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Astronomy 9620a / Physics 9302a – 2nd Problem List and Assignment

Your solution to problems 3, 7, and 11 has to be handed in on Tue., Nov. 19, 2013.

1. Paramagnetic materials. Let us consider an atom that has one unpaired electron, which moves in an orbit about the nucleus (classical model). a) Show that the magnetic dipole moment m of the electron can be expressed as

m = −q2M

L, (1.1)

where −q( ), M , and L are the electron’s charge, mass, and angular momentum respectively. b) This magnetic dipole is then subjected to an external magnetic induction B = Bez . Consider the torque felt by the magnetic dipole, and show that, under these conditions, L will precess about the z-axis (i.e., B ) with constant angle (let us call it θ ), and at the Larmor frequency ω L defined by

ω L =qB2M

. (1.2)

c) Paramagnetic materials generally consist of atoms, molecules, or ions that have a net nonzero angular momentum. Under the influence of an external magnetic induction, each components of a gas will have a potential energy U = −m ⋅B == mBcos θ( ). (1.3) The probability P θ( )dθ that the magnetic moment lies between the angles θ and θ + dθ is proportional to the Boltzmann factor and the solid angle 2π sin θ( )dθ P θ( )dθ ∝ eβmB cos θ( ) sin θ( )dθ, (1.4) where β = 1 kT . Show that the average value cos θ( ) is given by

cos θ( ) = coth βmB( ) − 1βmB

. (1.5)

[Hint. You will need the following relation

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∂∂αln eα xdx∫( ) = eα xx dx∫

eα xdx∫. (1.6)

] d) Assume that βmB≪1 , and show that

cos θ( ) =13βmB. (1.7)

Furthermore, since mz = m cos θ( ) is the only non-vanishing component of the mean dipole moment, show that if we define a paramagnetic susceptibility χm = M B , then

χm =nm2

3kT, (1.8)

with n the number density of molecules. (Please note that the magnetic susceptibility is usually defined by M = χmH .)

e) Using L ≈ ! , and your best guess for the other quantities involved in equation (1.8), provide an order-of-magnitude calculation for χm .

Solution. a) The magnetic dipole moment is given by

m =12x × J∫ d 3x. (1.9)

For an unpaired electron in orbit, we have J = −qvδ x − x0( ) , and equation (1.9) becomes

m =q2v × xδ x − x0( ) d 3x∫

= −q2x0 × v

= −q2M

x0 × Mv( )

= −q2M

L.

(1.10)

b) The torque on the dipole is given by N = m × B. (1.11) But we also know that N ≡ dL dt , combining this relation with equation (1.11) we write

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dLdt

= −q2M

L × B, (1.12)

any variation in L is then perpendicular to both B and L . If B = Bez , then

dLxdt

= −q2M

LyB

dLydt

=q2M

LxB

dLzdt

= 0,

(1.13)

or from the last equation Lz = cste . We now define a new variable η = Lx + iLy such that we can transform equations (1.13) to

dηdt

= i qB2M

η. (1.14)

The solution to equation (1.14) is straightforward with η = AeiωL t , (1.15) or

Lx = Acos ω Lt( )Ly = Asin ω Lt( ), (1.16)

where

ω L =qB2M

(1.17)

is the Larmor frequency. Since both A and Lz are constants, and that equations (1.16) are that of a circle of radius A , we see that the angular momentum vector precesses about the z- or B-axis at the Larmor frequency. The angle θ made by L relative to B is given by

θ = tan−1 ALz

⎛⎝⎜

⎞⎠⎟. (1.18)

c) Given that P θ( )dθ ∝ eβmB cos θ( ) sin θ( )dθ = eβmB cos θ( )d cos θ( )⎡⎣ ⎤⎦ , then

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cos θ( ) =

eβmB cos θ( ) cos θ( )d cos θ( )⎡⎣ ⎤⎦−1

1

∫eβmB cos θ( )d cos θ( )⎡⎣ ⎤⎦−1

1

∫=

∂∂ βmB( ) ln eβmB cos θ( )d cos θ( )⎡⎣ ⎤⎦−1

1

∫{ }. (1.19)

Since

eβmB cos θ( )d cos θ( )⎡⎣ ⎤⎦−1

1

∫ =1

βmBeβmB cos θ( )

−1

1

=2sinh βmB( )

βmB,

(1.20)

then

cos θ( ) =

βmB2sinh βmB( )

2cosh βmB( )βmB

−2sinh βmB( )

βmB( )2⎡

⎣⎢⎢

⎤

⎦⎥⎥

= coth βmB( ) − 1βmB

.

(1.21)

d) When βmB≪1 (or kT ≫ mB ) equation (1.21) simplifies to

cos θ( ) !1+ 1

2βmB( )2

βmB + 16

βmB( )3−

1βmB

!1

βmB1+ 1

2βmB( )2⎡

⎣⎢⎤⎦⎥1− 1

6βmB( )2⎡

⎣⎢⎤⎦⎥−1⎧

⎨⎩

⎫⎬⎭

!13βmB.

(1.22)

The total mean magnetization M = mz ez (the other components average to zero because of their cos ϕ( ) or sin ϕ( ) dependency) is given by

M = n mz =nm2B3kT

ez , (1.23)

and the paramagnetic susceptibility is

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χm =nm2

3kT. (1.24)

e) With L ≈ ! = 1.05 ×10−34 J ⋅ s , q = 1.6 ×10−19 C , and M = 9.1×10−31 kg , we have

m ≈

q!2M

= 9.2 ×10−24 C s-1m2. (1.25)

Furthermore, taking T = 300 K , n = 1025 m-3 (an approximate value for air), and k = 1.38 ×10−23 J K-1 , the paramagnetic susceptibility is χm ∼ 10

−1. (1.26) This value is higher than what can be calculated ( χm ∼ 10

−4 ) with a more careful analysis. 2. Diamagnetic materials. As we saw in the Problem 1, electrons moving around a nucleus in an atom will also precess with angular frequency

ω L =qB2M

, (2.1)

when subjected to an external magnetic induction B = Bez .

a) Show that the current I resulting from this precession is

I = −Zq2B4πM

, (2.2)

where Z is the number of electrons in the atom, and that the corresponding dipole moment is

m = −Zq2B4M

ρ2 , (2.3)

where ρ2 = x2 + y2 is the mean square radius of the electrons’ orbit when projected on the plane perpendicular to B . b) Assume spherical charge distributions for the electrons, and show that the diamagnetic susceptibility χ is given by

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χ = −n mB

= −nZq2 r2

6M, (2.4)

where n is still the number density, and r2 = x2 + y2 + z2 .

c) It is found that the magnitude of the diamagnetic susceptibility is in general much smaller than that of the paramagnetic susceptibility. Provide at least two more fundamental differences between the two quantities.

Solution. a) The precession will produce an electric current I = charge × revolutions per unit time , and if there are Z precessing electrons, then

I = −Zq( )ω L

2π

= −Zq( ) 12π

qB2M

= −Zq2B4πM

.

