In free space (ρ = 0 \rho = 0 ρ = 0 J = 0 \mathbf{J} = \mathbf{0} J = 0
∇ × ( ∇ × E ) = − ∂ ∂ t ( ∇ × B ) = − μ 0 ε 0 ∂ 2 E ∂ t 2 \nabla \times (\nabla \times \mathbf{E}) = -\frac{\partial}{\partial t}(\nabla \times \mathbf{B}) = -\mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} ∇ × ( ∇ × E ) = − ∂ t ∂ ( ∇ × B ) = − μ 0 ε 0 ∂ t 2 ∂ 2 E
Using the identity ∇ × ( ∇ × E ) = ∇ ( ∇ ⋅ E ) − ∇ 2 E \nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E} ∇ × ( ∇ × E ) = ∇ ( ∇ ⋅ E ) − ∇ 2 E ∇ ⋅ E = 0 \nabla \cdot \mathbf{E} = 0 ∇ ⋅ E = 0
∇ 2 E = μ 0 ε 0 ∂ 2 E ∂ t 2 \nabla^2 \mathbf{E} = \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} ∇ 2 E = μ 0 ε 0 ∂ t 2 ∂ 2 E
Similarly: ∇ 2 B = μ 0 ε 0 ∂ 2 B ∂ t 2 \nabla^2 \mathbf{B} = \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{B}}{\partial t^2} ∇ 2 B = μ 0 ε 0 ∂ t 2 ∂ 2 B
These are wave equations with wave speed c = 1 / μ 0 ε 0 ≈ 3 × 10 8 c = 1/\sqrt{\mu_0 \varepsilon_0} \approx 3 \times 10^8 c = 1/ μ 0 ε 0 ≈ 3 × 1 0 8
Theorem 5.1. Electromagnetic waves in free space are:
Transverse : E \mathbf{E} E B \mathbf{B} B Mutually perpendicular : E ⊥ B \mathbf{E} \perp \mathbf{B} E ⊥ B In phase : E = c B E = cB E = c B Linearly polarised (; other polarisations are superpositions).Energy. The energy density of an EM wave is u = 1 2 ( ε 0 E 2 + B 2 / μ 0 ) u = \frac{1}{2}(\varepsilon_0 E^2 + B^2/\mu_0) u = 2 1 ( ε 0 E 2 + B 2 / μ 0 )
The Poynting vector S = 1 μ 0 E × B \mathbf{S} = \frac{1}{\mu_0}\mathbf{E} \times \mathbf{B} S = μ 0 1 E × B
Problem. Show that E = E 0 cos ( k z − ω t ) x ^ \mathbf{E} = E_0 \cos(kz - \omega t)\,\hat{\mathbf{x}} E = E 0 cos ( k z − ω t ) x ^ B \mathbf{B} B
Solution. ∇ 2 E = ∂ 2 E x ∂ z 2 x ^ = − k 2 E 0 cos ( k z − ω t ) x ^ \nabla^2 \mathbf{E} = \frac{\partial^2 E_x}{\partial z^2}\hat{\mathbf{x}} = -k^2 E_0 \cos(kz - \omega t)\,\hat{\mathbf{x}} ∇ 2 E = ∂ z 2 ∂ 2 E x x ^ = − k 2 E 0 cos ( k z − ω t ) x ^
∂ 2 E ∂ t 2 = − ω 2 E 0 cos ( k z − ω t ) x ^ \frac{\partial^2 \mathbf{E}}{\partial t^2} = -\omega^2 E_0 \cos(kz - \omega t)\,\hat{\mathbf{x}} ∂ t 2 ∂ 2 E = − ω 2 E 0 cos ( k z − ω t ) x ^
The wave equation requires k 2 = μ 0 ε 0 ω 2 k^2 = \mu_0 \varepsilon_0 \omega^2 k 2 = μ 0 ε 0 ω 2 ω / k = c \omega/k = c ω / k = c
From Faraday’s law: ∇ × E = − ∂ B ∂ t \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} ∇ × E = − ∂ t ∂ B
( ∇ × E ) y = − ∂ E x ∂ z = k E 0 sin ( k z − ω t ) (\nabla \times \mathbf{E})_y = -\frac{\partial E_x}{\partial z} = k E_0 \sin(kz - \omega t) ( ∇ × E ) y = − ∂ z ∂ E x = k E 0 sin ( k z − ω t )
∂ B y ∂ t = − k E 0 sin ( k z − ω t ) ⟹ B y = k ω E 0 cos ( k z − ω t ) = E 0 c cos ( k z − ω t ) \frac{\partial B_y}{\partial t} = -k E_0 \sin(kz - \omega t) \implies B_y = \frac{k}{\omega} E_0 \cos(kz - \omega t) = \frac{E_0}{c}\cos(kz - \omega t) ∂ t ∂ B y = − k E 0 sin ( k z − ω t ) ⟹ B y = ω k E 0 cos ( k z − ω t ) = c E 0 cos ( k z − ω t )
So B = E 0 c cos ( k z − ω t ) y ^ \mathbf{B} = \frac{E_0}{c}\cos(kz - \omega t)\,\hat{\mathbf{y}} B = c E 0 cos ( k z − ω t ) y ^ ■ \blacksquare ■
Poynting’s theorem is the statement of energy conservation for electromagnetic fields.
