Electromagnetic Waves: Electromagnetic Momentum, Radiation Pressure, and Photon Momentum - Exercises
ElectromagneticWavesElectromagneticMomentumRadiationPressureAndPhotonMomentumExercises
2026-09-19 19:41:04
2026-09-19 19:44:37
Topic
added julia example code back in
electromagnetic momentum
momentum density
Poynting vector
Maxwell stress tensor
radiation pressure
absorption
reflection
transmission
oblique incidence
solar sail
photon momentum
photon flux
optical force
RF momentum transfer
exercises
worked solutions
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\section*{Electromagnetic Waves, Antennas, and RF: Electromagnetic Momentum, Radiation Pressure, and Photon Momentum - Exercises and Complete Worked Solutions}
EM18 established that electromagnetic fields carry momentum as well as energy. This companion article develops that result through worked problems ranging from local field momentum density to radiation pressure, the Maxwell stress tensor, solar-sail acceleration, photon momentum, and the agreement between classical and quantum momentum accounting.
The central vacuum relations are
\begin{equation}
\boxed{\mathbf g=\epsilon_0\mathbf E\times\mathbf B=\frac{\mathbf S}{c^2},}
\end{equation}
\begin{equation}
\boxed{\mathbf P_{\text{EM}}=\int_V\mathbf g\,dV,}
\end{equation}
and, for a plane wave,
\begin{equation}
\boxed{g=\frac{u}{c},\qquad P_{\text{EM}}=\frac{U}{c}.}
\end{equation}
At normal incidence, the radiation-pressure limits are
\begin{equation}
\boxed{p_{\text{abs}}=\frac{I}{c},\qquad p_{\text{refl}}=\frac{2I}{c}.}
\end{equation}
The photon description uses
\begin{equation}
\boxed{E_\gamma=h\nu=\frac{hc}{\lambda},\qquad p_\gamma=\frac{E_\gamma}{c}=\frac{h}{\lambda}.}
\end{equation}
The problems below are intended to make these formulas consequences of momentum conservation rather than isolated facts \cite{Griffiths2017,Jackson1999,OpenStaxV2,FeynmanV1,FeynmanV2}.
\section*{How to use this problem set}
Attempt all exercises in Part I before consulting Part II. In every radiation-pressure problem, identify three things before calculating: the direction of the incident momentum, what fraction of that momentum leaves after interaction, and whether the requested force is along the beam direction or normal to a material surface. This prevents most sign and factor-of-two errors.
\begin{center}
\includegraphics{EM18E1_fig01_momentum_chain.png}
\vspace{0.45em}
\textbf{Figure.} The electromagnetic momentum chain. The fields determine the Poynting vector, the Poynting vector determines momentum density, and momentum flux produces force and pressure when the radiation interacts with matter.
\end{center}
\section*{Part I: Exercises}
\section*{Exercise 1: momentum density of a vacuum plane wave}
At a particular point and instant, a vacuum plane wave has
\begin{equation}
\mathbf E=(300\,\text{V/m})\hat{\mathbf x}
\end{equation}
and propagates in the $+z$ direction.
Find:
\begin{enumerate}
\item[(a)] the corresponding magnetic-field vector $\mathbf B$;
\item[(b)] the instantaneous Poynting vector $\mathbf S$;
\item[(c)] the electromagnetic momentum density $\mathbf g$ using $\mathbf g=\mathbf S/c^2$;
\item[(d)] the total instantaneous energy density $u$;
\item[(e)] verify numerically that $g=u/c$.
\end{enumerate}
\section*{Exercise 2: total momentum in a finite electromagnetic pulse}
A short vacuum pulse carries total electromagnetic energy
\begin{equation}
U=2.4\,\text{mJ}
\end{equation}
through a beam of cross-sectional area
\begin{equation}
A=3.0\,\text{cm}^2.
\end{equation}
Its duration is
\begin{equation}
\tau=8.0\,\text{ns}.
\end{equation}
Assume a uniform rectangular pulse profile.
Find:
\begin{enumerate}
\item[(a)] the total electromagnetic momentum of the pulse;
\item[(b)] the pulse length $L=c\tau$;
\item[(c)] the pulse volume $V=AL$;
\item[(d)] the average energy density $u=U/V$ inside the pulse;
\item[(e)] the average momentum density $g=u/c$;
\item[(f)] verify that $gV=U/c$.
\end{enumerate}
\section*{Exercise 3: absorbing surface under a known intensity}
A normally incident electromagnetic beam has intensity
\begin{equation}
I=1200\,\text{W/m}^2
\end{equation}
and completely illuminates an absorbing plate of area
\begin{equation}
A=0.35\,\text{m}^2.
\end{equation}
Find:
\begin{enumerate}
\item[(a)] the radiation pressure;
\item[(b)] the force on the plate;
\item[(c)] the momentum transferred to the plate during $10\,\text{s}$.
\end{enumerate}
\section*{Exercise 4: perfect mirror and the factor of two}
A $8.0\,\text{W}$ laser beam is completely intercepted by a perfect mirror at normal incidence.
Find:
\begin{enumerate}
\item[(a)] the incident electromagnetic momentum arriving per second;
\item[(b)] the reflected electromagnetic momentum leaving per second, including its direction;
\item[(c)] the force on the mirror;
\item[(d)] explain from momentum conservation why the result is twice the absorbing-surface force for the same beam power.
\end{enumerate}
\section*{Exercise 5: partial absorption, reflection, and transmission}
A plane wave of intensity
\begin{equation}
I=2500\,\text{W/m}^2
\end{equation}
strikes a planar optical element at normal incidence. The power fractions are
\begin{equation}
A=0.25,
\qquad
R=0.60,
\qquad
T=0.15.
