Plane Waves and Math Review - Homework 01

This is the first homework assignment for Lasers and Optomechanics at Syracuse University.
It is due January 23, 2026

You will need to complete the questions in this jupyter notebook and submit it via gitlab

Tech References¶

%matplotlib widget
from ipywidgets import *
import numpy as np
import matplotlib as mpl
import matplotlib.pyplot as plt

plt.style.use('dark_background')

fontsize = 14
mpl.rcParams.update(
    {
        "text.usetex": True,
        "figure.figsize": (9, 6),
        "figure.autolayout": True,
        "font.family": "serif",
        "font.serif": "georgia",
        # 'mathtext.fontset': 'cm',
        "lines.linewidth": 1.5,
        "font.size": fontsize,
        "xtick.labelsize": fontsize,
        "ytick.labelsize": fontsize,
        "legend.fancybox": True,
        "legend.fontsize": fontsize,
        "legend.framealpha": 0.7,
        "legend.handletextpad": 0.5,
        "legend.labelspacing": 0.2,
        "legend.loc": "best",
        "axes.edgecolor": "#b0b0b0",
        "grid.color": "#707070",  # grid color"
        "xtick.color": "#b0b0b0",
        "ytick.color": "#b0b0b0",
        "savefig.dpi": 80,
        "pdf.compression": 9,
    }
)

1Approximations Review¶

In this course, you will need to remember and use some basic approximations. These approximations all come from taking the Taylor Expansion of a function $f(x)$ about some point $x = a$ :

f(x)\Bigr|_{x \rightarrow a} \approx f(a) + f'(a) (x-a) + \dfrac{1}{2!} f''(a) (x-a)^2 = \displaystyle \sum_{n=0}^\infty \dfrac{f^{(n)}(a)}{n!}(x - a)^n

(1)

1.1Question 1: Binomial Approximation¶

The binomial approximation to first order in $x$ is as follows:

\begin{align} (1 + x)^{n} \approx 1 + n x. \end{align}

(2)

Question 1A¶

Derive the binomial approximation using the Taylor Expansion to first order about $x = 0$

Question 1B¶

Find the second and third order terms of the binomial approximation

Question 1C¶

Plot the binomial function on $x \in [-1, 1]$ for $n = \dfrac{1}{2}$ .
Compare to plots of the first, second, and third order binomial approximation.
At what $x > 0$ does each approximation fail, becoming greater than 5% error?

Question 1A Solution: (This would be the cell you fill out)¶

Let

f(x) = (1 + x)^n,

(3)

then at $x = 0$ ,

f(0) = 1.

(4)

Then the first derivative $f'(x)$ is

f'(x) = n (1 + x)^{n-1}

(5)

and the derivative evaluated at $x = 0$ is

f'(0) = n

(6)

The Taylor Expansion to first order then becomes

\begin{align} f(x)\Bigr|_{x \rightarrow 0} &\approx f(0) + f'(0) x\\~\\ &= 1 + n x \end{align}

(7)

Question 1B Solution:¶

Taking the second and third derivatives, and evaluating at 0 yields

\begin{align} f''(x) &= n (n-1) (1 + x)^{n-2} \\ f'''(x) &= n (n-1) (n-2) (1 + x)^{n-3} \\~\\ f''(0) &= n (n-1)\\ f''(0) &= n (n-1) (n-2) \end{align}

(8)

The second order expansion is

\begin{align} f(x)\Bigr|_{x \rightarrow 0} &\approx f(0) + f'(0) x + \dfrac{1}{2!} f''(0) x^2\\~\\ &= 1 + n x + \dfrac{1}{2} n (n-1) x^2 \end{align}

(9)

The third order expansion is

\begin{align} f(x)\Bigr|_{x \rightarrow 0} &\approx f(0) + f'(0) x + \dfrac{1}{2!} f''(0) x^2 + \dfrac{1}{3!} f'''(0) x^3\\~\\ &= 1 + n x + \dfrac{1}{2} n (n-1) x^2 + \dfrac{1}{6} n (n-1) (n-2) x^3 \end{align}

(10)

def binom(xx:float, nn:float):
    """Binomial function (1 + xx)^nn
    
    Inputs:
    -------
    xx: float or array of floats
        binomial variable
    nn: float
        binomial exponent

    Output:
    -------
    binom: float or array of floats
        binomial expansion
    """
    return (1 + xx)**nn

# Parameter definitions.  Protip: never make single-letter variable names
nn = 0.5
xx = np.linspace(-1, 2, 100)

taylor0 = 1
taylor1 = taylor0 + nn * xx
taylor2 = taylor1 + 0.5 * nn * (nn - 1) * xx**2
taylor3 = taylor2 + (1/6) * nn * (nn - 1) * (nn - 2) * xx**3

