Poynting's theorem

4 min read

Poynting’s Theorem

If we have some charge and current configuration which, produces some fields E\mathbf{E} and B\mathbf{B}. After a while, the charges move around.

The question is, how much work dWdW is done by the electromagnetic forces in the interval dtdt?

To do this, we simply compute the work, which is

dW=Fdl=q(E+v×B)vdt=qEvdtdW = \mathbf{F} \cdot d \mathbf{l}=q(\mathbf{E}+\mathbf{v} \times \mathbf{B}) \cdot \mathbf{v} d t=q \mathbf{E} \cdot \mathbf{v} d t

We can rewrite this in terms of charge and current densities. Swap out qρdτq \rightarrow \rho d\tau and ρvJ\rho \mathbf{v} \rightarrow \mathbf{J}.

dWdt=V(EJ)dτ\frac{d W}{d t}=\int_{\mathcal{V}}(\mathbf{E} \cdot \mathbf{J}) d \tau

So EJ\mathbf{E} \cdot \mathbf{J} is the work done per time, per unit volume, or the power per volume. We would like to know what this quantity is.

Begin with the Ampere-Maxwell’s Law:

×B=μ0J+μ0ε0Et\nabla \times \mathbf{B}=\mu_{0} \mathbf{J}+\mu_{0} \varepsilon_{0} \frac{\partial \mathbf{E}}{\partial t}

and so we can dot both sides with E\mathbf{E}, using this equation to get rid of J\mathbf{J}:

EJ=1μ0E(×B)ϵ0EEt\mathbf{E} \cdot \mathbf{J}=\frac{1}{\mu_{0}} \mathbf{E} \cdot(\nabla \times \mathbf{B})-\epsilon_{0} \mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t}

We now have to deal with two terms. The first we can use the following vector calculus identity:

(A×B)=B(×A)A(×B)\nabla \cdot(\mathbf{A} \times \mathbf{B})=\mathbf{B} \cdot(\nabla \times \mathbf{A})-\mathbf{A} \cdot(\nabla \times \mathbf{B})

and now plugging in the fields, we have

(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})

using Faraday’s Law ((×E=B/t(\nabla \times \mathbf{E}=-\partial \mathbf{B} / \partial t), it follows that

E(×B)=BBt(E×B)\mathbf{E} \cdot(\nabla \times \mathbf{B})=-\mathbf{B} \cdot \frac{\partial \mathbf{B}}{\partial t}-\nabla \cdot(\mathbf{E} \times \mathbf{B})

Using another calculus identity, we have:

BBt=12t(B2), and EEt=12t(E2)\mathbf{B} \cdot \frac{\partial \mathbf{B}}{\partial t}=\frac{1}{2} \frac{\partial}{\partial t}\left(B^{2}\right), \quad \text { and } \quad \mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t}=\frac{1}{2} \frac{\partial}{\partial t}\left(E^{2}\right)

And so

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

And plugging it into our original expression for work, then calling the divergence theorem

V(v)dτ=Svda\int_{\mathcal{V}}(\nabla \cdot \mathbf{v}) d \tau=\oint_{S} \mathbf{v} \cdot d \mathbf{a}

on the second term allows to convert a volume integral into a surface integral. Finally, we have

dWdt=ddtV12(ϵ0E2+1μ0B2)dτ1μ0S(E×B)da\frac{d W}{d t}=-\frac{d}{d t} \int_{\mathcal{V}} \frac{1}{2}\left(\epsilon_{0} E^{2}+\frac{1}{\mu_{0}} B^{2}\right) d \tau-\frac{1}{\mu_{0}} \oint_{\mathcal{S}}(\mathbf{E} \times \mathbf{B}) \cdot d \mathbf{a}

Which is the work energy theorem of electrodynamics: The first term is the total energy stored in electromagnetic fields:

u=12(ϵ0E2+1μ0B2)u = \frac{1}{2} \left( \epsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right)

The second term is the rate at which energy is transported out of the surface. The mysterious second term is defined as the Poynting vector. It is interpreted as the energy per unit time, per unit area.

S1μ0(E×B)\boxed{\mathbf{S} \equiv \frac{1}{\mu_{0}}(\mathbf{E} \times \mathbf{B})}

So Sda\mathbf{S} \cdot d \mathbf{a} is the energy leaving surface dad \mathbf{a}.

Finally, we can write the above equation into a more compact form:

dWdt=ddtVudτSSda\frac{d W}{d t}=-\frac{d}{d t} \int_{\mathcal{V}} u d \tau-\oint_{\mathcal{S}} \mathbf{S} \cdot d \mathbf{a}

Now what is the meaning of this equation? Imagine we do work on some charge configuration. Either the energy stored in the fields had to have decreased, or the energy must have went outside the surface.

The second interpretation could use a little more work. What does it mean for energy to leave a surface? After all, we said that the volume V\mathcal{V} is arbitrary, and S\mathcal{S} is only required to be the boundary of such a volume.

To be concrete, let’s say our system is a battery, and pick V\mathcal{V} to be the volume of the battery. In a circuit, the battery is clearly doing work to drive say a lightbulb. (increasing dW/dtdW/dt)

2880px Poynting vectors of DC circuit svg

Image: Wikipedia

So where does the energy come from? If we say there aren’t any fields in the battery, then energy really is leaving the battery to drive the circuit, in order word the second term must decrease.

Finally, if dW/dt=0dW/dt = 0, then using the divergence theorem again gives

utdτ=Sda=(S)dτ\int \frac{\partial u}{\partial t} d \tau=-\oint \mathbf{S} \cdot d \mathbf{a}=-\int(\mathbf{\nabla} \cdot \mathbf{S}) d \tau

and removing the integrals gives us:

ut=S\frac{\partial u}{\partial t}=-\nabla \cdot \mathbf{S}

which is the continuity equation for energy! This says that energy is locally conserved!

If we compare this to the continuity equation for fluids, we see that the Poynting vector S\mathbf{S} really is the energy flux.

Image: Wikipedia. Dipole radiation of a dipole vertically in the page showing electric field strength (colour) and Poynting vector (arrows) in the plane of the page.

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Zhi Han