\end{equation}
Assume the incident, reflected, and transmitted beams are all in vacuum, with transmission continuing in the original direction.
Find:
\begin{enumerate}
\item[(a)] the incoming momentum flux;
\item[(b)] the outgoing reflected and transmitted momentum fluxes, with signs;
\item[(c)] the radiation pressure on the element;
\item[(d)] verify the equivalent formula
\begin{equation}
p_{\text{rad}}=\frac{(A+2R)I}{c}.
\end{equation}
\end{enumerate}
\begin{center}
\includegraphics{EM18E1_fig02_partial_surface_momentum.png}
\vspace{0.45em}
\textbf{Figure.} Momentum bookkeeping for a surface that absorbs, reflects, and transmits portions of an incident wave. The reflected momentum reverses sign, which is why reflection contributes twice its fractional power to the pressure.
\end{center}
\section*{Exercise 6: oblique reflection}
Sunlight with intensity
\begin{equation}
I=1360\,\text{W/m}^2
\end{equation}
strikes a perfectly reflecting flat sail of actual area
\begin{equation}
A=20\,\text{m}^2
\end{equation}
at an angle
\begin{equation}
\theta=35^\circ
\end{equation}
measured from the surface normal.
Find:
\begin{enumerate}
\item[(a)] the projected area seen by the beam;
\item[(b)] the incident power intercepted by the sail;
\item[(c)] the normal radiation pressure;
\item[(d)] the normal force on the sail.
\end{enumerate}
Explain physically why two factors of $\cos\theta$ appear in the normal force.
\section*{Exercise 7: Maxwell stress tensor for a plane wave}
A sinusoidal plane wave propagates in the $+z$ direction with peak electric-field amplitude
\begin{equation}
E_0=300\,\text{V/m}.
\end{equation}
At an instant when the electric field is at its positive peak, take
\begin{equation}
\mathbf E=E_0\hat{\mathbf x},
\qquad
\mathbf B=\frac{E_0}{c}\hat{\mathbf y}.
\end{equation}
Using
\begin{equation}
\sigma_{ij}
=
\epsilon_0\left(E_iE_j-\frac{1}{2}\delta_{ij}E^2\right)
+
\frac{1}{\mu_0}\left(B_iB_j-\frac{1}{2}\delta_{ij}B^2\right),
\end{equation}
find:
\begin{enumerate}
\item[(a)] the instantaneous total energy density $u$;
\item[(b)] $\sigma_{zz}$ at the field peak;
\item[(c)] the cycle-averaged magnitude $\langle|\sigma_{zz}|\rangle$;
\item[(d)] verify that $\langle|\sigma_{zz}|\rangle=I/c$.
\end{enumerate}
\section*{Exercise 8: stress tensor of a static electric field}
A uniform electrostatic field is
\begin{equation}
\mathbf E=(2.0\times10^6\,\text{V/m})\hat{\mathbf x},
\qquad
\mathbf B=\mathbf 0.
\end{equation}
Find the Maxwell stress tensor in Cartesian coordinates.
Then determine the traction vector
\begin{equation}
\mathbf t=\boldsymbol{\sigma}\cdot\hat{\mathbf n}
\end{equation}
for surfaces whose outward normals are:
\begin{enumerate}
\item[(a)] $\hat{\mathbf n}=\hat{\mathbf x}$;
\item[(b)] $\hat{\mathbf n}=\hat{\mathbf y}$.
\end{enumerate}
Interpret the signs as tension along the field direction and compression transverse to the field.
\begin{center}
\includegraphics{EM18E1_fig03_maxwell_stress_cube.png}
\vspace{0.45em}
\textbf{Figure.} A static electric field produces anisotropic electromagnetic stress: tensile along the field direction and compressive on transverse faces in the stated stress-tensor sign convention.
\end{center}
\section*{Exercise 9: impulse from reflecting a finite-energy pulse}
A light pulse of energy
\begin{equation}
U=0.75\,\text{J}
\end{equation}
reflects normally from a free mirror of mass
\begin{equation}
m=2.0\,\text{g}.
\end{equation}
Assume the mirror is initially at rest and its acquired speed is sufficiently small that the change in photon energy can be neglected to first order.
Find:
\begin{enumerate}
\item[(a)] the impulse delivered to the mirror;
\item[(b)] the mirror's change in speed.
\end{enumerate}
\section*{Exercise 10: idealized solar-sail acceleration}
At approximately $1\,\text{AU}$ from the Sun, take the solar intensity to be
\begin{equation}
I=1361\,\text{W/m}^2.
\end{equation}
An ideal perfectly reflecting sail has area
\begin{equation}
A=100\,\text{m}^2
\end{equation}
and total spacecraft mass
\begin{equation}
m=12\,\text{kg}.
\end{equation}
Assume normal incidence and neglect all other forces and the variation of solar intensity with distance.
Find:
\begin{enumerate}
\item[(a)] the radiation force;
\item[(b)] the resulting acceleration;
\item[(c)] the idealized change in speed after one day.
\end{enumerate}
\section*{Exercise 11: energy and momentum of a green photon}
A photon has wavelength
\begin{equation}
\lambda=532\,\text{nm}.
\end{equation}
Find:
\begin{enumerate}
\item[(a)] the frequency;
\item[(b)] the photon energy in joules;
\item[(c)] the photon momentum;
\item[(d)] verify numerically that $E_\gamma=p_\gamma c$.
\end{enumerate}
\section*{Exercise 12: photon counting reproduces laser radiation pressure}
A continuous laser has power
\begin{equation}
P=1.00\,\text{W}
\end{equation}
and wavelength
\begin{equation}
\lambda=632.8\,\text{nm}.
