#### Poynting's Theorem; the Poynting Vector
emd.ce.poy

Earlier: work done to assemble a static charge distribution: \[ W_e = \frac{\varepsilon_0}{2} \int d\tau ~E^2 \] Work necessary to get currents going: \[ W_m = \frac{1}{2\mu_0} \int d\tau ~B^2 \label{Gr(7.34)} \] Total energy should be sum of these two. Derivation from scratch.

Suppose that at time \(t\), we have fields \({\bf E}\) and \({\bf B}\) produced by some charge and current distributions \(\rho\) and \({\bf J}\). In an interval \(dt\), how much work is done by EM forces? From Lorentz force law: \[ {\bf F} \cdot d{\bf l} = q({\bf E} + {\bf v} \times {\bf B}) \cdot {\bf v} dt = q ~{\bf E} \cdot {\bf v} dt \] Really, we're looking at a small volume element \(d\tau\) carrying charge \(\rho d\tau\), moving at velocity \({\bf v}\) such that \({\bf J} = \rho {\bf v}\). Thus,

- Gr (8.6)

\[ \frac{dW}{dt} = \int_{\cal V} d\tau ~ {\bf E} \cdot {\bf J} \tag{dWdt_intEJ}\label{dWdt_intEJ} \] The integrand is the work done per unit time, per unit volume, i.e. the power delivered per unit volume. In terms of fields alone: use Ampère-Maxwell: \[ {\bf E} \cdot {\bf J} = \frac{1}{\mu_0} {\bf E} \cdot ({\boldsymbol \nabla} \times {\bf B}) - \varepsilon_0 {\bf E} \cdot \frac{\partial {\bf E}}{\partial t} \] Using div_xprod, \[ {\boldsymbol \nabla} \cdot ({\bf E} \times {\bf B}) = {\bf B} \cdot ({\boldsymbol \nabla} \times {\bf E}) - {\bf E} \cdot ({\boldsymbol \nabla} \times {\bf B}), \] Invoking Faraday \({\boldsymbol \nabla} \times {\bf E} = - \partial {\bf B}/\partial t\), \[ {\bf E} \cdot ({\boldsymbol \nabla} \times {\bf B}) = - {\bf B} \cdot \frac{\partial {\bf B}}{\partial t} - {\boldsymbol \nabla} \cdot ({\bf E} \times {\bf B}). \] But obviously, \[ {\bf B} \cdot \frac{\partial {\bf B}}{\partial t} = \frac{1}{2} \frac{\partial}{\partial t} B^2, \hspace{1cm} {\bf E} \cdot \frac{\partial {\bf E}}{\partial t} = \frac{1}{2} \frac{\partial}{\partial t} E^2 \label{Gr(8.7)} \] so we get \[ {\bf E} \cdot {\bf J} = -\frac{1}{2} \frac{\partial}{\partial t} \left( \varepsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) - \frac{1}{\mu_0} {\boldsymbol \nabla} \cdot ({\bf E} \times {\bf B}). \label{Gr(8.8)} \] Substituting this in dWdt_intEJ and using the divergence theorem, we obtain

Poynting's theorem

- Gr (8.9)

\[ \frac{dW}{dt} = -\frac{d}{d t} \int_{\cal V} d\tau \frac{1}{2} \left( \varepsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) - \frac{1}{\mu_0} \oint_{\cal S} d{\bf a} \cdot ({\bf E} \times {\bf B}) \tag{👉Thm}\label{👉Thm} \]

First term on RHS: total energy in EM fields. Second term: rate at which energy is carried by EM fields out of \({\cal V}\) across its boundary surface.

Energy per unit time, per unit area carried by EM fields: given by the

Poynting vector

- Gr (8.10)

\[ {\bf S} \equiv \frac{1}{\mu_0} ({\bf E} \times {\bf B}) \tag{PoyntingVec}\label{PoyntingVec} \]

We can thus express Poynting's theorem more compactly:

Poynting's theorem (integral form)

- Gr (8.11)

\[ \frac{dW}{dt} = - \frac{dU_{em}}{dt} - \oint_{\cal S} d{\bf a} \cdot {\bf S}. \tag{PoyntingThm_int}\label{PoyntingThm_int} \]

where we have defined the total

Energy in electromagnetic fields

- Gr (8.5)

\[ U_{em} \equiv \frac{1}{2} \int d\tau \left( \varepsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) \tag{Uem}\label{Uem} \]

We can rewrite the energies in terms of densities, \[ U_{em} = \int d\tau ~u_{em}, \hspace{1cm} u_{em} \equiv \frac{1}{2} \left( \varepsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) \label{Gr(8.13)} \] Then, \[ \frac{d}{dt} \int_{\cal V} d\tau ~u_{em} = -\oint_{\cal S} d{\bf a} \cdot {\bf S} = -\int_{\cal V} d\tau ~({\boldsymbol \nabla} \cdot {\bf S}) \] so we get the

Poynting theorem (differential form)

- Gr (8.14)

\[ \frac{\partial}{\partial t} u_{em} + {\boldsymbol \nabla} \cdot {\bf S} = 0 \tag{PoyntingThm}\label{PoyntingThm} \]

and has a similar for to the continuity equation (charge density goes into energy density, charge current goes into Poynting vector).

Example: Joule heating

Task: characterize the energy flow for a current-carrying wire.

Solution: the energy per unit time delivered to wire the wire can be obtained from Poynting's theorem.

Assuming that the field is uniform, the electric field parallel to the wire is \[ {\boldsymbol E} = \frac{V}{L} \hat{\boldsymbol x}, \] where \(V\) is the potential difference between the ends ald \(L\) is the length. Magnetic field is circumferential: wire of radius \(a\), \[ {\boldsymbol B} = \frac{\mu_0 I}{2\pi a} \hat{\boldsymbol \varphi} \] Poynting: \[ {\boldsymbol S} = \frac{1}{\mu_0} \frac{V}{L} \frac{\mu_0 I}{2\pi a} \hat{\boldsymbol x} \times \hat{\boldsymbol \varphi} = -\frac{VI}{2\pi a L} \hat{\boldsymbol r} \] and points radially inwards. Energy per unit time passing surface of wire: \[ \int d{\bf a} \cdot {\bf S} = S (2\pi a L) = -V I \] where the minus sign means energy is flowing in (the wire heats up), and the value is as expected.

Except where otherwise noted, all content is licensed under a Creative Commons Attribution 4.0 International License.

Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42