(2.5)

The corresponding dipole moment is the current time the area. So, if ρ2 = x2 + y2 is the mean square radius of the electrons’ orbit when projected on the plane perpendicular to B , then

m = Iπ ρ2

= −Zq2B4M

ρ2 . (2.6)

b) Assuming spherical orbits for the electrons, we set r2 = x2 + y2 + z2 as the mean square distance of the electrons from the nucleus so that

ρ2 = x2 + y2

=23r2 .

(2.7)

Substituting this relation into equation (2.6) yields

m = −Zq2

6Mr2 B, (2.8)

and the diamagnetic susceptibility is given by

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χ = −n mB

= −nZq2 r2

6M, (2.9)

with n is still the number density. c) Three more fundamental differences between paramagnetic and diamagnetic susceptibilites are: 1) The diamagnetic susceptibility is always negative, contrary to the paramagnetic susceptibility. 2) Diamagnetism is a property of all matter, whereas paramagnetism requires that elementary components have a net nonzero angular momentum. 3) The diamagnetic susceptibility does not depend on temperature. The 1 T dependence of the paramagnetic susceptibility expresses Curie’s Law. 3. Suppose that

E x,t( ) = −

14πε0

qr2H vt − r( )er

B x,t( ) = 0, (3.1)

where H x( ) is the Heaviside or step distribution (see equation (1.14) of the lecture notes), and r = x . Show that these fields satisfy all of Maxwell’s equations, and determine ρ and J . Describe the physical situation that gives rise to the fields.

Solution. This is the field of a point charge −q located at the origin, out to an expanding spherical shell of radius vt ; outside this shell the field is zero. It follows that the shell carries a total charge +q . We first calculate the charge density ρ with

∇ ⋅E = −q4πε0

H vt − r( )∇ ⋅1r2er

⎛⎝⎜

⎞⎠⎟+1r2er ⋅∇H vt − r( )⎡

⎣⎢⎤⎦⎥

= −q4πε0

−H vt − r( )∇2 1r

⎛⎝⎜

⎞⎠⎟+1r2er ⋅∇H vt − r( )⎡

⎣⎢⎤⎦⎥

= −q4πε0

4πH vt − r( )δ r( ) − 1r2δ vt − r( )⎡

⎣⎢⎤⎦⎥

= −q4πε0

4πH t( )δ r( ) − 1r2δ vt − r( )⎡

⎣⎢⎤⎦⎥,

(3.2)

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where equation (1.16) of the lecture notes was used. From Coulomb’s law, the charge density is

ρ = −qδ r( )H t( ) + q4πr2

δ vt − r( ). (3.3)

This equation mathematically verifies our physical interpretation of the fields. The current density J is easily evaluated from

J = ρv

=qv4πr2

δ vt − r( )er , (3.4)

since v = 0 at the origin for the (stationary) negative charge. Evidently, ∇ ⋅B = 0 , and ∇ × E + ∂B ∂t = 0 , since E has only one component along er that is independent of θ and ϕ . The only remaining law to verify is

∇ × B = µ0J + µ0ε0∂E∂t

= 0. (3.5)

For this, we have

∂E∂t

= −qv

4πε0r2 δ vt − r( )er , (3.6)

from equation (3.1), and −ε0 ∂E ∂t = J = ρv as a comparison with equation (3.4) readily shows. 4. A uniform magnetic induction field B t( ) = B t( )ez , fills a circular region of radius R located in the xy-plane . Let’s assume that the magnetic induction is increasing in time. Use Lenz’ law to qualitatively characterize the induced electric field, and then evaluate it. Solution. Lenz’ law states that the electric field induced (and the current potentially created) will oppose the change in magnetic induction. Using the right hand rule, the magnetic induction generated by the induced current must be oriented downward (i.e., along −ez ). The induced electric field must then run clockwise in the circumferential direction. Mathematically,

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E ⋅dl!∫ = R Eϕ dϕ =0

2π

∫ 2πR( )Eϕ = −dFdt

= −ddt

B ⋅ ez daS∫= −πR2

dB t( )dt

.

(4.1)

Therefore,

E = −R2dB t( )dt

eϕ . (4.2)

This equation verifies our interpretation using Lenz’ law. 5. A line of charge λ is uniformly glued onto the rim of a wheel of radius b , which is then suspended horizontally, as shown in Figure 1. Although it is originally at rest, the wheel is free to rotate (the spokes are made of some non-conducting material – wood, maybe). In the central region, out to a radius a < b( ) , there is initially a uniform magnetic induction B0 , pointing up. Now someone turns off the field off.

a) Use Lenz’ law, and whatever conservation law(s) that might be needed, to

qualitatively describe what happens to the wheel after the magnetic induction field is turned off. Does the wheel stay at rest or is it set in some sort of mechanical motion?

b) Now, quantify the answer you gave in a), again using the needed laws (either from conservation arguments, and/or electromagnetic considerations). Does the final state of the wheel depend on the details of how B0 is turned off?

c) If you determined that the wheel is set in some sort of motion by the disappearance of the magnetic induction, then specify exactly what was the source of the motion, and the agent responsible for this motion. Be specific.

Solution. a) The changing magnetic induction field will induce an electric field (Faraday’s law of

induction), curling around the axis of the wheel. This electric field exerts a force on the charges glued to the rim, and the wheel starts to turn. According to Lenz’ law, it will rotate in such a direction that its (magnetic induction) field tends to restore the initial upward flux. The rotational motion, then, is counterclockwise, as viewed from above.

b) Quantitatively, Faraday’s law states that

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E ⋅dl!∫ = −dFdt

= −ddt

B ⋅n daS∫

= −πa2 dBdt.

(5.1)

The torque on a segment of length dl along the rim of the wheel is given by

dN = r × Eλdl( )⎡⎣ ⎤⎦, (5.2)

where E is the induced electric field. Integrating over the rim, we get the total torque on the wheel

N ≡dLdt

= x × Eλdl( )!∫= ezbλ Edl!∫= −ezλbπa

2 dBdt.

(5.3)

The total angular momentum of the rim, after the initial magnetic induction has been turned off, is

L = Ndt∫= −ezλbπa

2 dBB0

0

∫= ezλbπa

2B0 ,

(5.4)

Figure 1 – The wheel of Problem 5. A line charge λ is uniformly glued to the wheel at a radius b , and a magnetic induction field B0 initially exists out to a radius a , until it is turned off. The wheel is free to rotate.

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and the wheel is turning clockwise when seen from above, as stated in a). Since we must also have L = Izω ez , with Iz the principal moment of inertia of the wheel about its symmetry axis, then

ω =λπa2bIz

B0 . (5.5)

That is, the speed of the wheel is independent of the way in which B0 is turned off.

c) There must be conservation of angular momentum for this system. Since the wheel has acquired a net quantity of angular momentum, the electromagnetic fields must have lost an equal amount. So, the source of the motion was the initial angular momentum present in the electromagnetic fields, and the agent for the motion of the source is the electric field induced by the cancellation of B0 .