\end{equation}
Find:
\begin{enumerate}
\item[(a)] the energy of one photon;
\item[(b)] the photon rate $\dot N$;
\item[(c)] the momentum of one photon;
\item[(d)] the force on a perfectly absorbing target from photon counting;
\item[(e)] the force on a perfect mirror;
\item[(f)] verify that the answers reduce to $P/c$ and $2P/c$.
\end{enumerate}
\section*{Exercise 13: wavelength dependence at fixed optical power}
Two lasers each have output power
\begin{equation}
P=2.0\,\text{W}.
\end{equation}
Laser A has wavelength $400\,\text{nm}$ and Laser B has wavelength $800\,\text{nm}$.
For each laser, calculate:
\begin{enumerate}
\item[(a)] photon energy;
\item[(b)] photon momentum;
\item[(c)] photon rate;
\item[(d)] momentum delivered per second to a perfectly absorbing target.
\end{enumerate}
Explain why the shorter-wavelength laser has larger momentum per photon but does not exert a larger force at the same total power.
\begin{center}
\includegraphics{EM18E1_fig04_photon_flux_bridge.png}
\vspace{0.45em}
\textbf{Figure.} At fixed beam power, shorter-wavelength photons carry more momentum individually but arrive at a lower photon rate. The product $\dot Np_\gamma$ remains $P/c$ for complete absorption.
\end{center}
\section*{Exercise 14: RF beam momentum transfer in the far field}
A transmitting antenna radiates
\begin{equation}
P_t=50\,\text{W}
\end{equation}
with linear gain
\begin{equation}
G=20
\end{equation}
in the direction of a receiving panel located
\begin{equation}
r=2.0\,\text{km}
\end{equation}
away. Assume free-space far-field spreading and use
\begin{equation}
I=\frac{P_tG}{4\pi r^2}.
\end{equation}
If an absorbing panel of area
\begin{equation}
A=1.2\,\text{m}^2
\end{equation}
is normal to the beam and intercepts the local radiation uniformly, find:
\begin{enumerate}
\item[(a)] the local intensity;
\item[(b)] the radiation pressure;
\item[(c)] the force on the panel.
\end{enumerate}
Comment on why momentum transfer is usually negligible in ordinary RF link budgets even though the same electromagnetic fields carry both energy and momentum.
\section*{Exercise 15: derive local electromagnetic momentum conservation}
Starting with the Lorentz force density
\begin{equation}
\mathbf f=\rho\mathbf E+\mathbf J\times\mathbf B,
\end{equation}
use Maxwell's equations to show that it can be written as
\begin{equation}
\boxed{
\mathbf f
=
\nabla\cdot\boldsymbol{\sigma}
-
\frac{\partial\mathbf g}{\partial t},}
\end{equation}
where
\begin{equation}
\mathbf g=\epsilon_0\mathbf E\times\mathbf B
\end{equation}
and
\begin{equation}
\sigma_{ij}
=
\epsilon_0\left(E_iE_j-\frac12\delta_{ij}E^2\right)
+
\frac{1}{\mu_0}\left(B_iB_j-\frac12\delta_{ij}B^2\right).
\end{equation}
State the physical meaning of each term in the final equation and write the corresponding integral momentum-balance equation over a fixed volume.
\section*{Exercise 16: Julia sweep of reflectivity and incidence angle}
Consider an opaque surface, so that
\begin{equation}
A+R=1,
\qquad
T=0.
\end{equation}
For incidence angle $\theta$ measured from the surface normal, the normal pressure is
\begin{equation}
p_n(R,\theta)=\frac{(1+R)I}{c}\cos^2\theta.
\end{equation}
Using
\begin{equation}
I=1000\,\text{W/m}^2,
\end{equation}
write a Julia program that evaluates $p_n$ for
\begin{equation}
R=0,\ 0.25,\ 0.50,\ 0.75,\ 1.00
\end{equation}
and
\begin{equation}
\theta=0^\circ,\ 30^\circ,\ 60^\circ,\ 80^\circ.
\end{equation}
Verify numerically that:
\begin{enumerate}
\item[(a)] at fixed angle the pressure is linear in $1+R$;
\item[(b)] at fixed reflectivity it scales as $\cos^2\theta$;
\item[(c)] the $R=0$, $\theta=0$ case gives $I/c$;
\item[(d)] the $R=1$, $\theta=0$ case gives $2I/c$.
\end{enumerate}
\section*{Part II: Complete Worked Solutions}
\section*{Solution 1: momentum density of a vacuum plane wave}
For a vacuum plane wave,
\begin{equation}
B=\frac{E}{c}.
\end{equation}
Therefore
\begin{align}
B
&=\frac{300}{2.99792458\times10^8}\\
&=1.00069\times10^{-6}\,\text{T}.
\end{align}
Because the propagation direction is $+z$ and
\begin{equation}
\hat{\mathbf x}\times\hat{\mathbf y}=\hat{\mathbf z},
\end{equation}
we have
\begin{equation}
\boxed{
\mathbf B=(1.00069\,\mu\text{T})\hat{\mathbf y}.}
\end{equation}
The instantaneous Poynting vector is
\begin{align}
\mathbf S
&=\frac{1}{\mu_0}\mathbf E\times\mathbf B\\
&=\frac{(300)(1.00069\times10^{-6})}{1.25663706\times10^{-6}}
\hat{\mathbf z}\\
&=238.90\,\hat{\mathbf z}\,\text{W/m}^2.
\end{align}
Thus
\begin{equation}
\boxed{\mathbf S=238.90\,\hat{\mathbf z}\,\text{W/m}^2.}
\end{equation}
The momentum density is
\begin{align}
\mathbf g
&=\frac{\mathbf S}{c^2}\\
&=\frac{238.90}{(2.99792458\times10^8)^2}\hat{\mathbf z}\\
&=2.6581\times10^{-15}\hat{\mathbf z}
\,\text{kg/(m}^2\text{s)}.
