Pre-Quantum Electrodynamics

• Gr 8.2.2

Total EM force on charges in volume \({\cal V}\): \[ {\boldsymbol F} = \int_{\cal V} \left( {\boldsymbol E} + {\boldsymbol v} \times {\boldsymbol B} \right) \rho ~d\tau = \int_{\cal V} \left( \rho {\boldsymbol E} + {\boldsymbol J} \times {\boldsymbol B} \right) d\tau \] Force per unit volume: \[ {\boldsymbol f} \equiv \rho {\boldsymbol E} + {\boldsymbol J} \times {\boldsymbol B}. \] Substitute for \(\rho\) and \({\boldsymbol J}\) using Maxwell (Gauss and Ampère-Maxwell): \[ {\boldsymbol f} = \varepsilon_0 ({\boldsymbol \nabla} \cdot {\boldsymbol E}) {\boldsymbol E} + \left( \frac{1}{\mu_0} {\boldsymbol \nabla} \times {\boldsymbol B} - \varepsilon_0 \frac{ \partial {\boldsymbol E}}{\partial t} \right) \times {\boldsymbol B}. \]

On the other hand we have

\begin{equation} \frac{\partial }{\partial t} \left( {\boldsymbol E} \times {\boldsymbol B} \right) = \frac{\partial {\boldsymbol E}}{\partial t} \times {\boldsymbol B} + {\boldsymbol E} \times \frac{\partial {\boldsymbol B}}{\partial t}. \end{equation}

Using Faraday to substitute for \(\frac{\partial {\boldsymbol B}}{\partial t}\),

\begin{equation} \frac{ \partial {\boldsymbol E}}{\partial t} \times {\boldsymbol B} = \frac{\partial }{\partial t} \left( {\boldsymbol E} \times {\boldsymbol B}\right) + {\boldsymbol E} \times \left({\boldsymbol \nabla} \times {\boldsymbol E} \right) \end{equation}

so \[ {\boldsymbol f} = \varepsilon_0 \left( \left( {\boldsymbol \nabla} \cdot {\boldsymbol E} \right) {\boldsymbol E} - {\boldsymbol E} \times \left( {\boldsymbol \nabla} \times {\boldsymbol E} \right) \right) - \frac{1}{\mu_0} \left( {\boldsymbol B} \times \left( {\boldsymbol \nabla} \times {\boldsymbol B} \right) \right) - \varepsilon_0 \frac{\partial}{\partial t} \left( {\boldsymbol E} \times {\boldsymbol B} \right). \] Since \({\boldsymbol \nabla} \cdot {\boldsymbol B} = 0\), we can symmetrize the expression in \({\boldsymbol E}\) and \({\boldsymbol B}\). Moreover, by grad_sprod, \[ \frac{1}{2}{\boldsymbol \nabla} \left( E^2 \right) = \left( {\boldsymbol E} \cdot {\boldsymbol \nabla} \right) {\boldsymbol E} + {\boldsymbol E} \times \left( {\boldsymbol \nabla} \times {\boldsymbol E} \right) \] so \[ {\boldsymbol E} \times \left( {\boldsymbol \nabla} \times {\boldsymbol E} \right) = \frac{1}{2} {\boldsymbol \nabla} E^2 - \left({\boldsymbol E} \cdot {\boldsymbol \nabla} \right) {\boldsymbol E} \] and similarly for \({\boldsymbol B}\). We thus get

\begin{align} {\boldsymbol f} =& \varepsilon_0 \left( \left( {\boldsymbol \nabla} \cdot {\boldsymbol E} \right) {\boldsymbol E} + \left( {\boldsymbol E} \cdot {\boldsymbol \nabla} \right) {\boldsymbol E} \right) + \frac{1}{\mu_0} \left( \left( {\boldsymbol \nabla} \cdot {\boldsymbol B} \right) {\boldsymbol B} + \left( {\boldsymbol B} \cdot {\boldsymbol \nabla} \right) {\boldsymbol B} \right) \\ &- \frac{1}{2} {\boldsymbol \nabla} \left( \varepsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) - \varepsilon_0 \frac{\partial}{\partial t} \left( {\boldsymbol E} \times {\boldsymbol B} \right). \end{align}

This expression can be greatly simplified by introducing the

The element \(T_{ij}\) represents the force per unit area in the \(i\) direction acting on a surface element oriented in the \(j\) direction. Diagonal elements are pressures, off-diagonal elements are shears.

We then obtain the

where \({\boldsymbol S}\) is the Poynting vector. Integrating, we obtain the