Physics 460
1
Fall 2006
Susan M. Lea
Conservation laws in E&M
Last semester we noted that charge is conserved, and we were able to write both
integral
Z
Z
dQ
d
=
½d¿ = ¡ ~j ¢ ^n dA
dt
dt
and di¤erential
@½ ~ ~
+r¢j= 0
@t
equations to express charge conservation. Our next task is to do the same thing
with energy, momentum and angular momentum.
1.1
Conservation of energy in E&M
We have already found expressions for the energy stored in electric and magnetic
¤elds.The total energy in a volume V is
Z µ
¶
1
1 B2
U=
" 0 E2 +
d¿
2
2 ¹0
V
The work done by electromagnetic forces on a charge q moving through a displacement d~l is
dW
=
=
=
~ + ~v £ B)
~ ¢ d~l
q(E
³
´
~ + ~v £ B
~ ¢ ~v dt
q E
~ ¢ ~v dt
qE
As usual, the magnetic force does no work. Thus the rate at which work is done
is
dW
~ ¢ ~v
= qE
dt
Now consider a volume element with charge dq = ½d¿ : The work done on this
charge is
dW
~ d¿ = ~j ¢ E
~ d¿
= ½~v ¢ E
dt
Now we sum up over the whole volume to ¤nd the total rate of doing work:
Z
dW
~j ¢ E
~ d¿
=
dt
V
We can express this result in terms of the ¤elds alone by using the AmpereMaxwell law for ~j :
~
~j = 1 r
~ £B
~ ¡ "0 @E
(1)
¹0
@t
1
Thus
dW
=
dt
Z
V
~ ¢
E
Ã
~
1 ~
~ ¡ "0 @ E
r£B
¹0
@t
!
d¿
Now from the front cover:
³
´
³
´
³
´
~ ¢ E
~ £B
~ =B
~ ¢ r
~ £E
~ ¡E
~¢ r
~ £B
~
r
The last term is what we have in our integrand, and we can use Faradayzs law
~ £E
~ to get:
for r
Ã
!
³
´
³
´
~
@
B
~¢ r
~ £B
~ =B
~¢ ¡
~ ¢ E
~£B
~
E
¡r
@t
Then
dW
=¡
dt
Z "
V
´
~
~
1 ~ @B
1 ~ ³~
~ +E
~ ¢ "0 @ E
B¢
+
r¢ E£B
¹0
@t
¹0
@t
#
d¿
Using the divergence theorem on the middle term, we have:
¸
Z ·
Z
´
dW
1
1 @ ³~ ~´
@ ³~ ~´
1 ³~
~ ¢n
=¡
B ¢ B + "0
E ¢E
d¿ ¡
E £B
^ dA
dt
2 V ¹0 @t
@t
S ¹0
Or, for a ¤xed volume V ;
¶
Z µ
d 1
B2
"0 E 2 +
d¿
dt 2
¹0
=
=
Z
1
¹
ZS 0
1
¡
S ¹0
¡
³
´
dW
~ £B
~ ¢n
E
^ dA ¡
dt
Z
³
´
~ £B
~ ¢n
~j ¢ E
~ d¿ (2)
E
^ dA ¡
V
This equation should be interpreted as follows:
rate of change of em energy stored in the volume = energy ¢ow
into the volume - rate at which energy is converted to non-em forms
as the ¤elds do work..
Remember that ^n points outward from the volume. Mathematically:
Z
dU
dW
= ¡ S~ ¢ ^n dA ¡
dt
dt
s
where
³
´
~= 1 E
~£B
~
S
¹0
is the Poynting vector that describes energy ¢ow in the ¤elds.
~ ¢n
J/m2 ¢s. S
^ is positive when energy ¢ows out of the volume.
2
(3)
(4)
Its units are
We can convert equation (2) to di¤erential form in the usual way, using the
divergence theorem.
