Pre-Quantum Electrodynamics

• PM 9.6
• Gr 9.2.3

Energy density in EM field: \[ u = \frac{1}{2} \left( \varepsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) \label{Gr(9.53)} \] For a monochromatic EM plane wave, \[ B^2 = \frac{1}{c^2} E^2 = \mu_0 \varepsilon_0 E^2 \label{Gr(9.54)} \] so the electric and magnetic contributions to energy density are equal and \[ u = \varepsilon_0 E^2 = \varepsilon_0 E_0^2 \cos({\boldsymbol k} \cdot {\boldsymbol r} - \omega t + \delta) \]

Poynting vector: \[ {\bf S} = \frac{1}{\mu_0} {\bf E} \times {\bf B}, \label{Gr(9.56)} \] so for a monochromatic EM plan wave, \[ {\bf S} = c \varepsilon_0 E_0^2 \cos^2 ({\boldsymbol k} \cdot {\boldsymbol r} - \omega t + \delta) ~\hat{\boldsymbol n} \times (\hat{\boldsymbol k} \times \hat{\boldsymbol n}) = \label{Gr(9.57)} \] or more succinctly:

This has a transparent physical interpretation: the energy density \(u\) flows with velocity \(c\) in direction of the wave.

Similary, we get the

Time averages: integrating over a (integer number of) cycle(s), we have \[ \langle u \rangle = \frac{\varepsilon_0}{2} E_0^2, \hspace{5mm} \langle {\boldsymbol S} \rangle = \frac{c \varepsilon_0}{2} E_0^2 ~\hat{\boldsymbol k}, \hspace{5mm} \langle {\boldsymbol g} \rangle = \frac{\varepsilon_0}{2c} E_0^2 ~\hat{\boldsymbol k}. \]

The average power per unit time per unit area transported by an EM wave is called the Intensity \[ I \equiv \langle S \rangle = \frac{c\varepsilon_0}{2} E_0^2 \]

The radiation pressure is the momentum transfer per unit area per unit of time \[ P = \frac{1}{A}\frac{\Delta p}{\Delta t} = \frac{\langle g \rangle A c \Delta t}{A \Delta t} = \frac{\varepsilon_0}{2} E_0^2 = \frac{I}{c}. \]