Solution Chap 13-14 Classical Elec

8/12/2019 Solution Chap 13-14 Classical Elec

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Phys 561 Classical Electrodynamics 2

Assume that we have no external charges and only a continuous polarization eld. The dipolemoment of a small cell of volume i can then be calculated to be (we choose to denote thetotal charge density by ):

p i =

i

d3 x x (x , t ) =

i

d3 x x P (x , t ) =

i

d3 x P (x , t ) (3)

Lets now return to the problem. Since we have no external charges, and effectively asingle dipole located at r0 (t), we can write the polarization eld as:

P (x , t ) = p (x r 0 (t)) (4)

We are also given that the dipole moment is not a function of position. Hence:

= P = (p ) = p p = (p ) (x r 0 (t)) (5)

For the current density we simply receive:

J = v = v(p ) (x r 0 (t)) (6)

(b)We are to calculate the magnetic dipole moment and the electric quadrupole moment. Theseare dened through the following equations, respectively:

m (x , t ) = 12c d3 x x J (x , t ) (7)

Qij (x , t ) = d3 x (3x ix j ij x 2 )(x , t ) (8)We will begin with the magnetic dipole. However, rst lets consider the following integral:

d3 x G f (x r 0 ) (9)where whether G is a scalar, vector or tensor function is immaterial. We rewrite the integralas:

d3 x G f i x i (x r 0 )Note that there are three terms in the integration. Consider just the ith term (i.e. no assumedsummation), and for convention let i denote (x i r 0 i):

dxidx j dxk G f i x i i j k == dx j dxk G f i i j k dxi x i (G f i) i j k=

x i

(G f i)x = r 0

Lets now switch the summation on, and also assume that the vector f is xed (i.e. no spacevariation, while time variation is still allowed):

x i

(G f i)x = r 0

= f i G x i

G f ix i x = r 0

= f i G x i x = r 0

= f G| x = r 0 (10)


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We have found an important result, which can immediately be tested on the electric dipolemoment. Recall that the electric dipole is dened in terms of the charge density in thefollowing way:

p (x , t ) =

d3 x x (x , t ) (11)

Using the result for the charge density obtained in part-a, and the above result for theintergral with G replaced by x and f replaced by p we get:

p (x , t ) = d3 x x p (x r 0 (t))= p x | x = r 0 = pi ix | x = r 0 = pi x i | x = r 0 = p

Condent in what we have so far been doing, we can now address the problem of calcu-lating the magnetic dipole moment:

m (x , t ) = 12c d

3

x

x

J (x

, t )=

12c d3 x x v(p ) (x r 0 (t))

= 1

2cp (x v)

x = r 0

= 1

2c pi i(x v)

x = r 0

= 1

2c pi( ix) v +

12c

pix ( iv )x = r 0

=

1

2c pi x i v + 1

2cx ( pi iv) x = r 0

= 1

2cp v +

12c

x [(p )v ]x = r 0

(12)

That the second term of this equation vanishes can be best seen by noting that at any instantin time r0 is xed in space (or has no gradient). This is true for the present case since r0 isa function of time only, and not a function of position. Thus:

x i

v(t) = x i

r 0 (t)t

= t

r 0 (t)x i

= 0

We therefore obtain the following result for the magnetic dipole:

m (t) = 12c

p v(t) (13)

The magnetic dipole is only time dependent through the velocity.We can see that this result is true by performing the calculations in completely different

way. We can treat a dipole as two particles of charge q and q , respectively, separated inspace by a vector d , pointing from the minus charge to the plus charge. Specically, supposethat the positive charge is located at R + d / 2 and that the negative charge is located atR d / 2, where R is the position of center of charge (whereby we do not require the particlesto posses equal amounts of mass). Then the charge density of the system can be convenientlywritten as a sum of Dirac delta functions:

= q (x R d / 2) q (x R + d / 2) (14)


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The electric dipole moment is simply:

p = d3 x x (x , t )= q (R + d / 2) q (R d / 2) = q d (15)

as one would expect. The magnetic dipole moment can be similarly calculated:

m (x , t ) = 12c d3 x x J (x , t )

= 12c d3 x x v(x , t )

= 12c

(q (R + d / 2) v q (R d / 2) v)

= 12c

p v (16)

We have only assumed that the particles are moving with the same velocity, an assumptionwhich is very proper for the present problem. We have made no specic requirements forthe separation distance, d between the charges; in fact it could be that the separation isinnitesimally small.