6. An alternating current I = I0 cos(ωt) flows down a long straight wire, and returns along a coaxial connecting tube of radius a . a) Using (among others) Ampère’s Law as an approximation, determine in what direction does the induced electric field point (radial, circumferential, or longitudinal)? b) Assuming that the electric field goes to zero as r→∞ , find E r,t( ) , where r is the radial distance from the wire. c) Using the result obtained in b) calculate the displacement current density Jd .

d) Integrate the equation for Jd to get the total displacement current Id .

e) Calculate the ratio of Id I . Set a = 1 mm ; can you advance a hypothesis as to why Faraday did not discover displacement currents? Solution. a) Using Ampère Law (in its integral form in the magnetostatics approximation) it is straightforward to determine that the magnetic induction field is circumferential (i.e., B = Beϕ ). Turning to Faraday’s Law of induction (again its integral form), we find that the induced electric field is longitudinal. b) We will be again using Ampère’s and Faraday’s Law. For the former, we define a surface of radius s that is perpendicular to the wire/tube pair. Whether we choose that s < a or s > a we find that the total current crossing the surface is I or zero, respectively. Then the magnetic induction field can be calculated to be

B =µ0I2π s

eϕ , s < a

0 s > a.

⎧⎨⎪

⎩⎪ (6.1)

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For Faraday’s Law, we choose a rectangular surface that has two sides of length l parallel to the wire; one side is inside the tube at a distance r from the wire, while the other side is very far away outside the tube ( r→∞ ). Applying Faraday’s Law we have

E ⋅dl!∫ = −ddt

B ⋅nda∫El = −

ddt

µ0I2π s

l dsr

a

∫ , (6.2)

which yields

E = −

µ02π

dIdtln a

r⎛⎝⎜

⎞⎠⎟ez

=µ0I0ω2π

sin ωt( ) ln ar

⎛⎝⎜

⎞⎠⎟ez , r ≤ a.

(6.3)

c) The displacement current density is given by

Jd = ε0

∂E∂t

=µ0ε02π

ω 2I0 cos ωt( ) ln ar

⎛⎝⎜

⎞⎠⎟ez

=µ0ε02π

ω 2I ln ar

⎛⎝⎜

⎞⎠⎟ez .

(6.4)

d) The total displacement current

Id = J ⋅nda∫ =µ0ε0ω

2I2π

ln ar

⎛⎝⎜

⎞⎠⎟2πr dr

0

a

∫

=ω 2Ic2

r ln a( ) − r ln r( )⎡⎣ ⎤⎦dr0

a

∫

=ω 2Ic2

ln a( ) r2

2−

r2

2ln r( ) − r

2

4⎡

⎣⎢

⎤

⎦⎥

⎧⎨⎩⎪

⎫⎬⎭⎪ 0

a

=ω 2Ia2

4c2.

(6.5)

e) The ratio of the two currents is

IdI=ω 2a2

4c2. (6.6)

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With a = 1 mm, we see that even if we set the angular frequency as high as ω = 2c = 600 ×106 rad/s , then Id I = 1×10−6 . Obviously, a frequency of the order of 100 MHz was not accessible to Faraday, and he therefore did not have a chance of experimentally discovering displacement currents. 7. Consider the Jefimenko formulae for the electromagnetic fields

E x,t( ) = 14πε0

eRR2

ρ ′x , ′t( )⎡⎣ ⎤⎦ret +eRcR

∂ρ ′x , ′t( )∂ ′t

⎡⎣⎢

⎤⎦⎥ret

−1c2R

∂J ′x , ′t( )∂ ′t

⎡⎣⎢

⎤⎦⎥ret

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪d 3 ′x∫

B x,t( ) = µ04π

J ′x , ′t( )⎡⎣ ⎤⎦ret ×eRR2

+∂J ′x , ′t( )

∂ ′t⎡⎣⎢

⎤⎦⎥ret

×eRcR

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪d 3 ′x∫

(7.1)

when applied to the case of point charge where

ρ ′x , ′t( ) = qδ ′x − r ′t( )⎡⎣ ⎤⎦J ′x , ′t( ) = qv ′t( )δ ′x − r ′t( )⎡⎣ ⎤⎦.

(7.2)

The position and velocity of the charge at time ′t are specified by r ′t( ) and v ′t( ) , respectively. In evaluating expressions involving the retarded time, one must put ′t = t − R c , where R = x − r ′t( ) (R = x − ′x inside delta functions within integrands).

a) As a preliminary to deriving the Heaviside-Feynman expressions for the electric and magnetic induction fields of a point charge, show that

δ ′x − r ′t( )⎡⎣ ⎤⎦ d3 ′x∫ =

1κ, (7.3)

where κ = 1− eR ⋅v c . Take note that κ is evaluated at the retarded time.

b) Starting with equations (7.1), use equations (7.2) and the result of part a) to show that

E x,t( ) = q

4πε0

eRκR2⎡⎣⎢

⎤⎦⎥ret

+1c∂∂t

eRκR⎡⎣⎢

⎤⎦⎥ret

−1c2

∂∂t

vκR⎡⎣⎢

⎤⎦⎥ret

⎧⎨⎩

⎫⎬⎭

B x,t( ) = q µ04π

v × eRκR2

⎡⎣⎢

⎤⎦⎥ret

+1c∂∂tv × eRκR

⎡⎣⎢

⎤⎦⎥ret

⎧⎨⎩

⎫⎬⎭.

(7.4)

c) Show that

∂ f[ ]ret∂t

=1κ

∂f∂ ′t

⎡⎣⎢

⎤⎦⎥ret, (7.5)

where f is any well-behaved function of r ′t( ) and ′t .

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d) Finally, use equations (7.4) and (7.5) to derive the Heaviside-Feynman expressions for the electric and magnetic induction fields of a point charge. That is, show that

E x,t( ) = q4πε0

eRR2

⎡⎣⎢

⎤⎦⎥ret

+R[ ]retc

∂∂t

eRR2

⎡⎣⎢

⎤⎦⎥ret

+1c2

∂2

∂t 2eR[ ]ret

⎧⎨⎩

⎫⎬⎭

B x,t( ) = q µ04π

v × eRκ 2R2

⎡⎣⎢

⎤⎦⎥ret

+1

c R[ ]ret∂∂tv × eRκ

⎡⎣⎢

⎤⎦⎥ret

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪.

