Download Lecture Notes 18: Magnetic Monopoles/Magnetic Charges; Magnetic Flux Quantization, Dirac Quantization Condition, Coulomb/Lorentz Force Laws for Electric/Magnetic Charges, Duality Transformations

UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 18
Prof. Steven Errede
LECTURE NOTES 18
MAGNETIC MONOPOLES – FUNDAMENTAL / POINTLIKE MAGNETIC CHARGES
No fundamental, point-like isolated separate North or South magnetic poles – i.e. N or S
magnetic charges have ever been conclusively / reproducibly observed. In principle, there is no
physical law, or theory, that forbids their existence. So we may well ask, why did “nature” not
“choose” to have magnetic monopoles in our universe – or, if so, why are they so extremely rare,
given that many people (including myself) have looked for / tried to detect their existence…
If magnetic monopoles / fundamental point magnetic charges did exist in nature, they would
obey a Coulomb-type force law (just as electric charges do):
⎛μ ⎞g g
Fm ( r ) = g mtest Bm ( r ) = ⎜ o ⎟ m 2 m rˆ
⎝ 4π ⎠ r
test
src
( )
Source point S ′ r ′
Where:
g msrc = magnetic charge of source particle
r = r − r′
g msrc
g mtest = magnetic charge of test particle
g mtest
ẑ
g m ≡ + g ( ≡ North pole )
r′
r
g m ≡ − g ( ≡ South pole )
SI units of magnetic charge g = Ampere-meters (A-m)
ϑ
ŷ
x̂
Field / observation point P ( r )
Units check:
2
⎛ μo ⎞ g m
Fm ( Newtons ) = ⎜
⎟ 2
⎝ 4π ⎠ r
μo = 4π × 10−7 Newtons
Ampere 2
(N A )
2
2
2
⎛ N ⎞ ( A − m) ⎛ N ⎞ A − m
=N
=
Newton = N = ⎜ 2 ⎟
⎜ 2 ⎟
2
2
⎝A ⎠ m
m
A
⎝
⎠
2
Then (the radial) B -field of a magnetic monopole is:
src
N
⎛μ ⎞g
)
Bm ( r ) = ⎜ o ⎟ m2 rˆ (SI units = Tesla =
A−m
⎝ 4π ⎠ r
⎛ N ⎞ A2 − m
N
N
=⎜ 2 ⎟
=
Units check: Tesla =
A − m ⎝ A ⎠ m2
A−m
1 Tesla ≡ 1
1T=1
Newton
Ampere − meter
N
A−m
gm = Ampere-meters (A-m)
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
1
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 18
Prof. Steven Errede
The magnetic flux associated with a magnetic monopole with magnetic charge g msrc is:
(
)
⎛ μo ⎞ g msrc
src
2
2
src
ˆ
B
r
nda
=
i
(
)
⎜
⎟ 2 4π r = μo g m or: Φ m = μo g m (Webers, or Tesla-m )
∫S m
⎝ 4π ⎠ r
rˆ = r , and r = r because the source charge g msrc is located at the origin ϑ :
Φm ≡
SI units of magnetic flux:
Tesla-meters2 ( Φ m = B i A )
⎛ N −m⎞
1 (T–m2) = 1 ⎜
⎟ = 1 Weber (Wb)
⎝ A ⎠
Units check:
N
N − m⎫ N
N −m
⎧
im 2 =
T − m 2 ⎨=
⎬= 2 i A −m =
A ⎭ A
A
⎩ A−m
gm
nˆ = rˆ
ẑ
r
area
element
da
ŷ
ϑ
x̂
We know that electric charge is quantized in units of e = 1.602x 10−19 Coulombs. Similarly
we would expect magnetic charges to also be quantized (if they do indeed exist in nature).