We have thus shown the validity of our previous calculations, and are now in a positionto attack the problem of the electric quadrupole moment. Lets rewrite the equation in theform:

Q = d3 x G (17)where Q and G are 3 3 tensors, and:

G = 3xx T I| x |2

=2x2 y2 z 2 3xy 3xz

3xy x2 + 2 y2 z 2 3yz 3xz 3yz x2 y2 + 2 z 2

(18)

Note the following in the above equation. First, I denotes the identity matrix. Second, therst term appearing in the denition of G involves a dyadic vector multiplication, where thesuperscript T stands for transpose.

By using the usual results, we receive:

Q = d3 x G (p ) (x r 0 )= ( p )G|x = r 0

Now consider the elements of Q separately:

Qij = ( p )Gij= pk k(3x ix j ij xlx l)= pk(3x i kx j + 3x j kxi 2 ij xl kx l)

= pk(3x i jk + 3x j ik 2 ij x l kl )= 3 x i p j + 3x j pi 2 ij xl pl (19)


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Calculating the terms and writing them in matrix form we get:

Q(t) = 3xp T + 3 px T 2 I xT p

=4xpx 2ypy 2zpz 3xpy + 3ypx 3zpx + 3 xpz

3xpy + 3ypx 2xpx + 4 ypy 2zpz 3ypz + 3zpy3zpx + 3 xpz 3ypz + 3zpy 2xpx 2ypy + 4zpz

(20)

where it is to be understood that the vector x is now actually r0 (t).

Problem 2A particle of charge e and mass m moves with speed v, v/c 1 in a uniform magnetic eld B . Suppose the motion is conned to the plane perpendicular to B.

a. Calculate the power radiated, P in terms of B and v, and show that:

P = dE dt

= E,

where E is the energy of the particle, and nd . This gives

E (t) = E (0)e t ,

1/ is the mean lifetime of the motion. For an electron, nd 1/ in seconds for a magnetic eld of 104 Gauss.

b. Describe qualitatively the trajectory of the particle as a function of time.

(a)We shall make use of the Lienard result (1898) 2 :

P = 2e2

3c31

(1 2 )3(v)2

1c2

(v v)2 (21)

Since we can expand the acceleration into parallel and perpendicular components with re-spect to the velocity, we can write:

(v)2 1

c2(v v)2 = ( v || )2 + ( v )2

1

c2(v )2 (v)2 = ( v || )2 + ( v )2 (1 2 ) (22)

whence the Lienard result becomes:

P = dE dt

= 2e2

3c31

(1 2 )3(v || )2 + ( v )2 (1 2 ) (23)

The minus sign appears since we choose to dene E as the energy of the particle, and P is just the rate of decrease of this energy. In the non-relativistic limit ( 1) and foracceleration perpendicular to velocity, as in the present case, we receive:

P = 2e2

3c3 (v )2

(24)2 See for instance Jackson [1], eq.14.26.


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The acceleration can be found to be:

a = v = Fm

= ev B

mc (25)

Since we assume that the motion is conned to a plane perpendicular to the magneticeld, the magnitude of acceleration is just:

a =eBmc

v (26)

It is interesting to note that the proportionality constant between speed and radius of trajectory is also the same as that between acceleration and speed:

v =eBmc

R (27)

The result for the power is then:

P = 2e4 B 2

3m2 c5v2 (28)

while the energy of the particle is given through:

E = 12

mv 2 (29)

We thus receive:

P = dE dt =

4e4 B 2

3m3 c5 E = E (30)

with an obvious denition for the constant . A quick dimensional check will certify that indeed has the proper dimensions of inverse time. The solution of this differential equationis trivial:

E (t) = E (0)e t (31)

For the specic values of:

e = 4.8 10 10 esuB = 104 Gaussm = 9 10 28 gc = 3 1010 cm/s

we get:

1

= 3m3 c5

4e4 B 2 = 2.50sec (32)


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Now, the second integral vanishes since it is the integral of an odd function (or rather theintegral of the multiplication of an odd function with an even function) over a symmetricinterval around the origin. Hence, reducing the interval of integration in half by making useof the symmetry we get:

c = 2z I l/2

0dz sin 1

2kl kz cos t R

ccos z cos

c

We now invoke a second good old friend (from the times when mathematics was simple, andso was the universe):

sin A cos B = 12

[sin(A + B) + sin( A B)] (40)

