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Poynting's Theorem and Poynting Vector 📂Electrodynamics

Poynting's Theorem and Poynting Vector

Theorem1

The work done by the electromagnetic force on a charge is equal to the decrease in energy stored in the electromagnetic field plus the energy that has leaked out through the boundary. This is called Poynting’s theorem.

dWdt=−ddt∫V12(ϵ0E2+1μ0B2)dτ−1μ0∮S(E×B)⋅da=−ddt∫Vudτ−∮SS⋅da \begin{align*} \dfrac{dW}{dt} &= -\dfrac{d}{dt} \int_{\mathcal{V}} \dfrac{1}{2} \left( \epsilon_{0} E^2 + \dfrac{1}{\mu_{0}} B^2 \right) d\tau - \dfrac{1}{\mu_{0}} \oint_{\mathcal{S}} (\mathbf{E} \times \mathbf{B}) \cdot d \mathbf{a} \\ &= -\dfrac{d}{dt} \int_{\mathcal{V}} u d\tau - \oint_{\mathcal{S}}\mathbf{S} \cdot d\mathbf{a} \end{align*}

S\mathcal{S} is the boundary of V\mathcal{V}. uu is the total energy stored in the electromagnetic field of a unit volume space. The integrand S\mathbf{S} on the right-hand side, which is the energy carried by the electromagnetic field passing through a unit area in a unit time, is also known as the Poynting vector.

S=1μ0(E×B) \mathbf{S} =\dfrac{1}{\mu_{0}} (\mathbf{E} \times \mathbf{B})

It is also referred to as energy flux density.

Description

From Poynting’s theorem, we can derive the continuity equation for energy.

∂u∂t=−∇⋅S \dfrac{\partial u}{\partial t} = -\nabla \cdot \mathbf{S}

This equation tells us that the energy stored in an electromagnetic field is conserved.

It is easy to mistake Poynting for pointing, but it is the name of a person, Poynting. It has nothing to do with the English word point or the act of pointing, so do not be confused.

Poynting’s theorem is a principle on how to calculate the work done by electromagnetic forces on a charge in a unit of time. In particular, the Poynting vector is a crucial concept that constantly appears later in electrodynamics, dealing with content on momentum and energy.

Proof

The work required to create a charge distribution (work done against the Coulomb force) and to induce a current flow (work done against the electromotive force) are as follows.

WE=ϵ02∫E2dτ,WB=12μ0∫B2dτ W_{E} =\dfrac{\epsilon_{0}}{2}\int E^2 d\tau , \quad \quad W_{B} =\dfrac{1}{2\mu_{0}}\int B^2 d\tau

E\mathbf{E} and B\mathbf{B} are the electromagnetic fields created by the respective charge and current distributions. Therefore, the total energy stored in the electromagnetic field of a unit volume of space is as follows.

u=12(ϵ0E2+1μ0B2) u =\dfrac{1}{2}\left( \epsilon_{0} E^2 +\dfrac{1}{\mu_{0}} B^2\right)

  • Part 1. Poynting’s theorem

Let’s say an electric field E\mathbf{E} and a magnetic field B\mathbf{B} are generated from a certain charge and current distribution at time tt. If the charge moved after time dtdt, the work done by the electromagnetic force on this charge during time dtdt can be expressed as follows, according to the definition of work and Lorentz force law.

dW=F⋅ds=q(E+v×B)⋅vdt=qE⋅vdt \begin{align*} dW &= \mathbf{F}\cdot d\mathbf{s} \\ &= q(\mathbf{E} +\mathbf{v}\times \mathbf{B}) \cdot \mathbf{v} dt \\ &= q\mathbf{E} \cdot \mathbf{v} dt \end{align*}

The last equality holds because v×B\mathbf{v}\times \mathbf{B} is perpendicular to v\mathbf{v}, or because magnetic forces do no work. The charge amount is q=∫ρdτq=\int \rho d\tau, and ρv=J\rho\mathbf{v}=\mathbf{J}, therefore

dW=∫VE⋅ρvdτdt=∫VE⋅Jdτdt  ⟹  dWdt=∫VE⋅Jdτ \begin{align} dW &= \int_{\mathcal{V}} \mathbf{E}\cdot \rho\mathbf{v} d\tau dt \nonumber \\ &= \int_{\mathcal{V}} \mathbf{E} \cdot \mathbf{J} d\tau dt \nonumber \\ \implies \dfrac{dW}{dt} &= \int_{\mathcal{V}} \mathbf{E} \cdot \mathbf{J} d\tau \end{align}

Thus, E⋅J\mathbf{E} \cdot \mathbf{J} can be considered as the work done per unit volume per unit time. In other words, it’s the power per unit volume. According to Maxwell’s equations (iv)\text{(iv)}, we have J=1μ0∇×B−ϵ0∂E∂t\mathbf{J}=\frac{1}{\mu_{0}}\nabla \times \mathbf{B} - \epsilon_{0} \dfrac{\partial \mathbf{E}}{\partial t}, therefore

E⋅J=1μ0E⋅(∇×B)−ϵ0E⋅∂E∂t \mathbf{E} \cdot \mathbf{J} = \dfrac{1}{\mu_{0}}\mathbf{E} \cdot (\nabla \times \mathbf{B}) - \epsilon_{0} \mathbf{E}\cdot \dfrac{\partial \mathbf{E}}{\partial t}

Using the product rule (d) E⋅(∇×B)=B⋅(∇×E)−∇⋅(E×B)\mathbf{E} \cdot (\nabla \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{E}) - \nabla \cdot(\mathbf{E} \times \mathbf{B}) and Faraday’s law ∇×E=−∂B∂t\nabla \times \mathbf{E}= -\dfrac{\partial \mathbf{B}}{\partial t}, we can simplify the first term even further.

