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 $3^\circ$ 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 $\lambda \sim 1.1$ 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 $\langle u_{_{EM}} \rangle = \epsz \langle E^2 \rangle = 4.0\times10^{-14}$ J/m³, then $E_{rms} \equiv \sqrt{\langle E^2 \rangle} = \sqrt{\langle u_{_{EM}} \rangle / \epsz } = \sqrt{4.0\times10^{-14} \over 0.88541878\times 10^{-11}} =$  . (The wavelength, while interesting, is irrelevant to the question.)

2. STANDING WAVES: Consider standing electromagnetic waves:
   
   $$\begin{array}{rcl} \Vec{E} &=& E_0 \left( \sin k z \; \sin \ . . . . . . t( \cos k z \; \cos \omega t \right) \Hat{y} \; . \end{array}$$
   
   
   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 $\nabla^2 \sin k z = -k^2 \sin k z$ and $\nabla^2 \cos k z = -k^2 \cos k z$; $\dbyd{}{t} \sin \omega t = -\omega^2 \sin \omega t$ and $\dbyd{}{t} \cos \omega t = -\omega^2 \cos \omega t$; so $\nabla^2 \Vec{E} - (1/c^2)\dbyd{\Vec{E}}{t}= (-k^2 + \omega^2/c^2)\Vec{E}$ and similarly for $\Vec{B}$. But $(-k^2 + \omega^2/c^2) = -k^2[1-(\omega^2/k^2)/c^2] = 0$, since $c = \omega/k$. Thus and similarly for $\Vec{B}$. 

   2. Show that we must also have $c = \omega/k$ and E₀ = c B₀. ANSWER:  Since $c = \omega/k$ is a universal property of all solutions of The Wave Equation (TWE), that's a given. Applying FARADAY'S LAW, $\Curl{E} = -\dbyd{\Vec{B}}{t}$, gives $E_0 \sin \omega t \; \Vec{\nabla} \times (\sin k z \; \Hat{x}) = - B_0 \cos k z \; \dbyd{(\cos \omega t)}{t} \; \Hat{y}$ or $k \; \Hat{y} \; E_0 \cos k z \; \sin \omega t \; = \omega \; \Hat{y} \; B_0 \cos k z \; \sin \omega t$. Dividing out the common factor $\Hat{y} \;\cos k z \; \sin \omega t$ gives $k E_0 = \omega B_0$ or (since $c = \omega/k$)  .

   3. Show that the time-averaged power flow across any area will be zero. ANSWER:  $\Vec{S} = \Vec{E} \times \Vec{H} = (\Hat{x} \times \Hat{y})$
      $(E_0 B_0/\mu) \; (\sin k z \; \sin \omega t) \; (\cos k z \; \cos \omega t) . . . . . . \; (E_0 B_0/\mu) \; (\sin k z \; \cos k z) \; (\sin \omega t \; \cos \omega t)$. Looking only at the t-dependence to get the time average, we note that $\sin \omega t \; \cos \omega t = {1\over2} \sin (2\omega t)$ 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, $\Vec{S} = (E_0 B_0/4\mu) \; \sin 2kz \cdot \sin 2\omega t \; \Hat{z}$. This is only zero where $\sin 2kz = 0$, i.e. at z=0 and $2kz = n\pi$ (where n is any integer). That is, for $z = n\lambda/4$. At any other position, $\Vec{S}$ oscillates in the $\pm\Hat{z}$ 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 $\Hat{x}$ direction. Prove that this must be so. [Hint: Let the polarization vectors of the reflected and transmitted waves be
   
   $$\begin{array}{rcl} \Hat{n}_T &=& \cos \theta_T \Hat{x} + \si . . . . . . &=& \cos \theta_R \Hat{x} + \sin \theta_R \Hat{y} \end{array}$$
   