Using this equation in the integrand, we receive:

c = z I cos t R

c l/ 2

0 dz sin12kl kz +

zcos c + sin

12kl kz

zcos c

We can now safely carry out the integration:

c = z I cos t R

ccos(12 kl kz + z cos c )k cos c +

cos(12 kl kz z cos c )k + cos cl/ 2

0

= z I cos t R

ccos(lk cos 2 )k(1 cos ) + cos(

lk cos 2 )k(1 + cos ) cos(

lk2 )k(1 cos ) cos(

lk2 )k(1 + cos )

= z I cos t R

c coslk cos

2 coslk2

1k(1 cos ) +

1k(1 + cos )

= z 2Ik sin2

cos t R

ccos

lk cos 2

coslk2

Lastly, lets remind the reader of a third old -and again good- equation:

cos A cos B = 2sinA + B

2sin

A B2

(41)

This equation allows us to nalize our result for c:

c = z 4Icsin2

cos t Rc

sin lk4

(1 + cos )sin lk4

(1 cos ) (42)

Since c has only z component, the product n c yields:

n c = |c|sin sin

cos sin 0

(43)

whence the radiated power per solid angle becomes:

dP d =

14c3 (n c)

2

= 14c3

|c| 2 sin2 (44)


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Lets take the time averge of the power over a full cycle. The only time dependence comesabout from the term:

cos2 t R

c

whose time average is obviously 1/ 2. We thus obtain the time-averaged power radiated persolid angle:

dP d

= 2I 2 /c

sin2 sin2

lk4

(1 + cos )sin 2 lk4

(1 cos ) (45)

(b)Note that the above expression for the power per solid angle vanishes whenever one of thesine functions goes to zero. This, on the other hand, happens whenever:

lk4 (1 cos ) = n, n = 1, 2,...

This we can rewrite as:

cos = 4n

kl 1

or, equivalently:

cos = 1 4n

kl (46)

which is the desired result. By denition the right hand side may not exceed the interval[ 1, 1]. That is:

1 4n

kl 1 (47)

This completes the solution.

Problem 4A particle of charge e moves in a circular path of radius R in the x y plane with a constant

angular velocity 0 .a. Show that the exact expression for the angular distribution of power radiated into the mth multiple of 0 is

dP md

= e2 40 R 2

2c3 m2

dJ m (m sin )d(m sin )

2

+ cot2

2 J 2m (m sin )

where = 0 R/c , and J m (x) is the Bessel function of order m.b. Assume non-relativistic motion and obtain an approximate result for dP m /d for the

value of m that is most important.c. Assume extreme relativistic motion and obtain the results of Jackson [1], eq.14.79 for

a relativistic particle in instantaneously circular motion. 33 This is problem 14.15 of Jackson [1].


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(a)We will make use of an important formula derived in problem 14.13 of Jackson [1]. Here wequote the result: ... if the motion of a radiating particle repeats itself with periodicity T ,the continuous frequency spectrum becomes a discrete spectrum containing frequencies that

are integral multiples of the fundamental ... a general expression for the time-averaged power radiated per unit solid angle in each multiple m of the fundamental frequency 0 = 2/T is: 4

dP md

= e2 40 m2

(2c)3 2/ 0

0v (t) n exp im0 t

n x(t)c

dt2

(48)

We begin by writing the components of the vectors involved in the above integral:

x(t) = Rcos 0 tsin 0 t

0

, v(t) = R0 sin 0 tcos 0 t

0

, n =cos sin sin sin

cos

(49)

The dot and cross products can easily be evaluated:

n x = R(cos 0 t cos sin + sin 0 t sin sin ) = R cos( 0 t)sin

v n = R0i l k

sin 0 t cos0 t 0cos sin sin sin cos

= R0cos 0 t cos sin 0 t cos

cos( 0 t)sin (50)

Since the power will be symmetrically distributed over the azimuthal angle we set, withoutloss of generality, = 0. Then we have an integral of the following form:

= 2/ 0

0R0

cos 0 t cos sin 0 t cos

cos 0 t sin exp im0 t

R cos 0 t sin c

dt

2

(51)

corresponding to the magnitude of a vector with three independent components. Switch tovariable = 0 t, and set = 0 R/c , and x = m sin . Also dene:

