Exercises — Electromagnetic waves — derivation from Maxwell's equations

WHAT we look for: the term that survives when there are no charges and no currents. The four vacuum equations are $\nabla \cdot \vec{E}=0$, $\nabla \cdot \vec{B}=0$, $\nabla \times \vec{E}=-{\partial }_t\vec{B}$, and $\nabla \times \vec{B}={\mu }_0{\epsilon }_0{\partial }_t\vec{E}$. Set $\vec{J}=0$: the ordinary Ampère term ${\mu }_0\vec{J}$ vanishes. The only thing left on the right of the fourth equation is ${\mu }_0{\epsilon }_0{\partial }_t\vec{E}$ — the displacement current term. Answer: the Ampère–Maxwell law, via its displacement-current term ${\mu }_0{\epsilon }_0{\partial }_t\vec{E}$. Without it, $\nabla \times \vec{B}=0$ in vacuum and no wave equation forms.

WHAT we do: compare the coefficient of ${\partial }_t^2$ in both equations. Our equation has coefficient ${\mu }_0{\epsilon }_0$; the template has $1/v^2$. WHY they must be equal: two equations of identical form describing the same field must have identical coefficients, so $\frac{1}{v^2}={\mu }_0{\epsilon }_0⟹v=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}=c.$ Answer: $v=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$.

WHAT we know: the triad $(\hat{E},\hat{B},\hat{k})$ is right-handed, meaning $\hat{E}\times \hat{B}=\hat{k}$. Here $\hat{E}=\hat{y}$ and $\hat{k}=\hat{x}$. We need $\hat{B}$ with $\hat{y}\times \hat{B}=\hat{x}$. Test $\hat{B}=\hat{z}$: $\hat{y}\times \hat{z}=\hat{x}$. ✓ Answer: $\vec{B}$ points along $+\hat{z}$. Look at the figure below — $\vec{E}$ up, $\vec{B}$ out of the page-plane, wave rolling right.

WHY $c=f\lambda$: from $v=\omega /k$ with $\omega =2\pi f$ and $k=2\pi /\lambda$, the $2\pi$ cancels leaving $c=f\lambda$. $\lambda =\frac{c}{f}=\frac{3.00\times 10^8}{2.45\times 10^9}=0.1224\text{m}\approx 12.2\text{cm}.$ Answer: $\lambda \approx 0.122\text{m}$ (about 12 cm — that's why microwave cavities are hand-sized).

WHY: Faraday's law for a plane wave gives $k{E}_0=\omega B_0$, i.e. $E_0/B_0=\omega /k=c$. $B_0=\frac{E_0}{c}=\frac{90}{3.00\times 10^8}=3.0\times 10^{-7}\text{T}.$ Answer: $B_0=3.0\times 10^{-7}\text{T}$. Notice $B_0$ is tiny numerically — that's a units effect, not a "weaker" field.

WHAT we do: multiply, take the reciprocal square root. ${\mu }_0{\epsilon }_0=(1.2566\times 10^{-6})(8.854\times 10^{-12})=1.1127\times 10^{-17}{\text{s}}^2/{\text{m}}^2.$ $\frac{1}{\sqrt{1.1127\times 10^{-17}}}=2.998\times 10^8\text{m/s}.$ Answer: $c\approx 3.00\times 10^8\text{m/s}$. Two lab constants, neither about light, give the speed of light.

$f=\frac{c}{\lambda }=\frac{3.00\times 10^8}{633\times 10^{-9}}=4.74\times 10^{14}\text{Hz}.$ $\omega =2\pi f=2\pi (4.74\times 10^{14})=2.98\times 10^{15}\text{rad/s}.$ Answer: $f\approx 4.74\times 10^{14}\text{Hz}$, $\omega \approx 2.98\times 10^{15}\text{rad/s}$.

WHAT the identity gives: $\nabla \times (\nabla \times \vec{E})=\nabla (\nabla \cdot \vec{E})-{\nabla }^2\vec{E}$. The assumption: Maxwell's first vacuum equation is $\nabla \cdot \vec{E}=0$ (Gauss's law with $\rho =0$). This makes $\nabla (\nabla \cdot \vec{E})=\nabla (0)=0$. So the left side collapses to $-{\nabla }^2\vec{E}$, giving the clean wave equation. If $\rho \ne 0$: Gauss's law reads $\nabla \cdot \vec{E}=\rho /{\epsilon }_0$, so the surviving term is $\nabla (\rho /{\epsilon }_0)$. The equation becomes ${\nabla }^2\vec{E}-{\mu }_0{\epsilon }_0{\partial }_t^2\vec{E}=\nabla \left(\frac{\rho }{{\epsilon }_0}\right),$ a driven (inhomogeneous) wave equation. Charges act as a source term and the pristine free wave is disturbed.

