Pre-Quantum Electrodynamics

• PM 9.6
• Gr 8.1.2

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,

\[ \frac{dW}{dt} = \int_{\cal V} d\tau ~ {\bf E} \cdot {\bf J} \tag{dWdt_intEJ}\label{dWdt_intEJ} \] 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

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

We can thus express Poynting's theorem more compactly:

where we have defined the total

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

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