Z ·
µ
¶
³
´¸
@ 1
B2
~ +r
~ ¢ 1 E
~ £B
~
" 0 E2 +
+ ~j ¢ E
d¿ = 0
@t 2
¹0
¹0
V
for any volume V; so
µ
¶
³
´
@ 1
B2
2
~ +r
~ ¢ 1 E
~ £B
~ =0
"0E +
+ ~j ¢ E
@t 2
¹0
¹0
or, equivalently,
@uE M
~ ¢S
~ = ¡~j ¢ E
~
+r
(5)
@t
It is interesting at this point to consider energy ¢ow in a simple circuit with
a battery and a resistor. The battery supplies energy to the circuit, and the
resistor converts that energy to non-electro-magnetic forms (primarily thermal
energy). So energy is transported from the battery to the resistor. But how?
First letzs look at the ¤eld con¤guration. There are both electric and magnetic
¤elds. After the switch is closed the battery quickly establishes the necessary
charge distribution to create the potential distribution and electric ¤eld that
drive current through the resistor. The current loop produces a magnetic ¤eld,
as shown in the diagram. Then there is a non-zero Poynting ¢ux that transmits
¯ ¯
¯ ~¯
energy across the empty space between the battery and the resistor. ¯ S
¯ is
greatest near (but not in ) the wires, where the ¤elds are greatest.
3
1.2
1.2.1
Conservation of momentum
Newtonzs 3rd law in E&M
It is easy to show that Newtonzs third law holds in electrostatics: The force on
charge A due to charge B is equal and opposite to the charge on B due to A¡
this comes directly from Coulombzs law. But things get more interesting when
magnetic ¤elds are involved. The current density due to a charge q moving
with speed ~v is
~j = ½~v = q± (~r ¡ ~rq (t)) ~v
and this current density produces a magnetic ¤eld according to the Biot-Savart
Law. Provided that v ¿ c; we may write
~ (~r ) =
B
=
=
Z ~ ~
j£R 0
~ = ~r ¡ ~r0
d¿
where R
R3
Z
~
¹0
~v £ R
q ± (~r ¡ ~rq (t))
d¿
3
4¼
R
~
¹0 ~v £ R
~ = ~r ¡ ~rq (t)
q
where R
4¼ R3
¹0
4¼
Now consider two charges moving at right angles, as shown:
~ 1 at the position of charge 2, with a
Charge 1 produces a magnetic ¤eld B
resulting force exerted on 2 by 1
F~o n
2 by 1
~1
= q2 ~v 2 £ B
that points downward. But the magnetic ¤eld produced by charge 2 is zero at
~ = 0; and so the force
the (instantaneous) position of charge 1 because ~v £ R
exerted on 1 by 2 is zero. This result clearly violates Newtonsz third law. It
then appears that momentum is not conserved, as
d~p2
= F~o n
dt
2 by 1 ;
d~p1
= F~on
dt
1 by 2
=0
But momentum conservation is one of the most fundamental principles in physics.
The problem is resolved by realising that there is momentum stored in the ¤elds
that is also changing.
It should not surprise us that momentum is stored in the ¤elds if energy is.
Consider a particle moving with velocity ~v : It has kinetic energy K = 12 mv 2
4
and momentum ~p = m~v : The ratio of the magnitudes is
K
v
=
j~p j
2
Later this semester, we will see that as the particlezs speed approaches the speed
of light, the ratio becomes
Ã
!
r
µ
¶
K
°¡1 c
v2 ³ c ´
=c
=c 1¡ 1¡ 2
! c as v ! c
j~p j
°
v
c
v
We have been using a classical ¤eld model of electricity and magnetism, but we
could also model the ¤elds using a particle picture. The photon is a massless
particle that travels at speed c; and so for the photon K=p = c: Thus we might
guess that for the EM ¤eld, the momentum ¢ux is given by
´
~
S
1 ³~
~
=
E £B
c
c¹0
and then the momentum density
³
´
~
~ EM = f lux = S = "0 E
~£B
~
P
2
c
c
(6)
Our next task is to prove this conjecture.
1.2.2
The Maxwell Stress tensor
We begin by looking at the force per unit volume f~ in a system of charged
particles and ¤elds. The charge in a volume d¿ is dq = ½d¿ ; and then
³
´
~ + ~v £ B
~
f~d¿ = dF~ = ½d¿ E
f~
~ + ~j £ B
~
= ½E
Now we use Maxwellzs equations to eliminate ½ and ~j :
~ ¢E
~
½ = "0 r
and ~j from equation (1):
³
´
~ ¢E
~ E
~ +
f~ = "0 r
Letzs work on the last term.
product matters!