(7.6)

Solution. a) Referring to Figure 2, let us consider an infinitesimal rectangle of area ab (we work in two dimensions for simplicity) that is not moving relative to the observer. If the same element is set in motion at a velocity v relative to the observer, then it will not appear as a rectangle anymore (as shown in the figure), since electromagnetic signals emanating from different points and arriving at the same time at the position of the observer must leave the source at different times (while it is continuously moving). More importantly, (again referring to Figure 2) a signal originating from point 1 must leave before another leaving point 2, if they are to arrive at the same time at the observer, and as the source is moving we will find that ′b ≠ b (while ′a = a , since it is along a direction perpendicular to v ). More precisely, during the time interval Δ ′t that elapses between the emissions of the two signals, the signal from the point 1 travels a distance cΔ ′t in the direction of the observer

cΔ ′t = ′bvv⋅ eR . (7.7)

Figure 2 – A infinitesimally small rectangle as seen from an observer located a long distance away along the eR unit vector, when the rectangle is not moving (left) or moving at a velocity v (right) relative to the observer.

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During the same time interval, the surface element will travel the distance v Δ ′t = ′b − b( ). (7.8) Combining equations (7.7) and (7.8), we get

′b =b

1− eR ⋅vc

. (7.9)

A generalization to three-dimension is straightforward, and it follows that if d 3 ′x is the infinitesimal volume element at rest, then d 3 ′x κ , with κ = 1− eR ⋅v c , is the corresponding apparent volume element seen by the observer when the volume element is moving at a velocity v . Notably,

δ ′x − r ′t( )⎡⎣ ⎤⎦ d3 ′x∫ =

1κ. (7.10)

There is at least one other way of deriving this result (from Rybicki & Lightman, pp. 77-79). Consider the following integral (for the total charge) Q x,t( ) = ρ ′x , ′t( )⎡⎣ ⎤⎦∫ ret

d 3 ′x . (7.11)

We are certainly free to rewrite this equation as Q x,t( ) = d 3 ′x d ′t ρ ′x , ′t( )δ ′t − t + x − ′x c( )∫∫ . (7.12) If we substitute the first of equations (7.2) into equation (7.12), and integrate over space, we have

Q x,t( ) = q δ ′t − t + x − r ′t( ) c( )∫ d ′t

= q δ ′t − t + R ′t( ) c( )∫ d ′t , (7.13)

where R ′t( ) = x − r ′t( ) and R ′t( ) = R ′t( ) . We now make the change of variable ′′t = ′t − t + R ′t( ) c in the time integral of equation (7.13), which implies that

d ′′t = d ′t +1cdR ′t( )d ′t

d ′t . (7.14)

But

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dR2 ′t( )d ′t

= 2R ′t( ) dR ′t( )d ′t

=dR2 ′t( )d ′t

= 2R ′t( ) ⋅ dR ′t( )d ′t

= −2R ′t( ) ⋅v ′t( ),

(7.15)

or, alternatively

dR ′t( )d ′t

= −eR ⋅v ′t( ). (7.16)

Inserting equations (7.16) and (7.14) in equation (7.13), we get

Q x,t( ) = qκ

δ ′′t( )∫ d ′′t =qκ, (7.17)

which yields the desired result, with κ as defined above. b) Before integrating in equations (7.1), it is important to realize that since ′t = t − R c , where R = x − ′x is not a function of the retarded time ′t (it only becomes a function of ′t after the integration, see equations (7.12) and (7.13)), then in the integrands

ddt![ ]ret =

dd ′t!( )⎡

⎣⎢⎤⎦⎥ret. (7.18)

From this equation, and the definitions for the current and the charge densities of equations (7.2), it is straightforward to obtain equations (7.4) from the Jefimenko formulae for the electromagnetic fields. c) It is now essential to realize that after the integrations R becomes a function of the retarded time. More precisely, R ′t( ) = x − r ′t( ) , and therefore equation (7.18) does not

apply anymore. So if t = ′t + R ′t( ) c , then we have that dt = 1+ c−1dR ′t( ) d ′t( )d ′t . It follows from equation (7.16) that

dt = 1− eR ⋅vc

⎛⎝⎜

⎞⎠⎟=κd ′t , (7.19)

and, therefore, that

17

ddt![ ]ret =

1κ

dd ′t!( )⎡

⎣⎢⎤⎦⎥ret. (7.20)

d) We start with the last term on the right-hand side of the first of equations (7.4)

1c2

∂∂t

vκR⎡⎣⎢

⎤⎦⎥ret

= −1c2

∂∂t

1κR

∂R∂ ′t

⎡⎣⎢

⎤⎦⎥ret

= −1c2

∂∂t

1κ∂eR∂ ′t

⎡⎣⎢

⎤⎦⎥ret

−∂∂tRκ

∂∂ ′t

1R

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥ret

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

= −1c2

∂2

∂t 2eR[ ]ret +

∂∂t

eRκR

∂∂ ′t

R( )⎡⎣⎢

⎤⎦⎥ret

⎧⎨⎩

⎫⎬⎭

= −1c2

∂2

∂t 2eR[ ]ret −

∂∂t

eRκR

eR ⋅v( )⎡⎣⎢

⎤⎦⎥ret

⎧⎨⎩

⎫⎬⎭

= −1c2

∂2

∂t 2eR[ ]ret + c

∂∂t

eRκR

κ −1( )⎡⎣⎢

⎤⎦⎥ret

⎧⎨⎩

⎫⎬⎭

= −1c2

∂2

∂t 2eR[ ]ret −

1c∂∂teRR

⎡⎣⎢

⎤⎦⎥ret

+1c∂∂t

eRκR⎡⎣⎢

⎤⎦⎥ret.

(7.21)

The last term of this equation will cancel with the second term on the right-hand side of the first of equations (7.4). We now transform the second term on the right-hand side of equation (7.21) with

1c∂∂teRR

⎡⎣⎢

⎤⎦⎥ret

=1c∂∂t

R eRR2

⎡⎣⎢

⎤⎦⎥ret

=1c

eRR2

⎡⎣⎢

⎤⎦⎥ret

∂∂t

R[ ]ret + R[ ]ret∂∂t

eRR2

⎡⎣⎢

⎤⎦⎥ret

⎧⎨⎩

⎫⎬⎭

=eRR2

κ −1( )κ

⎡⎣⎢

⎤⎦⎥ret

+R[ ]retc

∂∂t

eRR2

⎡⎣⎢

⎤⎦⎥ret

=eRR2

⎡⎣⎢

⎤⎦⎥ret

−eRκR2⎡⎣⎢

⎤⎦⎥ret

+R[ ]retc

∂∂t

eRR2

⎡⎣⎢

⎤⎦⎥ret.

