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\