Problem 24.7

24.7.
Equation 24.12 says that
$\frac{\partial^{2} E}{\partial x^{2}} ~=~ \mu_{0}\epsilon_{0} \frac{\partial^{2} E}{\partial t^{2}}$
We start with our equation for E.
$E(x,t) = E_{m} \cos (k x - \omega t)$
Taking two spatial derivatives,
$\frac{\partial E}{\partial x} ~=~ - E_{m} k \sin (k x - \omega t)$
$\frac{\partial^{2} E}{\partial x^{2}} = - E_{m} k^{2} \cos (k x - \omega t)$
Now, for the time derivatives
$\frac{\partial E}{\partial t} ~=~ - E_{m} (-\omega) \sin (k x - \omega t)$
$\frac{\partial^{2} E}{\partial t^{2}} = - E_{m} (\omega^{2}) \cos (k x - \omega t)$
To show that equation 24.12 is true, we must show that
$\frac{\partial^{2} E}{\partial x^{2}} = - \mu_{0}\epsilon_{0} E_{m} (\omega^{2}) \cos (k x - \omega t)$
This holds if
$- k^{2} = - \mu_{0}\epsilon_{0} \omega^{2}$
But since
$\mu_{0}\epsilon_{0} = \frac{1}{c^{2}}$
We only have to show that
$k^{2} = \frac{\omega^{2}}{c^{2}}$
But, $k = \frac{2\pi}{\lambda}$
and $\omega = 2\pi f$
So, our equation reduces to
$\frac{1}{\lambda^{2}} = \frac{f^{2}}{c^{2}}$
or $c = f\lambda$
But that is the definition of a wave, that the frequency times the wavelength is the speed of the wave.
So, we have proven equation 24.12. The version of this problem for the magnetic form of the wave equation is identical.