Lecture Notes 18: Magnetic Monopoles/Magnetic Charges

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 ggtest src test ommˆ Frmmm()== gBr () ⎜⎟ 2 r ⎝⎠4π r Source point Sr′( ′) src src test Where: gm = magnetic charge of source particle gm r =−rr′ gm test gm = magnetic charge of test particle zˆ gm ≡+ g() ≡ North pole r′ r

gm ≡− g() ≡ South pole SI units of magnetic charge g = Ampere-meters (A-m) ϑ yˆ xˆ Field / observation point Pr( )

2 ⎛⎞μomg −7 Newtons N Units check: Fm () Newtons = ⎜⎟2 μπo =×410 22 ⎝⎠4π r Ampere( A )

2 ⎛⎞NN()Am− ⎛⎞A2 − m2 Newton = N = ⎜⎟22= ⎜⎟ = N ⎝⎠Am ⎝⎠A2 m2

Newton Then (the radial) B -field of a magnetic monopole is: 1 Tesla ≡1 Ampere− meter ⎛⎞μ g src N N omˆ Brm ()= ⎜⎟2 r (SI units = Tesla = ) 1 T = 1 ⎝⎠4π r Am− Am− NN⎛⎞A2 − m N Units check: Tesla == = gm = Ampere-meters (A-m) ⎜⎟2 2 Am− ⎝⎠A m Am−

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 1 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 18 Prof. Steven Errede

src The magnetic flux associated with a magnetic monopole with magnetic charge gm is: src ⎛⎞μo gm 2 src src 2 Φ≡mmB() ri ndaˆ = 4π = μomg or: Φ=momμ g (Webers, or Tesla-m ) ∫S ⎜⎟2 ()r ⎝⎠4π r src rˆ = r , and r = r because the source charge gm is located at the originϑ :

SI units of magnetic flux: r zˆ nrˆˆ= 2 area Tesla-meters ( Φ=m BiA ) element da 2 ⎛⎞Nm− 1 (T–m ) = 1⎜⎟=1 Weber (Wb) gm yˆ ⎝⎠A ϑ

Units check: xˆ

22⎧⎫NNmN− Nm− Tm−=⎨⎬i m = =i A −=m ⎩⎭Am− A A 2 A

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 Φm is quantized - the quantum (i.e. the smallest unit) of magnetic flux is one flux quantum:

−34 o ⎛⎞h6.626×− 10 Joule sec −15 o Φ≡m ⎜⎟ =−19 =4.136 × 10 Webers −gm ⎝⎠e1.602× 10 coulombs

Where: h = Planck’s constant = 6.626 x 10−34 Joule-sec

o 2 ⎛⎞N 2 ⎛⎞Nm− o Units check: Φ=m Webers = Tesla − m =⎜⎟im = ⎜⎟ +gm ⎝⎠Am− ⎝⎠A

⎛⎞h Joule−−−sec Joules Newton meters ⎛ N m ⎞ oo⎛⎞h And: ⎜⎟=== = ⎜ ⎟ 1 Φ=mom⎜⎟ =μ g ⎝⎠e coulombs Amp Amp ⎝ A ⎠ ⎝⎠e

o Then the smallest integer unit of quantized magnetic charge gm is:

−15 oo oo1 1⎛⎞hWb 4.136× 10 −9 Φ=momμ g or: gmm=Φ=⎜⎟ =−72 =3.2914 × 10 Ampere-meters (A-m) μμoo⎝⎠eNA410 π×

Now, it’s possible that magnetic monopoles could exist with integer multiples of this smallest / o no quantized amount of magnetic charge gm , i.e. gngmm= where n = integer = ±1, ±2, ±3, . . . . oo and: +=gmm N ( North Pole ) and −= g S ( South Pole ) .