\end{align}
Therefore
\begin{equation}
\boxed{
\mathbf g=2.6581\times10^{-15}\hat{\mathbf z}\,\text{kg/(m}^2\text{s)}.}
\end{equation}
For a plane wave, the electric and magnetic energy densities are equal, so
\begin{equation}
u=\epsilon_0E^2.
\end{equation}
Hence
\begin{align}
u
&=(8.85418781\times10^{-12})(300)^2\\
&=7.96877\times10^{-7}\,\text{J/m}^3.
\end{align}
Thus
\begin{equation}
\boxed{u=7.96877\times10^{-7}\,\text{J/m}^3.}
\end{equation}
Finally,
\begin{align}
\frac{u}{c}
&=\frac{7.96877\times10^{-7}}{2.99792458\times10^8}\\
&=2.6581\times10^{-15}\,\text{kg/(m}^2\text{s)},
\end{align}
which agrees with the momentum density obtained from $\mathbf S/c^2$.
\section*{Solution 2: total momentum in a finite electromagnetic pulse}
For a vacuum plane-wave pulse,
\begin{equation}
P_{\text{EM}}=\frac{U}{c}.
\end{equation}
Therefore
\begin{align}
P_{\text{EM}}
&=\frac{2.4\times10^{-3}}{2.99792458\times10^8}\\
&=8.01\times10^{-12}\,\text{kg m/s}.
\end{align}
Thus
\begin{equation}
\boxed{P_{\text{EM}}=8.01\times10^{-12}\,\text{kg m/s}.}
\end{equation}
The pulse length is
\begin{align}
L
&=c\tau\\
&=(2.99792458\times10^8)(8.0\times10^{-9})\\
&=2.398\,\text{m}.
\end{align}
The area is
\begin{equation}
A=3.0\,\text{cm}^2=3.0\times10^{-4}\,\text{m}^2.
\end{equation}
Therefore the pulse volume is
\begin{align}
V
&=AL\\
&=(3.0\times10^{-4})(2.398)\\
&=7.195\times10^{-4}\,\text{m}^3.
\end{align}
The average energy density is
\begin{align}
u
&=\frac{U}{V}\\
&=\frac{2.4\times10^{-3}}{7.195\times10^{-4}}\\
&=3.336\,\text{J/m}^3.
\end{align}
The average momentum density is then
\begin{align}
g
&=\frac{u}{c}\\
&=1.113\times10^{-8}\,\text{kg/(m}^2\text{s)}.
\end{align}
Multiplying by the pulse volume gives
\begin{align}
gV
&=(1.113\times10^{-8})(7.195\times10^{-4})\\
&=8.01\times10^{-12}\,\text{kg m/s},
\end{align}
which is exactly the same result as $U/c$ to rounding.
\section*{Solution 3: absorbing surface under a known intensity}
For complete absorption at normal incidence,
\begin{equation}
p_{\text{abs}}=\frac{I}{c}.
\end{equation}
Thus
\begin{align}
p_{\text{abs}}
&=\frac{1200}{2.99792458\times10^8}\\
&=4.003\times10^{-6}\,\text{Pa}.
\end{align}
Therefore
\begin{equation}
\boxed{p_{\text{abs}}=4.00\,\mu\text{Pa}.}
\end{equation}
The force is
\begin{align}
F
&=pA\\
&=(4.003\times10^{-6})(0.35)\\
&=1.401\times10^{-6}\,\text{N}.
\end{align}
Hence
\begin{equation}
\boxed{F=1.40\,\mu\text{N}.}
\end{equation}
The momentum delivered in $10\,\text{s}$ is the impulse
\begin{align}
\Delta p
&=F\Delta t\\
&=(1.401\times10^{-6})(10)\\
&=1.401\times10^{-5}\,\text{kg m/s}.
\end{align}
So
\begin{equation}
\boxed{\Delta p=1.40\times10^{-5}\,\text{kg m/s}.}
\end{equation}
\section*{Solution 4: perfect mirror and the factor of two}
The incident momentum arriving per unit time is
\begin{equation}
\frac{dp_{\text{in}}}{dt}=\frac{P}{c}.
\end{equation}
For $P=8.0\,\text{W}$,
\begin{align}
\frac{dp_{\text{in}}}{dt}
&=\frac{8.0}{2.99792458\times10^8}\\
&=2.6685\times10^{-8}\,\text{N}.
\end{align}
Take the incident direction as positive. The reflected radiation carries momentum away in the negative direction at the rate
\begin{equation}
\frac{dp_{\text{out}}}{dt}=-\frac{P}{c}.
\end{equation}
Therefore the field momentum changes at the rate
\begin{align}
\frac{d}{dt}\Delta p_{\text{field}}
&=-\frac{P}{c}-\frac{P}{c}\\
&=-\frac{2P}{c}.
\end{align}
The mirror receives the opposite force:
\begin{align}
F_{\text{mirror}}
&=\frac{2P}{c}\\
&=5.337\times10^{-8}\,\text{N}.
\end{align}
Hence
\begin{equation}
\boxed{F_{\text{mirror}}=53.4\,\text{nN}.}
\end{equation}
An absorber removes the incident momentum $+P/c$ from the field. A mirror changes the field momentum from $+P/c$ to $-P/c$. The total change is therefore twice as large.
\section*{Solution 5: partial absorption, reflection, and transmission}
The incoming momentum flux is
\begin{equation}
\Pi_{\text{in}}=\frac{I}{c}.
