2.1: Maxwell's Equations

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Franz X. Kaertner
Massachusetts Institute of Technology via MIT OpenCourseWare

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Maxwell’s equations are given by

\[\vec{\nabla} \times \vec{H} = \vec{j} + \dfrac{\partial \vec{D}}{\partial t}, \nonumber \]

\[\vec{\nabla} \times \vec{E} = -\dfrac{\partial \vec{B}}{\partial t}, \label{eq2.1.2} \]

\[\vec{\nabla} \cdot \vec{D} = \rho, \nonumber \]

\[\vec{\nabla} \cdot \vec{B} = 0 \nonumber \].

The material equations accompanying Maxwell’s equations are:

\[\vec{D} = \epsilon_0 \vec{E} + \vec{P}, \nonumber \]

\[\vec{B} = \mu_0 \vec{H} + \vec{M}. \nonumber \]

Here, \(\vec{E}\) and \(\vec{H}\) are the electric and magnetic field, \(\vec{D}\) the dielectric flux, \(\vec{B}\) the magnetic flux, \(\vec{j}\) the current density of free carries, \(\rho\) is the free charge density, \(\vec{P}\) is the polarization, and \(\vec{M}\) the magnetization. By taking the curl of Equation \ref{eq2.1.2} and considering \(\vec{\nabla} \times (\vec{\nabla} \times \vec{E}) = \vec{\nabla} (\vec{\nabla} \vec{E}) - \Delta \vec{E}\), we obtain

\[\Delta \vec{E} - \mu_0 \dfrac{\partial}{\partial t} \left (\vec{j} + \epsilon_0 \dfrac{\partial \vec{E}}{\partial t} + \dfrac{\partial \vec{P}}{\partial t} \right ) = \dfrac{\partial}{\partial t} \vec{\nabla} \times \vec{M} + \vec{\nabla} (\vec{\nabla} \cdot \vec{E}) \nonumber \]

and hence

\[\left (\Delta - \dfrac{1}{c_0^2} \dfrac{\partial^2}{\partial t^2} \right ) \vec{E} = \mu_0 \left ( \dfrac{\partial vec{j}}{\partial t} + \dfrac{\partial^2}{\partial t^2} \vec{P} \right ) + \dfrac{\partial}{\partial t} \vec{\nabla} \times \vec{M} + \vec{\nabla} ( \vec{\nabla} \cdot \vec{E}).\label{eq2.1.8} \]

The vacuum velocity of light is

\[c_0 = \sqrt{\dfrac{1}{\mu_0 \epsilon_0}}. \nonumber \]