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UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 3
Prof. Steven Errede
LECTURE NOTES 3
Conservation Laws (continued): Angular Momentum Associated with EM Fields
We have learned that the macroscopic EM fields have associated with them:
• EM Energy:
EM Energy Density:
1
1 2
u EM r , t o E 2 r , t
B r,t
2
o
Joules/m
3
1
1 2
B r , t d
EM Energy: U EM t u EM r , t d o E 2 r , t
v
v 2
o
1
S r,t
E r ,t B r ,t
• Poynting’s Vector:
Joules
Watts Joules
2
2
m -sec
m
o
• EM Linear Momentum:
EM Linear Momentum Density:
1
EM r , t o o S r , t 2 S r , t o E r , t B r , t
c
EM Linear Momentum:
1
pEM t EM r , t d 2 S r , t d o E r , t B r , t d
v
v
c v
kg
m 2 -sec
kg-m
sec
The macroscopic EM fields can additionally have associated with them:
• EM Angular Momentum:
EM Angular Momentum Density:
1
EM r , t r EM r , t o o r S r , t 2 r S r , t
kg
c
m-sec
o r E r , t B r , t
EM Angular Momentum:
1
LEM t EM r , t d r EM r , t d 2 r S EM r , t d kg -m 2
v
v
c v
sec
o r E r , t B r , t d
v
• Note that even STATIC E & B fields can carry net linear momentum pEM fcn t and
net angular momentum LEM fcn t as long as E B is non-zero! Again, at the microscopic
level, virtual photons associated with the macroscopic EM fields carry angular momentum
L as well as linear momentum p and (kinetic) energy E!
• Only when the EM field contributions are included for the total linear momentum pTot and the
total angular momentum LTot , i.e. pTot pmech pEM and LTot Lmech LEM is conservation of
linear momentum and conservation of angular momentum separately, independently satisfied.
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
1
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 3
Prof. Steven Errede
Griffiths Example 8.4
EM Angular Momentum Associated with a Long Solenoid & Coaxial Cylindrical Capacitor.
Consider a long solenoid of radius R and length L R , with n turns per unit length
n NTot L carrying a steady/DC current of I Amperes.
Coaxial with the long solenoid is a cylindrical (i.e. coaxial) capacitor consisting of two
long cylindrical conducting tubes, one inside the solenoid, of radius a < R and one
outside the solenoid, of radius b > R. The cylinders are free to rotate about the ẑ -axis.
Both cylindrical conducting tubes have same length L with a & b .
The inner (outer) conducting cylindrical tube carries electric charge +Q (–Q) uniformly
distributed over their surfaces, respectively.
When the current I in the long solenoid is slowly/gradually reduced (see e.g. Griffiths Example
7.8, p. 306-7), the cylindrical conducting tubes begin to rotate - the inner (outer) conducting
cylindrical tube rotating counter-clockwise (clockwise), respectively as viewed from above!!!
Long Solenoid
ẑ
R
z 12 L
a R b L
Binsol zˆ
a
LR
L
Q
b
Q
cap
Ein
ˆ
z 12
a, b ; L
z 12
I
z 12 L
Cylindrical
Capacitor
QUESTION: From where/how does the mechanical angular momentum originate?
ANSWER: The mechanical angular momentum imparted/transferred to the cylindrical
conducting tubes was initially stored in the EM fields associated with this system:
sol
Binside
o nIzˆ R, z L 2 and
The B -field associated with the long solenoid:
cap
The E -field associated with the cylindrical capacitor: Einside
2
Q 1
ˆ
2 o
a b,
z 2
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 3
Prof. Steven Errede
cap
sol
n.b. Einside
is non-zero for { a b .and. z 2 }, Binside
is non-zero for { R
.and. z L 2
L }. Hence, in the region { a R &
R b
z 2 } only:
Poynting’s Vector (Energy Flux Density):
1 Q 1
nQI
S r E r B r
ˆ zˆ
ˆ o nIzˆ
2 o
o 2 o
ˆ
Watts
2
m
ẑ
Very useful table:
ˆ ˆ zˆ
ˆ ˆ zˆ
zˆ ˆ ˆ
ˆ zˆ ˆ
zˆ ˆ ˆ
ˆ zˆ ˆ
ˆ ẑ
S
̂
̂
b
Cylindrical
Coordinates
nQI
Watts
S
ˆ
2 o m 2
a R
ˆ ẑ ˆ
EM energy/EM power circulates in the ̂ direction in the region a R & z 2 :
EM field linear momentum density: EM r o o S r o E r B r
nQI
nQI
EM r o o
ˆ o
ˆ
2 o
2
EM angular momentum density:
EM r r EM r EM r o
n.b. ˆ in cylindrical coordinates, thus:
kg
2
m -sec
E r B r
kg
m-sec
nQI
nQI
nQI
kg
EM r EM r ˆ o
ˆ o
ˆ ˆ o
zˆ
2
2
2
m-sec
zˆ
nQI
constant!!! EM r points in ẑ direction.