We know that magnetic flux Φ m is quantized - the quantum (i.e. the smallest unit) of magnetic
flux is one flux quantum:
−34
⎛ h ⎞ 6.626 ×10 Joule − sec
Φ om ≡ ⎜ ⎟ =
= 4.136 × 10−15Webers
−19
⎝ e ⎠ 1.602 × 10 coulombs
Where: h = Planck’s constant = 6.626 x 10−34 Joule-sec
⎛ N ⎞ 2 ⎛ N −m⎞
Units check: Φ om = Webers = Tesla − m 2 = ⎜
⎟i m = ⎜
⎟
⎝ A ⎠
⎝ A− m ⎠
⎛ h ⎞ Joule − sec Joules Newton − meters ⎛ N − m ⎞
And: ⎜ ⎟ =
=
=
=⎜
⎟
Amp
Amp
⎝ e ⎠ coulombs
⎝ A ⎠
− g mo
+ g mo
⎛h⎞
1Φ om = ⎜ ⎟ = μo g mo
⎝e⎠
Then the smallest integer unit of quantized magnetic charge g mo is:
Φ om = μo g mo or: g mo =
1
μo
Φ om =
1 ⎛ h ⎞ 4.136 ×10−15Wb
= 3.2914 × 10−9 Ampere-meters (A-m)
⎜ ⎟=
−7
2
μo ⎝ e ⎠ 4π ×10 N A
Now, it’s possible that magnetic monopoles could exist with integer multiples of this smallest /
quantized amount of magnetic charge g mo , i.e. g mn = ng mo where n = integer = ±1, ±2, ±3, . . . .
and:
+ g mo = N ( North Pole) and − g mo = S ( South Pole) .
⎛h⎞
⎛h⎞
Then if Φ om = μo g mo = ⎜ ⎟ , then Φ nm = μo ng mo = μo g mn = n ⎜ ⎟ i.e. Φ nm = nΦ om .
⎝e⎠
⎝e⎠
2
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
⎛ h ⎞
⎛h⎞
Fall Semester, 2007
Lecture Notes 18
Prof. Steven Errede
1
μo g mn = n ⎜ ⎟ ⇒ eg mn = n ⎜ ⎟ but:
= ε o c 2 where c = 3 x 108 m/s (speed of light)
μ
μ
e
⎝ ⎠
o
⎝ o⎠
2
nε o hc
n
⎛ h ⎞
⎛ h ⎞ 2 n ⎛ 4πε o c ⎞
n
= 4πε o ⎜
Defining ≡ ⎜
⎟ then g m =
⎟c = ⎜
⎟ ec
e
2
2 ⎝ e2 ⎠
⎝ 2π ⎠
⎝ 2π ⎠
And:
≡ 1 α em
e2
1
n⎛ 1 ⎞
α
≡
=
ec
(A-m)
and
the
fine-structure
constant
i.e. g mn = ⎜
⎟
em
4πε o c 137.036...
2 ⎝ α em ⎠
2
1
(dimensionless quantity) and numerically, 2α em =
137.036 68.5
Then: g mn = 68.5n ( ec ) A-m
(n = ±1, ±2, ±3, . . . .)
Thus, we see that the relative strength of magnetic monopole (e.g. North-South pole) attraction is
huge in comparison to that associated with electric monopole (e.g. e+-e−) attraction:
(n)
m
Fm( n ) F
=
Fe
Fe
⎛ μo ⎞ g mn 2
2
⎜
⎟
r2
⎛ g mn ⎞
1
2
4π ⎠
⎝
2 2 2
=
= ε o μo ⎜
⎟ = 2 68.5 n c = ( 68.5n )
⎛ 1 ⎞ e2
c
⎝ e ⎠
⎜
⎟ r2
⎝ 4πε o ⎠
4700n 2
(n = ±1, ±2, ±3, . . . .)
Force of attraction between two
Force of magnetic attraction between N-S monopoles
(opposite) electric charges.
⎛h⎞
If Φ om = ⎜ ⎟ magnetic flux quantum (= 4.136 x 10-15 Wb)
⎝e⎠
Φ om
=
Then:
Φ oE
But:
∴
∫
∫
S
S
ˆ
B inda
ˆ
E inda
=
⎛ go ⎞ 1 ⎛ go ⎞
μo g mo
= ε o μo ⎜ m ⎟ = 2 ⎜ m ⎟
e εo
⎝ e ⎠ c ⎝ e ⎠
⎛ 1 ⎞
⎛ g mo ⎞
g = 68.5ec or: ⎜
⎟c
⎟ = 68.5c = ⎜
⎝ e ⎠
⎝ 2α em ⎠
o
m
where: α em =
e2
4πε o c
=
1
137.036...
