Chapter 8: Conservation Laws
8.1 Charge and Energy 8.1.1 The Continuity Equation
Conservation laws
in electrodynamics
EM
Tsun-Hsu Chang
清華物理 張存續
Charge
the paradigm
KK[ˋpærә͵daɪm]
範例
Energy
Momentum
Angular momentum
Global conservation of charge: the total charge in the universe
is constant.
Local conservation of charge: If the total charge in some
volume changes, then exactly that amount of charge must
have passed in or out through the surface.
dQtotal
Qenc
0
  J  da
S
dt
t
1
The Continuity Equation
 (r, t )


d


(


J
)
d
V t
V


   J
t
(invoking the divergence theorem)
the continuity equation
(in differential form)
This equation is a precise mathematical statement of the
local conservation of charge.
It can be derived from Maxwell’s equations. since   (  B)  0


E
1
1
  0 (  E)   0 (  )   0  ( J 
  B)
0
 0 0
t
t
t

   J (a consequence of the law of electrodynamics)
t
Q1: The energy and momentum density  analogous to .
Q2: The energy and momentum “current”  analogous to J.
2
8.1.2 Poynting’s Theorem (I)
in vacuum
The work necessary to assemble a static charge distribution
We 
0
2
E
d

(
against
the
Coulomb
repulsion
of
like
charges
)

2
The work required to get current going
1
2
(
against
the
back
emf
)

Wm 
B
d

2 0
The total energy stored in the electromagnetic fields might be
1
1 2
2
U em  We  Wm   ( 0 E 
B )d   uem d
2
0
1
1 2
2
where uem  ( 0 E 
B )
2
0
Q: Can we derive this equation in a more general and more
persuasive way?
3
Poynting’s Theorem (II)
Starting point: How much work, dW, is done by the
electromagnetic forces acting on these charges in the
interval dt? (using the Lorentz force law)
dW   F j  dl j   q j (E j  v j  B j )  v j dt   q j E j  v j dt
j
j
j
dW
  q j E j  v j   (E  v)  d   (E  J )d
V
V
dt
j
(the work done per unit time, per unit volume,
i.e., the power delivered per unit volume)
Q: Can we express this quantity in terms of the fields alone?
Yes, use the Ampere-Maxwell law to eliminate J,
analogous to the proof of the continuity equation.
1
E
  B  0
J
t
0
4
Poynting’s Theorem (III)
E
1
E
E  J  E  (   B  0
E  (  B)   0E 
)
t
t
0
0
1
E  (  B)  B  (  E)    (E  B) (product rule 6)
(Faraday’s law)
B
E  
t
E 1
B
1
E  J  ( 0E 

B )
  (E  B )
t 0
t
0
1 
1 2
1
2

( 0 E 
B )
  (E  B )
2 t
0
0
dW
d
1
1 2
1
2
  (E  J )d    ( 0 E 
B )d  
(E  B)  da
V
S
0
dt
dt V 2
0
(invoking the divergence theorem)
5
Poynting’s Theorem and Poynting Vector
dW
d
1
1 2
1
2
   ( 0 E 
B )d  
(E  B)  da
S
dt
dt V 2
0
0
(total energy stored in the field, Uem)
(the rate at which energy is carried out of
V, across its boundary surface S, by the
electromagnetic fields.)
Poynting’s theorem: “work-energy theorem” of electrodynamics.
Poynting vector: the energy per unit time, per unit area,
transported by the fields.
S
1
0
(E  B )
(S: the energy flux density)
6
Differential Form of Poynting’s Theorem
dW
d
1
1 2
2
   ( 0 E 
B )d   S  da
S
dt
dt V 2
0
dW d
  umech d (umech: the mechanical energy density)
dt
dt V
d
1
1 2
d
2
:
the
energy
density
(u
(
)

E

B
d


u
d

em
0
em


dt V 2
0
dt V
of the fields)
d
d
u
d
u
d
d





S

a
So
mech
em




V
V
S
dt
dt
(divergence theorem)
d
d
u
d
u
d
d





(


S
)

mech
em



V
V
V
dt
dt

(umech  uem )    S (the differential form of Poynting’s theorem)
t

d
Q: What’s the difference between
and
?
t
dt
7
Example 8.1
When current flows down a wire, work is
done, which shows up as Joule heating of
the wire. Find the energy per unit time
delivered to the wire using Poynting vector?
V
ˆ
E

z
Sol:
1
L
(E  B )
S
0 I ˆ
0
B(r  a ) 