´
@ ³~
~
E £B
@t
=
=
Ã
~
1 ~
~ ¡ "0 @E
r£B
¹0
@t
!
~
£B
Remember: the order of the vectors in a cross
~
~
@E
~ +E
~ £ @B
£B
@t
@t
³
´
~
@E
~ +E
~ £ ¡r
~ £E
~
£B
@t
5
(Faradayzs law)
Thus
³
´
³
´
~
@E
~ = @ E
~ £B
~ +E
~ £ r
~ £E
~
£B
@t
@t
Now we may expand the triple cross product, to get
h
³
´i
~£ r
~ £E
~
E
i
@
Em
@x l
=
"ijk E j " klm
=
(±il ±jm ¡ ±im ±jl ) E j
@
Em
@xl
@
@
Ej ¡ Ej
Ei
@x i
@x j
1 @ 2 ³~ ~´
=
E ¡ E ¢ r Ei
2 @xi
³
´
³
´
~ £B
~ £B
~ = ¡B
~ £ r
~ £B
~ : Putting all this
with a similar expansion for r
back into f~; we get
·
¸
·
¸
³
´
´
1 1 ~ 2 ³~ ~ ´ ~
1 ~ 2 ³~ ~ ´ ~
@ ³~
~
~
~
~
~
f =
"0 r ¢ E E ¡
rB ¡ B ¢ r B ¡ "0
rE ¡ E ¢ r E ¡ " 0
E £B
¹0 2
2
@t
µ 2
¶
´
@ ³~
B
~ ¡r
~
= ¡"0
E£B
+ "0 E 2
@t
¹0
h³
´
³
´ i
h³
´
³
´i
~ ¢E
~ E
~+ E
~ ¢r
~ E
~ + 1
~ ¢r
~ B
~+B
~ r
~ ¢B
~
+"0 r
B
¹0
³
´
~ r
~ ¢B
~ ; which is identically zero, to make
In the last line I added the term B
=
Ej
~ and B:
~ Tidying up a bit, we get
the last two terms symmetrical in E
h³
´
³
´ i 1 h³
´
³
´i
~
@S
~
~ ¢E
~ E
~+ E
~ ¢r
~ E
~ +
~ ¢r
~ B
~ +B
~ r
~ ¢B
~
f~ = ¡" 0¹0
¡ ru+"
r
B
0
@t
¹0
The ¤rst term is, from equation (6),
¡
~
~ EM
1 @S
@P
=
¡
c2 @t
@t
The remaining terms may be written as the divergence of a tensor T ij : A tensor
as a total of 3£3 = 9 components, labelled with the two indices i and j; each
of which ranges over the values 1-3, 1 = x; 2 = y; 3 = z: It is much easier to
6
write equations in index notation when we have tensors. So
·
¸
h³
´
³
´ i
h³
´
³
´i
@
@
E2
B2
~ ¢E
~ Ei + E
~ ¢r
~ Ei + 1
~ ¢r
~ Bi + Bi r
~ ¢B
~
Tij = ¡
"0
+
+ "0 r
B
@xj
@x i
2
¹0
¹0
·
¸
·
¸
·
¸
2
2
@
E
B
@E j
@
1
@
@Bj
= ¡±ij
"0
+
+ "0
Ei + Ej
Ei +
Bj
Bi + B i
@x j
2
¹0
@x j
@x j
¹0
@x j
@xj
·
¸
µ
¶
2
2
@
E
B
@
@
1
= ¡±ij
"0
+
+
("0 E j E i ) +
Bj Bi
@x j
2
¹0
@xj
@x j ¹0
·
·
¸
¸
@
E2
B2
1
=
¡± ij "0
+
+ " 0 Ej E i +
Bj Bi
(7)
@x j
2
¹0
¹0
We may write the components conveniently as a matrix. The ¤rst term contributes only when i = j; ie on the diagonal of the matrix.