Derivation. Start with the two Maxwell equations containing time derivatives:
∇ × E = − ∂ B ∂ t , ∇ × B = μ 0 J + μ 0 ε 0 ∂ E ∂ t \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}, \quad \nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t} ∇ × E = − ∂ t ∂ B , ∇ × B = μ 0 J + μ 0 ε 0 ∂ t ∂ E
Compute B ⋅ ( ∇ × E ) − E ⋅ ( ∇ × B ) \mathbf{B} \cdot (\nabla \times \mathbf{E}) - \mathbf{E} \cdot (\nabla \times \mathbf{B}) B ⋅ ( ∇ × E ) − E ⋅ ( ∇ × B )
B ⋅ ( ∇ × E ) = − B ⋅ ∂ B ∂ t = − ∂ ∂ t ( B 2 2 ) \mathbf{B} \cdot (\nabla \times \mathbf{E}) = -\mathbf{B} \cdot \frac{\partial \mathbf{B}}{\partial t} = -\frac{\partial}{\partial t}\left(\frac{B^2}{2}\right) B ⋅ ( ∇ × E ) = − B ⋅ ∂ t ∂ B = − ∂ t ∂ ( 2 B 2 )
− E ⋅ ( ∇ × B ) = − μ 0 E ⋅ J − μ 0 ε 0 E ⋅ ∂ E ∂ t = − μ 0 E ⋅ J − ∂ ∂ t ( ε 0 E 2 2 ) -\mathbf{E} \cdot (\nabla \times \mathbf{B}) = -\mu_0 \mathbf{E} \cdot \mathbf{J} - \mu_0 \varepsilon_0 \mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t} = -\mu_0 \mathbf{E} \cdot \mathbf{J} - \frac{\partial}{\partial t}\left(\frac{\varepsilon_0 E^2}{2}\right) − E ⋅ ( ∇ × B ) = − μ 0 E ⋅ J − μ 0 ε 0 E ⋅ ∂ t ∂ E = − μ 0 E ⋅ J − ∂ t ∂ ( 2 ε 0 E 2 )
Using the vector identity ∇ ⋅ ( E × B ) = B ⋅ ( ∇ × E ) − E ⋅ ( ∇ × B ) \nabla \cdot (\mathbf{E} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{E}) - \mathbf{E} \cdot (\nabla \times \mathbf{B}) ∇ ⋅ ( E × B ) = B ⋅ ( ∇ × E ) − E ⋅ ( ∇ × B )
∇ ⋅ ( E × B ) = − μ 0 J ⋅ E − μ 0 ε 0 ∂ ∂ t ( E 2 2 ) − ∂ ∂ t ( B 2 2 ) \nabla \cdot (\mathbf{E} \times \mathbf{B}) = -\mu_0 \mathbf{J} \cdot \mathbf{E} - \mu_0 \varepsilon_0 \frac{\partial}{\partial t}\left(\frac{E^2}{2}\right) - \frac{\partial}{\partial t}\left(\frac{B^2}{2}\right) ∇ ⋅ ( E × B ) = − μ 0 J ⋅ E − μ 0 ε 0 ∂ t ∂ ( 2 E 2 ) − ∂ t ∂ ( 2 B 2 )
Dividing by μ 0 \mu_0 μ 0
− ∇ ⋅ S = J ⋅ E + ∂ u ∂ t \boxed{-\nabla \cdot \mathbf{S} = \mathbf{J} \cdot \mathbf{E} + \frac{\partial u}{\partial t}} − ∇ ⋅ S = J ⋅ E + ∂ t ∂ u
Where S = 1 μ 0 E × B \mathbf{S} = \frac{1}{\mu_0}\mathbf{E} \times \mathbf{B} S = μ 0 1 E × B u = 1 2 ( ε 0 E 2 + B 2 μ 0 ) u = \frac{1}{2}\left(\varepsilon_0 E^2 + \frac{B^2}{\mu_0}\right) u = 2 1 ( ε 0 E 2 + μ 0 B 2 )
Interpretation: The rate of energy leaving a volume equals the work done on charges plus The rate of increase of field energy. In integral form:
