Themes > Science > Physics > Electromagnetism > Magnetostatics > Magnetic Field > Electromagnetic Waves

Just as conservation of charge is an immediate consequence of Maxwell's equations, so too is the existence of electromagnetic waves, the immediate interpretation of light as an electromagnetic wave. This elucidation of the true nature of light is one of the great triumphs of classical electrodynamics as embodied in Maxwell's equations.

First let us recall that the 1-dimensional wave equation is  
 \begin{displaymath} \frac{{\partial }^2 \psi }{\partial x^2} - \frac{1}{v^2}\frac{{\partial }^2 \psi }{\partial t^2} = 0\end{displaymath} (131)

where $\psi (x,t)$ represents the wave and v is the speed of the wave. Contrast this to several other well known equations such as the heat equation
 \begin{displaymath} \frac{{\partial }^2 \psi }{\partial x^2} - \frac{1}{v}\frac{\partial \psi }{\partial t} = 0\end{displaymath} (132)

or the Schrödinger equation 
 \begin{displaymath} - \frac{{\hbar }^2}{2m}\frac{{\partial }^2 \psi }{\partial x^2} + V \psi = i \hbar \frac{\partial \psi }{\partial t}\end{displaymath} (133)

or the Klein-Gordon equation 
 \begin{displaymath} ({\Box }^2 + m^2 ) \phi = 0\end{displaymath} (134)
where  
 \begin{displaymath} {\Box }^2 \equiv \frac{1}{v^2}\frac{{\partial }^2 }{\partial t^2} . - {\bigtriangledown }^2\end{displaymath} (135)

In 3-dimensions the wave equation is  
 \begin{displaymath} {\bigtriangledown }^2 \psi - \frac{1}{v^2}\frac{{\partial }^2 \psi }{\partial t^2} =0\end{displaymath} (136)

We would like to think about light travelling in the vacuum of free space far away from sources of charge or current (which actually do produce the electromagnetic waves in the first place). Thus we set $\rho = j = 0$ in Maxwell's equations, giving Maxwell's equations in free space as  
 \begin{displaymath} \bigtriangledown . {\bf E} = 0 ,\end{displaymath} (137)
 
 \begin{displaymath} \bigtriangledown . {\bf B} = 0 ,\end{displaymath} (138)
 
 \begin{displaymath} \bigtriangledown \times {\bf E} + g \frac{\partial {\bf B}}{\partial t} =0 ,\end{displaymath} (139)
 
 \begin{displaymath} \bigtriangledown \times {\bf B} - \frac{1}{g c^2 } \frac{\partial {\bf E}}{\partial t} = 0\end{displaymath} (140)
Taking the curl of Faraday's law gives $\bigtriangledown \times (\bigtriangledown \times {\bf E}) + g \frac{\partial }{\partial t} (\bigtriangledown \times {\bf B})=0 $ and substituting $\bigtriangledown \times {\bf B}$ from Ampère's law gives $\bigtriangledown \times (\bigtriangledown \times {\bf E}) + \frac{1}{c^2} \frac{{\partial }^2{\bf E}}{\partial t^2} = 0$. However $\bigtriangledown \times (\bigtriangledown \times {\bf E}) = \bigtriangledown (\... ...wn . {\bf E}) - {\bigtriangledown }^2 {\bf E} = - {\bigtriangledown }^2 {\bf E}$ because of . (do Problem 3.2). Thus we have  
 \begin{displaymath} {\bigtriangledown }^2 {\bf E} - \frac{1}{c^2} \frac{{\partial }^2{\bf E}}{\partial t^2} = 0\end{displaymath} (141)

and with the same analysis (do Problem 3.3)  
 \begin{displaymath} {\bigtriangledown }^2 {\bf B} - \frac{1}{c^2} \frac{{\partial }^2{\bf B}}{\partial t^2} = 0\end{displaymath} (142)

showing that the electric and magnetic fields correspond to waves propagating in free space at speed c. Note that the constant g cancels out, so that these wave equations look the same in all units. It turns out that the permeability and permittivity of free space have the values such that $1/\sqrt{{\mu }_0 {\epsilon }_0 }$ equals the speed of light ! Thus the identification was immediately made that these electric and magnetic waves are light.

Probably the physical meaning of Maxwell's equations is unclear at this stage. Don't worry about this yet. The purpose of this chapter was to give a brief survey of Maxwell's equations and some immediate consequences. We shall study the physical meaning and solutions of Maxwell's equations in much more detail in the following chapters.


Information provided by: http://www.uwm.edu