oo⎛⎞h non⎛⎞h no Then if Φ=momμ g =⎜⎟, then Φ=momomμμng = g = n⎜⎟ i.e. Φmm=Φn . ⎝⎠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 Fall Semester, 2007 Lecture Notes 18 Prof. Steven Errede

n ⎛⎞h n ⎛⎞h 1 2 8 And: μomgn= ⎜⎟ ⇒ egm = n⎜⎟ but: = ε oc where c = 3 x 10 m/s (speed of light) ⎝⎠e ⎝⎠μo μo 2 ⎛⎞h n nhcεπεoonhn⎛⎞2 ⎛⎞4 c Defining ≡ ⎜⎟ then gcecmo==4πε = 2π ⎜⎟ ⎜⎟2 ⎝⎠ ee22⎝⎠π 2⎝⎠ ≡ 1 αem 2 n n ⎛⎞1 e 1 i.e. gecm = ⎜⎟ (A-m) and the fine-structure constant αem ≡= 2 ⎝⎠αem 4πε oc 137.036... 21 (dimensionless quantity) and numerically, 2α = em 137.036 68.5

n Then: gnecm = 68.5 () 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 2 ⎛⎞μo gm ()nnF ()n ⎜⎟ 2 2 Fgmmm ⎝⎠4π r ⎛⎞ 1 22 2 2 2 == =εμoo⎜⎟ =68.5 nc = ()68.5nn 4700 Fe⎛⎞2 2 e Fe 1 e ⎝⎠ c ⎜⎟r 2 ⎝⎠4πε o (n = ±1, ±2, ±3, . . . .)

Force of magnetic attraction between N-S monopoles Force of attraction between two (opposite) electric charges.

o ⎛⎞h -15 If Φ=m ⎜⎟ magnetic flux quantum (= 4.136 x 10 Wb) ⎝⎠e

ooooBi ndaˆ ΦmS∫ μ omggg⎛⎞ m1 ⎛⎞ m Then: o ==== εμoo⎜⎟2 ⎜⎟ ΦEoEi ndaˆ eeceε ⎝⎠ ⎝⎠ ∫S

o 2 o ⎛⎞gm ⎛⎞1 e 1 But: gecm = 68.5 or: ⎜⎟==68.5cc⎜⎟ where: αem == ⎝⎠e ⎝⎠2αem 4πε oc 137.036...

Φoo111111⎛⎞g ⎛⎞⎛⎞ ∴ mm==c = ==68.5 o 22⎜⎟ ⎜⎟⎜⎟ c ΦEemememce⎝⎠ c⎝⎠⎝⎠222ααα c c

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 3 2005-2008. All Rights Reserved. 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:

⎛⎞he⎛⎞2 ⎛⎞h ⎛⎞eh h Φ=oo22ααcc Φ= = 2 c = 2 where ≡ e− Eemmem⎜⎟ ⎜⎟⎜⎟ ⎜⎟ ⎝⎠e ⎝⎠4πε o c ⎝⎠e ⎝⎠4πε o 2π ⎛⎞ ⎛⎞ o ⎜⎟eh 4π eh ⎛⎞e + o e Φ=E 2 =⎜⎟= ⎜⎟ e 1 Φ=E ⎜⎟h ε o ε o ⎜⎟4πε ⎝⎠4π ε o h ⎝⎠ ⎝⎠o ()2π Numerically: −19 2 o ⎛⎞e 1.602×− 10 Coulombs−8 N m Φ=E ⎜⎟ = Electric Flux Quantum ==×2 1.810 10 ⎝⎠ε o Coulombs Coulomb 8.85× 10−12 Nm− 2 Φ=ooμ g =h = Magnetic Flux Quantum = 4.136 x 10−15 Wb (T–m2) mom()()e

o Φm 1 n n o = and gecm = Ampere-meters where n = ±±1, 2, ±3 ΦEem2α c 2αem

Gauss’ Law:

oohN−15⎛⎞ 2⎛⎞ 2 ⎛⎞Fm Φ=momμ gWbTmmAreaBArea = =4.136 × 10 ⎜⎟ = − =⎜⎟ − =⎜⎟ii =m eAmg⎝⎠⎝⎠− ⎝⎠m

o eN−82⎛⎞⎛⎞ ⎛⎞FE Φ=E =1.810 × 10 ⎜⎟⎜⎟ −mAreaEArea = ⎜ ⎟ii = ε o ⎝⎠⎝⎠ce⎝⎠

4 © 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:

egm eg∗==μom 2 nh (SI units) Dirac Quantization Condition ε oc e = electric charge = 1.602 x 10−19 Coulombs

gm = magnetic charge (SI units of Ampere-meters) μπ=×410−7 Newtons (magnetic permeability of free space) o Ampere2 ⎛⎞Coulombs2 ε =×8.85 10−12 F (electric permittivity of free space) o m⎜⎟2 ⎝⎠Newton− meter 8 c =1 ε ooμ = speed of light (in free space/vacuum) = 3 × 10 meters / second 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 omˆ Brm ()= ⎜⎟2 r (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 inϕ , i.e. ψ ee(ϕπ= 20n)()== ψϕ 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:

iϕ in(2π ) ψ eee()r→=ψψθψθ′ () r( re,,) = e( re)