Φ om 1 ⎛ g mo ⎞ 1 ⎛ 1 ⎞
1⎛ 1 ⎞
1
=
=
=
= 68.5
c
⎜
⎟
⎜
⎟=
⎜
⎟
2
2
o
c
Φ E c ⎝ e ⎠ c ⎝ 2α em ⎠
c ⎝ 2α em ⎠ 2α em c
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
3
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 18
Prof. Steven Errede
We can rearrange this latter relation to obtain the electric flux quantum:
⎛ e2 ⎞ ⎛ h ⎞
⎛ eh ⎞
h
⎛h⎞
Φ oE = 2α em cΦ om = 2α em c ⎜ ⎟ = 2 ⎜⎜
e−
⎟⎟ c ⎜ ⎟ = 2 ⎜
⎟ where ≡
4
e
πε
2
π
4
πε
c
e
⎝ ⎠
o ⎠
⎝
o
⎝
⎠ ⎝ ⎠
⎛
⎞
eh
⎟ = ⎛ 4π e h ⎞ = ⎛⎜ e ⎞⎟
Φ oE = 2 ⎜
1Φ oE = e
e+
εo
⎜⎜ 4πε h
⎟⎟ ⎜⎝ 4π ε o h ⎟⎠ ⎝ ε o ⎠
o
2π ⎠
⎝
Numerically:
2
1.602 ×10−19 Coulombs
−8 N − m
Φ oE = ⎛⎜ e ⎞⎟
=
Electric
Flux
Quantum
=
=
1.810
×
10
⎝ εo ⎠
Coulombs 2
Coulomb
8.85 × 10−12
2
N −m
−15
2
Φ om = ( μo g mo ) = h
e = Magnetic Flux Quantum = 4.136 x 10 Wb (T–m )
(
)
( )
Φ om
1
n
=
ec Ampere-meters where n = ±1, ± 2, ± 3
and g mn =
o
Φ E 2α em c
2α em
Gauss’ Law:
Φ om = μo g mo =
Φ oE =
4
⎛ Fm ⎞
h
⎛
⎛ N ⎞
2⎞
= 4.136 × 10−15Wb ⎜ = T − m 2 = ⎜
⎟i Area = Bm i Area
⎟−m ⎟ =⎜
e
⎝ A−m ⎠
⎝
⎠ ⎝ gm ⎠
⎛⎛ N ⎞
⎞ ⎛F
= 1.810 × 10−8 ⎜ ⎜ ⎟ − m 2 ⎟ = ⎜ E
εo
⎝⎝ c ⎠
⎠ ⎝ e
e
⎞
⎟i Area = E i Area
⎠
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 18
Prof. Steven Errede
The Dirac Quantization Condition
In 1931, Paul Adrian Maurice Dirac showed (see P.A.M. Dirac, Proc. Roy. Soc., London, Ser.
A133, 60, (1931)) that quantization of electric charge (i.e. why e is e could be explained if
magnetic monopoles existed, because then:
e ∗ μo g m =
eg m
= nh (SI units)
ε oc2
Dirac Quantization Condition
e = electric charge = 1.602 x 10−19 Coulombs
g m = magnetic charge (SI units of Ampere-meters)
(magnetic permeability of free space)
μo = 4π ×10−7 Newtons
Ampere 2
⎛ Coulombs 2 ⎞
(electric permittivity of free space)
ε o = 8.85 ×10
m ⎜ Newton − meter 2 ⎟
⎝
⎠
c = 1 ε o μo = speed of light (in free space/vacuum) = 3 × 108 meters / second
−12
F
n = integer (≠ 0) n = ±1, ±2, ±3, . . .