   
   and prove from the boundary conditions that $\theta_T = \theta_R = 0$.] ANSWER:  We must have $\Vec{E}_\parallel$ continuous across the boundary. Since the normal direction is $\Hat{k} = \Hat{z}$, $\Vec{E}_\parallel$ is constituted of x and y components. Thus $\Vec{E}_I + \Vec{E}_R = \Vec{E}_T$ or $E_I + E_R \cos \theta_R = E_T \cos \theta_T$ [1] and $E_R \sin \theta_R = E_T \sin \theta_T$ [2]. Similarly, $\Vec{H}_\parallel$ must be continuous across the boundary, and, as always, $v \Vec{B} = \Hat{k} \times \Vec{E}$, giving ${E_I - E_R \cos \theta_R \over \mu_1 v_1} = {E_T \cos \theta_T \over \mu_2 v_2}$ [3] and ${E_R \sin \theta_R \over \mu_1 v_1} = - {E_T \sin \theta_T \over \mu_2 v_2}$ [4]. If $\beta \equiv {\mu_1 v_1 \over \mu_2 v_2}$, Eq. [4] reads $E_R \sin \theta_R = - \beta E_T \sin \theta_T$, which we can combine with Eq. [2] to conclude that $E_T \sin \theta_T = - \beta E_T \sin \theta_T$, which can be true only if E_T = 0 (trivial case) or (mod $2\pi$). Equation [2] then also requires (mod $2\pi$).

4. (p. 392, Problem 9.15) - COMPLEX ALGEBRA EXERCISE: Suppose that we have six nonzero constants A,B,C,a,b,c such than Ae^iax + Be^ibx = Ce^icx 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 $\mu_1 = \mu_2 = \muz$.) ANSWER:  FRESNEL'S EQUATIONS read
   
   $${\tilde{E}_0^R \over \tilde{E}_0^I} = \left( \alpha - \be . . . . . . over \tilde{E}_0^I} = \left( 2 \over \alpha + \beta \right)$$
   
   
   where ${\displaystyle \alpha \equiv {\cos \theta_T \over \cos \theta_I} = {\sqrt{1 - . . . . . . sqrt{1 - \left[{n_1 \over n_2} \sin \theta_I\right]^2} \over \cos \theta_I} }$ and ${\displaystyle \beta \equiv {\mu_1 v_1 \over \mu_2 v_2} = {\mu_1 n_2 \over \mu_2 n_1} }$. In this case $\beta = 2.42$ (we assume the light is entering the diamond rather than emerging) and ${\displaystyle \alpha = {\sqrt{1 - \left(\sin \theta_I / 2.42 \right)^2} \over \cos \theta_I} }$. 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 $\theta_I = 0$, $\alpha = 1$, giving ${\displaystyle \tilde{E}_0^R = {1-2.42 \over 1+2.42} \tilde{E}_0^I }$ or and ${\displaystyle \tilde{E}_0^T = {2 \over 1+2.42} \tilde{E}_0^I }$ or  .

   2. Brewster's angle; ANSWER:  ${\displaystyle \sin^2 \theta_B = {1-\beta^2 \over (n_1/n_2)^2 - \beta^2} }$ $= {1 - 5.8564 \over (1/5.8564) - 5.8564} = 0.85415$ or $\sin \theta_B = 0.9242$ $\Rightarrow$  .

   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 $\alpha - \beta = 2$ or ${\displaystyle \alpha = {\sqrt{1 - \left(\sin \theta_I / 2.42 \right)^2} \over \cos \theta_I} = 4.42 }$ or $1 - \left(\sin \theta_I / 2.42 \right)^2 = 19.5364 \cos^2 \theta_I$ or $5.8564 - 1 + \cos^2 \theta_I = 114.413 \cos^2 \theta_I$ or $4.8564 = 113.413 \cos^2 \theta_I$ or $\cos^2 \theta_I = 4.8564/113.413 = 0.04282$ or $\cos \theta_I = 0.20693$ $\Rightarrow$  .