− The quantum physics of the e gm system gives 22ππμnegh= ( om) where n = ±1, ±2, ±3, . . . egμ Or: n = om or: egμ = nh ⇐ Dirac Quantization Condition (in SI units) h om 2 2 1 e h Using: c = and: αem ≡ = fine structure constant (dimensionless) and ≡ ε ooμ 4πε oc 2π

n gm n This relation can be rewritten as: gecm = or: = cnc 68.5 2αem e 2αm

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 5 2005-2008. All Rights Reserved. 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 zˆ orbiting electron on the electron is: m Free( )()=− evBr e × me ve is everywhere ⎛⎞μ g omˆ tangent to =−eve ×⎜⎟2 r ⎝⎠4π r electron’s e orbit ve n.b. assume gm = North − ρ e magnet pole i.e. gm > 0 e ve re ze re ⎛⎞egμ v× rˆ θ ˆ om e gm y =−⎜⎟2 ⎝⎠4π r e ϑ ⎛⎞nh vre × ˆ =−⎜⎟2 ⎝⎠4π re nhmv× rˆ ⎛⎞ c2 ˆ ⎛⎞ee x =−⎜⎟ 22⎜⎟ ⎝⎠4π rmcee⎝⎠ 2 ⎛⎞nhpree× ⎛⎞ c Imaginary sphere of radius re =−⎜⎟32⎜⎟ ⎝⎠4π rmcee⎝⎠ 2 m ⎛⎞nh Lce The angular momentum of the electron is Lrpee= × e: ∴ Free()=+⎜⎟ 3 ⎝⎠4π mcr ee2 (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

ee C 22sinπρπeer θ I == τ orbit == = τ orbit()Cv/ e vvee v e Which in turn creates an orbital magnetic dipole moment:

22eve 2 mIaIee==πρ() −= zˆˆ πρ e() − z (SI units Amp-m ) ρe 2πρe 1 mevzeee=−2 ρ ()ˆ where ρee= r sinθ ze 11 mevzevrzeeeee=−22ρ ˆˆ =− sinθ 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 ee()r 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.

nCλne==2πρ 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. vvxee=− ˆ , 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 EBand at the microscopic / fundamental / elementary particle physics level, there (obviously) exists an intimate connection between EB and 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 physics of electromagnetism unchanged – it’s a “knob” that allows us to rotate space time!!!

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 2 e.g. ()EcB , ()ec gm , ⎜⎟ε o 2 n.b. c = invariant under a duality transform ⎝⎠μoc

Other duality transform pairs are electric and magnetic currents and/or charge densities:

1 1 JJ ρ ρ emc emc 1 1 KK σ σ emc emc 1 1 I I λ λ emc emc dQ I ≡ m m dt

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 7 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 18 Prof. Steven Errede

1 Duality Transform for EcB Duality Transform for eg and c m

1 1 cB′ cB E′ g′ g e′ c m c m

ϕ ϕ ϕ E ϕ e

1 EE′ =+cosϕ cB sinϕ ee′ =+cosϕ g sinϕ c m 11 cB′ =− cBcosθ E sinϕ gg′ =−cosϕ e sinϕ ccmm

Duality Transforms for: Duality Transforms for: 1 1 JJ ρ ρ emc emc 1 1 KK σ σ emc emc 1 1 I I λ λ emc emc