I 1 = 2

0d cos eim ix cos

I 2 =

2

0d sin eim ix cos (52)

Then the above expression (eq.51), written as the sum of the absolute squares of the indi-vidual components, looks like:

= R2 |I 1 |2 cos2 + R2 |I 2 |2 cos2 + R2 |I 1 |2 sin2 = R2 |I 1 |2 + R2 |I 2 |2 cos2 (53)

We therefore need to evaluate the integrals I 1 and I 2 alone. That we now do. But beforethat we remind the reader of a couple of equations regarding Bessel functions. The Besselfunction of order m is given through:

J m (x) = 12i m

2

0 deix cos im

(54)4 Jackson [1], p.702.


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Since Bessel functions are real, this is equivalent to:

J m (x) = J m (x) = 1

2( i)m 2

0deim ix cos (55)

Therefore:

I 1 = 2

0d cos eim ix cos =

2

0d

1 i

ddx

eim ix cos = 2( i)m

id

dxJ m (x) (56)

The second integral, I 2 is a little more involved but still relatively easy. Dene:

u = eim

dv = d sin e ix cos (57)

and use partial integration:

I 2 = udv = uv| vdu= eim

1ix

e ix cos 2

0

imix

2

0deim ix cos =

2m ( i)m

x J m (x) (58)

Eq.53 now becomes:

= R2 |I 1 |2 + R2 |I 2 |2 cos2

= (2 R )2 ddx

J m (x)2

+m cos

x J m (x)

2

(59)

Substituting this into the equation for the radiated power, and recalling that x = m sin ,we get:

dP md

= e2 40 R 2

2c3 m2

dJ m (m sin )d(m sin )

2

+ cot2

2 J 2m (m sin ) (60)

(b)In the non-relativistic limit 0, so that:

J m (m sin ) = 12i m 2

0deim sin cos im

12i m

2

0de im (1 + im sin cos ) (61)

since cos cos . The rst term integrates to zero over the interval. The second termis also zero as long as m = 1 since exp( im ) is a function of cosm and sin m, and cos mis orthogonal to cos unless m = 1, while sin m is always orthogonal to cos . We thereforeconclude that J m (x) is nonzero only for m = 1.

Actually, by far, the largest element of the Bessel sequence is J 0 (x); though m = 0 is notphysical, and therefore is discarded. See for yourself that the radiated power is zero in thatcase. On the other hand, we also assume that m cannot be innitely large for a physicalsystem. And, in fact, should be relatively small in the present case.


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The derivative J m (x) can be analyzed by invoking the following relation, that can befound in any standard mathematical handbook 5 :

dJ m (x)dx

= J m +1 (x) + mx

J m (x) = mx

J m (x) + J m 1 (x) (62)

We observe that Bessel functions can be written in terms of one another in recursion relations.However, we proved above that only J 1 (x) is of signicant magnitude for nonrelativisticspeeds, which also implies that the argument x is very small, while we do not expect m to betoo large. Then the greatest contribution to J m (x) comes from the mJ m (x)/x term, whichis largest again for m = 1.

We therefore nd that in the nonrelativistic limit, only the fundamental mode counts.Lets now calculate the approximate values of the Bessel functions and subsitute them in thepower equation:

J 1 ( sin ) 1

2i 2

0de i i sin cos

= 12

2

0d sin cos2 =

12

sin (63)

and:

J 1 ( sin ) 1 sin

J 1 ( sin ) = 12

(64)

whence:

dP 1d

e2 40 R2

2c31

2

2

+

1

2 cos

2

=

e2 40 R2

8c3 (1 + cos2

) (65)References[1] Jackson, John David. Classical Electrodynamics. John Wiley & Sons, 3rd edition,

1998. (ISBN 0-471-30932-X)

[2] Lecture notes and homeworks.

[3] Abramowitz, M. and I. A. Stegun. Handbook of Mathematical Functions. Dover Pub-lications. (ISBN 0-486-61272-4)

[4] Hildebrand, Francis B. Advanced Calculus for Applications. Prentice Hall, 2nd edi-tion, 1976. (ISBN 0-13-011189-9)

[5] Watson, G.N. Theory of Bessel Functions. Cambridge University Press, 2nd edition,1952.

5 See, for instance, Abramowitz & Stegun [3].

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Solution Chap 13-14 Classical Elec