− ∮ S S ⋅ d A = d d t ∫ V u d V + ∫ V J ⋅ E d V -\oint_S \mathbf{S} \cdot d\mathbf{A} = \frac{d}{dt}\int_V u\,dV + \int_V \mathbf{J} \cdot \mathbf{E}\,dV − ∮ S S ⋅ d A = d t d ∫ V u d V + ∫ V J ⋅ E d V
Intensity. For a plane wave, the time-averaged Poynting vector is:
⟨ S ⟩ = E 0 2 2 μ 0 c k ^ = 1 2 ε 0 c E 0 2 k ^ \langle\mathbf{S}\rangle = \frac{E_0^2}{2\mu_0 c}\,\hat{\mathbf{k}} = \frac{1}{2}\varepsilon_0 c E_0^2\,\hat{\mathbf{k}} ⟨ S ⟩ = 2 μ 0 c E 0 2 k ^ = 2 1 ε 0 c E 0 2 k ^
Example: Radiation pressure A plane wave normally incident on a perfectly absorbing surface exerts a radiation pressure. The momentum flux of the wave is ⟨ S ⟩ / c \langle S \rangle/c ⟨ S ⟩ / c
P a b s = ⟨ S ⟩ c = ε 0 E 0 2 2 P_{\mathrm{abs} = \frac{\langle S \rangle}{c} = \frac{\varepsilon_0 E_0^2}{2}} P abs = c ⟨ S ⟩ = 2 ε 0 E 0 2
For a perfectly reflecting surface, the momentum transfer is doubled:
P r e f = 2 ⟨ S ⟩ c = ε 0 E 0 2 P_{\mathrm{ref} = \frac{2\langle S \rangle}{c} = \varepsilon_0 E_0^2} P ref = c 2 ⟨ S ⟩ = ε 0 E 0 2
A 1 kW/m2 ^2 2 3.3 μ 3.3\ \mu 3.3 μ ■ \blacksquare ■
Example: Polarization of EM waves Linear polarization. E = E 0 cos ( k z − ω t ) x ^ \mathbf{E} = E_0\cos(kz - \omega t)\,\hat{\mathbf{x}} E = E 0 cos ( k z − ω t ) x ^
Circular polarization. Two orthogonal linear polarizations with a phase difference of π / 2 \pi/2 π /2
E = E 0 cos ( k z − ω t ) x ^ ± E 0 sin ( k z − ω t ) y ^ \mathbf{E} = E_0\cos(kz - \omega t)\,\hat{\mathbf{x}} \pm E_0\sin(kz - \omega t)\,\hat{\mathbf{y}} E = E 0 cos ( k z − ω t ) x ^ ± E 0 sin ( k z − ω t ) y ^
The tip of E \mathbf{E} E + + + − - −
Elliptical polarization. The general case with arbitrary amplitudes and phase:
E = E 0 x cos ( k z − ω t ) x ^ + E 0 y cos ( k z − ω t + δ ) y ^ \mathbf{E} = E_{0x}\cos(kz - \omega t)\,\hat{\mathbf{x}} + E_{0y}\cos(kz - \omega t + \delta)\,\hat{\mathbf{y}} E = E 0 x cos ( k z − ω t ) x ^ + E 0 y cos ( k z − ω t + δ ) y ^
■ \blacksquare ■
In a conductor with conductivity σ \sigma σ J = σ E \mathbf{J} = \sigma\mathbf{E} J = σ E
∇ × B = μ 0 σ E + μ 0 ε 0 ∂ E ∂ t \nabla \times \mathbf{B} = \mu_0\sigma\mathbf{E} + \mu_0\varepsilon_0\frac{\partial \mathbf{E}}{\partial t} ∇ × B = μ 0 σ E + μ 0 ε 0 ∂ t ∂ E
For a monochromatic wave E = E 0 e − i ω t \mathbf{E} = \mathbf{E}_0\,e^{-i\omega t} E = E 0 e − iω t
k ~ 2 = μ 0 ε 0 ω 2 + i μ 0 σ ω \tilde{k}^2 = \mu_0\varepsilon_0\omega^2 + i\mu_0\sigma\omega k ~ 2 = μ 0 ε 0 ω 2 + i μ 0 σ ω
