Poynting's theorem

Poynting's theorem is a result expressing conservation of energy in electrodynamics.

> where $u(\mathbf{r})$ is the electromagnetic energy density of [[electric field]] $\mathbf{E}$ and [[magnetic field]] $\mathbf{B}$ > $$u=\frac{1}{2}\left( \varepsilon_{0}E^{2}+ \frac{1}{\mu_{0}}B^{2} \right)

S\mathcal{S} is the bounding surface of V\mathcal{V} and S\mathbf{S} is the Poynting vector.

In differential form it reads

> [!quote]- Proof > We begin by looking at the work being done by the Lorentz force on a [[Electric charge|point charge]] $q$ moving at velocity $\mathbf{v}$ over a line element $d\mathbf{s}=\mathbf{v}dt$: > $$dW=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})\cdot d\mathbf{s}=q\mathbf{E}\cdot \mathbf{v}dt+q\cancel{ (\mathbf{v}\times \mathbf{B})\cdot \mathbf{v}dt }=q\mathbf{E}\cdot d\mathbf{s}

To go from point charges to charge distributions, we can pick a tiny volume dτd\tau with charge dq=ρdτdq=\rho d\tau. If this charge element moves at a velocity v\mathbf{v}, the work over a time dtdt is

> since $\rho \mathbf{v}=\mathbf{J}$. $\mathbf{E}\cdot \mathbf{J}$ must the work done per unit time, per unit volume. If we "divide" by $dt$ and integrate over the volume we get the [[power]] > $$P=\frac{dW}{dt}=\int_{\mathcal{V}}\mathbf{E}\cdot \mathbf{J}\ d\tau\tag{1}

We can elaborate on E⋅J\mathbf{E}\cdot \mathbf{J} by expressing J\mathbf{J} through the Ampere-Maxwell law:

\mathbf{E}\cdot \mathbf{J}&=\mathbf{E}\cdot \frac{\nabla\times\mathbf{B}}{\mu_{0}}-\varepsilon_{0}\mathbf{E}\cdot\frac{ \partial \mathbf{E} }{ \partial t } \\ \left( \mathbf{E}\cdot \frac{ \partial \mathbf{E} }{ \partial t } =\frac{1}{2}\frac{ \partial E^{2} }{ \partial t } \right)&=\frac{1}{\mu_{0}} \mathbf{E}\cdot(\nabla\times\mathbf{B})- \frac{\varepsilon_{0}}{2}\frac{ \partial E^{2} }{ \partial t } \\ &=\frac{1}{\mu_{0}} [\mathbf{B}\cdot(\nabla\times\mathbf{E})- \nabla\cdot(\mathbf{E}\times \mathbf{B})]- \frac{\varepsilon_{0}}{2} \frac{ \partial E^{2} }{ \partial t } \\ &=\frac{1}{\mu_{0}} \left[ -\mathbf{B}\cdot \frac{ \partial \mathbf{B} }{ \partial t } - \nabla \cdot(\mathbf{E}\times \mathbf{B}) \right]- \frac{\varepsilon_{0}}{2}\frac{ \partial E^{2} }{ \partial t } \\ \left( \mathbf{B}\cdot \frac{ \partial \mathbf{B} }{ \partial t } =\frac{1}{2}\frac{ \partial B^{2} }{ \partial t } \right)&=- \frac{1}{2\mu_{0}}\frac{ \partial B^{2} }{ \partial t } - \frac{\varepsilon_{0}}{2}\frac{ \partial E^{2} }{ \partial t } -\nabla\cdot \mathbf{S} \end{align}
> where we expanded > $$\nabla \cdot(\mathbf{E}\times \mathbf{B})=\mathbf{B}\cdot(\nabla\times \mathbf{E})-\mathbf{E}\cdot(\nabla\times \mathbf{B})

and rearranged to find E⋅(∇×B)\mathbf{E}\cdot(\nabla\times \mathbf{B}), used Faraday's law and defined the Poynting vector

> > If we extract and merge the time derivative from the first two terms, we are left with the electromagnetic energy density by volume $u$: > $$- \frac{1}{2\mu_{0}}\frac{ \partial B^{2} }{ \partial t } - \frac{\varepsilon_{0}}{2}\frac{ \partial E^{2} }{ \partial t }=- \frac{1}{2}\frac{ \partial }{ \partial t } \left( \frac{1}{\mu_{0}}B^{2} + \varepsilon_{0}E^{2} \right)=-\frac{ \partial u }{ \partial t }

Which means

> which is the differential expression of the theorem. If we plug this back into $(1)$, we get > $$P=-\frac{ \partial }{ \partial t } \int_{V}u\ d\tau-\int_{V}\nabla\cdot\mathbf{S}=-\frac{ \partial }{ \partial t } \int_{V}u\ d\tau-\oint_{S}\mathbf{S}\cdot d\mathbf{a}

where we used the divergence theorem. This completes the proof.

The two integrals can be interpreted individually: the first is the stored energy1, of which we take the time derivative of, whereas the second is the transfer of energy from inside the volume V\mathcal{V} to outside, as determined by checking what goes through the boundary S\mathcal{S}.

It is equivalent in nature to the mechanical work-energy theorem and binds stored energy to transported energy as a consequence of energy conservation: if something goes out, it must be removed from what's inside. It is in large part the same statement as the continuity equation for electric current ∇⋅J=−∂ρ∂t\nabla\cdot\mathbf{J}=-\frac{ \partial \rho }{ \partial t }, except instead of charge we have energy. In fact, if no work is being done on the charges, then E⋅J=0\mathbf{E}\cdot \mathbf{J}=0 and we get ∇⋅S=−∂u∂t\nabla\cdot \mathbf{S}=-\frac{ \partial u }{ \partial t }. This confirms that the Poynting vector is an "energy current": it represents the flow of energy and therefore plays a pivotal role in energy conservation.

Footnotes

  1. Where the energy is stored is a whole other problem. Electrodynamics alone does not give a really satisfactory answer and you could realistically place it anywhere, be it in fields, matter or charges, but since field theory relies on this, it is best to think of it as being stored in the fields. ↩