# At which x does the error become greater than 10%?
# First, we divide the approximation by the real function,
# Second, we subtract 1 from that ratio
# Third, we take the absolute value of the subtraction
# Fourth, we look for the first location where the final result is greater than 0.1
# Fifth, we find where x > 0
# Sixth, we take the intersection of the indices found
# Seventh, we find the first index where the error is large for plotting
error = 0.05
model = binom(xx, nn)

xx_errors = np.array([])
for taylor in [taylor1, taylor2, taylor3]:
    abs_errors = np.abs(taylor/model - 1) # final result
    indices_error = np.argwhere(abs_errors > error)
    indices_x = np.argwhere(xx > 0)
    
    indices_final = np.intersect1d(indices_error, indices_x)
    index = indices_final[0]
    
    xx_errors = np.append(xx_errors, xx[index])
print(xx_errors)

[0.87878788 1.3030303  1.42424242]

/var/folders/t1/fq8mx5mj0bx4kn1hlgb4hh840000gn/T/ipykernel_46996/3003632814.py:14: RuntimeWarning: divide by zero encountered in divide
  abs_errors = np.abs(taylor/model - 1) # final result

fig, s1 = plt.subplots(1)

s1.plot(xx, binom(xx, nn), label="Binomial Function")
s1.plot(xx, taylor1, ls="--", label="Taylor 1")
s1.plot(xx, taylor2, ls="--", label="Taylor 2")
s1.plot(xx, taylor3, ls="--", label="Taylor 3")

s1.axvline(x=0, label=f"$x = 0$")

for ii, xx_error in enumerate(xx_errors):
    s1.axvline(x=xx_error, color=f"C{ii+1}",ls=":", label=f"Taylor {ii+1} Error")

s1.set_title("Binomial Approximations about $x = 0$ for $n = " + f"{nn}" + "$")
s1.set_xlabel("$x$")
s1.set_ylabel("$f(x)$")
s1.legend()
s1.grid()
plt.show()

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1.2Question 2: Sine and Cosine Approximations¶

Repeat the Taylor Expansion approximations for
A. sine and
B. cosine
to second order about $x = 0$ .
Make the plots, but you don’t need to calculate the 5% error point.

Question 2A Solution: Sine Approximation¶

Set $f(x) = \sin(x)$ , then

f'(x) = \cos(x), \quad f''(x) = -\sin(x)

(11)

Evaluating the Taylor Series about $x=0$ :

\begin{align} \sin{x} &\approx f(0) + f'(0) x + \dfrac{1}{2!} f''(0) x^2\\ \sin{x} &\approx 0 + x + 0\\ \sin{x} &\approx x \end{align}

(12)

# Changing the plotting parameter from x -> yy
yy = np.linspace(-np.pi, np.pi, 100)

taylor_sine1 = yy
taylor_sine3 = taylor_sine1 - yy**3 / 6

fig, s1 = plt.subplots(1)

s1.plot(yy, np.sin(yy), label=r"$ \sin(x) $")
s1.plot(yy, taylor_sine1, ls="--", label=r"$x$")
s1.plot(yy, taylor_sine3, ls="--", label=r"$x - x^3/6$")

s1.set_title("Sine Approximations about $x = 0$")
s1.set_xlabel("$x$")
s1.set_ylabel(r"$\sin(x)$")
s1.legend()
s1.grid()
plt.show()

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Question 2B Solution: Cosine Approximation¶

Set $f(x) = \cos(x)$ , then

f'(x) = -\sin(x), \quad f''(x) = -\cos(x)

(13)

Evaluating the Taylor Series about $x=0$ :

\begin{align} \cos{x} &\approx f(0) + f'(0) x + \dfrac{1}{2!} f''(0) x^2\\ \cos{x} &\approx 1 + 0 + \dfrac{x^2}{2}\\ \cos{x} &\approx 1 + \dfrac{x^2}{2} \end{align}

(14)

# Changing the plotting parameter from x -> yy
zz = np.linspace(-np.pi, np.pi, 100)

taylor_cosine0 = np.ones_like(zz)
taylor_cosine2 = taylor_cosine0 - zz**2 / 2

fig, s1 = plt.subplots(1)

s1.plot(zz, np.cos(yy), label=r"$ \sin(x) $")
s1.plot(zz, taylor_cosine0, ls="--", label=r"$1$")
s1.plot(zz, taylor_cosine2, ls="--", label=r"$1 - x^2/6$")

s1.set_title("Cosine Approximations about $x = 0$")
s1.set_xlabel("$x$")
s1.set_ylabel(r"$\cos(x)$")
s1.legend()
s1.grid()
plt.show()

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1.3Question 3: Complex Number Review¶

Question 3A:¶

Plot the following complex function on a domain of $\phi \in [0, 2 \pi]$ :

\begin{align} z_1(\phi) &= 2 + e^{i \phi}\\ z_2(\phi) &= \dfrac{3}{2 - e^{i \phi}}\\ z_3(\phi) &= e^{(\sigma + i \omega) t} \end{align}

(15)

where for $z_4$ , $\sigma = -0.5$ , and $\omega = 1$ .