0
B
B
Ã
! B
B
T =B
B
B
@
¡
¢
"0
2
E
E y2 ¡ E z2 ¢
2
¡x ¡
1
2
+ 2¹ Bx ¡ By2 ¡ Bz2
0
" 0 Ex E y +
1
B B
¹0 x y
"0 E xE z +
1
B B
¹0 x z
1
¹ 0 Bx By
" 0 Ex E y +
¡ 2
¢
"0
2
2
E
y ¡ Ex ¡ E z ¢
2
¡
+ 2¹1 By2 ¡ Bx2 ¡ Bz2
0
"0 E y E z +
1
B B
¹0 y z
Note that this tensor is symmetric, that is:
"0 E xE z +
1
¹ 0 Bx Bz
"0 E y E z + ¹1 By Bz
0
¡ 2
¢
"0
2
2
z ¡ Ex ¡ E y ¢
2 E
¡
+ 2¹1 Bz2 ¡ Bx2 ¡ By2
0
(8)
Tij = T ji
Finally we write the force density equation as:
fi = ¡
and integrating over the volume:
Z
Z
fi d¿ = ¡
V
V
~E M ; i
@P
@T ij
+
@t
@x j
~ EM ; i
@P
d¿ +
@t
Z
V
(9)
@T ij
d¿
@xj
Using the divergence theorem, we convert the last term to a surface integral:
Z
Z
~ EM ; i
@P
Fi = ¡
d¿ +
T ij nj dA
(10)
@t
V
S
or equivalently
or
Z
Z
d
Ã
! ~
~
~
F =¡
P EM d¿ +
T ¢ dA
dt
S
Z
Z
d
Ã
! ~
~ EM d¿ =
P
T ¢ dA ¡ F~
dt
S
We may interpret this as:
7
(11)
1
C
C
C
C
C
C
C
A
Rate of change of stored EM momentum in the volume=-force
exerted by ¤elds in the volume on charges in the volume - force
exerted by ¤elds in the volume on the surface of the volume.
The physical interpretation of the tensor component T ij is:
T ij is the force per unit area in the ith direction exerted on an
area element with normal in the jth direction.
The diagonal elements such as T xx are pressures (normal force per unit area)
while the o¤-diagonal components are shears.
Letzs see what happens to equation 11 when there is no time dependence.
We have
Z
Ã
! ~
~
F =
T ¢ dA
S
This means that we can calculate the force equally well by calculating the force
per unit volume, and integrating over the volume, or the force per unit area and
integrating over the area. For example, Suppose we have two point charges
separated by a distance D. The volume integral is just Coulombzs law:
F~ =
q1 q2
r^
4¼"0 D2
But we should be able to get the result by integrating over any surface completely surrounding one of the charges.We need the total electric ¤eld due to
both charges:
µ
¶
q1
q2
~ =E
~1 + E
~2 = 1
E
r
^
+
r
^
1
2
4¼"0 r21
r22
Letzs put the x¡axis along the line joining the two charges, with the origin half
way between them.
The easiest surface to use is a rectangular box, with ¢at surfaces at x = 0;
x ! 1; y ! §1 and z ! §1: This box encloses charge 2, but if we choose
8
x ! ¡1; that box encloses charge 1. The ¤elds go to zero fast enough that the
only non-zero contribution is from the surface at ¤nite x: (The ¤elds decrease
like 1/distance2 ; so the product of two ¤eld components goes like 1/distance4 ;
but the area only increases as distance2 ; so the product! 0 as 1/distance2 : )
Since the box enclosing charge one has outward normal n
^ = x^ while the box
enclosing charge 2 has outward normal ^n = ¡^x; we can see immediately that the
forces each charge exerts on the other are equal and opposite. Letzs calculate
the force on charge 2.