(7.22)

Substituting equations (7.21) and (7.22) into the first of equations (7.4) yields the Feynman formula for the electric field due to a moving point charge

E x,t( ) = q4πε0

eRR2

⎡⎣⎢

⎤⎦⎥ret

+R[ ]retc

∂∂t

eRR2

⎡⎣⎢

⎤⎦⎥ret

+1c2

∂2

∂t 2eR[ ]ret

⎧⎨⎩

⎫⎬⎭. (7.23)

For the magnetic induction field, we start with

18

1c∂∂tv × eRκR

⎡⎣⎢

⎤⎦⎥ret

=1c

1R[ ]ret

∂∂tv × eRκ

⎡⎣⎢

⎤⎦⎥ret

+v × eRκ

⎡⎣⎢

⎤⎦⎥ret

∂∂t

1R

⎡⎣⎢

⎤⎦⎥ret

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

=1

c R[ ]ret∂∂tv × eRκ

⎡⎣⎢

⎤⎦⎥ret

−1cv × eRκR2

⎡⎣⎢

⎤⎦⎥ret

∂∂t

R[ ]ret

=1

c R[ ]ret∂∂tv × eRκ

⎡⎣⎢

⎤⎦⎥ret

−v × eRκR2

κ −1( )κ

⎡⎣⎢

⎤⎦⎥ret

=1

c R[ ]ret∂∂tv × eRκ

⎡⎣⎢

⎤⎦⎥ret

+v × eRκ 2R2

⎡⎣⎢

⎤⎦⎥ret

−v × eRκR2

⎡⎣⎢

⎤⎦⎥ret.

(7.24)

Inserting this result in the second of equations (7.4) yields the Heaviside formula for the magnetic induction field due to a moving point charge

B x,t( ) = q µ04π

v × eRκ 2R2

⎡⎣⎢

⎤⎦⎥ret

+1

c R[ ]ret∂∂tv × eRκ

⎡⎣⎢

⎤⎦⎥ret

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪. (7.25)

8. Conservation of angular momentum. Show that the differential and integral forms of the law of conservation of angular momentum are

∂∂t

Lmech +Lfield( ) +∇ ⋅!M = 0, (8.1)

and

ddt

Lmech +Lfield( )V∫ d 3x + n ⋅

!M da

S∫ = 0, (8.2)

where the field angular momentum density is

Lfield = x × g =

1c2x × E ×H( ), (8.3)

and the flux of angular momentum is described by the tensor

!M =

!T × x. (8.4)

Solution. The mechanical force density F on a charge density ρ (of current density J ) subjected to electromagnetic fields is

19

F = ρE + J × B. (8.5) The corresponding density of the mechanical torque N is

N ≡∂Lmech

∂t= x × F

= x × ρE + J × B( ). (8.6)

Referring to equations (4.78) to (4.85) (pp. 97-98) of the lecture notes, it can be shown that equation (8.6) can be transformed to

∂Lmech

∂t= x × ∇ ⋅

!T −

∂g∂t

⎛⎝⎜

⎞⎠⎟, (8.7)

where

!T and g are the Maxwell stress tensor, and the electromagnetic momentum

density vector, respectively. If we define the field angular momentum density with equation (8.3), then

∂∂t

Lmech +Lfield( ) = −∇ ⋅!T × x. (8.8)

We now evaluate the term on the right-hand side of equation (8.8)

∇ ⋅!T × x( )i = εijk∂mTjmxk

= ∂m εijkTjmxk( ) − εijkTjm∂mxk

= ∂m εijkTjmxk( ) − εijkTjmδkm

= ∂m εijkTjmxk( ) − εijkTjk

= ∂m εijkTjmxk( ),

(8.9)

since εijkTjk = 0 , from the symmetry of the Maxwell stress tensor (i.e., Tij = Tji ; see equation (4.82) of the lecture notes). Inserting equation (8.9) in equation (8.8) yields

∂∂t

Lmech +Lfield( ) +∇ ⋅!M = 0, (8.10)

with

!M =

!T × x (according to equation (8.9),

!M has two indices i and j , and is therefore

a second rank tensor). Finally, integrating equation (8.10) over space will give us

ddt

Lmech +Lfield( )V∫ d 3x + ∇ ⋅

!M d 3x

V∫ = 0, (8.11)

20

or, using the divergence theorem,

ddt

Lmech +Lfield( )V∫ d 3x + n ⋅

!M da

S∫ = 0. (8.12)

9. Suppose you had an electric charge qe and a magnetic monopole qm (as far as we know, they don’t exist; see Jackson sections 6.11 and 6.12). The field of the electric charge is

E r( ) = 14πε0

qer2er , (9.1)

with the electron located at the origin, and the field of the magnetic monopole is

B r( ) = µ04π

qm′r 2 e ′r , (9.2)

with ′r = r − dez (the magnetic monopole is located a distance d away from the electron and on the z-axis ). Find the total angular momentum stored in the fields. This is a static problem.

Solution. With the problem as stated

E r( ) = qe4πε0

rr3

B r( ) = qmµ04π

′r′r 3 = qm

µ04π

r − dez( )r2 + d 2 − 2rd cos θ( )⎡⎣ ⎤⎦

32, (9.3)

with z = r cos θ( ) . The electromagnetic momentum density is

g = ε0 E × B( )

= −qeqmdµ016π 2

r × ez( )r3 r2 + d 2 − 2rd cos θ( )⎡⎣ ⎤⎦

32. (9.4)

The angular momentum density is

21

L = r × g

= −qeqmdµ016π 2

r × r × ez( )r3 r2 + d 2 − 2rd cos θ( )⎡⎣ ⎤⎦

32

= −qeqmdµ016π 2

r2 cos θ( )er − ez⎡⎣ ⎤⎦r3 r2 + d 2 − 2rd cos θ( )⎡⎣ ⎤⎦

32.

(9.5)

But since er has components along ex and ey , and that are proportional, respectively, to sin θ( )cos ϕ( ) and sin θ( )sin ϕ( ) , then

Lx = Lxr2 sin θ( ) dr dθ dϕ∫ ∝ cos ϕ( ) dϕ

0

2π

∫ = 0

Ly = Lyr2 sin θ( ) dr dθ dϕ∫ ∝ sin ϕ( ) dϕ

0

2π

∫ = 0. (9.6)

We are left with

Lz = Lzr2 sin θ( ) drdθdϕ∫

= −qeqmdµ016π 2 dϕ dθ dr

r sin θ( ) cos2 θ( ) −1⎡⎣ ⎤⎦r2 + d 2 − 2rd cos θ( )⎡⎣ ⎤⎦

320

∞

∫0

π

∫0

2π

∫ . (9.7)

We make the change u ≡ cos θ( ) , then

Lz = qeqmdµ08π

du 1− u2( ) dr r

r2 + d 2 − 2rdu( )320

∞

∫−1

1

∫ . (9.8)

The radial integral gives

dr r

r2 + d 2 − 2rdu( )320

∞

∫ =ru − d( )

d 1− u2( ) r2 + d 2 − 2rdu0

∞

=1

d 1− u( ) , (9.9)

and

22

Lz = qeqmµ08π

1− u2

1− u⎡

⎣⎢

⎤

⎦⎥du−1

1

∫

= qeqmµ08π

1+ u( )du−1

1

∫= qeqm

µ04π.