Note that: EM r o
2
We then compute the EM angular momentum LEM by integrating EM r over the volume v
corresponding to the region a R & z 2 :
R 2 z 12 nQI
LEM EM r d
zˆ d d dz
o
v
a 0 z 12
2
constant vector!
nQI
o
2
zˆ
dz
1 d d
a 0 z 2
d
= volume v of region { a R .and. z 2 }
nQI
ẑ R 2 a 2 12 o nQI R 2 a 2 zˆ
o
2
R
2
z 12
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
3
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 3
Thus, the EM angular momentum is LEM 12 o nQI R 2 a 2 zˆ
kg-m
2
Prof. Steven Errede
sec
When the current I in the long solenoid is slowly/gradually reduced, the changing magnetic field
d
induces a changing circumferential electric field, by Faraday’s Law: E d Bda
C
dt S
sol
ẑ
o nIzˆ R then
Since: Binside
R
for contour C1 R :
Bin o nIzˆ
IN
1
dI
1
dI
2
E
R
o n ˆ o n ˆ
1 2
dt
2
dt
2
S2 S1
C1 C
2
for contour C2 R :
I
E OUT R
1
dI
1
dI R 2
2
ˆ
n
R
n
ˆ
o
o
2
dt
dt
2
The {instantaneous} mechanical torque exerted on the inner conducting cylinder
(of radius a) by the tangential E -field E IN R is:
dI t
1
a ˆ o nQ
aˆ
dt
2
dI t 2
dI t 2
kg-m 2
N amech a, t o nQ
a
a zˆ N-m=
ˆ ˆ o nQ
dt
dt
sec 2
zˆ
d L t
But torque (by its definition) = time rate of change of angular momentum, i.e. N t
dt
The corresponding {increase} in the mechanical angular momentum the inner cylinder
acquires in the time it takes the current in the solenoid to decrease from I t 0 I to
mech
d Lmech t
kg-m 2
I t t final 0 is given by: N
d Lmech t N mech t dt
t
and:
dt
sec
mech
mech
mech
Lmech
t t final
final a t t final
La mech
d Lmech = Lmech
N amech t dt
final a t t final Linit a t 0 = L final a t t final
Linit a t 0 0
0
t
N amech r , t r FE r , t
a
r QE IN , t
a
0
t t final
t t final
dI t 2
1
N amech dt
a zˆ dt
Lamech Lmech
o nQ
final a t t final t 0
t 0
dt
2
t t final dI t
I t 0
1
1
1
o nQa 2
dt o nQa 2
dI t o nQa 2 zˆ 0 I
0
t
I
t
I
2
2
2
dt
mech
kg-m 2
1
2
ˆ
t
t
L
nQa
Iz
Thus: Lmech
final a
final
a
o
n.b. points in ẑ direction.
2
sec
the inner conducting cylinder {viewed from above} rotates counter-clockwise @ t t final !
4
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 3
Prof. Steven Errede
Similarly, the {instantaneous} mechanical torque exerted on the outer conducting cylindrical
tube (of radius b) by the tangential E -field E OUT R is:
N bmech r , t r FE r , t
b
r QE OUT , t
1
dI t R 2
ˆ
b
nQ
ˆ
b
0
dt
2
dI t 2
dI t 2
1
1
N bmech b, t o nQ
R ˆ ˆ o nQ
R zˆ
dt
dt
2
2
kg-m 2
N-m=
sec 2
zˆ
The corresponding {increase} in the mechanical angular momentum the outer cylinder
acquired in the time it takes the current in the solenoid to decrease from I t 0 I
to I t t final 0 is given by:
mech
mech
mech
Lmech
t t final
final b t t final
mech
Lb mech
d Lbmech = Lmech
t
t
L
t
0
L
t
t
final b
final
init b
final b
final
t 0 Nb t dt
Linit b t 0 0
0
t t final
t t final dI t
dI t 2
1
1
Lbmech Lmech
R zˆ dt o nQR 2 zˆ
dt
o nQ
final b t t final
t 0
t
0
2
dt
dt
2
I t t final 0
1
1
o nQR 2 zˆ
dI t o nQR 2 zˆ 0 I
I t 0 I
2
2
mech
kg-m 2
1
2
ˆ
t
t
L
nQIR
z
Thus: Lmech
final b
final
b
o
n.b. points in ẑ direction.