\end{equation}
Numerically,
\begin{align}
\Pi_{\text{in}}
&=\frac{2500}{2.99792458\times10^8}\\
&=8.339\times10^{-6}\,\text{Pa}.
\end{align}
The reflected momentum flux is negative because it travels opposite the incident direction:
\begin{equation}
\Pi_R=-\frac{RI}{c}.
\end{equation}
The transmitted momentum flux remains positive:
\begin{equation}
\Pi_T=+\frac{TI}{c}.
\end{equation}
Thus the net outgoing momentum flux is
\begin{equation}
\Pi_{\text{out}}
=-\frac{RI}{c}+\frac{TI}{c}.
\end{equation}
The momentum delivered to the element per unit area per unit time is
\begin{align}
p_{\text{rad}}
&=\Pi_{\text{in}}-\Pi_{\text{out}}\\
&=\frac{(1+R-T)I}{c}.
\end{align}
Using the given fractions,
\begin{equation}
1+R-T=1+0.60-0.15=1.45.
\end{equation}
Hence
\begin{align}
p_{\text{rad}}
&=\frac{(1.45)(2500)}{2.99792458\times10^8}\\
&=1.209\times10^{-5}\,\text{Pa}.
\end{align}
Therefore
\begin{equation}
\boxed{p_{\text{rad}}=12.1\,\mu\text{Pa}.}
\end{equation}
Since
\begin{equation}
A=1-R-T,
\end{equation}
we have
\begin{align}
A+2R
&=(1-R-T)+2R\\
&=1+R-T,
\end{align}
so
\begin{equation}
\boxed{p_{\text{rad}}=\frac{(A+2R)I}{c}}
\end{equation}
is identical to the momentum-flux result.
\section*{Solution 6: oblique reflection}
The projected area normal to the beam is
\begin{align}
A_\perp
&=A\cos\theta\\
&=(20)\cos35^\circ\\
&=16.38\,\text{m}^2.
\end{align}
The intercepted power is therefore
\begin{align}
P_{\text{inc}}
&=IA\cos\theta\\
&=(1360)(20)\cos35^\circ\\
&=2.228\times10^4\,\text{W}.
\end{align}
For ideal specular reflection, the normal component of momentum reverses. The normal pressure on the actual surface is
\begin{equation}
p_n=\frac{2I}{c}\cos^2\theta.
\end{equation}
Thus
\begin{align}
p_n
&=\frac{2(1360)}{2.99792458\times10^8}\cos^2 35^\circ\\
&=6.088\times10^{-6}\,\text{Pa}.
\end{align}
Hence
\begin{equation}
\boxed{p_n=6.09\,\mu\text{Pa}.}
\end{equation}
The normal force is
\begin{align}
F_n
&=p_nA\\
&=(6.088\times10^{-6})(20)\\
&=1.218\times10^{-4}\,\text{N}.
\end{align}
Therefore
\begin{equation}
\boxed{F_n=1.22\times10^{-4}\,\text{N}.}
\end{equation}
One factor of $\cos\theta$ comes from the projected collecting area $A\cos\theta$. The second comes from taking the normal component of the incident and reflected momentum.
\section*{Solution 7: Maxwell stress tensor for a plane wave}
At the electric-field peak, a vacuum plane wave has equal electric and magnetic energy densities, so the total instantaneous energy density is
\begin{equation}
u=\epsilon_0E_0^2.
\end{equation}
Thus
\begin{align}
u
&=(8.85418781\times10^{-12})(300)^2\\
&=7.96877\times10^{-7}\,\text{J/m}^3.
\end{align}
For propagation along $z$, both $E_z$ and $B_z$ vanish. Hence
\begin{align}
\sigma_{zz}
&=-\frac12\epsilon_0E^2-\frac{B^2}{2\mu_0}\\
&=-u.
\end{align}
At the peak,
\begin{equation}
\boxed{\sigma_{zz}=-7.96877\times10^{-7}\,\text{Pa}.}
\end{equation}
For sinusoidal fields, the cycle average of $E^2$ is $E_0^2/2$. Therefore
\begin{align}
\left\langle|\sigma_{zz}|\right\rangle
&=\frac12\epsilon_0E_0^2\\
&=3.98438\times10^{-7}\,\text{Pa}.
\end{align}
The average wave intensity is
\begin{equation}
I=\frac12\epsilon_0cE_0^2.
\end{equation}
Therefore
\begin{equation}
\frac{I}{c}=\frac12\epsilon_0E_0^2,
\end{equation}
which is exactly the same average stress magnitude.
Numerically,
\begin{equation}
I=119.45\,\text{W/m}^2,
\end{equation}
so
\begin{equation}
\boxed{\frac{I}{c}=3.98438\times10^{-7}\,\text{Pa}.}
\end{equation}
\section*{Solution 8: stress tensor of a static electric field}
With $\mathbf B=\mathbf 0$, the stress tensor is
\begin{equation}
\sigma_{ij}
=\epsilon_0\left(E_iE_j-\frac12\delta_{ij}E^2\right).
\end{equation}
For
\begin{equation}
\mathbf E=E\hat{\mathbf x},
\end{equation}
we have
\begin{equation}
E_x=E,
\qquad
E_y=E_z=0.
\end{equation}
Therefore
\begin{equation}
\boldsymbol{\sigma}
=
\frac12\epsilon_0E^2
\begin{bmatrix}
1&0&0\\
0&-1&0\\
0&0&-1
\end{bmatrix}.
\end{equation}
The stress scale is
\begin{align}
\frac12\epsilon_0E^2
&=\frac12(8.85418781\times10^{-12})(2.0\times10^6)^2\\
&=17.708\,\text{Pa}.