WHAT we compute: $\nabla \cdot \vec{E}={\partial }_x{E}_x+{\partial }_y{E}_y+{\partial }_z{E}_z$. For a wave of form $e^{i(kx-\omega t)}$, the only spatial dependence is on $x$, and ${\partial }_x$ brings down a factor $ik$. So $\nabla \cdot \vec{E}=ik{E}_x$ (only the $x$-component's own $x$-derivative can be nonzero here). Set it to zero: $ik{E}_{0x}=0$. Since $k\ne 0$, we need $E_{0x}=0$. Meaning: the component of $\vec{E}$ along the travel direction $\hat{x}$ must vanish. The field lives only in the $y$–$z$ plane, perpendicular to $\vec{k}$ — the wave is transverse. The same argument on $\vec{B}$ (using $\nabla \cdot \vec{B}=0$) makes $\vec{B}$ transverse too.

Differentiate twice in space: ${\partial }_{x}E=E_{0}k\cos (kx-\omega t)$, ${\partial }_x^{2}E=-E_0{k}^2\sin (kx-\omega t)=-k^{2}E$. Differentiate twice in time: ${\partial }_{t}E=E_0\omega \cos (kx-\omega t)\cdot (-1)\cdot (-1)$... carefully: ${\partial }_t(kx-\omega t)=-\omega$, so ${\partial }_{t}E=-E_0\omega \cos (kx-\omega t)$, and ${\partial }_t^{2}E=-E_0{\omega }^2\sin (kx-\omega t)=-{\omega }^{2}E$. Substitute: $-k^{2}E={\mu }_0{\epsilon }_0(-{\omega }^{2}E)$. Cancel $-E$: $k^2={\mu }_0{\epsilon }_0{\omega }^2⟹\frac{\omega }{k}=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}=c.$ Answer: it is a solution iff $\omega =ck$. This is the dispersion relation — in vacuum it is a straight line ($\omega$ proportional to $k$), meaning every frequency travels at the same speed $c$ (no spreading).

(a) $B_0=E_0/c=200/(3.00\times 10^8)=6.667\times 10^{-7}\text{T}$. (b) $u_{\max }=\frac{1}{2}{\epsilon }_0{E}_0^2=\frac{1}{2}(8.854\times 10^{-12})(200{)}^2=1.771\times 10^{-7}{\text{J/m}}^3$. (c) $\frac{1}{2{\mu }_0}{B}_0^2=\frac{(6.667\times 10^{-7}{)}^2}{2(1.2566\times 10^{-6})}=1.768\times 10^{-7}{\text{J/m}}^3$. Answer: the two energy densities agree (to rounding, $\approx 1.77\times 10^{-7}{\text{J/m}}^3$), confirming the electric and magnetic parts store equal energy — the L2-trap resolved with numbers. See Poynting Vector and EM Energy.

Solve for $E_0$: $E_0=\sqrt{\frac{2I}{c{\epsilon }_0}}$. $E_0=\sqrt{\frac{2(1000)}{(3.00\times 10^8)(8.854\times 10^{-12})}}=\sqrt{7.53\times 10^5}=868\text{V/m}.$ Then $B_0$: $B_0=E_0/c=868/(3.00\times 10^8)=2.9\times 10^{-6}\text{T}$. Answer: $E_0\approx 8.7\times 10^2\text{V/m}$, $B_0\approx 2.9\times 10^{-6}\text{T}$. Sunlight's magnetic field is comparable to Earth's own field ($\sim 5\times 10^{-5}$ T) but oscillating at $\sim 10^{14}$ Hz.

Key fact: in vacuum both travel at $c$, so $f=c/\lambda$ and the ratio depends only on wavelengths. $\frac{f_2}{f_1}=\frac{c/{\lambda }_2}{c/{\lambda }_1}=\frac{{\lambda }_1}{{\lambda }_2}=\frac{700}{400}=\mathrm{1.75.}$ Answer: violet's frequency is $1.75\times$ the red's. See Electromagnetic Spectrum for where these sit.

