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Physics 506 Winter 2006

Homework Assignment #10 Solutions

Textbook problems: Ch. 12: 12.15, 12.16, 12.19, 12.20

12.15 Consider the Proca equations for a localized steady-state distribution of current thathas only a static magnetic moment. This model can be used to study the observ-able effects of a finite photon mass on the earths magnetic field. Note that if themagnetization is M(x ) the current density can be written as J = c( M).

a) Show that if M = mf(x ), where m is a fixed vector and f(x ) is a localized scalarfunction, the vector potential is

A(x ) = m

f(x )e|xx

|

|x x | d3x

In the static limit, the Proca equation

[ + 2]A =

4

cJ

takes the form

[2 2]A = 4c

J

This admits a time independent Greens function solution

A(x) =1

c

J(x

)G(x, x)d3x

where

G(x, x) = e|xx

|

|x x |Taking J = c( M) with M = mf(x ) gives

J = c ( mf(x )) = cf m = c m fThen

A = m

f(x ) e|xx |

|x x | d3x

Integration by parts (assuming the surface term vanishes since the source is lo-calized) gives

A = m

f(x )

e|xx

|

|x x |

d3x

= m

f(x )

e|xx|

|x x |

d3x

= m

f(x )e|xx

|

|x x | d3x

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where we made use of the fact that G(x, x) = G(x, x).b) If the magnetic dipole is a point dipole at the origin [f(x ) = (x )], show that

the magnetic field away from the origin is

B(x ) = [3r(r m) m]1 + r + 2r2

3 er

r3 23

2

m

er

r

For f(x ) = (x ) the resulting vector potential is

A = m

er

r

= (1 + r)

er

r3m r

The magnetic field is then

B = A = (1 + r)er

r3 ( m r ) + (1 + r)er

r3 ( m r )

= (3 + 3r + 2r2) er

r3r ( m r)

+ (1 + r)er

r3( m( r) ( m )r )

= (3 + 3r + 2r2) er

r3( m r(r m )) + (2 + 2r) e

r

r3m

= (3r(r m) m)

1 + r +2r2

3

er

r3 232 m

er

r

c) The result of part b) shows that at fixed r = R (on the surface of the earth), theearths magnetic field will appear as a dipole angular distribution, plus an addedconstant magnetic field (an apparently external field) antiparallel to m. Satelliteand surface observations lead to the conclusion that this external field is lessthan 4 103 times the dipole field at the magnetic equator. Estimate a lowerlimit on 1 in earth radii and an upper limit on the photon mass in grams fromthis datum.

At the magnetic equator we have r m = 0. Hence

Bdipole =

m(1 + R +

2R2

3

)eR

R3

, Bexternal =

m(23

2R2)eR

R3

Setting | Bdipole|/| Bexternal| < 4 103 gives23 (R)

2 < 4 103(1 + R + 13 (R)2)

or R < 0.08. The lower limit on 1 is then

1 > 12.5R = 8.0 109 cm

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where we have used the radius of the earth R = 6.38 108 cm. This correspondsto an upper limit on the photon mass

m =h

c=

1.05 1027 ergs(8.0 109 cm)(3 1010cm/s) = 4.4 10

48 gm

12.16 a) Starting with the Proca Lagrangian density (12.91) and following the same pro-cedure as for the electromagnetic fields, show that the symmetric stress-energy-momentum tensor for the Proca fields is

=1

4

gFF

+1

4gFF

+ 2

AA 12

gAA

The Proca Lagrangian density is

L=

1

16

FF +

1

8

2AA

Since

T =L

A A L

we find

T = 14

FA +1

16F2 1

82A2

where we have used a shorthand notation F2 FF and A2 AA. Inorder to convert this canonical stress tensor to the symmetric stress tensor, wewrite A = F

+ A

. Then

T = 14

[FF 14F2 + 122A2] 14 FA

= 14

[FF 14F2 + 122A2 (F)A]1

4(F

A)

Using the Proca equation of motion F + 2A = 0 then gives

T = + S

where

= 14 F

F 14F2 2(AA 12A2) (1)is the symmetric stress tensor and S = (1/4)FA is antisymmetric on thefirst two indices.

b) For these fields in interaction with the external source J, as in (12.91), show thatthe differential conservation laws take the same form as for the electromagneticfields, namely

=

JF

c

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Taking a 4-divergence of the symmetric stress tensor (1) gives

= 1

4

F

F + FF

12FF

2(AA + AAAA)= 1

4

F

F +12F(2

F F) + 2A(A A)

= 14

(F

+ 2A)F +12F(

F + F + F)