\end{align}
Thus
\begin{equation}
\boxed{
\boldsymbol{\sigma}
=
\begin{bmatrix}
17.708&0&0\\
0&-17.708&0\\
0&0&-17.708
\end{bmatrix}\text{Pa}.}
\end{equation}
For a surface normal to $+x$,
\begin{equation}
\mathbf t_x=\boldsymbol{\sigma}\cdot\hat{\mathbf x}
=(17.708\,\text{Pa})\hat{\mathbf x}.
\end{equation}
For a surface normal to $+y$,
\begin{equation}
\mathbf t_y=\boldsymbol{\sigma}\cdot\hat{\mathbf y}
=-(17.708\,\text{Pa})\hat{\mathbf y}.
\end{equation}
The field therefore produces a tensile stress along its own direction and a compressive stress transverse to that direction in this sign convention.
\section*{Solution 9: impulse from reflecting a finite-energy pulse}
The incident pulse momentum is
\begin{equation}
p_i=\frac{U}{c}.
\end{equation}
Ideal reflection reverses the electromagnetic momentum, so the mirror receives an impulse
\begin{equation}
J=\frac{2U}{c}.
\end{equation}
Thus
\begin{align}
J
&=\frac{2(0.75)}{2.99792458\times10^8}\\
&=5.0035\times10^{-9}\,\text{N s}.
\end{align}
Therefore
\begin{equation}
\boxed{J=5.00\times10^{-9}\,\text{N s}.}
\end{equation}
The mirror mass is
\begin{equation}
m=2.0\,\text{g}=2.0\times10^{-3}\,\text{kg}.
\end{equation}
Hence
\begin{align}
\Delta v
&=\frac{J}{m}\\
&=\frac{5.0035\times10^{-9}}{2.0\times10^{-3}}\\
&=2.502\times10^{-6}\,\text{m/s}.
\end{align}
So
\begin{equation}
\boxed{\Delta v=2.50\,\mu\text{m/s}.}
\end{equation}
\section*{Solution 10: idealized solar-sail acceleration}
For a perfectly reflecting sail at normal incidence,
\begin{equation}
F=\frac{2IA}{c}.
\end{equation}
Therefore
\begin{align}
F
&=\frac{2(1361)(100)}{2.99792458\times10^8}\\
&=9.080\times10^{-4}\,\text{N}.
\end{align}
Thus
\begin{equation}
\boxed{F=0.908\,\text{mN}.}
\end{equation}
The acceleration is
\begin{align}
a
&=\frac{F}{m}\\
&=\frac{9.080\times10^{-4}}{12}\\
&=7.566\times10^{-5}\,\text{m/s}^2.
\end{align}
Hence
\begin{equation}
\boxed{a=7.57\times10^{-5}\,\text{m/s}^2.}
\end{equation}
One day is
\begin{equation}
\Delta t=86400\,\text{s}.
\end{equation}
Under the stated constant-acceleration approximation,
\begin{align}
\Delta v
&=a\Delta t\\
&=(7.566\times10^{-5})(86400)\\
&=6.54\,\text{m/s}.
\end{align}
Therefore
\begin{equation}
\boxed{\Delta v\approx6.54\,\text{m/s per day}}
\end{equation}
under the idealized assumptions. A real trajectory requires solar gravity, sail orientation, reflectivity, thermal effects, and the variation of solar flux with heliocentric distance.
\section*{Solution 11: energy and momentum of a green photon}
The frequency is
\begin{equation}
\nu=\frac{c}{\lambda}.
\end{equation}
With
\begin{equation}
\lambda=532\times10^{-9}\,\text{m},
\end{equation}
we obtain
\begin{align}
\nu
&=\frac{2.99792458\times10^8}{532\times10^{-9}}\\
&=5.6352\times10^{14}\,\text{Hz}.
\end{align}
The photon energy is
\begin{align}
E_\gamma
&=h\nu\\
&=(6.62607015\times10^{-34})(5.6352\times10^{14})\\
&=3.7339\times10^{-19}\,\text{J}.
\end{align}
The photon momentum is
\begin{align}
p_\gamma
&=\frac{h}{\lambda}\\
&=\frac{6.62607015\times10^{-34}}{532\times10^{-9}}\\
&=1.2455\times10^{-27}\,\text{kg m/s}.
\end{align}
Thus
\begin{equation}
\boxed{
\nu=5.6352\times10^{14}\,\text{Hz},
\quad
E_\gamma=3.7339\times10^{-19}\,\text{J},
\quad
p_\gamma=1.2455\times10^{-27}\,\text{kg m/s}.}
\end{equation}
Finally,
\begin{align}
p_\gamma c
&=(1.2455\times10^{-27})(2.99792458\times10^8)\\
&=3.7339\times10^{-19}\,\text{J},
\end{align}
so $E_\gamma=p_\gamma c$ is verified numerically.
\section*{Solution 12: photon counting reproduces laser radiation pressure}
The photon energy is
\begin{equation}
E_\gamma=\frac{hc}{\lambda}.
\end{equation}
For $\lambda=632.8\,\text{nm}$,
\begin{align}
E_\gamma
&=\frac{(6.62607015\times10^{-34})(2.99792458\times10^8)}{632.8\times10^{-9}}\\
&=3.1391\times10^{-19}\,\text{J}.
\end{align}
The photon arrival rate is
\begin{align}
\dot N
&=\frac{P}{E_\gamma}\\
&=\frac{1.00}{3.1391\times10^{-19}}\\
&=3.1856\times10^{18}\,\text{s}^{-1}.
\end{align}
The momentum per photon is
\begin{align}
p_\gamma
&=\frac{h}{\lambda}\\
&=1.0471\times10^{-27}\,\text{kg m/s}.
