Poynting's theorem

Poynting's theorem states the rate of energy going into charged matter in a volume $Ω$ is equal to the decrease in energy stored in the fields in $Ω$ minus the rate of energy transported out of $Ω$ by the fields¹

𝑃 = ∭ Ω ⃗ 𝐄 \cdot ⃗ 𝐉 𝑑 𝜏 = - 𝑑 𝑑 𝑡 ∭ Ω 𝑢 𝑑 𝜏 - \subset \supset \iint 𝜕 Ω ⃗ 𝐒 \cdot 𝑑 ⃗ 𝐚

or equivalently by Острогра́дский's divergence theorem

⃗ 𝐄 \cdot ⃗ 𝐉 = - 𝜕 𝑢 𝜕 𝑡 - ⃗ \nabla \cdot ⃗ 𝐒

where

𝑢 = 12 (1 𝜇 0 𝐵 2 + 𝜖 0 𝐸 2) ⃗ 𝐒 = 1 𝜇 0 ⃗ 𝐄 \times ⃗ 𝐁

are the Electromagnetic energy density and Poynting vector respectively. In the case $⃗ 𝐄 \cdot ⃗ 𝐉 = 0$ everywhere we get a continuity equation for electromagnetic energy.

Derivation from Maxwell's equations

First note that by the Lorentz force law, the force density acting on the charge density it $⃗ 𝐟 \pm = 𝜌 \pm (⃗ 𝐄 + ⃗ 𝐯 \pm \times ⃗ 𝐁)$ whence the work done is

𝑑 𝑊 \pm = ⃗ 𝐟 \pm \cdot 𝑑 ⃗ ℓ \pm = 𝜌 \pm (⃗ 𝐄 + ⃗ 𝐯 \pm \times ⃗ 𝐁) \cdot (⃗ 𝐯 \pm 𝑑 𝑡) = 𝜌 \pm ⃗ 𝐄 \cdot ⃗ 𝐯 \pm 𝑑 𝑡

and the power per unit volume is

𝑑 𝑊 𝑑 𝑡 = ⃗ 𝐄 \cdot (𝜌 + ⃗ 𝐯 + + 𝜌 - ⃗ 𝐯 -) = ⃗ 𝐄 \cdot ⃗ 𝐉

Applying ^Differential to eliminate $⃗ 𝐉$

⃗ 𝐄 \cdot ⃗ 𝐉 = 1 𝜇 0 ⃗ 𝐄 \cdot (⃗ \nabla \times ⃗ 𝐄) - 𝜖 0 ⃗ 𝐄 \cdot 𝜕 ⃗ 𝐄 𝜕 𝑡

Now by a Product rule and Faraday's law of induction

⃗ \nabla \cdot (⃗ 𝐄 \times ⃗ 𝐁) = ⃗ 𝐁 \cdot (⃗ \nabla \times ⃗ 𝐄) - ⃗ 𝐄 \cdot (⃗ \nabla \times ⃗ 𝐁) = ⃗ 𝐁 \cdot 𝜕 ⃗ 𝐁 𝜕 𝑡 - ⃗ 𝐄 \cdot (⃗ \nabla \times ⃗ 𝐁)

⃗ 𝐄 \cdot ⃗ 𝐉 = - 1 𝜇 0 ⃗ 𝐁 \cdot 𝜕 ⃗ 𝐁 𝜕 𝑡 - 𝜖 0 ⃗ 𝐄 \cdot 𝜕 ⃗ 𝐄 𝜕 𝑡 - 1 𝜇 0 ⃗ \nabla \cdot (⃗ 𝐄 \times ⃗ 𝐁) = - 12 𝜕 𝜕 𝑡 (1 𝜇 0 𝐵 2 + 𝜖 0 𝐸 2) - 1 𝜇 0 ⃗ \nabla \cdot (⃗ 𝐄 \times ⃗ 𝐁)

giving the expressions above.

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2023. Electromagnetism and special relativity, p. 73 ↩

Poynting's theorem

Footnotes