Question 3B:¶

Calculate the magnitude $r(\phi)$ and argument $\theta(\phi)$ for each $z_i$ .

Question 3C:¶

Calculate the velocity of the phasors with respect to $\phi$ , and draw them for each $z_i$ evaluated at $\phi = \left\{0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2} \right\}$

Question 3D:¶

What is the primary difference between $z_1$ and $z_2$ ?

Question 3E:¶

For $z_3$ , substitute $t$ for $\phi$ , and calculate the normalized time derivatives : $\dfrac{\dot{z_3}}{z_3}$ , $\dfrac{\ddot{z_3}}{z_3}$

and find expressions for the normalized real polar coordinates $\dfrac{\dot{r}}{r}, \dfrac{\ddot{r}}{r}, \dot{\theta}, \ddot{\theta}$ .

Discuss how the expressions you found for the polar coordinates relate to the path you plotted for $z_3$ in part A.

What happens if $\sigma = +0.5$ ?

# For Part A, define the complex functions for plotting
def z1(phi):
    return 2 + np.exp(1j * phi)

def z2(phi):
    return 3 /(2 - np.exp(1j * phi))

def z3(phi, sigma=-0.5, omega=1):
    return np.exp((sigma + 1j * omega) * phi)

# For Part C, define the derivatives of the complex functions for plotting
def dz1(phi):
    return 1j * np.exp(1j * phi)

def dz2(phi):
    return 3 * 1j * np.exp(1j * phi) / (2 - np.exp(1j * phi))**2

def dz3(phi, sigma=-0.5, omega=1):
    return (sigma + 1j * omega) * np.exp((sigma + 1j * omega) * phi)

phis = np.linspace(0, 2*np.pi, 100)
aa = 2
bb = 1

sigma = -0.5
omega = 1

# Make convenient ploting vectors
plot_z1_re = np.real(z1(phis))
plot_z1_im = np.imag(z1(phis))
plot_z2_re = np.real(z2(phis))
plot_z2_im = np.imag(z2(phis))
plot_z3_re = np.real(z3(phis, sigma, omega))
plot_z3_im = np.imag(z3(phis, sigma, omega))

Question 3A Solution¶

fig, s1 = plt.subplots(1)

line1, = s1.plot(plot_z1_re, plot_z1_im, label=r"$ z_1 $")
line2, = s1.plot(plot_z2_re, plot_z2_im, lw=2, ls=":", label=r"$ z_2 $")
line3, = s1.plot(plot_z3_re, plot_z3_im, ls="-", label=r"$ z_3 $")

s1.set_xlim([-3,3])
s1.set_ylim([-3,3])

s1.set_aspect('equal')

s1.set_title("Complex Number Plots")
s1.set_xlabel("Real")
s1.set_ylabel(r"$i$")
s1.legend()
s1.grid()
plt.show()

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Question 3B Solution:¶

$z_1$ :

\begin{align} z_1(\phi) &= 2 + e^{i \phi}\\ z_1(\phi) &= 2 + \cos\phi + i \sin\phi \end{align}

(16)

\begin{align} r_1(\phi) &= \sqrt{z_1 z_1^*}\\ r_1(\phi) &= \sqrt{(2 + e^{i \phi})(2 + e^{-i \phi})}\\ r_1(\phi) &= \sqrt{ 5 + 4 \cos\phi }\\ \theta_1(\phi) &= \arctan2(\sin\phi, 2 + \cos\phi) \end{align}

(17)

$z_2$ :

\begin{align} z_2(\phi) &= \dfrac{3}{2 - e^{i \phi}}\\ z_2(\phi) &= \dfrac{3}{2 - e^{i \phi}} \dfrac{ 2 - e^{-i \phi} }{ 2 - e^{-i \phi} }\\ z_2(\phi) &= \dfrac{ 3 (2 - e^{-i \phi}) }{ 5 - 4 \cos\phi }\\ z_2(\phi) &= \dfrac{ 6 - 3 \cos\phi }{ 5 - 4 \cos\phi } + i \dfrac{ 3 \sin\phi }{ 5 - 4 \cos\phi } \end{align}