Step 1 — curl the Ampère–Maxwell law. Start from $\nabla \times \vec{B}={\mu }_0{\epsilon }_0{\partial }_t\vec{E}$ and take curl of both sides: $\nabla \times (\nabla \times \vec{B})={\mu }_0{\epsilon }_0\nabla \times ({\partial }_t\vec{E}).$ Why: curling brings a $\nabla \times \vec{E}$ onto the right, and Faraday gives us a formula for exactly that. Step 2 — left side via the identity. $\nabla \times (\nabla \times \vec{B})=\nabla (\nabla \cdot \vec{B})-{\nabla }^2\vec{B}$. Since $\nabla \cdot \vec{B}=0$, the first term dies: $\nabla \times (\nabla \times \vec{B})=-{\nabla }^2\vec{B}.$ Step 3 — right side, swap order and insert Faraday. Space-curl and time-derivative commute for smooth fields: ${\mu }_0{\epsilon }_0\nabla \times ({\partial }_t\vec{E})={\mu }_0{\epsilon }_0{\partial }_t(\nabla \times \vec{E})={\mu }_0{\epsilon }_0{\partial }_t\left(-{\partial }_t\vec{B}\right)=-{\mu }_0{\epsilon }_0{\partial }_t^2\vec{B}.$ Step 4 — equate. $-{\nabla }^2\vec{B}=-{\mu }_0{\epsilon }_0{\partial }_t^2\vec{B}$; minus signs cancel: $\boxed{{\nabla }^2\vec{B}={\mu }_0{\epsilon }_0{\partial }_t^2\vec{B}}$ Same form, same speed $c=1/\sqrt{{\mu }_0{\epsilon }_0}$. $\vec{E}$ and $\vec{B}$ obey identical wave equations — they must, since they are two faces of one wave.

(a) $c^{\prime }=1/\sqrt{2{\mu }_0{\epsilon }_0}=\frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}=c/\sqrt{2}$. Light would be slower by a factor $1/\sqrt{2}\approx 0.707$, i.e. $c^{\prime }\approx 2.12\times 10^8\text{m/s}$. (b) At fixed $f$, $\lambda =c/f$, so ${\lambda }^{\prime }=c^{\prime }/f=(c/\sqrt{2})/f=\lambda /\sqrt{2}$. Wavelength shrinks by the same factor $\approx 0.707$. Answer: speed and wavelength both drop by $1/\sqrt{2}$; frequency (set by the source) is unchanged. This is exactly how a dielectric medium behaves — larger effective $\epsilon$ slows light and shortens $\lambda$.

Set ${\partial }_t\vec{E}=0$: the right side is zero, leaving ${\nabla }^2\vec{E}=0$ — Laplace's equation. Interpretation: this is the equation of static fields (electrostatics), not waves. There is no propagation because nothing changes in time — the seesaw isn't moving, so no push is transmitted. The lesson: a wave requires time variation. The displacement-current coupling only acts when ${\partial }_t\vec{E}\ne 0$. Zero time-derivative is the degenerate limit where the wave equation gracefully reduces to the static case — no contradiction, just no wave.

(a) Direction of $\vec{B}$: need $\hat{E}\times \hat{B}=\hat{k}$, i.e. $\hat{y}\times \hat{B}=\hat{x}$, giving $\hat{B}=\hat{z}$ (since $\hat{y}\times \hat{z}=\hat{x}$). Amplitude $B_0=E_0/c$. In phase with $\vec{E}$: $\vec{B}=\frac{E_0}{c}\hat{z}\cos (kx-\omega t).$ (b) Poynting magnitude with, say, $E_0=100\text{V/m}$: $B_0=100/(3.00\times 10^8)=3.333\times 10^{-7}\text{T}$. $S=\frac{E_0{B}_0}{{\mu }_0}=\frac{(100)(3.333\times 10^{-7})}{1.2566\times 10^{-6}}=26.5{\text{W/m}}^2.$ Direction: $\vec{S}\propto \vec{E}\times \vec{B}=\hat{y}\times \hat{z}=\hat{x}$ — energy flows in the propagation direction, as it must. See Poynting Vector and EM Energy and Polarization. Answer: $\vec{B}=\frac{E_0}{c}\hat{z}\cos (kx-\omega t)$; peak $S\approx 26.5{\text{W/m}}^2$ along $+\hat{x}$.

Did you get: 2.1 ≈ 0.122 m ::: yes / 2.2 = $3.0\times 10^{-7}$ T ::: yes / 2.4 $f\approx 4.74\times 10^{14}$ Hz ::: yes / 4.2 $E_0\approx 868$ V/m ::: yes / 5.2 factor $1/\sqrt{2}$ ::: yes / 5.4 $S\approx 26.5$ W/m² ::: yes