2
sec
outer conducting tube rotates clockwise at @ t t final viewed from above!
2
mech
mech
Lmech
L
L
Now note that, for t t final :
Lmech
final Tot
final a
final b
final i
i 1
1
1
1
Lmech
o nQIa 2 zˆ o nQIR 2 zˆ o nQI R 2 a 2 zˆ
final Tot
2
2
2
1
But this is precisely the EM field angular momentum, for t 0 : LEM o nQI R 2 a 2 zˆ
2
mech
1
i.e. LEM t 0 L final Tot t t final o nQI R 2 a 2 zˆ kg-m 2 sec
2
Thus, we explicitly see that angular momentum is conserved – angular momentum that was
originally stored in the EM fields of this device is converted to mechanical angular momentum as
the current in the long solenoid is slowly/steadily decreased!
Again, microscopically, angular momentum is carried by virtual photons associated with the
macroscopic E & B fields in this region of space. The angular momentum, as initially carried by
the EM fields in the region a R and z 2 is transferred to the two charged conducting
inner/outer cylindrical tubes as the current flowing conducting in the solenoid is slowly
decreased from I 0 , the cylinders acquiring non-zero mechanical angular momentum,
the total of which = initial EM angular momentum! Note also the time-reversed situation
(I increasing, i.e. from 0 I ) also does the time-reversed thing – because the EM
force/interaction (microscopically & macroscopically) obeys time-reversal invariance!!!
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
5
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 3
Prof. Steven Errede
The EM Field Energy Density uEM , Poynting’s Vector S , Linear Momentum Density EM
and Angular Momentum Density EM Associated with a Point Electric Charge qe
and a Point Magnetic Monopole g m
n.b. This is a static problem – i.e. it has no time dependence!
1 qe
1
Point electric charge at origin: E r
2 rˆ
4 o r
4 o
Point magnetic monopole e.g. located at d dzˆ :
B r o
4
ẑ
d dzˆ
d
x̂
r r rˆ
P observation/field point
Vectorially:
r r dzˆ r r dzˆ
r
ŷ
qe @
origin
r rrˆ
0 g m
gm
3 r
2 rˆ
r
4 0 r
r
gm
qe
3r
r
r r dzˆ & r r d
Law of Cosines: r 2 r 2 d 2 2rd cos
r r 2 d 2 2rd cos
o g m o
g m r dzˆ
B r
3 r
3
4 r
4 r 2 d 2 2rd cos 2
EM field energy density:
1 2 1
1
1
u EM r o E 2 r
B r o E r E r
B r B r
2
o
o
2
uEM
u EM
u EM
6
1 1
r
o
2 4 o
2
r 2
qe
gm
1 o
r r
2
2
3
2
2
4
r
o
r d 2rd cos
2
2
1
1 1 qe
r o
2
2 4 o r o
1
qe2
1
r o 2 2 4
2 16 o r o
gm
o 2
2
4 r d 2rd cos
2
r 3
2
g m2
o
but: o o 1 c 2
2
2
2
2
16 r d 2rd cos
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
u EM
r
Fall Semester, 2015
Lect. Notes 3
Prof. Steven Errede
2
g m2 c 2
g m2 c 2
1 1 2
qe
qe
2
32 2 o r 4 r 2 d 2 2rd cos 2 32 2 o r 4
1 dr 2 dr cos
1
n.b. for r d (also true for d = 0): u EM r
1 2
2
2
qe g m c
32 o r 4
1
2
2
Joules
3
m
1
S r
E r B r
Watts
2
o
m
g m r dzˆ
1 1 qe o
S r
3 r
3
2
2
2
4
4
r
o
o
r d 2rd cos
In spherical coordinates: r r 0 , zˆ cos rˆ sin ˆ , r zˆ r sin ˆ and r̂ ˆ ˆ , r rrˆ .