On the plane
0
1
µ
¶
µ
¶
q1
D
q2
¡D
C
~ = 1 B
E
x^ + y y^ + z z^ + ³¡ ¢
x^ + y y^ + z^z A
@ ³¡ ¢
´3=2
´3=2
4¼"0
2
2
2
2
D
D
+ y2 + z 2
+ y2 + z 2
2
2
Fx
= ¡
Z
+1
dy
¡1
Z +1
Z
+1
¡1
Z +1
dzTxx
¢
"0 ¡ 2
E x ¡ E y2 ¡ E 2z
2
¡1
¡1
Z +1
Z +1
µ
¶
E2
2
= ¡
dy
dz"0 E x ¡
2
¡1
¡1
0 h
i2 1
Z +1
Z +1
q 1 D=2
q2 D=2
¡ (D 2=4+y 2+z2) 3=2
1
2
2
2 3=2
B
C
= ¡
dy
dz @ (D =4+y
h +z )
i2
A
2
q 1r^1
q 1r^2
1
(4¼) "0 ¡1
¡1
¡ 2 D 2=4+y 2+z2 + D 2=4+y 2+z 2
= ¡
dy
dz
We may use polar coordinates on the plane, where y 2 + z 2 = s 2
Z
2¼
0
Z
+1
sdsdµ
0
q21 D 2 =4
(D2 =4 + s2 )3
=
=
and
Z
0
2¼
Z
0
+1
sdsdµ
q21
¼q12 D2
4
Z
+1
du
D2
where
u
=
+ s2
3
4
D 2=4 u
µ
¶¯1
¼q12 D2
1 ¯
2¼q12
¡ 2 ¯¯
=
4
2u
D2
D 2=4
2
(D2 =4 + s 2 )
=
¼q12
µ
¶ ¯1
1 ¯¯
4¼q21
¡ 2 ¯
=
u
D2
D 2 =4
Thus the ¤rst term in each of the brackets sum to give
2¼q12
1 4¼q21
¡
=0
D2
2 D2
Of course we expect this because the result should involve only the product of
the two di¤erent charges. The same thing happens with the last term in each
9
of the two brackets. Thus we are left with the cross terms.
Z 2¼ Z +1
Z
¡q1 q2 D 2=2
D 2 1 du
D2
sdsdµ
=
¡¼q
q
where
u
=
+ s2
1
2
2 D 2=4 u3
4
(D2 =4 + s2 )3
0
0
µ
¶¯ 1
D2
1 ¯
= ¡¼q1 q2
¡ 2 ¯¯
2
2u
D 2=4
q1 q2
= ¡4¼ 2
D
and
Z 2¼ Z
0
+1
sdsdµ
0
2q1 q2 ^r1 ¢ r^2
Z
=
2
(D2 =4 + s2 )
2¼
0
=
q1 q2
Z
Z
+1
2sdsdµ
0
2¼
0
=
q1 q2
Z
2¼
0
=
q1 q2
Z
0
2¼
¡
¢
q1q2 r2y + rz2 ¡ r2x
2
(D2 =4 + s2 ) r 2
¡ 2
¢
+1
s ¡ D 2=4
2sdsdµ
3
(D2 =4 + s2 )
0
¡
¢
Z +1
s2 + D2 =4 ¡ D2 =2
2sdsdµ
3
(D2 =4 + s2 )
0
"
#
Z +1
1
D2 =2
2sdsdµ
2 ¡
3
(D2 =4 + s2 )
(D 2 =4 + s 2)
0
Z
We havre already done the second integral, so letzs work on the ¤rst:
Z
2¼
0
Z
+1
2sdsdµ
0
1
(D2 =4 +
s 2 )2
= 2¼
Z
1
D 2=4
1
8¼
du = 2
2
u
D
Thus
Fx
=
=
µ
q1 q2
¡
2
(4¼) "0
q1 q2
4¼"0 D 2
4¼
1
¡ 2 ¡
D
2
·
8¼
8¼
¡ 2
D2
D
¸¶
as expected.
Finally letzs look at Fy :
Fy
= ¡
= ¡
= ¡
Z
+1
dy
¡1
Z +1
dy
¡1
1
2
(4¼) "0
Z
+1
dzT yx
¡1
Z +1
Z
dz" 0E x E y
¡1
+1
dy
¡1
Z
+1
dz
¡1
"
£
q1 D=2
(D2 =4 + y 2 + z2 )
"
q1 y
3=2
3=2
(D 2 =4 + y 2 + z 2 )
10
¡
¡
q2 D=2
3=2
(D2 =4 + y 2 + z 2 )
#
q2 y
(D2 =4 + y 2 + z2 )
3=2
#
The terms look like
Z
+1
dy
¡1
Z
+1
dz
¡1
q12yD=2
3
(D2 =4 + y 2 + z 2 )
=0
because the integral over y has an odd integrand integrated over an even interval.