2 a
0 I ˆ
VI
1 V
( zˆ 
S
)  
rˆ (point radially inward)
So
2 a
2 aL
0 L
The energy per unit time passing through the surface of the
wire is:
dU em
dW
  S  da  S (2 aL)  VI 
 0 (static fields)
S
dt
dt
8
8.2 Momentum
8.2.1 Newton’s Third Law in Electrodynamics
Suppose two charges, q1 and q2, travel in
along x axis and y axis, respectively. They
can only slide on the axes with velocities
v1 and v2 as shown in the figure.
Q: Is Newton’s third law valid?
EM
Tsun-Hsu Chang
清華物理 張存續
B1   B1zˆ
B 2   B2 zˆ
The electric force between them satisfies the third law, but
the magnetic force does not hold (same magnitudes, but
their directions are not opposite).
The proof of conservation of the momentum, however, rests
on the cancellation of the internal forces, which follows from
the third law. In electrodynamics the third law does not hold.
Q: How to rescue the momentum conservation?
The fields themselves carry momentum. (Surprise!)
9
The Fields of a Moving Charge
The electric field of a
moving charge is not
given by Coulomb’s law.
The magnetic field of a moving
charge does not constitute a
steady current. Thus it is not
given by Biot-Sarvart law.
10
8.2.2 Maxwell’s Stress Tensor
The total electromagnetic force on the charges in volume V:
F    (E  v  B)d   (  E  J  B)d   fd
V
V
V
where f denotes the force per unit volume.
f  E  J  B
Eliminate  and J by using Maxwell’s equations.
   0 (  E)
E
  B  0
J
t
0
1
E
f   0 (  E)E  (   B   0
)B
0
t
1
E
  0 (  E)E 
(  B)  B   0 (  B)
0
t
1
11
Maxwell’s Stress Tensor (II)
E
f   0 (  E)E 
(  B)  B   0 (  B)
0
t
1

E
B
(E  B ) 
B  E
t
t
t
B
(Faraday’s law)
   E
t
1

f   0 (  E)E 
(  B)  B   0E  (  E)   0 (E  B)
0
t
1
2
E  (  E)  E  (E )E
2
1
2
(  B)  B  B  (  B)   B  (B )B
2
( A  B )  A  (   B )  B  (   A )  ( A   ) B  ( B   ) A
( B  B )  B  (   B )  B  (   B )  ( B   ) B  ( B   ) B
2
 B  2B  (  B)  2( B  )B
Chap. 1 p.23
12
Maxwell’s Stress Tensor (III)
1

2
f   0 (  E)E 
B 
(B )B   0 E   0 (E )E   0 (E  B)
2 0
2
t
0
0
1
  0 [(  E)E  (E )E]  [(B )B  (  B)B]
1
1
2
0
0


B  E   0 ( E  B )
2 0
2
t
1
2
2
It can be simplified by introducing the Maxwell stress tensor.
1
1
1
2
2
( Bi B j   ij B )
Tij   0 ( Ei E j   ij E ) 
2
2
0
(the Kronecker delta, another example see Prob. 3.45)
1 if i  j
 ij  
0 if i  j
13
Maxwell’s Stress Tensor (IV)
1
1
1
1 if i  j
2
2
( Bi B j   ij B )  ij  
Tij   0 ( Ei E j   ij E ) 
2
2
0
0 if i  j
0 2
1
2
2
2
2
2
( Bx  B y  Bz )
Txx  ( Ex  E y  Ez ) 
2
2 0
0 2
1
2
2
2
2
2
( B y  Bz  Bx )
Tyy  ( E y  Ez  Ex ) 
2
2 0
0 2
1
2
2
2
2
2
( Bz  B y  Bx )
Tzz  ( Ez  E y  Ex ) 
2
2 0
1
( Bx B y )
Txy  Tyx   0 ( Ex E y ) 
Because Tij carries two indices,
0
1
( B y Bz ) it is sometimeswritten with a
Tyz  Tzy   0 ( E y Ez ) 
0
double arrow T .
1
( Bz Bx )
Tzx  Txz   0 ( Ez Ex ) 
0
Q: How does the tensor operate?
See, “Vector Analysis”, Chap.8, M. E. Spiegel, McGRAW-HILL.
14
Maxwell’s Stress Tensor (V)

On can form the dot product of tensor T with a vector a:
(row vector)

(a  T) j 

i x, y, z
aiTij
(column vector)