ẑ
r rrˆ
rˆ zˆ
n.b. rˆ zˆ sin ˆ
ŷ
is to rˆ and to zˆ
Poynting’s Vector:
̂
x̂
S r
S r
1
qe g m dr zˆ
r sin ˆ
16 2 o r 3 r 2 d 2 2rd cos 3 2
qe g m
d
16 2 o r 3 r 2 d 2 2rd cos 3 2
r zˆ
r sin ˆ
d
r zˆ
qe g m
16 o r 3 r 2 d 2 2rd cos
2
3
2
d
qe g m
16 o r 2 r 2 d 2 2rd cos 3 2
2
sin ˆ
Note that Poynting’s vector S r ˆ
i.e. EM energy is circulating in the ̂ (azimuthal)
direction in a static problem! Note also that S r vanishes when d = 0 (i.e. monopole g m is on top
of electric charge qe ) and also vanishes whenever rˆ
or anti- to zˆ
(then rˆ zˆ sin ˆ 0 )!
kg
EM Field Linear momentum density: EM r o o S r S r c 2 o E r B r 2
m -s
r sin ˆ
d
qe g m
d
qe g m
EM r o 2
r zˆ o 2
sin ˆ
3
16 r 2 r 2 d 2 2rd cos 2
16 r 2 r 2 d 2 2rd cos 3 2
Here again, note that EM r ˆ i.e. EM linear momentum density is circulating in the ̂
(azimuthal) direction in a static problem! Note also thatEM r vanishes when d = 0
and also vanishes whenever rˆ
or anti-
to zˆ (then rˆ zˆ sin ˆ vanishes)!
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
7
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 3
Prof. Steven Errede
EM Field Angular momentum density: EM r r EM r
d
qe g m
r rˆ zˆ
EM r o 2
3
16 r 2 r 2 d 2 2rd cos 2
d
qe g m
EM r o 2
rˆ rˆ zˆ
16 r r 2 d 2 2rd cos 3 2
kg
m-s
but: r rrˆ
Now: rˆ rˆ zˆ rˆ rˆ zˆ zˆ rˆrˆ rˆ rˆ zˆ zˆ rˆ cos zˆ rˆ cos rˆ cos ˆ sin ˆ sin
where: rˆ zˆ rˆ rˆ cos ˆ sin cos and: zˆ rˆ cos ˆ sin
d
qe g m
sin ˆ
EM r o 2
16 r r 2 d 2 2rd cos 3 2
u EM
EM Field Energy Density:
r
2
g m2 c 2
Joules
1 qe
2 2 2
2
2
3
2
32 o r r
m
2
cos
r
d
rd
1
S r
Poynting’s Vector:
EM Linear Momentum Density:
kg
m-s
sin ˆ
Watts
2
m
d
qe g m
sin ˆ
EM r o 2
16 r 2 r 2 d 2 2rd cos 3 2
kg
2
m -s
qe g m
d
16 o r 2 r 2 d 2 2rd cos
2
3
2
d
qe g m
sin ˆ
EM Angular Momentum Density: EM r o 2
16 r r 2 d 2 2rd cos 3 2
kg
m-s
The Total EM Field Energy: U EM u EM r d Joules because E r 0 and B r r
v
both diverge/are both singular (at r 0 and r r respectively) – so this is not a surprise!!!
However, the EM Power flowing through/crossing the enclosing surface S is zero (!!!):
PEM S r da
S
d
16 o
2
qe g m S
1
r 2 r 2 d 2 2rd cos
3
2
sin ˆ da 0 Watts !!!
But: S r 0 !!! The EM Power PEM 0 because da danˆ darˆ and ˆ rˆ 0 , i.e. ̂ is always
to rˆ !!! EM field energy associated with electric charge – magnetic monopole qe g m
system circulates! (i.e. is fully contained within enclosing surface S !!!)
8
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 3
Prof. Steven Errede
Total EM Field Linear Momentum:
d
pEM EM r d o 2
v
16
Note that:
S
v
qe g m
r 2 r 2 d 2 2rd cos
3
sin ˆ d
2
kg-m
sec
EM r da 0 because EM r ˆ daˆ rˆ EM Field Linear Momentum
circulates (i.e. EM field linear momentum is also fully contained within the enclosing surface S)!