h = Planck’s Constant = 6.626 x 10−34 Joule-sec = (N-m-s)
Dirac originally obtained this relation by considering the motion of an electron circling a
magnetic monopole of magnetic charge gm, with radial magnetic field
⎛μ ⎞g
Bm ( r ) = ⎜ o ⎟ m2 rˆ (SI units: Tesla = N / A-m)
⎝ 4π ⎠ r
Quantum mechanically, the wave function ψ e ( r ) of the electron circling the magnetic monopole
(assumed to be infinitely heavy) must be single-valued in ϕ , i.e. ψ e (ϕ = 2π n ) = ψ e (ϕ = 0 )
in analogy e.g. to the Bohr model of the Hydrogen atom (e− bound to proton, p)
In other words for the electron-monopole system, Dirac demanded:
ψ e ( r ) → ψ e′ ( r ) = ψ e ( r ,θ ) eiϕ = ψ e ( r , θ ) ei( 2π n )
The quantum physics of the e−gm system gives 2π n = 2π ( eμo g m ) h where n = ±1, ±2, ±3, . . .
eμ o g m
or: eμo g m = nh ⇐ Dirac Quantization Condition (in SI units)
h
1
e2
and: α em ≡
= fine structure constant (dimensionless) and
Using: c 2 =
ε o μo
4πε o c
gm
n
n
ec or:
=
c 68.5nc
This relation can be rewritten as: g m =
2α em
2α m
e
Or:
n=
≡
h
2π
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
5
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 18
Prof. Steven Errede
Classically, the motion of an electron circling a magnetic monopole is shown below, at a height
zc above it (at the origin).
Classical path of
The “Lorentz” force acting
ẑ
orbiting electron
on the electron is:
Fem ( re ) = −eve × Bm ( re )
v is
e
everywhere
tangent to
electron’s
orbit
ve
e−
ρe
ve
⎛μ ⎞g
= −eve × ⎜ o ⎟ m2 rˆ
⎝ 4π ⎠ re
n.b. assume g m = North
re ze
gm
magnet pole i.e. gm > 0
re
θ
ŷ
ϑ
⎛ eμ g ⎞ v × rˆ
= −⎜ o m ⎟ e 2
⎝ 4π ⎠ re
⎛ nh ⎞ ve × rˆ
= −⎜
⎟ 2
⎝ 4π ⎠ re
2
⎛ nh ⎞ me ve × rˆ ⎛ c ⎞
= −⎜
⎜
⎟
2
2 ⎟
⎝ 4π ⎠ re
⎝ me c ⎠
x̂
2
⎛ nh ⎞ pe × re ⎛ c ⎞
= −⎜
⎟ 3 ⎜
2 ⎟
⎝ 4π ⎠ re ⎝ me c ⎠
Imaginary sphere of radius re
The angular momentum of the electron is Le = re × pe :
2
⎛ nh ⎞ Le c
m
=
+
F
r
∴ e ( e)
⎜
⎟
3
⎝ 4π ⎠ me c2 re
(n.b. the direction of Le is not constant here..)
The magnetic force acting on the electron bends it around in the orbit as shown in above figure.
We can also view this from a different perspective: the electron circling the magnetic monopole
in this manner creates an electric current I
e
e
C 2πρe 2π re sin θ
=
=
τ orbit = =
I=
τ orbit ( C / ve )
ve
ve
ve
Which in turn creates an orbital magnetic dipole moment:
eve
πρe2 ( − zˆ ) (SI units Amp-m2)
me = Ia = I πρe2 ( − zˆ ) =
ρe
2πρe
me = 12 eve ρe ( − zˆ )
where
ρe = re sin θ
me = − 12 eve ρ e zˆ = − 12 eve re sin θ zˆ
ze
θ
re
Reminder: I = conventional current, which for a circulating e− electron, flows in the direction
opposite to the electron’s orbital motion (see above figure).
The orbital magnetic moment of the electron then interacts with the magnetic monopole.
6
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 18
Prof. Steven Errede
Quantum mechanically, the e− behaves as a wave, not a point particle and thus the wavefunction
ψ e ( re ) of the electron spreads out along its orbit as a periodic wave in such a way that an integer
number of deBroglie wavelengths fit around the classical circumferential path, i.e.
nλn = C = 2πρe with n = 1, 2, 3, . . ..