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:
   
   $$\begin{eqnarray*} \Vec{E}(z,t) &=& E_0 \cos(kz - \omega t + \delta) \Hat{x} \cr . . . . . . }(z,t) &=& {E_0\over c} \cos(kz - \omega t + \delta) \Hat{y} \cr \end{eqnarray*}$$
   
   ANSWER:  Recall Eq. (8.19) on p. 352:
   
   $$T_{ij} = \epsz \left( E_iE_j - \delta_(ij) E^2/2 \right) + \left( B_iB_j - \delta_(ij) B^2/2 \right)/\muz \; .$$
   
   
   Here $E_i = \delta_{i1} E$ where $E \equiv E_0 \cos(kz - \omega t + \delta)$ and $B_i = \delta_{i2} B$ where $B \equiv {E_0 \over c} \cos(kz - \omega t + \delta)$ = E/c, so all off-diagonal elements are zero. We have $T_{11} = \epsz \left( E^2 - E^2/2 \right) - B^2/2\muz = \epsz \left( E^2/2 - E^2/2\epsz\muz c^2 \right) = \epsz \left( E^2/2 - E^2/2 \right)$ or T₁₁ = 0, $T_{22} = - \epsz E^2/2 + \left( B^2 - B^2/2 \right) \muz = \epsz \left( - E^2/2 + E^2/2\epsz\muz c^2 \right) = \epsz \left( - E^2/2 + E^2/2 \right)$ or T₂₂ = 0 and $T_{33} = - \epsz E^2/2 - B^2/2\muz$ 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 T_ij represents the force per unit area acting in the $\Hat{x}_i$ direction on a surface whose normal is in the $\Hat{x}_j$ direction, then the diagonal elements are pressures and T₃₃ is the radiation pressure on a surface normal to $\Hat{z}$. In the same way -T₃₃ represents the the momentum current density transported by the fields, and is (as expected) in the same direction as $\Hat{k}$ and is, in fact, equal to $\Vec{S}/c$.

7.

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

$$\Vec{E}(r,\theta,\phi,t) = {A \sin\theta \over r} \left[ . . . . . . ight) \sin \left( k r - \omega t \right) \right] \Hat{\phi}$$

with $c = \omega/k$, as usual. [This is, incidentally, the simplest possible spherical wave.  For notational convenience, let $(k r - \omega t) \equiv u$ in your calculations.]

1. Show that $\Vec{E}$ obeys all four of Maxwell's equations, in vacuum, and find the associated magnetic field. ANSWER:  Since $\Vec{E} = E \Hat{\phi}$ and E does not depend on $\phi$, GAUSS' LAW reads (in spherical coordinates)
   

   $$\Div{E} = {1\over r\sin\theta}\DbyD{E}{\phi} = 0 \; . \quad \hbox{\sf\chk}$$ (1)

   
   
   
   

   $$% \Curl{E} = {1\over r\sin\theta} \DbyD{}{\theta}\left(E\sin . . . . . . {1\over r} \left( E + r \DbyD{E}{r} \right) \Hat{\theta} \; .$$ (2)

   
   
   
   

   $$\hbox{\sf Now, ~ } \DbyD{E}{\theta} = {A\cos\theta \over r} . . . . . . r k r } \sin u \right] = E \; {\cos\theta \over \sin\theta}$$ (3)

   
   
   
   

   $$\hbox{\sf and ~ } \DbyD{E}{r} - {A\sin\theta \over r^2} \left(\cos u - {\sin u \over kr} \right . . . . . . over r} \left( -k \sin u + {\sin u \over kr^2} -{k \cos u \over kr} \right) \cr$$ = (4)

   
   
   
   

   $$\hbox{\sf so ~ } \Curl{E} = {A\over r^2} \left\{ 2\cos\theta . . . . . . \over kr} \right) \sin u \right] \Hat{\theta} \right\} \; .$$ (5)

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

   $$\Vec{B} = - \int \left( \Curl{E} \right) dt = - {A\over r . . . . . . ( kr - {1 \over kr} \right) S \right] \Hat{\theta} \right\}$$ (6)

   
   
   
   

   $$\hbox{\sf where ~ } C \equiv \int \cos u \; dt = - {\sin . . . . . . r \omega} \; . \qquad (\hbox{\sf Note: } \omega = c k \; . )$$ (7)

