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}}
= $ \fbox{ 0.0672~V/m } . (The wavelength, while interesting, is irrelevant to the question.)

  2. STANDING WAVES: Consider standing electromagnetic waves:

    \begin{displaymath}\begin{array}{rcl}
\Vec{E} &=& E_0 \left( \sin k z \; \sin \ . . . 
 . . . t( \cos k z \; \cos \omega t \right) \Hat{y} \; .
\end{array} \end{displaymath}

    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 \fbox{ $\Box^2 \Vec{E} = 0$\ } 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 $c = \omega/k$) \fbox{ $E_0 = c B_0$\ } .

    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{displaymath}\begin{array}{rcl}
\Hat{n}_T &=& \cos \theta_T \Hat{x} + \si . . . 
 . . . &=& \cos \theta_R \Hat{x} + \sin \theta_R \Hat{y}
\end{array} \end{displaymath}

    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 ET = 0 (trivial case) or \fbox{ $\theta_T = 0$\ } (mod $2\pi$). Equation [2] then also requires \fbox{ $\theta_R = 0$\ } (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 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 $\mu_1 = \mu_2 = \muz$.) ANSWER FRESNEL'S EQUATIONS read

    \begin{displaymath}
{\tilde{E}_0^R \over \tilde{E}_0^I}
= \left( \alpha - \be . . . 
 . . . over \tilde{E}_0^I}
= \left( 2 \over \alpha + \beta \right)
\end{displaymath}

    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.)

    \epsfbox{images/FresnelDiamond.ps}

    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 \fbox{ $\tilde{E}_0^R = - 0.4152 \tilde{E}_0^I$\ } and ${\displaystyle
\tilde{E}_0^T = {2 \over 1+2.42} \tilde{E}_0^I }$ or \fbox{ $\tilde{E}_0^T = 0.5848 \tilde{E}_0^I$\ } .

    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 $\sin \theta_B = 0.9242$ $\Rightarrow$ \fbox{ $\theta_B = 67.55^\circ$\ } .

    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 $\cos \theta_I = 0.20693$ $\Rightarrow$ \fbox{ $\theta_C = 78.06^\circ$\ } .

  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:

    \begin{displaymath}
T_{ij} = \epsz \left( E_iE_j - \delta_(ij) E^2/2 \right)
+ \left( B_iB_j - \delta_(ij) B^2/2 \right)/\muz \; .
\end{displaymath}

    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!) \fbox{ $T_{33} = - \epsz E^2 = - u_{_{EM}}$\ } .

    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 $\Hat{x}_i$ 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

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

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


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


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


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


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

    In order to satisfy FARADAY'S LAW we must therefore have (within a constant of integration)
    \begin{displaymath}
\Vec{B} = - \int \left( \Curl{E} \right) dt
= - {A\over r . . . 
 . . . ( kr - {1 \over kr} \right) S \right]
\Hat{\theta} \right\}
\end{displaymath} (6)


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


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

    or $\Vec{B} = B_r \; \Hat{r} + B_\theta \; \Hat{\theta}$ where
    \begin{displaymath}
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] \; .
\end{displaymath} (9)

    This should satisfy GAUSS' LAW too:  ${\displaystyle \Div{B} = {1\over r^2}\DbyD{}{r}\left(r^2 B_r\right)
+ {1\over r\sin\theta}\DbyD{}{\theta}\left(\sin\theta B_\theta\right) }$
      = $\displaystyle {2A\cos\theta \over c k r^2}\DbyD{}{r}\left(\sin u
+ { 1 \over k  . . . 
 . . . k r} + k r \right) \cos u \right]
\DbyD{}{\theta}\left(\sin^2\theta \right) \cr$ (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
    $\displaystyle \Curl{B}$ = $\displaystyle {1\over r} \left\{ {A\sin\theta \over c k r^2} \left[
\sin u - \left( kr - {1 \over kr} \right) \cos u \right] \right. \cr$ (11)

    Now, if we're to get any joy from this, it had better be equal to ${\displaystyle {1 \over c^2} \DbyD{\Vec{E}}{t}
= {k^2 \over \omega^2} \DbyD{\Vec{E}}{t}
= {k \over c \omega} \DbyD{\Vec{E}}{t} }$
      = $\displaystyle {k \over c \omega} {A \sin\theta \over r} \DbyD{}{t} \left[
\cos  . . . 
 . . .  \omega} \left[
\omega \sin u + \omega {\cos u \over kr} \right]
\Hat{\phi} \cr$ (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
    $\displaystyle \Vec{S}$ = $\displaystyle {\Vec{E} \times \Vec{B} \over \muz}
= {1\over\muz} \left( E \; \H . . . 
 . . . - {1\over\muz} \left( E B_r \; \Hat{\theta}
+ E B_\theta \; \Hat{r} \right) \cr$ (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
    \begin{displaymath}
\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} \; ,
\end{displaymath} (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

    \begin{displaymath}
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)
\end{displaymath}


    \begin{displaymath}
\hbox{\sf or ~ }
P = {\pi A^2 \over \muz c} \int_{-1}^{+1} . . . 
 . . . {\displaystyle
P = {4\pi A^2 \over 3\muz c} }$ }~. ~ \chk }
\end{displaymath}

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