(9.10)

The total angular momentum is therefore

L = qeqmµ04πez . (9.11)

[Note: This result is independent of the separation between the charges. From quantum mechanics, we know that the angular momentum comes in half-integer multiples of ! . So, this result suggests that if magnetic monopoles exist, electric and magnetic charges must be quantized. That is, qeqmµ0 4π = n! 2 , for n = 1, 2, 3,… ; an idea proposed by Dirac in 1931. If even only one monopole exists somewhere in the Universe, this would explain why the electric charge comes in distinct units (from Griffiths, p. 362).] 10. A classical atomic electron of charge −q circles about a nucleus of charge Q on a stable orbit of radius r . The centripetal acceleration provided by the Coulomb attraction between opposite charges is counterbalanced by the centrifugal force mv2 r due to the motion of the electron; m and v are the mass and the speed of the electron, respectively. a) Show that the kinetic energy of the electron is given by

T ≡12mv2 = 1

8πε0

qQr. (10.1)

b) A very small magnetic induction field dB is slowly turned on, perpendicular to the plane of the orbit. Assume that this change causes the electron to move at a new speed v1 = v + dv on a new orbit of radius r1 = r + dr with dv≪ v and dr ≪ r , show that the kinetic energy becomes

T1 ≡12mv1

2

!18πε0

qQr

1− drr

⎛⎝⎜

⎞⎠⎟+12qvrdB.

(10.2)

c) Use Faraday’s Law to show that the increase in kinetic energy dT ≡ T1 − T is just right to sustain the circular motion at the same radius. That is, show that dr = 0 .

23

[Hints: i) Keep your calculations to first order, while assuming that dB is of first order (e.g., dvdB ! 0 ).

ii) When considering a change dr on the orbital radius r , you can assume that it is infinitesimal such that, for example, 1+ dr r( )−1 ! 1− dr r . The same would be true for the speed.]

Solution. a) Since the system is initially in equilibrium, we must have from Lorentz force

mv2

r=

14πε0

qQr2, (10.3)

where m and v are, respectively, the mass and the speed of the electron, and therefore the kinetic energy T of the electron is

T =12mv2 = 1

8πε0

qQr. (10.4)

b) After the magnetic induction field is turned on, the electron is on a new orbit r1 = r + dr , with the assumption that dr ≪ r , and moves at a new velocity v1 . Therefore, the Lorentz force tells us that

mv12

r1=

14πε0

qQr12 + qv1dB, (10.5)

and to first order

T1 =12mv1

2

=18πε0

qQr1

+12qv1r1dB

=18πε0

qQr + dr( ) +

12q v + dv( ) r + dr( )dB

!18πε0

qQr

1− drr

⎛⎝⎜

⎞⎠⎟+12qvrdB.

(10.6)

c) The change in kinetic energy is

dT = T1 − T !

qvr2dB −

18πε0

qQr2

dr. (10.7)

24

However, according to the Faraday law of induction we have for the induced electric field

E ⋅dl!∫ = 2πrE

= −ddt

B ⋅n da∫= −πr2 dB

dt,

(10.8)

or

E = −r2dBdt. (10.9)

So, the force imparted on the electron by the induced electric field is

m dvdt

= −qE =qr2dBdt, (10.10)

or

mdv = qr2dB. (10.11)

The increase in kinetic energy is therefore

dT = d 12mv2⎛

⎝⎜⎞⎠⎟= mvdv = qvr

2dB, (10.12)

and comparing equations (10.7) and (10.12) we find that dr = 0. (10.13) Therefore, the size of the orbit remains the same after the apparition of the (infinitesimal) magnetic induction field dB . 11. Electromagnetic Fields Inside a Conductor. To make a current flow in a conductor a force must be applied. It is usually found that for most substance the current density J and the force per unit charge f are related through J = σ f , (11.1)

25

with σ the conductivity of the substance. But when the force driving the current is electromagnetic in nature, equation (11.1) takes the form of the so-called Ohm’s Law J = σ E + v × B( ). (11.2) It is normally the case that the mean velocity of the charges within the conductor is small enough that Ohm’s Law is approximated to J = σE. (11.3) a) We know that in electrostatic the electric field and the charge density within a conductor are zero, but this is not necessarily the case for time varying fields. This, in fact, should be obvious from Ohm’s Law. Assume that we are given a charge density ρ and a current density J , and use Gauss’ Law, Ohm’s Law, and the continuity equation to show that for a homogeneous medium of permittivity ε the time evolution of the charge density is

ρ t( ) = ρ 0( )e−σεt. (11.4)

Equation (11.4) implies that any charge density will eventually dissipate in a time inversely proportional to the conductivity. b) Since we know that any charge density will eventually dissipate, set ρ = 0 in Maxwell’s equations and use Faraday’s Law and the Ampère-Maxwell Law to show that the medium allows the propagation of attenuated waves such as E x,t( ) = E0e−κ⋅xei k⋅x−ω t( ), (11.5) where k and κ are collinear and

k =ω µε2

1+ σωε

⎛⎝⎜

⎞⎠⎟2

+1⎡

⎣⎢⎢

⎤

⎦⎥⎥

1 2

κ =ω µε2

1+ σωε

⎛⎝⎜

⎞⎠⎟2

−1⎡

⎣⎢⎢

⎤

⎦⎥⎥

1 2

.

(11.6)

Equations (11.5) and (11.6) show that electromagnetic fields can only exist inside a conductor for a distance of the order of the so-called skin depth δ , with

δ ≡1κ. (11.7)

26

Also, show that the electric and magnetic fields are transverse to the direction of propagation and perpendicular to each other. c) Assume that a plane wave propagating with a wave vector ′k is incident on the conductor’s surface, whose outward normal unit vector is n . This plane wave will excite current and charge motions in the conductor such that electromagnetic fields (as expressed by equation (11.5)) will be induced. Consider the boundary conditions stemming from Maxwell’s equations, and show that for a good conductor (i.e., where σ ≫ωε ) the electric and magnetic fields E1 and H1 outside the conductor are related at its surface by E1 surface = ZsH1 surface , (11.8) with Zs the surface impedance of the conductor defined with

Zs =1− i( )δσ

. (11.9)

Solution. a) Using Gauss’ Law, equation (11.3) for Ohm’s Law, and the continuity equation we find

∇ ⋅E =1σ∇ ⋅ J = −

1σ∂ρ∂t

=ρε, (11.10)

or

ρ t( ) = ρ 0( )e−σεt. (11.11)

b) Faraday’s Law and the Ampère-Maxwell Law (using Ohm’s Law) can be written as

∇ × E = −

∂B∂t

∇ × B = µσE + µε ∂E∂t. (11.12)

Taking the curl of both equations we get

∇ ∇ ⋅E( ) − ∇2E = −µσ ∂E

∂t− µε ∂

2E∂t 2

∇ ∇ ⋅B( ) − ∇2B = −µσ ∂B∂t

− µε ∂2B∂t 2

. (11.13)

27

But since ∇ ⋅B = ∇ ⋅E = 0 (from equation (11.11) for the electric field) then

∇2E = µε ∂

2E∂t 2

+ µσ ∂E∂t

∇2B = µε ∂2B∂t 2

+ µσ ∂B∂t. (11.14)

