electrodynamics / Companion to J.D. Jackson's Classical Electrodynamics 3rd ed. - R. Magyar

Problem 9.3

A radiating

V (t) = V0 cos(ωt). There should be a diagram showing a sphere split across the equator. The top half is kept at a potential V (t) while the bottom half is −V (t). From Jackson 9.9,

If kd = k2R << 1 (as it is), the higher order terms in this expansion fall o rapidly. In our case, it is su cient to consider just the first term.

~

Integrating by parts and substituting · J = iωρ, we find

I solved the static situation (but neglected to include it) earlier.

So

|p0| = 32 4πǫ0V0R2

The time dependent dipole moment is

p~(t) = 6πǫ0V0R2 cos(ωt)ˆz

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The power radiated per solid angle can be obtained from the Poynting vector.

Notice how the complex conjugation and absolute signs get rid of the pesky wave factors.

Integrate over all solid angles to find the total radiated power.

Putting all this together, the final result is

3πǫ0V 2R4ω4

PT otal = 0

c3

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Problem 9.10

The transitional charge and current densities for the radiative transition from the excited state m = 0, 2p in hydrogen to the ground state 1s, i.e.

|H; 2pi → |1si

are, in the notation of (9.1) and with the neglect of spin, the matrix elements, h2p|ρˆ|1si:

the Bohr orbit speed.

~

a. Find the e ective transitional “magnetization”, calculate · M , and evaluate all the non-vanishing radiation multi-poles in the longwavelength limit.

~

~r × J = −ia0ρv0 (tan θ sin φxˆ − tan θ cos φyˆ)

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Obviously,

Z ∞

ue−f (u)du = 0

−∞

if f (u) is even. Thus, the integrals over the x and y coordinates vanish. We are left with

Plug in κ explicitly.

p~ = 1.49qa0zˆ

Now, for the magnetic moment,

Well, ρ is even but y and x are odd so m~ is zero. The magnetic dipole and electric quadrapole terms vanish because of their dependence on m.

We suspect that electric octo-pole and every other pole thereafter might persist because of symmetry, but we won’t worry about that.

b.In the electric dipole approximation calculate the total time-

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c.Interpreting the classically calculated power as the photon energy times the transition probability, evaluate numerically the transition probability in units of reciprocal seconds.

hω¯ = P . Using numbers, = 6.3 × 108 seconds−1 .

d.If, instead of the semi-classical charge density used above, the electron in the 2p state was described by a circular Bohr orbit of radius 2a0, rotating with the transitional frequency ω0, what would the predicted power be? Express your answer in the same units as in part b and evaluate the ratio of the two powers numerically.

For a Bohr transition, a dipole transition,

p~ = q(2a0 − a0)ˆze−iωt = qa0e−iωtzˆ

which gives an emitted power of PBohr = 0.018(¯hω0) α4c . And the ratio:

a0

PBohr 0.45

PJ ackson

The grader claims that this is incorrect citing a correct value of 0.55. You decide, and tell me what you conclude.

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Figure 1:

Problem 9.16

A thin linear antenna of length d is excited in such a way that the sinusoidal current makes a full wavelength of oscillation as shown in some Jacksonian figure.

a. Calculate exactly the power radiated per unit solid angle and plot the

¯

angular distribution of radiation. Assume the antenna is center fed.

Note J(±d2 ) = 0 as makes sense. Jackson makes some arguments to justify this current density for a center fed antenna. I’ll take his word for it, but if you’re not convinced, consult Jackson page 416 in the third edition.

The vector potential due to an oscillating current is

The vector potential with the current density can be explicitly written

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Bear in mind that this expression for the vector potential includes all multipoles. The integral can be done quite easily. Use Euler’s theorem, e−ix =

cos x − i sin x → sin x = ex−e−x , to write:

2i

Each integral can be solved quite easily by “ u ” substitution.

The result is a mess. Use Maple or have patience. It takes a bit of algebra to get the neat result,

Who cares about the vector potential? We want E and B fields. Fortunately, we know how to write the E and B fields in the radiation zone in terms of the vector potential.

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The time averaged angular distribution of power is

In this problem λ = d so kd = 2λπ d = 2π.

Well, cos π = −1, and of course, cos α + 12 = 2 cos2( α2 ).

b. Determine the total power radiated and find a numerical value for the radiation resistance.

Integrate the result from part a over all solid angles.

Obviously3 , the integral equals about 0.84.

I2µ0c P = (0.84) 0

π

We learned in high school that P = I2R and it does take much to show R = IP2 = µ80πc (6.7) ≈ 100Ω. Here, Ω stands for Ohms. Actually, Jackson seems to define the radiative resistance as 2 times this, but typically Jackson is hard to follow so I’ll ignore this factor without a better explanation about its origin.

3Solve the integral numerically.

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