= 1c

JF =1

cJF

Note that in the second line we have used the fact that A = 0, which is

automatic for the Proca equation. To obtain the last line, we used the Bianchiidentity 3[F] = 0 as well as the Proca equation of motion.

c) Show explicitly that the time-time and space-time components of

are

00 =1

8[E2 + B2 + 2(A0A0 + A A)]

i0 =1

4[( E B)i + 2AiA0]

Given the explicit form of the Maxwell tensor, it is straightforward to show that

F2 FF = 2(E2 B2), A2 AA = (A0)2 A 2

Thus

= 14

FF +

12

(E2 B2) 2(AA 12((A0)2 A 2))

The time-time component of this is

00 = 14

F0iF0i +

12 (E

2 B2) 2((A0)2 12 ((A0)2 A 2))

= 14

12 (E2 + B2) 122((A0)2 + A 2)

= 18

E2 + B2 + 2((A0)2 + A 2)

Similarly, the time-space components are

0i = 14

F0jF

ij 2A0Ai = 14

Ej(ijkBk) 2A0Ai

= 1

4

ijkEjBk 2A0Ai = 14

( E B)i + 2A0Ai

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12.19 Source-free electromagnetic fields exist in a localized region of space. Consider thevarious conservation laws that are contained in the integral of M

= 0 over allspace, where M is defined by (12.117).

a) Show that when and are both space indices conservation of the total fieldangular momentum follows.

Note thatM = xx

Hence

M0ij = 0ixj 0jxi = c(gixj gjxi) = cijk(g x )k = cijk(x g )k

where g is the linear momentum density of the electromagnetic field. Since xgis the angular momentum density, integrating M0ij over 3-space gives the fieldangular momentum

Mij

M0ijd3x = cijk

(x g )k d3x = cijkLk

The conservation law Mij = 0 then corresponds to the conservation of angular

momentum in the electromagnetic field.

b) Show that when = 0 the conservation law is

d X

dt=

c2 PemEem

where X is the coordinate of the center of mass of the electromagnetic fields,defined by

X

u d3x =

xu d3x

where u is the electromagnetic energy density and Eem and Pem are the totalenergy and momentum of the fields.

In this case, we have

M0i M00id3x = (00xi 0ix0) d3x=

(uxi cgix0) d3x =

(uxi c2tgi) d3x

Making use of the definition

uxid3x = EXi where E =

u d3x is the total fieldenergy, we have simply

M0i = EXi c2tPi

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where P =

g d3x is the (linear) field momentum. Since M0i is a conservedcharge, its time derivative must vanish. This gives

0 =d

dt(EX) c2 d

dt(t P) = E

d X

dt c2 P

(where we used the fact that energy and momentum are conserved, namelydE/dt = 0 and d P/dt = 0). The result d X/dt = c2 P /E then follows.

12.20 A uniform superconductor with London penetration depth L fills the half-space x > 0.The vector potential is tangential and for x < 0 is given by

Ay = (aeikx + beikx)eit

Find the vector potential inside the superconductor. Determine expressions for theelectric and magnetic fields at the surface. Evaluate the surface impedance Zs (inGaussian units, 4/c times the ratio of tangential electric field to tangential magnetic

field). Show that in the appropriate limit your result for Zs reduces to that given inSection 12.9.

The behavior of the vector potential inside the superconductor may be describedby the massive Proca equation

2 1c2

2

t2 2

A = 0

Working with a harmonic time behavior eit, the Proca equation may be rewrit-ten as

[

2 + (2/c2

2)] A = 0

This has a generic solution of the form

A(x, t) = A0eikxit

where|k| =

2/c2 2 = i

2 2/c2

The second form of the square root is appropriate for sufficiently low frequencies.Since the vector potential outside the superconductor (x < 0) only points in they direction, and since the wave is normally incident (ie only a function of x), itis natural to expect the solution inside the superconductor to be of the form

Ay = (e

22/c2x + e

22/c2x)eit

for appropriate constants and . To avoid an exponentially growing behavior,we take = 0. Then it is straightforward to see that matching at x = 0 gives

Ay(x, t) =

(aeikx + beikx)eit x < 0

(a + b)e

22/c2xeit x > 0

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In the absence of a scalar potential, the electric and magnetic fields are

E(x = 0+) = 1c

tA

x=0+

=i

c(a + b)yeit

and B(x = 0+) = Ax=0+

=

2 2/c2(a + b)zeit

The surface impedance is given by

Zs =4

c

EyBz

= 4ic2

2 2/c2

Setting = 1/L and = 2c/ finally yields

Zs = 82i

c

L

(1 (2L/)2

)

1/2

This reduces in the long wavelength limit ( L) to the expected result

Zs = 82i

c

L