Problem 5.6.1: Reading a monochromatic wave

The amplitude is $A=10V/m$, along $a_y$. $\omega =6\times 10^8\pi rad/s$ and $\varphi (z)=3\pi z$, so $\beta =3\pi rad/m$. The argument is $\omega t-\beta z$ with a minus sign: a crest $\omega t=\beta z$ moves towards $+z$.

$\varphi (z)=3\pi z$, so $\nabla \varphi =3\pi a_z$ and $|\nabla \varphi |=3\pi rad/m$:

Write the cosine as the real part of an exponential and separate the time factor:

The factor that multiplies $e^{j\omega t}$ is the phasor:

$\lambda =2\pi /\beta =2\pi /3\pi =2/3\mathrm{m}$, the same as $v_p/f$. In Figure 5.6.12a every quarter period ($T/4=0.83\mathrm{ns}$) moves the whole pattern a quarter wavelength ($\lambda /4=16.7\mathrm{cm}$) towards $+z$, which is $2\times 10^{8}m/s$ again. In the surface of the statement the ridges are the crests, and their slope in the $(t,z)$ plane is the phase velocity.