(18)

\begin{align} r_2(\phi) &= \sqrt{z_2 z_2^*}\\ r_2(\phi) &= \sqrt{\left( \dfrac{3}{2 - e^{i \phi}} \right)\left( \dfrac{3}{2 - e^{-i \phi}} \right)}\\ r_2(\phi) &= \sqrt{ \dfrac{9}{5 - 4 \cos\phi } }\\ \theta_2(\phi) &= \arctan2\left(\dfrac{ 3 \sin\phi }{ 5 - 4 \cos\phi }, \dfrac{ 6 - 3 \cos\phi }{ 5 - 4 \cos\phi }\right)\\ \theta_2(\phi) &= \arctan2\left( \sin\phi, 2 - \cos\phi \right) \end{align}

(19)

$z_3$ :

\begin{align} z_3(\phi) &= e^{(\sigma + i \omega) \phi}\\ z_3(\phi) &= e^{\sigma \phi} e^{i \omega \phi}\\ z_3(\phi) &= e^{\sigma \phi} \cos(\omega \phi) + i e^{\sigma \phi} \sin(\omega \phi) \end{align}

(20)

\begin{align} r_3(\phi) &= \sqrt{z_3 z_3^*}\\ r_3(\phi) &= \sqrt{ (e^{\sigma \phi} e^{i \omega \phi}) (e^{\sigma \phi} e^{-i \omega \phi}) }\\ r_3(\phi) &= e^{\sigma \phi}\\ \theta_3(\phi) &= \arctan2(\sin(\omega \phi), \cos(\omega \phi))\\ \theta_3(\phi) &= \omega \phi \end{align}

(21)

Question 3C Solution¶

\begin{align} \dfrac{dz_1}{d\phi} &= ie^{i\phi}\\ \dfrac{dz_2}{d\phi} &= \dfrac{3ie^{i\phi}}{(2-e^{i\phi})^2}\\ \dfrac{dz_3}{d\phi} &= (\sigma + i\omega) e^{(\sigma + i\omega)\phi} \end{align}

(22)

fig, [ax1,ax2,ax3] = plt.subplots(1, 3, figsize=(12,4))

for ii, func, dfunc, ax in zip(np.arange(1,4),[z1,z2,z3],[dz1,dz2,dz3],[ax1,ax2,ax3]):
    text = f"$ z_{ii} $"
    dtext = f"$ dz_{ii} $"
    color = f"C{ii}"

    zz_re = np.real(func(phis))
    zz_im = np.imag(func(phis))

    line1, = ax.plot(zz_re, zz_im, color=color, label=text)
    
    for phi in np.linspace(0, 3*np.pi/2, 4):
        zz0 = func(phi)
        dz0 = dfunc(phi)
        ax.arrow(np.real(zz0), np.imag(zz0), np.real(dz0), np.imag(dz0), shape='full', color=f"C{ii+3}", lw=1, length_includes_head=True, head_width=.10, zorder=2, label=dtext)
        dtext = ""

    if ii < 3:
        ax.set_xlim([-3,4])
        ax.set_ylim([-3,4])
    else:
        ax.set_xlim([-1,2])
        ax.set_ylim([-1,2])        
    
    ax.set_aspect('equal')
    
    ax.set_title(text)
    ax.set_xlabel("Real")
    ax.set_ylabel(r"$i$")
    ax.legend()
    ax.grid()
plt.show()

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Question 3D Solution¶

The primary difference between $z_1$ and $z_2$ is the evolution rate, i.e. their derivatives. Both $z_1$ and $z_2$ trace out perfect circles centered at $z_0 = 2$ with radius $r_0 = 1$ .
However, the rate at which the circles are traced out are vastly different, which can be seen in the velocities plotted in the solution to Question 3C.

$z_1$ traces out the circle at a constant velocity. $z_2$ starts fast at $\phi = 0$ , and slows down immediately, yielding a minimum velocity at $\phi = \pi$ .

Question 3E Solution¶

Subbing in $\phi \rightarrow t$ , and taking the time derivatives:

\begin{align} z_3(t) &= e^{(\sigma + i \omega) t}\\~\\ \dot{z_3}(t) &= (\sigma + i \omega)e^{(\sigma + i \omega) t}\\~\\ \ddot{z_3}(t) &= (\sigma + i \omega)^2 e^{(\sigma + i \omega) t} \end{align}

(23)

which implies

\begin{align} \dfrac{\dot{z_3}}{z_3} &= \sigma + i \omega\\~\\ \dfrac{\ddot{z_3}}{z_3} &= (\sigma + i \omega)^2 = \sigma^2 - \omega^2 + i 2 \sigma \omega \end{align}

(24)

