaRl

nIQdlL oo

emem ˆ2

ˆ2

2222

When the current is turned off the changing m-field induces an e-filed given by

Rssdt

dIn

dt

dsE o

22 Rsdt

dInsE o ˆ

21

RsRdt

dIn

dt

dsE o

22 Rsdt

dI

s

RnE o ˆ

2

21

This e-filed exerts a force on the two charged cylinders and so a torque. For the

outer cylinder we have

ˆ2

21

dt

dI

b

QRnEQF obsb

zdt

dIQnR

dt

dI

b

QRnsbFrN oob ˆˆˆ

221

2

21

For the inner cylinder we have

ˆ21

dt

dInaQEQF oasa

zdt

dIQna

dt

dInaQsaFrN ooa ˆˆˆ

221

21

Using the relation: dt

LdN

zIQnRzdtdt

dIQnRL o

Iob ˆˆ

221

02

21

zIQnazdtdt

dIQnaL o

Ioa ˆˆ

221

02

21

The total mechanical angular momentum is then

zIQRanLLL oabmech ˆ22

21

Which is exactly equal to total em-angular momentum, i.e., the angular

momentum lost by the field is exactly equal to the angular momentum gained by

the cylinders, and the total angular momentum (em+mech) is conserved.