2 r
d
r 2 dr sin 2 d d
ˆ
pEM o 2 qe g m
3
0 0 r 0
2
2
2
2
16
r r d 2rd cos
2 r
o d
dr sin 2 d d
ˆ
q
g
e m 0 0 r 0
3
2
2
2
16 2
r d 2rd cos
Let’s do the -integral first (trivial – get 2 ):
r
d
dr sin 2 d
ˆ
pEM o qe g m
3
0 r 0
2
2
2
8
r
d
rd
2
cos
Next, let’s do the r-integral:
2 2r 2d cos sin 2 d
d
pEM o qe g m
ˆ
2
0
8
4d 2 2d cos
r 2 d 2 2rd cos
4 r d cos sin 2 d
d
o qe g m
0
8
4 d 2 1 cos 2 r 2 d 2 2rd cos
r d cos sin 2 d
o d
qe g m 0 2
8 d 2
sin r 2 d 2 2rd cos
o
r d cos d
qe g m 0 2 2
8 d
r d 2rd cos
o qe g m 1 cos d ˆ
0
8 d
r
ˆ
r 0
r
ˆ
r 0
r
ˆ
r 0
Finally, we carry out the -integral:
pEM o qe g m sin 0 ˆ o qe g m 0 0 0 ˆ o qe g m ̂
8 d
8 d
8 d
qg
kg-m
pEM o e m ˆ
or:
8d
sec
Note that the EM field linear momentum pEM is finite as long as the electric charge-magnetic
monopole separation distance d 0 . When the electric charge qe is on top of the monopole g m ,
then d 0 and pEM becomes infinite.
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
9
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Total EM Field Angular Momentum: LEM EM r d
kg-m
v
LEM
o d qe g m
16 2
v
sin ˆ
rˆ cos zˆ
r r 2 d 2 2rd cos
3
d
2
Lect. Notes 3
o d qe g m
16 2
2
Prof. Steven Errede
sec
2
0
0
r
r 0
rˆ cos zˆ r 2 sin d d
r r 2 d 2 2rd cos
3
2
Let’s work this out in Cartesian coordinates: rˆ sin cos xˆ sin sin yˆ cos zˆ
2
o d qe g m 2 r sin cos xˆ sin sin yˆ cos zˆ cos zˆ r sin d d dr
LEM
3
16 2 0 0 r 0
r r 2 d 2 2rd cos 2
LEM
dqg
o e2 m
16
2
0
0
2
Now note that:
0
sin cos cos xˆ sin cos sin yˆ cos 1 zˆ r
2
r
r r 2 d 2 2rd cos
r 0
3
2
sin d d dr
2
2
cos d 0 sin d
2
0
sin 0
cos d
2
0
2
sin d 0
2
cos 0
0
Thus, the integrals over the -variable for both the xˆ and yˆ terms explicitly vanish,
x̂ - and ŷ - components of LEM both vanish due to manifest axial/azimuthal symmetry
(rotational invariance) of this problem about the ẑ -axis; only the zˆ-term remains:
LEM
dqg
o e2 m
16
2
0
0
cos 1 zˆ
2
r
r 0
r r d 2rd cos
2
2
3
r 2 dr sin d d
2
Now: 1 sin 2 cos 2 sin 2 cos 2 1
dqg
LEM o e2 m
16
2
2
0
0
sin 2 zˆ
r
r 0
r r 2 d 2 2rd cos
3
r 2 dr sin d d
2
Let’s do the -integral first - (trivial), since integrand has no explicit -dependence, get:
d q g 2 r
sin 2 zˆ
r 2 dr sin d
LEM o e m
3
r
0
0
2
2
2
8
r r d 2rd cos
Next, let’s do the r-integral – noting that:
r
r 0
2
r dr
r r 2 d 2 2rd cos
3
2
r
r 0
rdr
r
2
d 2 2rd cos
3
2
r cos d
d 1 cos r 2 d 2 2rd cos
2
1 cos
1 cos
cos
d
1
2
2
2
d 1 cos d 1 cos d d 1 cos d 1 cos 1 cos d 1 cos
10
r
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
r 0
UIUC Physics 436 EM Fields & Sources II
LEM
d qe g m
o
8
0
Fall Semester, 2015
qg
sin 3 d
zˆ o e m
8
d 1 cos
0
Lect. Notes 3
Prof. Steven Errede
sin 2 sin d
zˆ
1 cos
Finally, let’s do the -integral:
Let u cos and du d cos sin d , and sin 2 1 cos 2 1 u 2
For 0 u 1 and for u 1
Then: LEM
1 u du zˆ
qg
o e m
8
qg
o e m
8
qg
qg
1 2
o e m 2 zˆ o e m zˆ
u 2 u zˆ
8
4
u 1
u 1
2
but: 1 u 2 1 u 1 u
1 u
q g u 1 1 u 1 u
q g u 1
duzˆ o e m 1 u duzˆ
LEM o e m
u 1
u 1
8
8
1 u
LEM
u 1
LEM o qe g m zˆ
4
u 1
kg-m
2
sec
Note that the EM field angular momentum associated with the electric charge-magnetic monopole
system is independent of the qe g m separation distance, d !!!
