Exercise 5.1.2: The wave equation and the speed of light

Taking the curl of $\nabla \times E=-\partial B/\partial t$ and using Ampère–Maxwell with $J=0$,

Since $\nabla \cdot E=\rho /\epsilon =0$, the identity gives $\nabla \times \nabla \times E=-{\nabla }^{2}E$, hence

For $E_x=f(u)$ with $u=t-z/v$: ${\partial }^2{E}_x/\partial z^2=f^{\prime \prime }/v^2$ and ${\partial }^2{E}_x/\partial t^2=f^{\prime \prime }$, so the equation holds if and only if $1/v^2=\mu \epsilon$. The shape $f$ is arbitrary: every disturbance travels undistorted at $v=1/\sqrt{\mu \epsilon }$. (Home exercise 5.1.1 is the case $f=E_0\cos \omega u$.)

With the values given in the statement, ${\mu }_0{\epsilon }_0=1.112650\times 10^{-17}{\mathrm{s}}^2/{\mathrm{m}}^2$, so

With the rounded constants of this book, ${\epsilon }_0=10^{-9}/(36\pi )F/m$ and ${\mu }_0=4\pi \times 10^{-7}H/m$, the same formulae give exactly $c=3\times 10^{8}m/s$ and ${\eta }_0=120\pi =376.99\Omega$, within $0.07\%$ of the values above. ${\epsilon }_0$ comes from Coulomb's law and ${\mu }_0$ from the force between currents, two table-top experiments, and their product gives the speed of light to all digits. Since 1983 the logic is reversed: $c$ is defined exactly and fixes the metre, and ${\epsilon }_0=1/({\mu }_0{c}^2)$ follows.