Recalling from class the generalized time-derivative expressions for polar coordinates:

\begin{align} \dfrac{\dot{z}}{z} &= \dfrac{\dot{r}}{r} + i \dot{\theta}\\ \dfrac{\ddot{z}}{z} &= \left[ \dfrac{\ddot{r}}{r} - \dot{\theta}^2 \right] + i \left[ \ddot{\theta} + \dfrac{2 \dot{\theta} \dot{r}}{r} \right] \end{align}

(25)

We can see a direct relation between the real and imaginary parts of the above equations:

\begin{align} \dfrac{\dot{r_3}}{r_3} &= \sigma\\ \dot{\theta_3} &= \omega\\ \dfrac{\ddot{r_3}}{r_3} &= \sigma^2\\ \ddot{\theta_3} &= 0 \end{align}

(26)

This can also be seen in another way:

\begin{align} z_3(t) &= r_3(t) e^{i \theta_3(t) }\\~\\ z_3(t) &= e^{(\sigma + i \omega) t}\\ z_3(t) &= e^{\sigma t} e^{i \omega t}\\~\\ \rightarrow r_3(t) &= e^{\sigma t}, \quad \theta_3(t) = \omega t \end{align}

(27)

then taking the derivatives from there.

From the polar derivatives in Eq. (26), we can make some statements about the evolution of the path traced by $z_3$ over time. First, the sign of $\sigma$ is of paramount importance to whether $r_3$ decays to zero or grows to infinity.

If $\sigma > 0$ , then $r_3$ grows to infinity.

If $\sigma < 0$ , then $r_3$ decays to zero.

If $\sigma = 0$ , then $r_3$ remains constant, and $z_3$ proceeds in an infinite circle.

For the angular term, $\omega$ is simply the angular frequency of the oscillations intrinsic to $z_3$ .

A negative $\omega$ term is possible here, it flips the propogation of the spiral from counterclockwise to clockwise.

A negative $\omega$ has no impact on the decay rate $\dot{r_3}$ . It does flip the sign of the angular acceleration term $2 \sigma \omega$ , which is necessary to ensure decay for a negative $\sigma$ or growth for a positive $\sigma$ .

1.4Question 4: Electric field propogating in 2D¶

In class, we assumed that an plane wave was propogating in the $\hat{k} = \hat{z}$ direction, with the electric field oscillating in the $\hat{x}$ direction. Suppose now that the is oscillating in the $\dfrac{1}{\sqrt{2}} (\hat{x} + \hat{y})$ direction:

\begin{align} \boldsymbol{E} = E_0 \cos(\vec{k} \cdot \vec{r} - \omega t) \dfrac{1}{\sqrt{2}} (\hat{x} + \hat{y}) \end{align}

(28)

Question 4A:¶

What direction of propogation $\hat{k}$ and magnetic field vector $\boldsymbol{B}$ are now possible?
Draw a diagram of the electric field vector and the plane of propogation.

Question 4B:¶

What are the expressions for $\hat{k}$ and $\boldsymbol{B}$ if we constrain the direction of propogating to be (partially) in the positive $\hat{x}$ direction?

Question 4A Solution¶

The electric field oscillates in the $\hat{n} = \dfrac{1}{\sqrt{2}}(\hat{x} + \hat{y})$ direction.

The direction of propogation of the wave must be perpendicular to that vector, in other words, anywhere in the plane normal to $\hat{n}$ .

The equation for a plane is

A (x - x_0) + B (y - y_0) + C (z - z_0) = 0,

(29)

where $\vec{p} = \{x_0, y_0, z_0\}$ is a point in the plane, and $\hat{n} = \{A, B, C\}$ is the normal vector out of the plane.

We set $\vec{p} = \{0, 0, 0\}$ for simplicity, and $\hat{n} = \dfrac{1}{\sqrt{2}}\{1, 1, 0\}$ , then solve. We find that

y = - \dfrac{A}{B} x = - x

(30)

for all $z$ , so the plane of propogation is simply

x - y = 0 \quad \forall z.

(31)

The diagram of this solution is plotted below.

import matplotlib.pyplot as plt
import numpy as np

fig = plt.figure()
ax = fig.add_subplot(projection='3d')

# Define the plane (e.g., normal vector and a point)
normal = 1/np.sqrt(2)*np.array([1, 1, 0])
point = np.array([0, 0, 0])

# Create a meshgrid for the plane
size = 2
xx, zz = np.meshgrid(np.arange(-size, size, 0.5), np.arange(-size, size, 0.5))

# Calculate the Z values for the plane
# Equation of a plane: ax + by + cz = d, where d = normal . point
d = np.dot(normal, point)
yy = -normal[1] / normal[0] * xx

# Plot the plane
ax.plot_surface(xx, yy, zz, alpha=0.5, color='blue',label="Propogation Plane")

ax.quiver(0, 0, 0, normal[0], normal[1], normal[2], length=1.0, normalize=False, color='red', label=r"$\vec{E}$")

ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')
ax.set_title('3D Plane and Arrow')
ax.legend()

# ax.set_axis_off()

plt.show()

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Question 4B Solution¶

If we constrain the wave to propogate in the positive $\hat{x}$ direction, then from our result above,

\hat{k} = \dfrac{1}{\sqrt{2}} (\hat{x} - \hat{y}).