\end{align}
For complete absorption, each photon transfers $p_\gamma$. Therefore
\begin{align}
F_{\text{abs}}
&=\dot Np_\gamma\\
&=(3.1856\times10^{18})(1.0471\times10^{-27})\\
&=3.3356\times10^{-9}\,\text{N}.
\end{align}
Thus
\begin{equation}
\boxed{F_{\text{abs}}=3.336\,\text{nN}.}
\end{equation}
For perfect reflection, each photon reverses momentum and transfers $2p_\gamma$:
\begin{equation}
\boxed{F_{\text{refl}}=6.671\,\text{nN}.}
\end{equation}
Algebraically,
\begin{align}
\dot Np_\gamma
&=\frac{P}{E_\gamma}\frac{E_\gamma}{c}\\
&=\frac{P}{c}.
\end{align}
Thus photon counting gives exactly the classical result
\begin{equation}
\boxed{F_{\text{abs}}=\frac{P}{c},\qquad F_{\text{refl}}=\frac{2P}{c}.}
\end{equation}
\section*{Solution 13: wavelength dependence at fixed optical power}
For each laser,
\begin{equation}
E_\gamma=\frac{hc}{\lambda},
\qquad
p_\gamma=\frac{h}{\lambda},
\qquad
\dot N=\frac{P}{E_\gamma}.
\end{equation}
For the $400\,\text{nm}$ laser,
\begin{align}
E_{\gamma,400}
&=4.9661\times10^{-19}\,\text{J},\\
p_{\gamma,400}
&=1.6565\times10^{-27}\,\text{kg m/s},\\
\dot N_{400}
&=4.0273\times10^{18}\,\text{s}^{-1}.
\end{align}
Therefore
\begin{equation}
\dot N_{400}p_{\gamma,400}
=6.6713\times10^{-9}\,\text{N}.
\end{equation}
For the $800\,\text{nm}$ laser,
\begin{align}
E_{\gamma,800}
&=2.4831\times10^{-19}\,\text{J},\\
p_{\gamma,800}
&=8.2826\times10^{-28}\,\text{kg m/s},\\
\dot N_{800}
&=8.0546\times10^{18}\,\text{s}^{-1}.
\end{align}
Therefore
\begin{equation}
\dot N_{800}p_{\gamma,800}
=6.6713\times10^{-9}\,\text{N}.
\end{equation}
The $400\,\text{nm}$ photon has twice the momentum of the $800\,\text{nm}$ photon, but only half as many $400\,\text{nm}$ photons are needed per second to carry the same power. The two effects cancel:
\begin{equation}
\boxed{F_{\text{abs}}=\frac{P}{c}=6.6713\,\text{nN}}
\end{equation}
for both beams.
\section*{Solution 14: RF beam momentum transfer in the far field}
The far-field intensity is
\begin{align}
I
&=\frac{P_tG}{4\pi r^2}\\
&=\frac{(50)(20)}{4\pi(2000)^2}\\
&=1.9894\times10^{-5}\,\text{W/m}^2.
\end{align}
Thus
\begin{equation}
\boxed{I=19.9\,\mu\text{W/m}^2.}
\end{equation}
For complete absorption,
\begin{align}
p_{\text{rad}}
&=\frac{I}{c}\\
&=6.636\times10^{-14}\,\text{Pa}.
\end{align}
Therefore
\begin{equation}
\boxed{p_{\text{rad}}=6.64\times10^{-14}\,\text{Pa}.}
\end{equation}
The force on the $1.2\,\text{m}^2$ panel is
\begin{align}
F
&=p_{\text{rad}}A\\
&=(6.636\times10^{-14})(1.2)\\
&=7.96\times10^{-14}\,\text{N}.
\end{align}
Hence
\begin{equation}
\boxed{F=7.96\times10^{-14}\,\text{N}.}
\end{equation}
This is extraordinarily small mechanically. In most RF systems the energy transfer and signal-to-noise ratio matter enormously, while the corresponding radiation force is far below ordinary mechanical disturbances.
\section*{Solution 15: derive local electromagnetic momentum conservation}
Begin with the Lorentz force density
\begin{equation}
\mathbf f=\rho\mathbf E+\mathbf J\times\mathbf B.
\end{equation}
Use Gauss's law,
\begin{equation}
\rho=\epsilon_0\nabla\cdot\mathbf E,
\end{equation}
and the Ampere--Maxwell law,
\begin{equation}
\mathbf J
=\frac{1}{\mu_0}\nabla\times\mathbf B
-\epsilon_0\frac{\partial\mathbf E}{\partial t}.
\end{equation}
Substitution gives
\begin{align}
\mathbf f
&=\epsilon_0(\nabla\cdot\mathbf E)\mathbf E
+\frac{1}{\mu_0}(\nabla\times\mathbf B)\times\mathbf B
-\epsilon_0\frac{\partial\mathbf E}{\partial t}\times\mathbf B.
\end{align}
Differentiate the cross product:
\begin{equation}
\frac{\partial}{\partial t}(\mathbf E\times\mathbf B)
=
\frac{\partial\mathbf E}{\partial t}\times\mathbf B
+
\mathbf E\times\frac{\partial\mathbf B}{\partial t}.
\end{equation}
Hence
\begin{equation}
-\epsilon_0\frac{\partial\mathbf E}{\partial t}\times\mathbf B
=
-\frac{\partial}{\partial t}
\left(\epsilon_0\mathbf E\times\mathbf B\right)
+
\epsilon_0\mathbf E\times\frac{\partial\mathbf B}{\partial t}.
