How does the solution

$latex E_x = f(z – v_p t)$

behave? Well, we know that subtracting a number from the argument of a function shifts that function to the right. For example, suppose that $latex g(x)$ is a Gaussian function of $latex x$. Replacing $latex x$ with $latex (x – d)$ will shift the function to the right by a distance $latex d$:

So when we’re considering $latex f(z – v_p t)$, if we were to plot this function vs. $latex z$ at $latex t = 0$ and also at $latex t = t_1$, the $latex t = t_1$ version will be the same as the $latex t = 0$ version except shifted to the right by a distance $latex v_p t_1$. If $latex f(x, t = 0)$ is a Gaussian, then, the plot of $latex f(z, t)$ at the two times would look like:

Since the function has moved in the positive $latex z$ direction by a distance $latex v_p t_1$ in time $latex t_1$, it must be moving at a speed of $latex v_p$. In fact, we call the quantity $latex v_p$ the phase velocity. The solutions to the wave equation are waves, and $latex v_p$ is the speed at which the wave is moving.

I mentioned briefly that there is another possible solution to the scalar wave equation, namely:

$latex E_x = f(z + v_p t)$

Hopefully you can see that this will be a wave moving to the left (negative $latex z$ direction) with speed $latex v_p$. In fact, if we went back to the more general form of the Helmholtz equation, we would find solutions moving in all possible directions in 3d space at speed $latex v_p$.

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