(32)

And since we know that

\boldsymbol{B} = \dfrac{1}{c}(\hat{k} \times \boldsymbol{E})

(33)

we can calculate the curl of the field like

\begin{align} \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\[6pt] \dfrac{1}{\sqrt{2}} & -\dfrac{1}{\sqrt{2}} & 0 \\[12pt] \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}} & 0 \end{vmatrix} \end{align}

(34)

which comes out to be exactly equal to $+\hat{z}$ :

\boldsymbol{B} = \dfrac{E_0}{c}\cos{(\vec{k} \cdot \vec{r} - \omega t)} \hat{z}

(35)

1.5Question 5: Spherical Plane Wave Intensity and Radiation Pressure¶

Suppose you have a sinusoidal spherical plane wave source a distance $d$ away along the $\hat{z}$ axis from a cylindrical mirror with radius $a$ .
Use the center of the spherical wave as the origin, and the distance from that center as the variable $r$ .
Assume that the cylinder is in the $xy$ plane.
Also assume that the spherical wave is emitting total power $P_\mathrm{total}$ in all directions.

Question 5A:¶

What is the Poynting vector $\boldsymbol{S}$ for the spherical waves?
Hint: Equation 9.49 of Griffith’s E&M may be helpful here

Question 5B:¶

What is the Poynting vector $\boldsymbol{S}$ incident on the mirror center?
What about the mirror edge?
Write an expression for the Poynting vector incident anywhere on the mirror’s surface.

Question 5C:¶

Using your result from Question 5B, find the intensity $I$ incident on the mirror.

Question 5D:¶

Find the total power $P$ incident on the mirror.
Compare to the total power emitted by the spherical plane wave.

Question 5E:¶

Calculate the radiation pressure $p_\mathrm{rad}$ incident on the mirror.
Also find the radiation pressure force $F_\mathrm{rad}$ .
Assume the mirror is a perfect reflector.
If the mirror has a mass $m$ , what is its acceleration?

1.6Question 5A Solution¶

First, we consider the direction the wave is propogating in. A spherical wave is conveniently propogating in the radial $\hat{r}$ direction. For ease of the next few questions, we define the $\hat{r}$ vector with respect to two angles $\theta, \phi$ , such that if $\theta = 0$ and $\phi = 0$ , $\hat{r} = \hat{z}$ :

\begin{align} \hat{r} = \cos\theta \cos\phi~\hat{z} + \cos\theta \sin\phi~\hat{x} + \sin\theta~\hat{y} \end{align}

(36)

From this definition, if $\theta = 0$ and $\phi = \dfrac{\pi}{2}$ , $\hat{r} = \hat{x}$
and if $\theta = \dfrac{\pi}{2}$ , $\hat{r} = \hat{y}$ .

Next, we write what the electric and magnetic fields are for a sinusoidal spherical plane wave. The electric field will be polarized in some $\hat{n}$ direction, which must be orthogonal to the radial direction $\hat{r}$ . We don’t bother to define $\hat{n}$ , although we could using Eq. (36).

\begin{align} \boldsymbol{E} &= \dfrac{E_0}{r} \cos(\boldsymbol{k} \cdot \boldsymbol{r} - \omega t) \hat{n}\\ \boldsymbol{B} &= \dfrac{E_0}{c r} \cos(\boldsymbol{k} \cdot \boldsymbol{r} - \omega t) (\hat{r} \times \hat{n}) \end{align}

(37)

where the magnetic field direction $(\hat{r} \times \hat{n})$ is just the remaining orthogonal direction to $\hat{r}$ and $\hat{n}$ . We also note that the phase term $\boldsymbol{k} \cdot \boldsymbol{r} = k r$ , since $\hat{k} = \hat{r}$ .