Quantum mechanically LEM is quantized in integer (or even half-integer) units of h 2 ,
where h = Planck’s constant,
i.e.
LEM
h o
qe g m
2 4
qe o g m 2h !!!
However, recall the Dirac Quantization Condition (P435 Lecture Notes 18) which arose from
insisting on the single-valued nature of the electron’s wavefunction circling/orbiting a {presumed}
heavy magnetic monopole:
e o g m
eg m
nh (SI units)
oc2
Dirac Quantization Condition
These two formulae agree if 2 n or n 2 , thus if n = 1,2,3,… then 12 ,1, 32 , 2... and
LEM 12 ,1, 32 , 2...
h o
qe g m
2 4
kg-m
2
sec
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
11
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 3
Prof. Steven Errede
The EM Field Energy Density uEM , Poynting’s Vector S , Linear Momentum Density EM
and Angular Momentum Density EM Associated with a Point Electric Charge qe
and a Point/Pure Magnetic Dipole Moment m mzˆ
n.b. This is {again} a static problem – has no time dependence!
1 qe
1 qe
1) This time, we locate the point charge qe at d dzˆ : E r
2 rˆ
3 r
4 o r
4 o r
2) We locate the pure/point magnetic dipole moment m mzˆ at the origin:
m
8
B r o 3 3 mrˆ rˆ m m 3 r
in coordinate-free form
3
4 r
m
8
B r o 3 2 cos rˆ sin ˆ m 3 r
in spherical coordinates
3
4 r
ẑ
P r = observation/field point
r
Vectorially:
r rrˆ , r r rˆ and m mzˆ
qe
r
d dzˆ
d
m mzˆ
r r d dzˆ
r r d r dzˆ
r r d
ŷ
(origin)
Law of cosines: r 2 r 2 d 2 2rd cos
x̂
r r 2 d 2 2rd cos r
EM Energy Density:
1 2 1
1
1
u EM r o E 2 r
B r o E r E r
B r B r
2
o
o
2
2
2
qe2
1 o m 2
1 1
2
2
4 cos
uEM r o
sin
2 4 o r 2 d 2 2rd cos 2 o 4 r 6
3cos 2 1
2
12
m 2 8
ˆ
ˆ 3 8 2 3 2
2 cos rˆ sin cos rˆ sin r m r
3
r 3
3
2
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
u EM
u EM
u EM
Fall Semester, 2015
Lect. Notes 3
Prof. Steven Errede
2
2
qe2
1 o m 2
1 1
2
r o
2
6 3cos 1
2
2
2 4 o r d 2rd cos o 4 r
2
16 m 2
8 2 3 2
2
2
3
sin r m r
2 cos
3 r 3
3
3cos 2 1
2
2
qe2
1 o m 2
1 1
2
r
o
2
6 3cos 1
2
2
2 4 o r d 2rd cos o 4 r
2
16 m 2
8 2 3 2
2
3
3cos 1 r 3 m r
3 r3
1 qe2
1
r 2 2 2
32 o r d 2rd cos 2
m2
16 3
8
2
o 6 3cos 2 1
r 3cos 2 1 3 r r 6 3 r
3
3
r
2
n.b. If r d (or d 0) then for r 0 : uEM r
qe2
m2
2
o 2 3cos 1
4
32 r o
r
1
1
S r
E r B r
Poynting’s Vector:
o
1 1 qe o m
8
3
ˆ
ˆ
ˆ
r
mz
r
r
S r
2
cos
sin
o 4 o r 3 4 r 3
3
qe m
8
r 2 cos rˆ sin ˆ r 3 zˆ 3 r
3 3
16 o r r
3
1
2
but: r r d r dzˆ rrˆ dzˆ and zˆ rˆ cos ˆ sin then:
r 2 cos rˆ sin ˆ rrˆ dzˆ 2 cos rˆ sin ˆ
2r cos rˆ rˆ r sin rˆ ˆ 2d cos zˆ rˆ d sin zˆ ˆ
0
sin ˆ
ˆ
cosˆ
r sin 2d sin cos d sin cos ˆ r sin 3d sin cos ˆ
r zˆ rrˆ dzˆ zˆ r rˆ zˆ d zˆ zˆ r rˆ zˆ
and:
rrˆ rˆ cos ˆ sin r rˆ rˆ cos r rˆ ˆ sin r sin ˆ
ˆ
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
13
UIUC Physics 436 EM Fields & Sources II
r
Lect. Notes 3
Prof. Steven Errede
In spherical coordinates:
r̂ ˆ ˆ and zˆ cos rˆ sin ˆ
ẑ
Fall Semester, 2015
̂
ˆ
r̂
ŷ
̂
zˆ rˆ cos rˆ sin ˆ rˆ
cos rˆ rˆ sin ˆ rˆ sin ˆ
0
zˆ ˆ cos rˆ sin ˆ ˆ
cos rˆ ˆ sin ˆ ˆ cos ˆ
ˆ
0
x̂
S r
qe m
8
3 3 r sin 3d sin cos
16 o r r
3
1
2
4
3
r sin r ˆ
qe m
8 4 3
3 3 r 3d cos r r sin ˆ
16 o r r
3
1
2
But: r r 2 d 2 2rd cos
8
r 3d cos r 4 3 r
1
3
sin ˆ
S r
qe m
3
2
16 o
r 3 r 2 d 2 2rd cos 2
Note that:
1.) S r points in the ̂ -direction!!!