⎛⎞′ ⎛⎞⎛⎞JJee′ ⎛ J m ⎞ ρρee⎛⎞ ⎛ ρ m ⎞ ⎜⎟⎜⎟1 ⎜ ⎟ ⎜⎟⎜⎟1 ⎜ ⎟ KK′ =+cosϕ K sinϕ ⎜⎟σ ′ =+σϕσϕcos sin ⎜⎟⎜⎟eec ⎜ m ⎟ ee⎜⎟c ⎜ m ⎟ ⎜⎟⎜⎟II′ ⎜ I ⎟ ⎜⎟λλ′ ⎜⎟ ⎜ λ ⎟ ⎝⎠⎝⎠ee ⎝ m ⎠ ⎝⎠ee⎝⎠ ⎝ m ⎠ ⎛⎞′ ⎛⎞JJmm′ ⎛⎞ ⎛⎞ J e ρρmm⎛⎞ ⎛⎞ ρ e 11⎜⎟ ⎜⎟ ⎜⎟ 11⎜⎟⎜⎟ ⎜⎟ KK′ =−cosϕ K sinϕ ⎜⎟σ ′ =−σϕσϕcos sin cc⎜⎟mm ⎜⎟ ⎜⎟ e ccmm⎜⎟ ⎜⎟ e ⎜⎟II′ ⎜⎟ ⎜⎟ I ⎜⎟λλ′ ⎜⎟ ⎜⎟ λ ⎝⎠mm ⎝⎠ ⎝⎠ e ⎝⎠mm⎝⎠ ⎝⎠ e

8 © 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

1 1 ′ Duality Transform for 2 2 ε o μo′c μoc 1 ε o and 2 ϕ μoc ϕ

ε o 1 11 ′ ε oo=+εϕcos2 sin ϕ and 22=−cosϕ εϕo sin μoc μμoo′cc

cosϕ + sinϕ We define the 2 × 2 duality transform rotation matrices as: R(ϕ ) ≡ ( −sinϕ cosϕ ) and its inverse R−1()ϕ ≡ cosϕ -sinϕ Then RR−−11=== R R 1 1 0 ()+sinϕ cosϕ (0 1) unit matrix EE′ EE−1 ′ Then: = R ϕ or: = R ϕ ()cB′ ()() cB ( cB) ()( cB′ ) ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ′ ′ ⎜⎟εεoo= R()ϕ ⎜⎟ or: ⎜⎟εεoo= R−1 ()ϕ ⎜⎟ ⎜⎟11 ⎜⎟ ⎜⎟11 ⎜⎟ ′ 22 22′ ⎝⎠μμoocc ⎝⎠ ⎝⎠μμoocc ⎝⎠ ee′ = R ϕ or: ee= R−1 ϕ ′ etc…. ()gmm′ cgc()() ( gmmcgc) ()( ′ )

EE′ An Example of the Use / Application of the Duality Transform = R ϕ ( cB′ ) ()() cB

Convert the solenoidal magnetic field associated with the motion of an electric charge ()vce into the solenoidal electric field associated with the motion of a magnetic charge vc ! ()gm ⎛⎞μoevr× ˆ o o 0 1 Start with Bre ()= ⎜⎟ q 2 choose: ϕ = 90 then R(90 )(= −1 0 ) ⎝⎠4π r ⎛⎞μ vr× ˆ 1 1 o ′ ′ ′ Multiply both sides by c: cB= ⎜⎟ cq 2 EcB= ε o = 2 eg= m ⎝⎠4π r μ c c o ⎛⎞μ c vr× ˆ 1 1 ′ ′ o ′ ′ ′ Change cB→ E : Ee=−⎜⎟() − 2 cB= − E 2 = −ε o gem = − ⎝⎠4π r μ′c c o 1 ⎛⎞μμc 1 vr× ˆˆ ⎛⎞ vr× ′ ′′oo⎛⎞ ′ Change −→egm : Eg=−⎜⎟⎜⎟mm22 =− ⎜⎟ g c ⎝⎠44ππ⎝⎠cr ⎝⎠ r 1 11⎛⎞vr× ˆ ′′ Change μo → 2 : Eg=− 22⎜⎟m ε o′c cr⎝⎠4πε o All EM quantities - everything electromagnetic - now duality-transformed Now, drop primes everywhere (i.e. can’t tell the difference after ϕ -rotation!!)

11⎛⎞vrm × ˆ ⎛⎞μoevr× ˆ Ermm()=− 22⎜⎟ g Bre ()= ⎜⎟ q 2 cr⎝⎠4πε o ⎝⎠4π r This is where / how the minus sign arises!!