📂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.

$$ \begin{align*} \dfrac{dW}{dt} &= -\dfrac{d}{dt} \int_{\mathcal{V}} \dfrac{1}{2} \left( \epsilon_{0} E^2 + \dfrac{1}{\mu_{0}} B^2 \right) d\tau - \dfrac{1}{\mu_{0}} \oint_{\mathcal{S}} (\mathbf{E} \times \mathbf{B}) \cdot d \mathbf{a} \\ &= -\dfrac{d}{dt} \int_{\mathcal{V}} u d\tau - \oint_{\mathcal{S}}\mathbf{S} \cdot d\mathbf{a} \end{align*} $$

$\mathcal{S}$ is the boundary of $\mathcal{V}$. $u$ is the total energy stored in the electromagnetic field of a unit volume space. The integrand $\mathbf{S}$ 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.

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

It is also referred to as energy flux density.

Description

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

$$ \dfrac{\partial u}{\partial t} = -\nabla \cdot \mathbf{S} $$

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.

$$ W_{E} =\dfrac{\epsilon_{0}}{2}\int E^2 d\tau , \quad \quad W_{B} =\dfrac{1}{2\mu_{0}}\int B^2 d\tau $$

$\mathbf{E}$ and $\mathbf{B}$ 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.

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

• Part 1. Poynting’s theorem

Let’s say an electric field $\mathbf{E}$ and a magnetic field $\mathbf{B}$ are generated from a certain charge and current distribution at time $t$. If the charge moved after time $dt$, the work done by the electromagnetic force on this charge during time $dt$ can be expressed as follows, according to the definition of work and Lorentz force law.

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

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

$$ \begin{align} dW &= \int_{\mathcal{V}} \mathbf{E}\cdot \rho\mathbf{v} d\tau dt \nonumber \\ &= \int_{\mathcal{V}} \mathbf{E} \cdot \mathbf{J} d\tau dt \nonumber \\ \implies \dfrac{dW}{dt} &= \int_{\mathcal{V}} \mathbf{E} \cdot \mathbf{J} d\tau \end{align} $$

Thus, $\mathbf{E} \cdot \mathbf{J}$ 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 $\text{(iv)}$, we have $\mathbf{J}=\frac{1}{\mu_{0}}\nabla \times \mathbf{B} - \epsilon_{0} \dfrac{\partial \mathbf{E}}{\partial t}$, therefore

$$ \mathbf{E} \cdot \mathbf{J} = \dfrac{1}{\mu_{0}}\mathbf{E} \cdot (\nabla \times \mathbf{B}) - \epsilon_{0} \mathbf{E}\cdot \dfrac{\partial \mathbf{E}}{\partial t} $$

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

$$ \mathbf{E} \cdot (\nabla \times \mathbf{B}) = -\mathbf{B} \cdot \dfrac{\partial \mathbf{B}}{\partial t} -\nabla \cdot (\mathbf{E} \times \mathbf{B}) $$

Thus, the original equation becomes

$$ \mathbf{E} \cdot \mathbf{J} = -\left( \epsilon_{0} \mathbf{E} \cdot \dfrac{\partial \mathbf{E} } {\partial t} + \dfrac{1}{\mu_{0}}\mathbf{B}\cdot\dfrac{\partial \mathbf{B}}{\partial t} \right) -\nabla \cdot (\mathbf{E} \times \mathbf{B}) $$

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

$$ \dfrac{\partial}{\partial t} A^2=\dfrac{\partial}{\partial t}(\mathbf{A} \cdot \mathbf{A})=2\mathbf{A} \cdot \dfrac{\partial \mathbf{A} }{\partial t} \\ \implies \mathbf{A}\cdot \dfrac{ \partial \mathbf{A}}{\partial t} = \dfrac{1}{2}\dfrac{\partial}{\partial t} A^2 $$

Applying this to the original equation gives

$$ \mathbf{E} \cdot \mathbf{J} = -\dfrac{1}{2}\dfrac{\partial }{\partial t} \left( \epsilon_{0} E^2 + \dfrac{1}{\mu_{0}}B^2 \right) -\dfrac{1}{\mu_{0}}\nabla \cdot (\mathbf{E} \times \mathbf{B}) $$

Finally, substituting this into $(1)$ gives

$$ \begin{align*} \dfrac{dW}{dt} &= -\dfrac{d}{dt} \int_{\mathcal{V}} \dfrac{1}{2}\left( \epsilon_{0} E^2 + \dfrac{1}{\mu_{0}}B^2 \right)d\tau -\dfrac{1}{\mu_{0}} \int_{\mathcal{V}} \nabla \cdot (\mathbf{E} \times \mathbf{B}) d\tau \\ &= -\dfrac{d}{dt} \int_{\mathcal{V}} \dfrac{1}{2}\left( \epsilon_{0} E^2 + \dfrac{1}{\mu_{0}}B^2 \right)d\tau -\dfrac{1}{\mu_{0}} \oint_{\mathcal{S}} (\mathbf{E} \times \mathbf{B}) \cdot d\mathbf{a} \end{align*} $$

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 $u$ stored in the electromagnetic field. The Poynting vector is defined as $\mathbf{S} \equiv \frac{1}{\mu_{0}}( \mathbf{E} \times \mathbf{B})$. By $u$ and $\mathbf{S}$, we can simply express Poynting’s theorem as

$$ \dfrac{dW}{dt} =-\dfrac{d}{dt} \int_{\mathcal{V}} u d\tau - \oint_{\mathcal{S}}\mathbf{S} \cdot d\mathbf{a} $$

• Part 2. Continuity equation

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

$$ \int \dfrac{\partial u}{\partial t} d\tau =- \oint \mathbf{S} \cdot d\mathbf{a}=-\int \nabla \cdot \mathbf{S} d\tau $$

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

$$ \dfrac{\partial u}{\partial t} = -\nabla \cdot \mathbf{S} $$

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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