The cross terms are zero for the same reason.
Z
+1
dy
¡1
Z
+1
dz
¡1
q1 q2 yD=2
(D2 =4 +
y 2 + z 2 )3
=0
Thus the force only has an x¡component, as expected.
1.2.3
Conservation of momentum
Letzs return to the full force law (11), Since the force acting on the charges in
the volume changes their mechanical momentum, we may rewrite the equation
as:
Z ³
Z
´
Ã
! ~
d
~ EM + P
~ M ec h d¿ =
P
T ¢ dA
dt
S
or, in di¤erential form,
´
!
@ ³~
~ Me ch = r
~ ¢Ã
PEM + P
T
@t
1.3
(12)
Angular momentum
If the ¤elds carry momentum, they must also carry angular momentum:
~`EM
~ EM
= ~r £ P
³
´
~ £B
~
= " 0~r £ E
As an example, consider a con¤guration of charges and currents as follows.
We have two concentric cylindrical shells of charge with radii a and c > a. The
11
cylinders have equal but opposite charges, so the surface charge densities are ¾
and ¡¾c=a respectively. There is an electric ¤eld
~ = ¾a s^
E
" 0s
for c > s > a. A solenoid with radius b; c > b > a; carries azimuthal current I;
so a magnetic ¤eld
~ = ¹0 nI z^
B
exists in the region s < b: Thus for a < s < b we have both electric and
magnetic ¤elds.
2
~ EM = "0 ¾a ^s £ ¹ nI z^ = ¡ ¾a ¹ nI ^µ
P
0
"0 s
2s 0
and the angular momentum density about the origin is:
³
´
~`EM = ~r £ ¡ ¾a ¹ nI ^µ
0
s
¾a
= ¡ ¹0 nI (s^s + z z^) £ ^µ
s
³
z ´
= ¡¾a¹0 nI z^ ¡ s^
s
The total angular momentum contained in a cylinder of height h is
Z
~`EM d¿
~LEM =
Z
Z
Z
n
³
z ´o
sds ¾a¹0 nI z^ ¡ s^
s
¡h=2
0
a
¡2
¢
2
= ¡¼¾a¹0 nI b ¡ a h z^
=
h =2
dz
2¼
dÁ
b
~ ! 0 and L
~ E M ! 0 also. Since there are no
Now if we turn the current o¤, B
~
~ produces an
external torques, L should be conserved. Aha, but the changing B
~
azimuthal E ; by Faradayzs law, and the resulting torque on the cylinders causes
them to rotate. We have
d
2¼sE µ = ¡ ¼s2 B
dt
s
dI
E µ = ¡ ¹0 n
2
dt
and the torque on a height h of the cylinder of radius a < b is
~¿
=
a¾ (2¼ah) E µ ^z
~ c y lin de r
a
dI
dL
¡2¼¾a2 h ¹0n z^ =
2
dt
dt
Thus the inner cylinder gains angular momentum:
Z t2 ~
dLcy lin d er
t
~Linn e r =
dt = ¡¼¾a3 h¹0 n Ijt21 z^
dt
t1
=
=
¼¾a3 h¹0 nI z^
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The initial ¢ux through a circle of radius c is ¼b2 B; since B is zero for b < s < c;
so we have
b2
dI
E µ (c) = ¡ ¹0 n
2c
dt
and, since the outer cylinder has a negative charge density,
giving
³ a´
b2
dI
~¿ = 2¼ ¾
c2 h ¹0 n z^
c
2c
dt
~ ou te r = ¡¼achb 2 ¹0nI z^
L
The total angular momentum gained by the two cylinders is
¡
¢
~ m e ch = ¡¼¾a¹0 nIh^
~ EM
L
z b2 ¡ a2 z^ = L
The total electromagnetic angular momentum that disappears exactly equals
the total mechanical angular momentum gained by the cylinders. Angular
momentum is conserved.
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