It is natural to test whether these equations allow harmonic fields of the type E x,t( ) = E0ei

!k⋅x−ω t( ) (11.15) as a solution. Inserting equation (11.15) into the first of equations (11.14) we find

!k2E = µ ω 2ε + iωσ( )E, (11.16)

which will yield an acceptable solution if

!k2= µ ω 2ε + iωσ( ). (11.17)

If we define !k = k + iκ, (11.18) then

!k2= !k ⋅ !k = k2 −κ 2 + 2ik ⋅ κ

=ω 2µε + iωσµ. (11.19)

But since 2k ⋅ κ =ωµσ for any frequency, and that µ and σ are assumed constant and isotropic, then k and κ must be collinear (that is, k ⋅ κ = kκ ) and

k ≡ k =ω µε2

1+ σωε

⎛⎝⎜

⎞⎠⎟2

+1⎡

⎣⎢⎢

⎤

⎦⎥⎥

1 2

κ ≡ κ =ω µε2

1+ σωε

⎛⎝⎜

⎞⎠⎟2

−1⎡

⎣⎢⎢

⎤

⎦⎥⎥

1 2

.

(11.20)

It follows that the electric field can be expressed as

28

E x,t( ) = E0e−κ⋅xei k⋅x−ω t( ), (11.21) with k and κ collinear. But because Gauss’ Law specifies that !k ⋅E = 0 , and since furthermore ∇ ⋅B = !k ⋅B = 0 , then it must also be that the fields are transverse to the direction of propagation. Finally, the Faraday and Maxwell-Ampère Law dictate that

!k × E =ωB!k × B = −µ ωε + iσ( )E,

(11.22)

which imply that E and B are also perpendicular to one another. c) We now assume that a plane wave propagating with a wave vector ′k is incident on the conductor’s surface, whose outward normal unit vector is n . This plane wave excites currents and charge motions in the conductor such that electromagnetic fields like expressed by equation (11.21) are induced. If the conductor is not perfect (although it is good), then there will not be any surface current on its surface. Instead a current density J = σE will be induced, with E the electric field inside the conductor. We first need to work out the relationship between the wave vectors k and ′k that respectively define the directions of propagation for the induced and incident waves, and n the unit normal at the boundary of the conductor. If we denote the electric and magnetic fields outside the conductor by E1 and H1 (made of the incident fields and whatever other fields that might be scattered by the conductor), then the boundary condition brought up by Faraday’s Law states that n × E1 − E( ) = 0. (11.23) The tangential components of the fields are continuous. If we call θ and ′θ the respective angles made by k and ′k with n , then at a point on the surface of the conductor some distance rb away (from the arbitrary origin in the direction of a unit tangential vector t ) the phases of the fields inside and outside the surface of the conductor will respectively be krb sin θ( ) and ′k rb sin ′θ( ) , which must be equal everywhere on the boundary at all times. This will only be achieved if k sin θ( ) = ′k sin ′θ( ). (11.24) For a good conductor such that σ ≫ωε we have

k !κ ! ωµσ

2!1δ, (11.25)

and at a given frequency ω we find that

29

limσ→∞

′kk=ωc

2ωµσ

→ 0. (11.26)

That is, the wave induced in the conductor must travel in a direction almost normal to the boundary (of course, this procedure is equivalent to applying Snell’s Law for the refraction of waves). In other words, we can to a good degree of precision make the following substitution k→ k + ′k sin ′θ( )t, (11.27) and the amplitude of the induced electric and magnetic fields E0 and B0 induced in the conductor are practically parallel to the boundary. So, if E! is the amplitude of the tangential component of the electric field at the exterior of the conductor’s surface, then the electric field within the conductor must be (from equation (11.23)) E x,t( ) ! E"e−κ⋅xei k⋅x+ ′k sin ′θ( )t ⋅x−ω t⎡⎣ ⎤⎦ . (11.28) Now, since for a good conductor the fields are confined to a very small skin depth δ , it is tempting to assume that a surface current density K is induced on the conductor. We find this surface current density by integrating the component of the current density J = σE (which is basically tangential to the surface of the conductor) along a direction perpendicular to the boundary. That is,

K = σ E!e−κ⋅xei k⋅x+ ′k sin ′θ( )t ⋅x−ω t⎡⎣ ⎤⎦dxn0

∞

∫=

σκ − ik( )E!e

i ′k sin ′θ( )t ⋅x−ω t⎡⎣ ⎤⎦

=δσ1− i( )E!e

i ′k sin ′θ( )t ⋅x−ω t⎡⎣ ⎤⎦ ,

(11.29)

where the notation dxn was used to make it apparent that the integration is along a path normal to the boundary. Alternatively, with E1 the electric field exterior to the conductor, we can invert equation (11.29) to write that at the surface of the conductor E1 surface = ZsK, (11.30) where we introduced the surface impedance of the conductor

Zs =1− i( )δσ

= 1− i( ) ωµ2σ. (11.31)

Finally, the boundary condition stemming from the Ampère-Maxwell Law states that

30

n ×H1 = n ×H = K, (11.32) or E1 surface = ZsH1 surface . (11.33)

12. Find all the elements of the Maxwell stress tensor !T for a monochromatic plane wave

traveling in the z direction and linearly polarized in the x direction. Does your result make sense, considering that

!T is the momentum flux density, and that

!T ⋅da is the rate

at which momentum crosses an area da ? How is the momentum flux related to the energy density? Solution. The components of the Maxwell stress tensor are given by

Tij = ε0 EiEj + c2BiBj −

12δ ij EkEk + c

2BkBk( )⎡⎣⎢

⎤⎦⎥. (12.1)

With the electric field given by E x,t( ) = exE0ei kz−ω t( ), (12.2) then it must be that

B x,t( ) = eyE0c2ei kz−ω t( ). (12.3)

With these fields, all the “off-diagonal” components (i.e., for i ≠ j ) in equation (12.1) will be zero. The diagonal components are

Txx = ε0 Ex2 −12

E2 +c2

c2E2⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ = 0

Tyy = ε0c2

c2Ey2 −12

E2 +c2

c2E2⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ = 0

Tzz = −ε02

E2 +c2

c2E2⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ = −ε0E

2 = −u,

(12.4)

where u is the energy density (see equation (4.71) of the lecture notes). So, we have

31

Tzz = −ε0E02 cos2 kz −ωt( ), (12.5)

with all the other components zero. The momentum of these fields is in the z direction, and it is being transported in the same direction. So, it makes sense that Tzz should be the only not zero component of the Maxwell stress tensor. We also know that −

!T ⋅da is the

rate at which the momentum crosses (i.e., leaving) an area da . Here we have no momentum transported in the x and y directions. The momentum per unit time per unit area flowing across the surface oriented in the z direction is −Tzz = u = g c (see equations (4.81) and (4.82) of the lecture notes), so Δp = g cAΔt , and hence Δp Δt = g cA = momentum per unit time crossing area A . Evidently, the momentum flux is the energy density.