(T  a)i   Tij a j
 x1
 1
 x
 x 2
(A1 A 2 A3 )  (A1 A 2 A3 )  1
 x
 3
 x
 x 1

x1
x 2
x
2
x
2
x3
x 2
x1 
3
x 
2
x 
3
x
3
x 
x 3 
 x1
 1

x

 A1 

  x1
 A2    2

  x
 A3   x1

 x 3

x 2
x 1
x
2
x
2
x 2
x 3
x3 
1
x 
A
3  1 
x  

A
2  2 
x 

A
3  3 
x 
x 3 
j  x, y, z
The divergence of the Maxwell stress tensor is:

1
2
(  T) j   0 [(  E) E j  (E ) E j   j E ]
2
1
1
2
 [(B ) B j  (  B) B j   j B ]
2
0
15
Maxwell’s Stress Tensor (VI)
The force per unit volume:

S
f    T   0 0
t
The total force on the charges in V is:

d
F   T  da   0 0  Sd
dt
S
V
Physically, the Maxwell stress tensor is the force per unit area
acting on the surface.
16
Example 8.2
A uniformly charged solid sphere of radius R
and charge Q is cut into two hemispheres.
Find the force required to prevent the
hemispheres from separating.
Sol: This is an electrostatics, no magnetic field involved.

1 Qr
ˆ
E
r
F

T

d
a
4 0 R3
S
The boundary surface consists of two parts---bowl and disk.

The net force is obviously in the z direction.

dFz  (T  da) z  Tzx dax  Tzy da y  Tzz daz
Express the electric component in Cartesian coordinates.
17
Example 8.2 (ii)
On the bowl, r  R.
E
Q
4 0 R
ˆ
r
2
rˆ  sin  cos  xˆ  sin  sin  yˆ  cos  zˆ
Tzx   0 Ez Ex   0 (
Tzy   0 Ez E y   0 (
Tzz 
0
2
2
( Ez
2
 Ey
Q
4 0 R
Q
2
)
sin

cos

cos

2
4 0 R
2
)
sin

cos

sin

2
0
2
 Ex ) 
(
Q
2 4 0 R
2
2
2
)
(cos


sin

)
2
dax  da  xˆ  R 2 sin  sin  cos  d d

2
2
da  R sin  d d rˆ  da y  da  yˆ  R sin  sin  sin  d d

2
 daz  da  zˆ  R sin  cos  d d
18
Example 8.2 (iii)
The force on the bowl is:
dFz  Tzx dax  Tzy da y  Tzz daz
 2sin 2  sin  cos  cos 2 



0
Q
2 2
2
2
R
d
d






)
2sin
sin
cos
sin
 (



2
2 4 0 R

2
2 




sin
cos
(cos
sin
)




0
Q 2
) sin  cos  d d
 (
2 4 0 R
 /2 2
Fbowl 
 
 0  0
 /2


 0
0
2 4 0 R
 0 2
2
(
Q
2
) sin  cos  d d
2
1
1 Q
(
) sin 2 d 
2
4 0 R 2
4 0 8R
Q
2
19
Example 8.2 (iv)
Contd.:
The force on the disk is:
R 2
Fdisk 
 
r 0  0
0
(
Q
2 4 0 R
2 3
)
r
drd


3
1
Q
2
4 0 16 R
2
The net force on the northern hemisphere is:
F
1
3Q
2
4 0 16 R
2
Q: Can we solve this problem using a simpler approach?
Yes, we can use the potential energy to find the net force.
20
8.2.3 Conservation of Momentum
Newton’s second law  the force on an object is equal to
the rate of change of its momentum.

dp mech
d
  T  da   0 0  Sd
F
dt
dt
S
V
where pmech is the total (mechanical) momentum of the
particles contained in the volume V.
 dp mech d
g
d


(an analogous interpretation,
mech

 dt
dt

not a rigorous proof)
V
 dp
d
em

  0 0  Sd , where pem   ( 0 0S)d   g em d
 dt
dt
V
V
V

(momentum stored in the
electromagnetic fields themselves)
(momentum density)
21
Conservation of Momentum (II)

(the momentum per unit time flowing in through the surface.
T

d
a

S
Conservation of momentum in electrodynamics:
Any increase in the total momentum (mechanical plus
electromagnetic) is equal to the momentum brought in
by the fields.


(g mech  g em )    (T)
t
(in differential form)
(momentum flux density, playing the role of J in
continuity equation, or S in Poynting’s theorem)
22
Conservation of Momentum (III)
The roles of Poynting’s vector:
the energy per unit area, per unit time, transported by
S
electromagnetic fields.
0 0S the momentum per unit volume stored in those fields.
The roles of momentum stress tensor:

T the electromagnetic stress acting on a surface.