   
   
   
   

   $$\hbox{\sf Thus ~ } \Vec{B} = {A\over c k r^2} \left\{ 2\c . . . . . . - {1 \over kr} \right) \cos u \right] \Hat{\theta} \right\}$$ (8)

   
   
   or $\Vec{B} = B_r \; \Hat{r} + B_\theta \; \Hat{\theta}$ where
   

   $$B_r = {2A\cos\theta \over c k r^2} \left[ \sin u + { 1 \ov . . . . . . sin u - \left( kr - {1 \over kr} \right) \cos u \right] \; .$$ (9)

   
   
   This should satisfy GAUSS' LAW too:  ${\displaystyle \Div{B} = {1\over r^2}\DbyD{}{r}\left(r^2 B_r\right) + {1\over r\sin\theta}\DbyD{}{\theta}\left(\sin\theta B_\theta\right) }$
   

   $${2A\cos\theta \over c k r^2}\DbyD{}{r}\left(\sin u + { 1 \over k . . . . . . k r} + k r \right) \cos u \right] \DbyD{}{\theta}\left(\sin^2\theta \right) \cr$$ = (10)

   
   
   It remains only to check AMPÈRE'S LAW:  ${\displaystyle \Curl{B} = {1\over r}\left[\DbyD{}{r}\left(rB_\theta\right) - \DbyD{B_r}{\theta}\right]\Hat{\phi} }$ or
   

   $$\Curl{B} {1\over r} \left\{ {A\sin\theta \over c k r^2} \left[ \sin u - \left( kr - {1 \over kr} \right) \cos u \right] \right. \cr$$ = (11)

   
   
   Now, if we're to get any joy from this, it had better be equal to ${\displaystyle {1 \over c^2} \DbyD{\Vec{E}}{t} = {k^2 \over \omega^2} \DbyD{\Vec{E}}{t} = {k \over c \omega} \DbyD{\Vec{E}}{t} }$
   

   $${k \over c \omega} {A \sin\theta \over r} \DbyD{}{t} \left[ \cos . . . . . . \omega} \left[ \omega \sin u + \omega {\cos u \over kr} \right] \Hat{\phi} \cr$$ = (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 $\Vec{S}$ over a full cycle to get the intensity vector $\Vec{I}$.  Does $\Vec{I}$ point in the expected direction?  Does it fall off like r^-2, as it should? ANSWER: 
   

   $$\Vec{S} {\Vec{E} \times \Vec{B} \over \muz} = {1\over\muz} \left( E \; \H . . . . . . - {1\over\muz} \left( E B_r \; \Hat{\theta} + E B_\theta \; \Hat{r} \right) \cr$$ = (13)

   
   
   The fact that $\Vec{S}$ has a non-radial component may seem alarming, but let's check the time average: all of  $\sin u \; \cos u$, $\sin^2 u$ and  $\cos^2 u$ oscillate in time, but only the first averages to zero; the other two average to ${1\over2}$, but their difference does average to zero. Thus
   

   $$\Vec{I} \equiv \langle \Vec{S} \rangle = {A^2 \sin^2 \the . . . . . . {A^2 \sin^2 \theta \over 2\muz c} \; {\Hat{r} \over r^2} \; ,$$ (14)

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

3. Integrate $\Vec{I}\cdot d\Vec{a}$ over a spherical surface to determine the total power radiated.
   [You should get $P = 4 \pi A^2 / 3\muz c$.] ANSWER: 
   
   $$P = \SurfInt \Vec{I} \cdot d\Vec{a} = {A^2 \over 2\muz c} . . . . . . \int_1^{-1} \left( 1 - \cos^2 \theta \right) d(\cos \theta)$$
   
   
   
   
   $$\hbox{\sf or ~ } P = {\pi A^2 \over \muz c} \int_{-1}^{+1} . . . . . . {\displaystyle P = {4\pi A^2 \over 3\muz c} }$ }~. ~ \chk }$$
   
   

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!