Then the Poynting vector $\boldsymbol{S}$ becomes

\begin{align} \boldsymbol{S} &= \dfrac{1}{\mu_0} (\boldsymbol{E} \times \boldsymbol{B})\\ \boldsymbol{S} &= \dfrac{1}{\mu_0} \left( \left(\dfrac{E_0}{r} \cos(\boldsymbol{k} \cdot \boldsymbol{r} - \omega t) \hat{n} \right) \times \left( \dfrac{E_0}{c r} \cos(\boldsymbol{k} \cdot \boldsymbol{r} - \omega t) (\hat{r} \times \hat{n})\right) \right)\\ \boldsymbol{S} &= \dfrac{E_0^2}{c \mu_0} \dfrac{1}{r^2} \cos(k r - \omega t)^2 (\hat{n} \times (\hat{r} \times \hat{n}))\\ \boldsymbol{S} &= \epsilon_0 E_0^2 \dfrac{1}{r^2} \cos(k r - \omega t)^2 (\hat{n} \times (\hat{r} \times \hat{n})) \end{align}

(38)

where we’ve used that $c^2 = \dfrac{1}{\epsilon_0 \mu_0}$ in the last line.

Remembering BAC CAB:

\begin{align} \boldsymbol{A} \times (\boldsymbol{B} \times \boldsymbol{C}) = \boldsymbol{B}(\boldsymbol{A} \cdot \boldsymbol{C}) - \boldsymbol{C}(\boldsymbol{A} \cdot \boldsymbol{B}) \end{align}

(39)

yields our solution

\begin{align} \boxed{ \boldsymbol{S} = \epsilon_0 E_0^2 \dfrac{1}{r^2} \cos(k r - \omega t)^2 \hat{r} } \end{align}

(40)

%load_ext jupyter_tikz

%%tikz
\begin{tikzpicture}[scale=1.5]
  % Define parameters
  \def\r{4}        % wavefront radius
  \def\d{2}        % distance to cylinder center
  \def\a{0.8}      % cylinder radius
  
  % Wavefront center at origin
  \coordinate (O) at (0, 0);
  \coordinate (C) at (\d, 0);  % cylinder center
  
  % Draw cylinder with front ellipse (facing origin) and back ellipse
  % Front ellipse (facing toward origin)
  \draw[red, thick] (C) ellipse ({0.15*\a} and {\a});
  
  % Back ellipse (receded, smaller horizontal scale)
  \draw[red, thick] (\d + 1.2, -\a) arc(-90:90:{0.15*\a} and {\a});
  % \draw[red, dashed, thick] (\d + 1.2, -\a) -- (\d + 1.2, \a);  % hidden edge (dashed)
  
  % Connecting lines (top and bottom)
  \draw[red, thick] (\d, \a) -- (\d + 1.2, \a);
  \draw[red, thick] (\d, -\a) -- (\d + 1.2, -\a);
  
  % Draw concentric spherical wavefronts (circular contours)
  \draw[cyan, thin] (0, 0) circle (0.5);
  \draw[cyan, thin] (0, 0) circle (1.0);
  \draw[cyan, thin] (0, 0) circle (1.5);
  \draw[cyan, thin] (0, 0) circle (2.0);
  \draw[cyan, thin] (0, 0) circle (2.5);
  \draw[cyan, thin] (0, 0) circle (3.0);

  % Mark origin
  \filldraw[white] (O) circle (2pt);
  \node[white, below left] at (O) {O};
  
  % Mark cylinder center
  \filldraw[white] (C) circle (1pt);
  
  % Add dimension labels
  \draw[|-|, gray, thin] (0, 0) -- (\d, 0);
  \node[below, gray] at (\d/2, 0) {$d$};
  
  \draw[|-|, gray, thin] (\d + 0.3, 0) -- (\d + 0.3, \a);
  \node[right, gray] at (\d + 0.3, \a/2) {$a$};
  
  % Label wavefront radius
  \draw[gray, thin] (0, 0) -- (\d, \a);
  \node[above, gray] at (\d/2, \a/2) {$r$};
  
\end{tikzpicture}

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1.7Question 5B Solution¶

At the center of the cylinder, $\vec{r} = d \hat{z}$ .
At the edge of the cyclinder in the y-direction, $\vec{r} = d \hat{z} + a \hat{y}$ .
We can generally write the magnitude of the $\vec{r} = r \hat{r}$ as

r = \sqrt{d^2 + x^2 + y^2}.

(41)

for any location $(x,y)$ on the mirror’s surface.