T the flow of momentum transported by the fields.
23
Optional
Example 8.3 (hidden momentum)
A long coaxial cable, of length l, consists of an inner conductor
(radius a) and an outer conductor (radius b). It is connected to
a battery at one end and a resistor at the other. The inner
conductor carries a uniform charge per unit length λ, and a
steady current to the right; the outer conductor has the
opposite charge and current. What is the electromagnetic
momentum stored in the fields.
24
Optional
Sol: The fields are
1 
E
sˆ
2 0 s
0 I ˆ
B

2 s
I
(E  B)  2 2 zˆ
S
0
4  0 s
1
(an astonishing result!)
The momentum in the fields is:
pem
0  I
  0 0Sd 
2
4
b
a
0 Il b
ˆ
ˆ
l

sds
z
z
2
ln(
)
2
a
2
s
1
In fact, if the center of mass of a localized system is at rest,
its total momentum must be zero.
There is “hidden” mechanical momentum associated with the
flow of current, and this exactly cancels the momentum in the
fields. (This is a relativistic effect: See Example 12.12)
25
8.2.4 Angular Momentum
The electromagnetic fields carry energy and momentum, not
merely mediators of forces between charges.
1
1 2
2
uem  ( 0 E 
B )
2
0
g em   0 0S   0 (E  B)
How about the angular momentum?
 em  r  g em   0 [r  (E  B)]
(again, not a rigorous proof)
Even perfectly static fields can harbor momentum and
angular momentum. See the following example.
26
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 the 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, as we found in Ex. 7.8. Where does
the angular momentum come from?
27
Sol: The fields are
1 Q
E
sˆ (a  s  b)
2 0 ls
B  0 nIzˆ ( s  R)
The momentum density is:
g em
0 nIQ ˆ
 0 0S   0E  B  
 (a  s  R)
2 ls
The angular momentum density is:
 em  r  g em
0 nIQ
  0 [r  (E  B)]  
zˆ (a  s  R)
2 l
The total angular momentum in the fields is:
0 nIQ 2 2
Lem    em d  
( R  a )zˆ
2
(an astonishing result!)
28
Conservation Laws
Conservation laws
in electrodynamics
Charge
Energy
Momentum
Angular momentum

   J
t

(umech  uem )    S
t


(g mech  g em )    (T)
t
What is the conservation law for angular momentum ?
Extra bonus (send your answer to me).
 em  r  g em   0 [r  (E  B)]
29
Homework of Chap.8
Problem 8.1 Calculate the power (energy per unit time) transported down the
cables of Ex. 7.13 and Prob. 7.62, assuming the two conductors are held at potential
difference V , and carry current I (down one and back up the other).
Problem 8.4
(a) Consider two equal point charges q, separated by a distance 2a. Construct the
plane equidistant from the wo charges. By integrating Maxwell's stress tensor
over this plane, determine the force of one charge on the other.
(b) Do the same for charges that are opposite in sign.
Problem 8.6 A charged parallel-plate capacitor (with uniform electric field
E = Ezˆ ) is placed in a uniform magnetic field B = Bxˆ , as shown in Fig.8.6.
(a) Find the electromagnetic momentum in the space between the plates.
(b) Now a resistive wire is connected between the plates, along the z axis, so that
the capacitor slowly discharges. The current through the wire will experience
a magnetic force; what is the total impulse delivered to the system, during the
discharge?
7
17
Problem 8.16 A sphere of radius R carries a uniform polarization P and a uniform
magnetization M (not necessarily in the same direction). Find the electromagnetic
3
momentum of this configuration. [Answer: (4/9)0 R (M  P )]
30
Homework of Chap.8
19
Problem 8.19
Suppose you had an electric charge q e and a magnetic monopole
1 qe
ˆ.
qm . The field of the electric charge is E=
r
4 0 r 2
0 qm
ˆ.
(of course), and the field of the magnetic monopole is B =
r
4 r 2
Find the total angular momentum stored in the fields, if the two charges are separated by a distance d . [ Answer : ( 0 / 4 )qe qm .]20
Problem 8.23
(a) Carry through the argument in Sect. 8.1.2, starting with Eq. 8.6, but using J f in
place of J. Show that the Poynting vector becomes
S  E  H,
(8.46)
and the rate of change of the energy density in the fields is
u
D
B
 E
 H

t
t
t
24
For linear media, show that
1
(8.47)
u  (E  D+B  H).
2
(b) In the same spirit, reproduce the argument in Sect. 8.2.2, starting with Eq. 8.15,
with  f and J f in place of  and J. Don't bother to construct the Maxwell stress
tensor, but do show that the momentum density is
g  D  B.
25
(8.48)
31