13. A localized electric charge distribution produces an electrostatic field, E = −∇Φ . Into this field is placed a small, localized time-independent current density J x( ) , which generates a magnetic field H .

a) Show that the momentum of these electromagnetic fields can be transformed to

Pfield =1c2

ΦJd 3x∫ (13.1)

provided that the product ΦH falls off rapidly enough at large distances. How rapidly is “rapidly enough”? b) Assuming that the current distribution is localized to a region small compared to the scale of variation of the electric field, expand the electrostatic potential in a Taylor series and show that

Pfield =1c2E 0( ) ×m, (13.2)

where E 0( ) is the electric field at the location of the current density and m is the magnetic moment caused by the current density. c) Suppose the current distribution is placed instead in a uniform electric field E0 (filling all space). Show that, no matter how complicated is the localized current density J , the result in part a) is augmented by a surface integral contribution from infinity equal to minus one-third the result of b), yielding

Pfield =23c2E0 ×m. (13.3)

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Solution. a) From Equation (4.86) of the Lecture Notes we have for the electromagnetic linear momentum

Pfield =1c2

E ×H d 3x∫= −

1c2

∇Φ ×H d 3x∫= −

1c2

∇ × ΦH( ) − Φ∇ ×H⎡⎣ ⎤⎦ d3x∫ .

(13.4)

Upon using Equation (1.31) of the Lecture Notes the first integral can be transformed into a surface integral such that

Pfield = −1c2

n × ΦH( ) da∫ +1c2

Φ∇ ×H d 3x∫=1c2

Φ∇ ×H d 3x∫=1c2

ΦJ d 3x∫ ,

(13.5)

since the surface integral will vanish at a large distance r if ΦH falls faster than r−2 when r→∞ and there is no displacement current because the electric field is time-independent. b) Expanding the scalar potential in a Taylor Series about x = 0 (where the current density is located) we have

Φ x( ) ! Φ 0( ) + x ⋅∇Φ x( ) x=0! Φ 0( ) − x ⋅E 0( )

(13.6)

when the scale of variation for E x( ) is large compared to the size of J x( ) . Inserting Equation (13.6) into Equation (13.5) we find

Pfield =Φ 0( )c2

J d 3x∫ −Ei 0( )c2

xiJ d3x∫ . (13.7)

However, we know from the discussion following Equation (3.33) of the Lecture Notes that J d 3x∫ = 0 (13.8)

33

and

Ei 0( ) xiJ d

3x∫ = −12E 0( ) × x × J( )d 3x∫

= −E 0( ) ×m, (13.9)

where the magnetic dipole moment is given by Equation (3.40) of the Lecture Notes. It follows from equations (13.7) to (13.9) that

Pfield =1c2E 0( ) ×m. (13.10)

c) If the electric field is uniform, then the scalar potential must be linear in x and Equation (13.6) is exact (when E 0( ) is replaced by E0 ). We must, however, start with Equation (13.4) instead of the last of Equation (13.5) since there is no guarantee that ΦH will fall faster than r−2 at large distances (in fact, at large distances the field due to the magnetic dipole moment m , which goes as r−3 , should dominate and therefore ΦH ∝ r−2 ). 1st way (as required) We then proceed with Equation (13.4)

Pfield = −1c2

∇ × ΦH( ) d 3x∫ +1c2

ΦJ d 3x∫= −

1c2

Φ 0( ) ∇ ×H d 3x∫ − E0i ∇ × xiH( ) d 3x∫⎡⎣

⎤⎦ +

1c2

ΦJ d 3x∫=E0ic2

n × xiH( ) da∫ +1c2

ΦJ d 3x∫ .

(13.11)

But at a very large distance r , we have

H x( ) = 14πr3

3n n ⋅m( ) −m⎡⎣ ⎤⎦ (13.12)

and therefore

n × xiH( ) = xi n ×H( )

=xi4πr3

n × 3n n ⋅m( ) −m⎡⎣ ⎤⎦{ }= −

xi4πr3

n ×m( ).

(13.13)

34

We can then write

E0i n × xiH( ) da∫ = −E0i

xi4πr3

n ×m( ) r2 dΩ∫= m × E0i

xi4πr

n dΩ∫⎡⎣⎢⎤⎦⎥.

(13.14)

By choosing the surface of integration to be a sphere centered at the origin (and of radius r→∞ ) we can write n = er , and since dΩ = sin θ( )dθdϕ the only term from E0 ⋅x that will yield a non-zero integral is E0zz = E0zr cos θ( ) . Moreover, we can also write E0zz = E0r cos θ( ) − E0θ sin θ( )⎡⎣ ⎤⎦r cos θ( ), (13.15) which when inserted in Equation (13.14) yields

E0i n × xiH( ) da∫ =12m × er( ) E0r cos

2 θ( ) − E0θ sin θ( )cos θ( )⎡⎣ ⎤⎦sin θ( )dθ0

π

∫=13E0r m × er( )

= −13E0 ×m.

(13.16)

From Equations (13.5), (13.9), (13.11), and (13.16) we have

Pfield =23c2E0 ×m. (13.17)

2nd way We proceed from and transform Equation (13.4) to

Pfield = −1c2

∇ × ΦH( ) d 3x∫ +1c2

ΦJ d 3x∫= −

1c2

Φ 0( ) ∇ ×H d 3x∫ − E0i ∇ × xiH( ) d 3x∫⎡⎣

⎤⎦ +

1c2

ΦJ d 3x∫=E0ic2

∇ × xiH( ) d 3x∫ +1c2

ΦJ d 3x∫

(13.18)

where Equation (13.8) was used. Instead of transforming the first integral into a surface integral evaluated, we will keep it as a volume integral and transform it as follows

35

∇ × xiH( ) d 3x∫⎡⎣ ⎤⎦k = εkmn ∂m xiHn( )d 3x∫

= εkmn δmiHn + xi∂mHn( )d 3x∫= εkin Hn d

3x∫ + εkmn xi∂mHn d3x∫ ,

(13.19)

and E0i ∇ × xiH( ) d 3x∫ = E0 × H d 3x∫ + E0i xi∇ ×H d 3x∫ . (13.20) But we also know from the discussion leading to, and from, Equation (3.53) that

Hd 3x∫ =23m, (13.21)

which implies that (also using Equation (13.9))

E0i ∇ × xiH( ) d 3x∫ = E0 ×23m⎛

⎝⎜⎞⎠⎟+ E0i xiJ d

3x∫

=23E0 ×m − E0 ×m

= −13E0 ×m.

(13.22)

It follows from the result obtained in b) that

Pfield =23c2E0 ×m. (13.23)

3rd way Equation (13.3) could have been derived directly from the first of Equation (13.4) and Equation (13.21) since the electric field is uniform over all space.