1 \newcommand{\Em}{E_\text{max}}
2 \newcommand{\Bm}{B_\text{max}}
3 \newcommand{\ctrig}{\cos(kx-\omega t)}
4 \newcommand{\strig}{\sin(kx-\omega t)}
6 Verify by substitution that the following equations are solutions to
7 Equations 24.15 and 24.16 respectively:
13 \npderiv{2}{x}{E} &= \epsilon_0\mu_0 \npderiv{2}{t}{E} \tag{24.15} \\
14 \npderiv{2}{x}{B} &= \epsilon_0\mu_0 \npderiv{2}{t}{B} \tag{24.16}
16 \end{problem*} % problem 24.9
19 This is just an excercise in partial derivatives.
21 \pderiv{x}{E} &= -\Em\strig\cdot k \\
22 \npderiv{2}{x}{E} &= -\Em k\ctrig\cdot k = -k^2 E\\
23 \pderiv{t}{E} &= -\Em\strig\cdot (-\omega) \\
24 \npderiv{2}{t}{E} &= \Em\omega\ctrig\cdot (-\omega) = -\omega^2 E \\
25 \frac{k\U{rad/m}}{\omega\U{rad/s}} &= \frac{1}{c}
26 = \sqrt{\epsilon_0\mu_0} \label{eq.c_to_e_mu} \\
27 \npderiv{2}{x}{E} &= \frac{k^2}{\omega^2} \npderiv{2}{t}{E}
28 = \epsilon_0\mu_0 \npderiv{2}{t}{E}
30 which is what we set out to show. Note that we used Equation 24.17 in
31 Equation \ref{eq.c_to_e_mu}. The situation for $B$ is exactly the
32 same with the replacement $E\rightarrow B$.
34 \pderiv{x}{B} &= -\Bm\strig\cdot k \\
35 \npderiv{2}{x}{B} &= -\Bm k\ctrig\cdot k = -k^2 B\\
36 \pderiv{t}{B} &= -\Bm\strig\cdot (-\omega) \\
37 \npderiv{2}{t}{B} &= \Bm\omega\ctrig\cdot (-\omega) = -\omega^2 B \\
38 \frac{k\U{rad/m}}{\omega\U{rad/s}} &= \frac{1}{c}
39 = \sqrt{\epsilon_0\mu_0} \\
40 \npderiv{2}{x}{B} &= \frac{k^2}{\omega^2} \npderiv{2}{t}{B}
41 = \epsilon_0\mu_0 \npderiv{2}{t}{B}