\end{equation}
Use Faraday's law,
\begin{equation}
\frac{\partial\mathbf B}{\partial t}=-\nabla\times\mathbf E,
\end{equation}
so
\begin{equation}
\epsilon_0\mathbf E\times\frac{\partial\mathbf B}{\partial t}
=-\epsilon_0\mathbf E\times(\nabla\times\mathbf E).
\end{equation}
Therefore
\begin{align}
\mathbf f
&=
\epsilon_0(\nabla\cdot\mathbf E)\mathbf E
-\epsilon_0\mathbf E\times(\nabla\times\mathbf E)\\
&\quad
+\frac{1}{\mu_0}(\nabla\times\mathbf B)\times\mathbf B
-\frac{\partial}{\partial t}
\left(\epsilon_0\mathbf E\times\mathbf B\right).
\end{align}
Now use
\begin{equation}
\mathbf A\times(\nabla\times\mathbf A)
=
\frac12\nabla(A^2)
-(\mathbf A\cdot\nabla)\mathbf A,
\end{equation}
and
\begin{equation}
(\nabla\times\mathbf B)\times\mathbf B
=
(\mathbf B\cdot\nabla)\mathbf B
-\frac12\nabla(B^2),
\end{equation}
where $\nabla\cdot\mathbf B=0$ has been used in the magnetic identity.
Then
\begin{align}
\mathbf f
&=
\epsilon_0\left[
(\nabla\cdot\mathbf E)\mathbf E
+(\mathbf E\cdot\nabla)\mathbf E
-\frac12\nabla(E^2)
\right]\\
&\quad
+\frac{1}{\mu_0}\left[
(\mathbf B\cdot\nabla)\mathbf B
-\frac12\nabla(B^2)
\right]\\
&\quad
-\frac{\partial}{\partial t}
\left(\epsilon_0\mathbf E\times\mathbf B\right).
\end{align}
The spatial terms are precisely the divergence of the Maxwell stress tensor,
\begin{equation}
\boldsymbol{\sigma}
=
\epsilon_0\left(
\mathbf E\mathbf E-\frac12E^2\mathbf 1
\right)
+
\frac{1}{\mu_0}\left(
\mathbf B\mathbf B-\frac12B^2\mathbf 1
\right).
\end{equation}
Define
\begin{equation}
\mathbf g=\epsilon_0\mathbf E\times\mathbf B.
\end{equation}
Thus
\begin{equation}
\boxed{
\mathbf f
=
\nabla\cdot\boldsymbol{\sigma}
-
\frac{\partial\mathbf g}{\partial t}.}
\end{equation}
Equivalently,
\begin{equation}
\boxed{
\frac{\partial\mathbf g}{\partial t}
+\mathbf f
-\nabla\cdot\boldsymbol{\sigma}
=0.}
\end{equation}
The terms have direct physical meanings:
\begin{itemize}
\item $\partial\mathbf g/\partial t$ is the local rate of change of electromagnetic momentum density;
\item $\mathbf f$ is the force density exerted by the fields on matter;
\item $\nabla\cdot\boldsymbol{\sigma}$ accounts for momentum transported through electromagnetic stresses.
\end{itemize}
Integrate over a fixed volume $V$:
\begin{equation}
\int_V\mathbf f\,dV
=
\int_V\nabla\cdot\boldsymbol{\sigma}\,dV
-
\frac{d}{dt}\int_V\mathbf g\,dV.
\end{equation}
Using the tensor form of the divergence theorem,
\begin{equation}
\int_V\nabla\cdot\boldsymbol{\sigma}\,dV
=
\oint_S\boldsymbol{\sigma}\cdot d\mathbf A,
\end{equation}
so the integral momentum balance is
\begin{equation}
\boxed{
\mathbf F_{\text{matter}}
=
\oint_S\boldsymbol{\sigma}\cdot d\mathbf A
-
\frac{d\mathbf P_{\text{EM}}}{dt}.}
\end{equation}
This equation is the momentum analogue of Poynting's energy theorem.
\section*{Solution 16: Julia sweep of reflectivity and incidence angle}
For an opaque surface,
\begin{equation}
A=1-R,
\end{equation}
so
\begin{align}
A+2R
&=1-R+2R\\
&=1+R.
\end{align}
The normal pressure is therefore
\begin{equation}
\boxed{
p_n(R,\theta)
=\frac{(1+R)I}{c}\cos^2\theta.}
\end{equation}
Note Julia verbatim code did not render so opening issue to tackle and leaving out here.
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\bibitem{Griffiths2017}
David J. Griffiths,
\emph{Introduction to Electrodynamics},
4th ed., Cambridge University Press, 2017,
sections on electromagnetic momentum and the Maxwell stress tensor.
\bibitem{Jackson1999}
John David Jackson,
\emph{Classical Electrodynamics},
3rd ed., Wiley, 1999,
sections on electromagnetic conservation laws, momentum, and stress.
\bibitem{OpenStaxV2}
Samuel J. Ling, Jeff Sanny, and William Moebs,
\emph{University Physics, Volume 2},
OpenStax, 2016,
sections on electromagnetic waves, momentum, and radiation pressure.
\bibitem{FeynmanV1}
Richard P. Feynman, Robert B. Leighton, and Matthew Sands,
\emph{The Feynman Lectures on Physics, Volume I},
Addison-Wesley, 1963,
chapters on radiation, photons, and momentum transfer.
\bibitem{FeynmanV2}
Richard P. Feynman, Robert B. Leighton, and Matthew Sands,
\emph{The Feynman Lectures on Physics, Volume II},
Addison-Wesley, 1964,
chapters on electromagnetic energy, momentum, and stress.
\bibitem{Barnett2010}
Stephen M. Barnett,
``Resolution of the Abraham--Minkowski Dilemma,''
\emph{Physical Review Letters},
Vol. 104, 070401, 2010.
\end{thebibliography}