Download - Chapter VIII Conservation Laws
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.