Note also that the electron is not actually bound to the magnetic monopole – its orbit is stable.
∃ no binding energy between these two particles; given an initial electron velocity, e.g.
ve = −ve xˆ , with the electron initially at height ze above the magnetic monopole, the radial
magnetic field of the magnetic monopole will bend the electron’s path into orbit shown. Recall
also that magnetic forces do no work…
The Duality Transformation for Electromagnetism
Because of the intimate connection between E and B at the microscopic / fundamental /
elementary particle physics level, there (obviously) exists an intimate connection between
E and B at the macroscopic level.
A duality transformation is a simultaneous rotation in an abstract mathematical space by an angle
ϕ of all electric and magnetic phenomena, which leaves all of the laws associated with the
time!!!
physics of electromagnetism unchanged – it’s a “knob” that allows us to rotate space
Electromagnetism is invariant under a duality transformation.
By carrying out a duality transformation, we simultaneously rotate all electric and magnetic
phenomena by an angle ϕ in this abstract mathematical space, thus we can change electric fields
magnetic fields, electric charges
magnetic charges and we would never know the
difference!
Note: In carrying out duality transformations on all electric and magnetic phenomena, in order
for all of these quantities to transform properly, each duality transform pair must have the same
physical units.
⎛
1 ⎞
e.g.
E
cB , ( ec
gm ) , ⎜ ε o
n.b. c2 = invariant under a duality transform
2 ⎟
μo c ⎠
⎝
(
)
Other duality transform pairs are electric and magnetic currents and/or charge densities:
Je
Ke
Ie
1
Jm
c
1
Km
c
1
Im
c
ρe
σe
λe
Im ≡
1
ρm
c
1
σm
c
1
λm
c
dQm
dt
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
7
UIUC Physics 435 EM Fields & Sources I
Duality Transform for E
cB′
Fall Semester, 2007
1
g m′
c
E′
ϕ
1
gm
c
1
gm
c
e′
ϕ
ϕ
E ′ = E cos ϕ + cB sin ϕ
cB′ = cB cos θ − E sin ϕ
Duality Transforms for:
1
Je
Jm
c
1
Ke
Km
c
1
Ie
Im
c
⎛ J e′ ⎞ ⎛ J e ⎞
⎛ Jm ⎞
⎜ ⎟ ⎜ ⎟
⎟
1⎜
⎜ K e′ ⎟ = ⎜ K e ⎟ cos ϕ + c ⎜ K m ⎟ sin ϕ
⎜ I e′ ⎟ ⎜ I e ⎟
⎜ Im ⎟
⎝ ⎠ ⎝ ⎠
⎝
⎠
⎛ J m′ ⎞
⎛ Jm ⎞
⎛ Je ⎞
⎟ 1⎜
⎟
⎜ ⎟
1⎜
K m′ ⎟ = ⎜ K m ⎟ cos ϕ − ⎜ K e ⎟ sin ϕ
⎜
c⎜
⎟ c ⎜ Im ⎟
⎜ Ie ⎟
⎝ I m′ ⎠
⎝
⎠
⎝ ⎠
8
Prof. Steven Errede
Duality Transform for e and