Writing k ~ = k + i κ \tilde{k} = k + i\kappa k ~ = k + iκ k k k κ \kappa κ
E ( z , t ) = E 0 e − κ z cos ( k z − ω t ) \mathbf{E}(z,t) = \mathbf{E}_0\,e^{-\kappa z}\cos(kz - \omega t) E ( z , t ) = E 0 e − κ z cos ( k z − ω t )
The field decays exponentially. The skin depth is the distance over which the amplitude Falls by a factor of 1 / e 1/e 1/ e
δ = 1 κ \delta = \frac{1}{\kappa} δ = κ 1
For a good conductor (σ ≫ ε 0 ω \sigma \gg \varepsilon_0\omega σ ≫ ε 0 ω
δ = 2 μ 0 σ ω \delta = \sqrt{\frac{2}{\mu_0\sigma\omega}} δ = μ 0 σ ω 2
Example: Skin depth in copper at 60 Hz and 1 MHz Copper: σ = 5.96 × 10 7 \sigma = 5.96 \times 10^7 σ = 5.96 × 1 0 7 μ r ≈ 1 \mu_r \approx 1 μ r ≈ 1
At f = 60 f = 60 f = 60 ω = 2 π × 60 \omega = 2\pi \times 60 ω = 2 π × 60
δ = 2 4 π × 10 − 7 × 5.96 × 10 7 × 2 π × 60 ≈ 8.5 m m \delta = \sqrt{\frac{2}{4\pi \times 10^{-7} \times 5.96 \times 10^7 \times 2\pi \times 60}} \approx 8.5\ \mathrm{mm} δ = 4 π × 1 0 − 7 × 5.96 × 1 0 7 × 2 π × 60 2 ≈ 8.5 mm
At f = 1 f = 1 f = 1 ω = 2 π × 10 6 \omega = 2\pi \times 10^6 ω = 2 π × 1 0 6
δ = 2 4 π × 10 − 7 × 5.96 × 10 7 × 2 π × 10 6 ≈ 65 μ m \delta = \sqrt{\frac{2}{4\pi \times 10^{-7} \times 5.96 \times 10^7 \times 2\pi \times 10^6}} \approx 65\ \mu\mathrm{m} δ = 4 π × 1 0 − 7 × 5.96 × 1 0 7 × 2 π × 1 0 6 2 ≈ 65 μ m
The skin depth decreases as 1 / f 1/\sqrt{f} 1/ f ■ \blacksquare ■
Electromagnetic waves can be guided by hollow conducting pipes (waveguides). Consider a Rectangular waveguide with dimensions a a a b b b
TE modes (transverse electric, E z = 0 E_z = 0 E z = 0 B z ≠ 0 B_z \neq 0 B z = 0 \mathrm{TE_}{10} With fields:
E y = E 0 sin ( π x a ) cos ( k g z − ω t ) E_y = E_0 \sin\!\left(\frac{\pi x}{a}\right)\cos(k_g z - \omega t) E y = E 0 sin ( a π x ) cos ( k g z − ω t )
B x = − k g ω E 0 sin ( π x a ) cos ( k g z − ω t ) B_x = -\frac{k_g}{\omega}E_0 \sin\!\left(\frac{\pi x}{a}\right)\cos(k_g z - \omega t) B x = − ω k g E 0 sin ( a π x ) cos ( k g z − ω t )
B z = π ω a E 0 cos ( π x a ) sin ( k g z − ω t ) B_z = \frac{\pi}{\omega a}E_0 \cos\!\left(\frac{\pi x}{a}\right)\sin(k_g z - \omega t) B z = ω a π E 0 cos ( a π x ) sin ( k g z − ω t )
Where the guide wave number is k g = ( ω / c ) 2 − ( π / a ) 2 k_g = \sqrt{(\omega/c)^2 - (\pi/a)^2} k g = ( ω / c ) 2 − ( π / a ) 2
Cutoff frequency. Waves propagate only when ω > ω c \omega \gt \omega_c ω > ω c
