THE UNIVERSITY OF BRITISH COLUMBIA

Physics 401 Assignment # 6:

Electromagnetic Waves

SOLUTIONS:

Wed. 8 Feb. 2006 - finish by Wed. 22 Feb.
1. CMBR: Most of the electromagnetic energy in the universe is in the cosmic microwave background radiation (CMBR), sometimes referred to as the Kelvin background. Penzias and Wilson discovered the CMBR in 1965 using a radio telescope, and subsequently received the Nobel Prize for this discovery. This background radiation has wavelength  mm. The energy density of the CMBR is about 4.0 x 10-14 J/m3. What is the rms electric field strength of the CMBR? ANSWER If  J/m3, then  . (The wavelength, while interesting, is irrelevant to the question.)

2. STANDING WAVES: Consider standing electromagnetic waves:

1. Show that these satisfy the wave equation (9.2). ANSWER When we're taking the spatial derivatives, the t-dependent factor is just part of the amplitude, and vice versa. Thus and ; and ; so and similarly for . But , since . Thus and similarly for

2. Show that we must also have and E0 = c B0. ANSWER Since is a universal property of all solutions of The Wave Equation (TWE), that's a given. Applying FARADAY'S LAW, , gives or . Dividing out the common factor gives or (since )  .

3. Show that the time-averaged power flow across any area will be zero. ANSWER
. Looking only at the t-dependence to get the time average, we note that which averages to zero.

4. Show that the Poynting vector will also be zero, i.e. there is no net energy flow. ANSWER I must apologize for a defective question. [The hazards of using someone else's problem!] As explained above, . This is only zero where , i.e. at z=0 and (where n is any integer). That is, for . At any other position, oscillates in the direction, averaging to zero.

3. (p. 386, Problem 9.14) - REFLECTED & TRANSMITTED POLARIZATION: In Eqs. (9.76) and (9.77) it was tacitly assumed that the reflected and transmitted waves have the same polarization as the incident wave, namely along the direction. Prove that this must be so. [Hint: Let the polarization vectors of the reflected and transmitted waves be

and prove from the boundary conditions that .] ANSWER We must have continuous across the boundary. Since the normal direction is , is constituted of x and y components. Thus or [1] and [2]. Similarly, must be continuous across the boundary, and, as always, , giving [3] and [4]. If , Eq. [4] reads , which we can combine with Eq. [2] to conclude that , which can be true only if ET = 0 (trivial case) or (mod ). Equation [2] then also requires (mod ).

4. (p. 392, Problem 9.15) - COMPLEX ALGEBRA EXERCISE: Suppose that we have six nonzero constants A,B,C,a,b,c such than Aeiax + Beibx = Ceicx for all x. Prove that a=b=c and A+B=C. ANSWER The first part is easy: if it were not true that a=b=c then even if the equation were satisfied at some position in x, it would not be satisfied at some nearby x. So a=b=c. The second part is even easier: at x=0, A+B=C. Done.

5. (p. 392, Problem 9.17) - DIAMOND: The index of refraction of diamond is 2.42. Construct the graph analogous to Figure 9.16 for the air/diamond interface. (Assume .) ANSWER FRESNEL'S EQUATIONS read

where and . In this case (we assume the light is entering the diamond rather than emerging) and . You can use your favourite spreadsheet or other plotting software to produce the graph below. (I used http://musr.org/muview/, a free Java spreadsheet applet we built at TRIUMF.)

In particular, calculate
1. the amplitudes at normal incidence; ANSWER For , , giving or and or  .

2. Brewster's angle; ANSWER or  .

3. and the "crossover" angle at which the reflected and transmitted amplitudes are equal. ANSWER Rather than try to read this off the graph, let's calculate it exactly: The condition is or or or or or or  .

6. PLANE WAVE STRESS TENSOR: Find all the elements of the Maxwell stress tensor of a monochromatic plane wave traveling in the z-direction, polarized in the x-direction:

ANSWER Recall Eq. (8.19) on p. 352:

Here where and where = E/c, so all off-diagonal elements are zero. We have or T11 = 0, or T22 = 0 and or (only nonzero element!)  .

In what direction does this EM wave transport momentum?  Does this agree with the form of the Maxwell stress tensor you just deduced? ANSWER If Tij represents the force per unit area acting in the direction on a surface whose normal is in the direction, then the diagonal elements are pressures and T33 is the radiation pressure on a surface normal to . In the same way -T33 represents the the momentum current density transported by the fields, and is (as expected) in the same direction as and is, in fact, equal to .

7.
(p. 412, Problem 9.33) - SPHERICAL WAVES: Suppose that

with , as usual. [This is, incidentally, the simplest possible spherical wave.  For notational convenience, let in your calculations.]

1. Show that obeys all four of Maxwell's equations, in vacuum, and find the associated magnetic field. ANSWER Since and E does not depend on , GAUSS' LAW reads (in spherical coordinates)
 (1)

 (2)

 (3)

 = (4)

 (5)

In order to satisfy FARADAY'S LAW we must therefore have (within a constant of integration)
 (6)

 (7)

 (8)

or where
 (9)

This should satisfy GAUSS' LAW too:
 = (10)

It remains only to check AMPÈRE'S LAW or
 = (11)

Now, if we're to get any joy from this, it had better be equal to
 = (12)

Thus the proposed function does satisfy all of MAXWELL'S EQUATIONS as advertised and is therefore also a valid solution of TWE (The Wave Equation). And this is the simplest possible spherical wave! (Don't you just love curvilinear coordinates?)

2. Calculate the Poynting vector. Average over a full cycle to get the intensity vector .  Does point in the expected direction?  Does it fall off like r-2, as it should? ANSWER
 = (13)

The fact that has a non-radial component may seem alarming, but let's check the time average: all of   and   oscillate in time, but only the first averages to zero; the other two average to , but their difference does average to zero. Thus
 (14)

which points radially outward and falls off like 1/r2, as expected.

3. Integrate over a spherical surface to determine the total power radiated.
[You should get .] ANSWER

This was a tedious problem; it took me all day to get it right. I will be duly impressed if you managed to grind through it successfully. Now you know why we like our plane waves so much, Huygens' principle notwithstanding!

Jess H. Brewer
2006-02-20