E⋅(∇×B)=−B⋅∂B∂t−∇⋅(E×B) \mathbf{E} \cdot (\nabla \times \mathbf{B}) = -\mathbf{B} \cdot \dfrac{\partial \mathbf{B}}{\partial t} -\nabla \cdot (\mathbf{E} \times \mathbf{B})

Thus, the original equation becomes

E⋅J=−(ϵ0E⋅∂E∂t+1μ0B⋅∂B∂t)−∇⋅(E×B) \mathbf{E} \cdot \mathbf{J} = -\left( \epsilon_{0} \mathbf{E} \cdot \dfrac{\partial \mathbf{E} } {\partial t} + \dfrac{1}{\mu_{0}}\mathbf{B}\cdot\dfrac{\partial \mathbf{B}}{\partial t} \right) -\nabla \cdot (\mathbf{E} \times \mathbf{B})

To simplify the first term on the right-hand side, we can apply the following relationship:

∂∂tA2=∂∂t(A⋅A)=2A⋅∂A∂t  ⟹  A⋅∂A∂t=12∂∂tA2 \dfrac{\partial}{\partial t} A^2=\dfrac{\partial}{\partial t}(\mathbf{A} \cdot \mathbf{A})=2\mathbf{A} \cdot \dfrac{\partial \mathbf{A} }{\partial t} \\ \implies \mathbf{A}\cdot \dfrac{ \partial \mathbf{A}}{\partial t} = \dfrac{1}{2}\dfrac{\partial}{\partial t} A^2

Applying this to the original equation gives

E⋅J=−12∂∂t(ϵ0E2+1μ0B2)−1μ0∇⋅(E×B) \mathbf{E} \cdot \mathbf{J} = -\dfrac{1}{2}\dfrac{\partial }{\partial t} \left( \epsilon_{0} E^2 + \dfrac{1}{\mu_{0}}B^2 \right) -\dfrac{1}{\mu_{0}}\nabla \cdot (\mathbf{E} \times \mathbf{B})

Finally, substituting this into (1)(1) gives

dWdt=−ddt∫V12(ϵ0E2+1μ0B2)dτ−1μ0∫V∇⋅(E×B)dτ=−ddt∫V12(ϵ0E2+1μ0B2)dτ−1μ0∮S(E×B)⋅da \begin{align*} \dfrac{dW}{dt} &= -\dfrac{d}{dt} \int_{\mathcal{V}} \dfrac{1}{2}\left( \epsilon_{0} E^2 + \dfrac{1}{\mu_{0}}B^2 \right)d\tau -\dfrac{1}{\mu_{0}} \int_{\mathcal{V}} \nabla \cdot (\mathbf{E} \times \mathbf{B}) d\tau \\ &= -\dfrac{d}{dt} \int_{\mathcal{V}} \dfrac{1}{2}\left( \epsilon_{0} E^2 + \dfrac{1}{\mu_{0}}B^2 \right)d\tau -\dfrac{1}{\mu_{0}} \oint_{\mathcal{S}} (\mathbf{E} \times \mathbf{B}) \cdot d\mathbf{a} \end{align*}

The second equality holds due to the divergence theorem. The first term on the right-hand side is, as seen at the top of this document, the total energy uu stored in the electromagnetic field. The Poynting vector is defined as S≡1μ0(E×B)\mathbf{S} \equiv \frac{1}{\mu_{0}}( \mathbf{E} \times \mathbf{B}). By uu and S\mathbf{S}, we can simply express Poynting’s theorem as

dWdt=−ddt∫Vudτ−∮SS⋅da \dfrac{dW}{dt} =-\dfrac{d}{dt} \int_{\mathcal{V}} u d\tau - \oint_{\mathcal{S}}\mathbf{S} \cdot d\mathbf{a}

  • Part 2. Continuity equation

When no work is done on the charge within volume V\mathcal{V}, then dWdt=0\dfrac{dW}{dt}=0, thus Poynting’s theorem is

∫∂u∂tdτ=−∮S⋅da=−∫∇⋅Sdτ \int \dfrac{\partial u}{\partial t} d\tau =- \oint \mathbf{S} \cdot d\mathbf{a}=-\int \nabla \cdot \mathbf{S} d\tau

The second equality holds due to the divergence theorem. Hence,

∂u∂t=−∇⋅S \dfrac{\partial u}{\partial t} = -\nabla \cdot \mathbf{S}

If the energy stored in the electromagnetic field within the volume changes (left-hand side), it means that an equivalent amount of energy in the electromagnetic field has escaped through the boundary of the volume (right-hand side). In other words, local energy is conserved. However, the energy of the electromagnetic field itself is not always conserved. Only when calculating the energy of both the electromagnetic field and the objects interacting with it is the energy conserved.

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  1. David J. Griffiths, 기초전자기학(Introduction to Electrodynamics, 김진승 역) (4th Edition1 2014), p381-385 ↩︎