Poynting Theorem

Poynting Theorem

1. What It States — Energy Conservation for EM Fields

Poynting theorem = work-energy theorem for EM: rate of change of field energy in volume $V$ + power flowing out through surface $S$ + power dissipated (Joule heating) =0.

In differential form, starting from $∇·(E×H) = H·(∇×E) - E·(∇×H)$ and Maxwell $∇×E=-∂B/∂t$, $∇×H=J+∂D/∂t$, one gets:

where

is energy density (J/m³) and

is Poynting vector (W/m²) — power per area flowing in direction $E×H$ (direction of wave propagation for plane wave).

Integral over volume $V$ bounded by $S$:

• First term: field energy decrease.
• Second: flux out through $S$.
• Third: $J·E$ power delivered to charges (heat, work).

If $J=0$, energy change = - flux out.

2. For Plane Wave (time-averaged)

For $E=E_0 cos(kz-ωt) \hat{x}$, $H=H_0 cos(kz-ωt) \hat{y}$, $H_0=E_0/Z_0$:

instantaneous

Time-averaged over cycle $⟨cos²⟩=1/2$ ⇒

which equals

and magnitude is intensity $I =⟨S⟩$ (W/m²).

3. Worked Examples

(a) Laser intensity I=1 mW/mm² =1000 W/m²: ⟨S⟩=I ⇒ E0=√(2 Z0 I)=√(2·377·1000)=868 V/m, B0=2.9 µT ✓

(b) Sunlight at Earth I≈1360 W/m² ⇒ ⟨S⟩ ⇒ E0≈1010 V/m (peak)?? Actually time-averaged gives 1010? Check: E0=√(2·377·1360)=1012 V/m peak, rms 716 V/m.

(c) Show S direction is E×H → propagation: \hat{x}×\hat{y}=\hat{z} matches wave k.

(d) For static fields? S still defined but may circulate with zero net flux.

4. Exam Points

• State Poynting vector $S=E×H$, energy density $u=1/2(εE²+B²/μ)$ and theorem $-∂u/∂t=∇·S+J·E$ — 5-7 marks, very frequent (almost certain).
• Derive from Maxwell (sketch: start $∇·(E×H)$) — 7 marks, very frequent (asked as proof).
• For plane wave $⟨S⟩=E_0²/(2Z_0)$, intensity = $⟨S⟩$ — 5 marks, frequent.
• Mention $S$ direction is power flow — 3 marks.