ω c , m n = c π ( m a ) 2 + ( n b ) 2 \omega_{c,mn} = c\pi\sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2} ω c , mn = c π ( a m ) 2 + ( b n ) 2
For the \mathrm{TE_}{10} mode: f c = c 2 a f_c = \frac{c}{2a} f c = 2 a c
Phase and group velocities. In a waveguide, the phase velocity exceeds c c c
v p = ω k g = c 1 − ( ω c / ω ) 2 > c v_p = \frac{\omega}{k_g} = \frac{c}{\sqrt{1 - (\omega_c/\omega)^2}} \gt c v p = k g ω = 1 − ( ω c / ω ) 2 c > c
The group velocity (signal velocity) is less than c c c
v g = d ω d k g = c 1 − ( ω c / ω ) 2 < c v_g = \frac{d\omega}{dk_g} = c\sqrt{1 - (\omega_c/\omega)^2} \lt c v g = d k g d ω = c 1 − ( ω c / ω ) 2 < c
They satisfy v p v g = c 2 v_p\,v_g = c^2 v p v g = c 2
special relativity. No information or energy travels faster than c c c v g < c v_g \lt c v g < c
An oscillating electric dipole is the simplest source of electromagnetic radiation.
Consider a dipole p ( t ) = p 0 cos ( ω t ) z ^ \mathbf{p}(t) = p_0\cos(\omega t)\,\hat{\mathbf{z}} p ( t ) = p 0 cos ( ω t ) z ^ radiation zone (r ≫ λ r \gg \lambda r ≫ λ
E = − μ 0 p 0 ω 2 4 π sin θ r cos [ ω ( t − r / c ) ] θ ^ \mathbf{E} = -\frac{\mu_0 p_0 \omega^2}{4\pi}\frac{\sin\theta}{r}\cos[\omega(t - r/c)]\,\hat{\boldsymbol{\theta}} E = − 4 π μ 0 p 0 ω 2 r s i n θ cos [ ω ( t − r / c )] θ ^
B = − μ 0 p 0 ω 2 4 π c sin θ r cos [ ω ( t − r / c ) ] ϕ ^ \mathbf{B} = -\frac{\mu_0 p_0 \omega^2}{4\pi c}\frac{\sin\theta}{r}\cos[\omega(t - r/c)]\,\hat{\boldsymbol{\phi}} B = − 4 π c μ 0 p 0 ω 2 r s i n θ cos [ ω ( t − r / c )] ϕ ^
The fields fall off as 1 / r 1/r 1/ r 1 / r 2 1/r^2 1/ r 2
Radiation pattern. The intensity varies as sin 2 θ \sin^2\theta sin 2 θ θ = π / 2 \theta = \pi/2 θ = π /2 θ = 0 , π \theta = 0, \pi θ = 0 , π
Total radiated power. Integrating the Poynting vector over a sphere:
P = μ 0 p 0 2 ω 4 12 π c P = \frac{\mu_0 p_0^2 \omega^4}{12\pi c} P = 12 π c μ 0 p 0 2 ω 4
Larmor formula. For a point charge q q q a a a
P = q 2 a 2 6 π ε 0 c 3 P = \frac{q^2 a^2}{6\pi\varepsilon_0 c^3} P = 6 π ε 0 c 3 q 2 a 2
This is the non-relativistic limit and is valid whenever v ≪ c v \ll c v ≪ c
Derivation: Power radiated by an oscillating dipole The time-averaged Poynting vector magnitude in the radiation zone:
⟨ S ⟩ = 1 2 μ 0 ∣ E θ ∣ ∣ B ϕ ∣ = μ 0 p 0 2 ω 4 32 π 2 c sin 2 θ r 2 \langle S \rangle = \frac{1}{2\mu_0}\lvert E_\theta\rvert\,\lvert B_\phi\rvert = \frac{\mu_0 p_0^2\omega^4}{32\pi^2 c}\frac{\sin^2\theta}{r^2} ⟨ S ⟩ = 2 μ 0 1 ∣ E θ ∣ ∣ B ϕ ∣ = 32 π 2 c μ 0 p 0 2 ω 4 r 2 s i n 2 θ
