PlanetPhysics/Derivation of Wave Equation From Maxwell's Equations

Maxwell was the first to note that Amp\`ere's Law does not satisfy conservation of charge (his corrected form is given in Maxwell's equation). This can be shown using the equation of conservation of electric charge:

$\nabla \cdot \mathbf {J} +{\frac {\partial \rho }{\partial t}}=0$

Now consider Faraday's Law in differential form:

$\nabla \times \mathbf {E} =-{\frac {\partial \mathbf {B} }{\partial t}}$

Taking the curl of both sides:

$\nabla \times (\nabla \times \mathbf {E} )=\nabla \times (-{\frac {\partial \mathbf {B} }{\partial t}})$

The right-hand side may be simplified by noting that

$\nabla \times ({\frac {\partial \mathbf {B} }{\partial t}})=-{\frac {\partial }{\partial t}}(\nabla \times \mathbf {B} )$

Recalling Amp\`ere's Law,

$-{\frac {\partial }{\partial t}}(\nabla \times \mathbf {B} )=-\mu _{0}\epsilon _{0}{\frac {\partial ^{2}\mathbf {E} }{\partial t^{2}}}$

Therefore

$\nabla \times (\nabla \times \mathbf {E} )=-\mu _{0}\epsilon _{0}{\frac {\partial ^{2}\mathbf {E} }{\partial t^{2}}}$

The left hand side may be simplified by the following Vector Identity:

$\nabla \times (\nabla \times \mathbf {E} )=-\nabla ^{2}\mathbf {E}$

Hence

$\nabla ^{2}\mathbf {E} =\mu _{0}\epsilon _{0}{\frac {\partial ^{2}\mathbf {E} }{\partial t^{2}}}$

Applying the same analysis to Amp\'ere's Law then substituting in Faraday's Law leads to the result

$\nabla ^{2}\mathbf {B} =\mu _{0}\epsilon _{0}{\frac {\partial ^{2}\mathbf {E} }{\partial t^{2}}}$

Making the substitution $\mu _{0}\epsilon _{0}=1/c^{2}$ we note that these equations take the form of a transverse wave travelling at constant speed $c$. Maxwell evaluated the constants $\mu _{0}$ and $\epsilon _{0}$ according to their known values at the time and concluded that $c$ was approximately equal to 310,740,000 ${\mbox{ms}}^{-1}$, a value within ~3\