2.) S r vanishes (for r > 0) when: 1 3 dr cos 0 !!!
Watts
2
m
i.e. when: cos 13 dr 3dr equation for a line-curve (corresponds to a surface in !!)
3.) S r also vanishes (for r > 0) when: sin 0 i.e. at 0 and i.e. @ N/S poles!
4.) Note also that {here} S r does not vanish when d = 0 (i.e. when point electric charge
qe and point magnetic dipole moment m mzˆ are on top of/coincident with each other!!
1 qe m
Watts
5.) For r d (or d 0 , with r > 0): S r
5 sin ˆ
2
2
16 o r
m
EM Field Linear Momentum Density: EM r o o S r
8 4 3
r
d
3
cos
r r
o
3
sin ˆ
EM r o o S r
qm
3
2 e
16
r 3 r 2 d 2 2rd cos 2
Same comments made above for S r apply here for EM r .
14
kg
2
m -sec
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 3
Prof. Steven Errede
EM Field Angular Momentum Density: Em r r EM r where r rrˆ and r̂ ˆ ˆ
Very Useful Table:
8 4 3
r 3d cos r r
3
sin ˆ
EM r o 2 qe m
3
2
2
2
2
16
r r d 2rd cos
rˆ ˆ ˆ
ˆ ˆ rˆ
ˆ rˆ ˆ
ˆ rˆ ˆ
ˆ ˆ rˆ
rˆ ˆ ˆ
ˆ direction for 0 2
Note that for 0 r 3d cos that EM r points in:
ˆ direction for 2
Energy in EM Field:
U EM u EM r d
v
1
32 2
2
1
q
v eo r 2 d 2 2rd cos 2
2
m2
16 3
8 6 3 2
2
2
3
o 6 3cos 1
r 3cos 1 r r r d
3
3
r
U EM E diverges at r =d , B diverges at r =0
d r 2 dr sin d d
Power in EM Field crossing/passing through enclosing surface S: PEM S r da 0
S
because da danˆ darˆ but S points in the ̂ -direction. EM energy circulates within volume
v, enclosed by surface S !!!
Total Linear Momentum in EM Field:
pEM EM
v
8 4 3
r 3d cos r r
3
sin ˆ d
r d o 2 qe m v
3
3
2
2
2
16
r r d 2rd cos
Carrying out the 3-D volume integral is ~ tedious. We do not explicitly wade through this
here. The contributions from each of the 3 terms associated with the numerator in the integrand
are a.) finite, b.) logarithmically-divergent, and c.) zero respectively. Thus pEM here, and
also note that each of these 3 terms is proportional to o2 qe m , which is strongly divergent as
d
the electric charge qe – point/pure magnetic dipole m separation distance d 0 .
Note again that EM r da 0 i.e. EM field linear momentum circulates in ̂ -direction.