The total power through a sphere of radius r r r
P = ∫ 0 2 π ∫ 0 π ⟨ S ⟩ r 2 sin θ d θ d ϕ = μ 0 p 0 2 ω 4 32 π 2 c ⋅ 2 π ∫ 0 π sin 3 θ d θ P = \int_0^{2\pi}\!\!\int_0^\pi \langle S \rangle\, r^2\sin\theta\,d\theta\,d\phi = \frac{\mu_0 p_0^2\omega^4}{32\pi^2 c} \cdot 2\pi \int_0^\pi \sin^3\theta\,d\theta P = ∫ 0 2 π ∫ 0 π ⟨ S ⟩ r 2 sin θ d θ d ϕ = 32 π 2 c μ 0 p 0 2 ω 4 ⋅ 2 π ∫ 0 π sin 3 θ d θ
Using ∫ 0 π sin 3 θ d θ = 4 / 3 \int_0^\pi \sin^3\theta\,d\theta = 4/3 ∫ 0 π sin 3 θ d θ = 4/3
P = μ 0 p 0 2 ω 4 32 π 2 c ⋅ 2 π ⋅ 4 3 = μ 0 p 0 2 ω 4 12 π c P = \frac{\mu_0 p_0^2\omega^4}{32\pi^2 c} \cdot 2\pi \cdot \frac{4}{3} = \frac{\mu_0 p_0^2\omega^4}{12\pi c} P = 32 π 2 c μ 0 p 0 2 ω 4 ⋅ 2 π ⋅ 3 4 = 12 π c μ 0 p 0 2 ω 4
■ \blacksquare ■
A[5_Electromagnetic Waves] --> B[Key Concepts]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
D --> G[Real-world usage]
Electromagnetic waves are self-sustaining ripples in the electric and magnetic fields that regenerate each other as they propagate. A changing electric field creates a magnetic field, which in turn creates an electric field, and this feedback loop allows the wave to travel through empty space. The energy carried by the wave is split equally between the electric and magnetic components, stored in the field configuration itself. Radiation from an accelerating charge becomes more intense at higher frequencies because the charge’s acceleration changes direction faster, producing stronger field ripples. The inverse-square law for intensity reflects how the wave’s energy spreads over an expanding sphere.
Mistake 1: Assuming E \mathbf{E} E B \mathbf{B} B In an electromagnetic wave, E \mathbf{E} E B \mathbf{B} B E \mathbf{E} E B \mathbf{B} B B = k ^ × E / c \mathbf{B} = \hat{\mathbf{k}} \times \mathbf{E}/c B = k ^ × E / c
Mistake 2: Confusing phase velocity with group velocity In free space, phase velocity v p = ω / k = c v_p = \omega/k = c v p = ω / k = c v g = d ω / d k = c v_g = d\omega/dk = c v g = d ω / d k = c v p > c v_p > c v p > c v g < c v_g < c v g < c c c c
Mistake 3: Forgetting that the Poynting vector represents energy flux The Poynting vector S = E × B / μ 0 \mathbf{S} = \mathbf{E} \times \mathbf{B}/\mu_0 S = E × B / μ 0 u = 1 2 ( ε 0 E 2 + B 2 / μ 0 ) u = \frac{1}{2}(\varepsilon_0 E^2 + B^2/\mu_0) u = 2 1 ( ε 0 E 2 + B 2 / μ 0 )