cB
cB
Lecture Notes 18
E
ϕ
e
1
e′ = e cos ϕ + g m sin ϕ
c
1
1
g m′ = g m cos ϕ − e sin ϕ
c
c
Duality Transforms for:
1
ρe
ρm
c
1
σe
σm
c
1
λe
λm
c
⎛ρ ′⎞ ⎛ρ ⎞
⎛ ρm ⎞
⎜ e ⎟ ⎜ e⎟
1⎜ ⎟
⎜ σ e′ ⎟ = ⎜ σ e ⎟ cos ϕ + ⎜ σ m ⎟ sin ϕ
c⎜ ⎟
⎜⎜ λ ′ ⎟⎟ ⎜ λ ⎟
e
e ⎠
⎝
⎝ λm ⎠
⎝ ⎠
⎛ρ ′⎞
⎛ρ ⎞
⎛ ρe ⎞
m
⎟ 1⎜ m ⎟
1⎜
⎜ ⎟
⎜ σ m′ ⎟ = ⎜ σ m ⎟ cos ϕ − ⎜ σ e ⎟ sin ϕ
c⎜
c⎜ ⎟
⎜λ ⎟
⎜ λm′ ⎟⎟
⎝ λm ⎠
⎝ e⎠
⎝
⎠
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
1
μo′ c 2
Duality Transform for
ε o and
Lecture Notes 18
1
μo c 2
1
μo c 2
Prof. Steven Errede
ε o′
ϕ
ϕ
εo
ε o′ = ε o cos ϕ +
1
1
1
sin ϕ and
cos ϕ − ε o sin ϕ
=
2
2
μo c
μo′ c
μo c 2
ϕ + sin ϕ
We define the 2 × 2 duality transform rotation matrices as: R (ϕ ) ≡ ( cos
− sin ϕ cos ϕ ) and its inverse
ϕ
R −1 (ϕ ) ≡ ( cos
+ sin ϕ
Then:
-sin ϕ
cos ϕ
)
Then RR −1 = R −1 R = 1 = ( 10 01 )
unit
matrix
( ) = R (ϕ ) ( )
E′
c B′
or:
E
cB
⎛ ε′ ⎞
⎛
⎞
⎜ 1 o ⎟ = R (ϕ ) ⎜ 1ε o ⎟ or:
⎜ μ ′ c2 ⎟
⎜ μ c2 ⎟
⎝ o ⎠
⎝ o ⎠
( ) = R (ϕ ) ( )
e′
g m′ c
e
gm c
or:
( )=R
E
cB
−1
(ϕ ) ( c EB′′ )
⎛ ε ⎞
⎛
⎞
⎜ 1 o ⎟ = R −1 (ϕ ) ⎜ 1ε o′ ⎟
⎜ μ c2 ⎟
⎜ μ ′ c2 ⎟
⎝ o ⎠
⎝ o ⎠
( ) = R (ϕ ) ( )
e
gm c
−1
e′
g m′ c
etc….
An Example of the Use / Application of the Duality Transform
( ) = R (ϕ ) ( )
E′
c B′
E
cB
Convert the solenoidal magnetic field associated with the motion of an electric charge ( ve
(
into the solenoidal electric field associated with the motion of a magnetic charge vgm
⎛ μ ⎞ v × rˆ
Start with Be ( r ) = ⎜ o ⎟ q e 2
⎝ 4π ⎠ r
)
c !
choose: ϕ = 90o then R ( 90o ) = ( −01 01 )
1
⎛ μ ⎞ v × rˆ
cB = ⎜ o ⎟ cq 2
E ′ = cB
ε o′ =
r
μo c 2
⎝ 4π ⎠
v × rˆ
1
⎛μ c⎞
E ′ = − ⎜ o ⎟ ( −e ) 2
cB′ = − E ′
= −ε o
r
μo′ c 2
⎝ 4π ⎠
⎛ μ c ⎞ ⎛ 1 ⎞ v × rˆ
⎛ μ ⎞ v × rˆ
E ′ = − ⎜ o ⎟ ⎜ g m′ ⎟ 2 = − ⎜ o ⎟ g m′ 2
r
⎝ 4π ⎠ ⎝ c ⎠ r
⎝ 4π ⎠
1
gm
c
Multiply both sides by c:
e′ =
Change cB → E ′ :
1
g m′ = −e
c
1
Change −e → g m′ :
c
c)
1
1 ⎛ 1 ⎞ v × rˆ
: E′ = − 2 ⎜
⎟ g m′ 2
2
ε o′ c
c ⎝ 4πε o ⎠
r
All EM quantities - everything electromagnetic - now duality-transformed
Now, drop primes everywhere (i.e. can’t tell the difference after ϕ -rotation!!)
Change μo →
Em ( r ) = −
1⎛ 1
⎜
c 2 ⎝ 4πε o
⎞ vm × rˆ
⎛ μ ⎞ v × rˆ
Be ( r ) = ⎜ o ⎟ q e 2
⎟ gm 2
r
⎝ 4π ⎠ r
⎠
This is where / how the minus sign arises!!
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
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