S
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
15
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 3
Prof. Steven Errede
Total Angular Momentum in EM Field:
8
r 3d cos r 4 3 r
3
sin ˆ d
LEM EM r d o 2 qe m
3
v
v
16
r 2 r 2 d 2 2rd cos 2
8 4 3
r 3d cos r r
3
r 2 dr sin 2 d d ˆ
o 2 qe m
3
v
16
r 2 r 2 d 2 2rd cos 2
8
r 3d cos r 4 3 r
3
dr sin 2 d d ˆ d r 2 dr sin d d
LEM o 2 qe m
3
v
16
r 2 d 2 2rd cos 2
We again choose to work this out in Cartesian coordinates, so sin ˆ rˆ cos zˆ
8
r 3d cos r 4 3 r
3
rˆ cos zˆ dr sin d d
LEM o 2 qe m
3
v
16
r 2 d 2 2rd cos 2
Then rˆ sin cos xˆ sin sin yˆ cos zˆ , thus:
rˆ cos zˆ sin sin cos cos xˆ sin cos sin yˆ cos2 zˆ zˆ sin
sin 2 cos cos xˆ sin 2 cos sin yˆ sin cos 2 zˆ sin zˆ
sin 2 cos cos xˆ sin 2 cos sin yˆ sin 1 cos 2 zˆ
sin 2 cos cos xˆ sin 2 cos sin yˆ sin 3 zˆ
Again, the integrals for the x̂ and ŷ components of LEM will contribute nothing when the
integrals of
2
0
2
... cos d and 0 ... sin d are carried out – only the
ẑ term survives the
-integration:
8
r 3d cos r 4 3 r
r
3
dr sin 3 d zˆ
LEM o qe m
3
0
0
r
8
r 2 d 2 2rd cos 2
Carrying out the remainder of the integration is ~ somewhat tedious, so we don’t explicitly
wade through this here, but interestingly enough, it yields a finite result (for d 0 ):
LEM o qe m 4 zˆ , which diverges as the electric charge qe – point magnetic dipole
8 d
moment m separation distance d 0 , which coincides with that of a real/physical electron
e
– i.e. a point electric charge e with point magnetic dipole moment of magnitude =
.
2me
16
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 3
Prof. Steven Errede
The main purpose of the above example, aside from its instructional use as academic exercise to
illustrate a simple static electromagnetic system in which energy, linear momentum and angular
momentum are all involved, is also to emphasize/underscore the important point that real/physical
electrons simultaneously have both a point electric charge and a point magnetic dipole moment –
both of which are necessary ingredients in order to be able to transfer {apparently} arbitrarily large
amounts of energy, linear and angular momentum to other such particles via the electromagnetic
interaction. Without the simultaneous presence of both an electric charge and a magnetic dipole
moment, transfer of linear & angular momentum could not occur!
It is not surprising that “classical” macroscopic electrodynamics “fails” here to correctly
quantitatively explain the physics operative at the microscopic scale – the domain of quantum
mechanics (and beyond – i.e. the structure of space-time itself at the smallest distance scales).
Despite more than 100 years of collective effort, since explicit discovery the electron by
J.J. Thompson in 1897, and the discovery of electron spin and the electron’s magnetic dipole
moment by first observed experimentally by O. Stern & W. Gerlach in 1922 and subsequently
explained theoretically by W. Pauli and S. Goudsmit and G. Uhlenbeck in 1925, today, we still
have gained no fundamental insight as to what precisely electric charge is, nor do we understand
the physics origins of intrinsic spin angular momentum (associated with either spin-½ fermions
{and the accompanying Pauli exclusion “principle”} or integer spin bosons, such as the photon
{and their accompanying “gregarious” nature at the quantum level – the opposite of that for
fermions!}, nor any fundamental explanation of the existence of the intrinsic magnetic dipole
moment(s) associated with all of the fundamental, point-like electrically-charged particles – three
generations of integer-charged point-like leptons e, , and six point-like quarks 2 3 : u , c, t
and 1 3 : d , s, b . Note that the W boson – the spin-1 electrically-charged mediator of the weak
interactions also has a magnetic dipole moment, as well as an electric quadrupole moment. These
same fundamental particles also interact via the weak interaction and thus have weak charges and
weak magnetic moments {the W boson also additionally has a weak quadrupole moment}. The
spin-½ quarks additionally interact via the strong interactions, and hence have strong “chromoelectric” charges (“red”, “green” & “blue”) as well as strong “chromo-magnetic” dipole moments.
Thus, point “charge” and point magnetic dipole moments, etc. associated with the all of the
fundamental particles we know and love transcends each of the individual forces, and in fact
points to/hints at a single common explanation. We do know that intrinsic spin and the
accompanying magnetic dipole moments of these particles are indeed manifestly fully-relativistic
phenomena, and thus “hint” at an explanation operative only at the smallest conceivable distance
scale, where the quantum behavior of space-time itself becomes manifest – i.e. the so-called
Planck distance scale, also known as the Planck length: LP GN c 3 1.61624 1035 meters,
with corresponding Planck time t P LP c GN c 5 5.39056 1044 seconds! It may seem
surprising that Newton’s gravitational constant GN enters here – however, Einstein’s general
theory of relativity tells us that there is an intrinsic link between the gravitational “force” as we
understand it macroscopically in the every-day world and the curvature of space-time!
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
17