This yields

\begin{align} \mathrm{Center:}& \qquad \boldsymbol{S}(r = d) = \epsilon_0 E_0^2 \dfrac{1}{d^2} \cos(k d - \omega t)^2 \hat{z} \\~\\ \mathrm{Mirror~Edge:}& \qquad \boldsymbol{S}(r = \sqrt{d^2 + a^2}) = \epsilon_0 E_0^2 \dfrac{1}{d^2 + a^2} \cos(k \sqrt{d^2 + a^2} - \omega t)^2 (\dfrac{d}{r} \hat{z} + \dfrac{a}{r} \hat{y})\\~\\ \mathrm{Anywhere~on~Mirror~Surface:}& \qquad \boldsymbol{S}(r = \sqrt{d^2 + x^2 + y^2}) = \epsilon_0 E_0^2 \dfrac{1}{d^2 + x^2 + y^2} \cos(k \sqrt{d^2 + x^2 + y^2} - \omega t)^2 (\dfrac{d}{r} \hat{z} + \dfrac{y}{r} \hat{y} + \dfrac{x}{r} \hat{x}) \end{align}

(42)

This means we have a Poynting vector which oscillates spatially across the mirror surface.
These spatial wave front oscillations are very important for raw phase estimates, and are strongly related to the Gouy Phase Accumulation, which will become imporant when calculate higher order mode resonance conditions in optical cavities.

1.8Question 5C Solution¶

For the intensity $I$ anywhere on the mirror’s surface $(x,y)$ , we take our final Poynting vector magnitude $S$ and integrate over one phase cycle in time $t$ .

To find the Poynting vector magnitude $S$ , we identify the normal vector to the cylinder surface as $\hat{z}$ ,
then isolate the $\hat{z}$ direction by dotting with the Poynting vector $\boldsymbol{S}$ for Anywhere on the Mirror Surface:

\begin{align} S &= \boldsymbol{S} \cdot \hat{z}\\ S &= \epsilon_0 E_0^2 \dfrac{d}{r^3} \cos(k r - \omega t)^2 \end{align}

(43)

This appropriately scales for the power incident on mirror at an angle $\hat{r} \cdot \hat{z} = \cos\theta \cos\phi = \dfrac{d}{r}$ .

Now calculating the intensity over the surface $I(x,y)$

\begin{align} I(x,y) &= \langle S \rangle\\ I(x,y) &= \dfrac{1}{T} \int_0^{T} \epsilon_0 E_0^2 \epsilon_0 E_0^2 \dfrac{d}{r^3} \cos(k r - \omega t)^2 dt\\ I(x,y) &= \dfrac{1}{2} \epsilon_0 E_0^2 \dfrac{d}{r^3}\\ I(x,y) &= \dfrac{1}{2} \epsilon_0 E_0^2 \dfrac{d}{(d^2 + x^2 + y^2)^{3/2}} \end{align}

(44)

The intensity peaks at $x = 0$ and $y = 0$ , the nearest locations to the center of the cylinder.

1.9Question 5D Solution¶

The power incident on the mirror $P$ is the double integral over the mirror surface out to radius $a$ :

\begin{align} P = \int \int I(x,y) dx dy \end{align}

(45)

We define a radius on the mirror surface

\vec{s} = x \hat{x} + y \hat{y}, \text{where}~s^2 = x^2 + y^2,~\text{and}~dx dy = s ds d\varphi

(46)

to make our integral easier.

$I(x,y)$ will become a single variable $I(s)$ :

\begin{align} I(x,y) &= \dfrac{1}{2} \epsilon_0 E_0^2 \dfrac{d}{(d^2 + x^2 + y^2)^{3/2}}\\ I(s) &= \dfrac{1}{2} \epsilon_0 E_0^2 \dfrac{d}{(d^2 + s^2)^{3/2}} \end{align}

(47)

For the power $P$ calculation, we have

\begin{align} P &= \int_0^{2\pi} \int_0^{a} I(s)~s ds d\varphi\\ P &= 2\pi \int_0^{a} I(s)~s ds\\ P &= 2\pi \int_0^{a} \dfrac{1}{2} \epsilon_0 E_0^2 \dfrac{s d}{(d^2 + s^2)^{3/2}}~ds\\ P &= \pi \epsilon_0 E_0^2 d \int_0^{a} \dfrac{s}{(d^2 + s^2)^{3/2}}~ds\\ P &= \pi \epsilon_0 E_0^2 d \left[ \dfrac{1}{d} - \dfrac{1}{\sqrt{d^2 + a^2}} \right] \\ \end{align}

(48)

\begin{align} \boxed{ P = \pi \epsilon_0 E_0^2 \left[ 1 - \dfrac{d}{\sqrt{d^2 + a^2}} \right] } \end{align}

(49)

If we let the cylinder radius $a$ go to infinity, then the right hand term goes to zero and we recover

\boxed{ P(a\rightarrow\infty) = \pi \epsilon_0 E_0^2 }

(50)

To get the total power $P_\mathrm{total}$ , we integrate (40) over one cycle $T$ to get total intensity $I_\mathrm{total}(r)$ , then spatially over a spherical surface.