let's consider a generic wave that propagates in space and time, expressed in frequency domain:

$$V(x)=A\cdot e^{-kx} + B\cdot e^{kx}$$

It may represent the voltage along a transmission line, or any physical quantity that appears in a second order differential equation that satisfies the oscillation conditions.

In time - domain it becomes something like that:

$$V(x)=A\cdot cos(-kx+\omega t) + B\cdot cos(kx+\omega t)$$

The first term is usually called direct or incident wave, while the second one is called reverse or reflected wave. They are two waves that moves in space and time, and precisely the first one moves towards x axis, while the second one moves in the opposite direction to the x axis.

My question is about this last statement: why are the term with "-kx" the direct wave and the term with "kx" the reverse wave?

I tried to answer my question by using the concept of phase velocity. The phase velocity of a wave is defined as the velocity an observer should have in order not to see any difference of phase, i.e. such that:

$$d\cdot (-kx+\omega t) = 0$$ for the first term;

$$d\cdot (kx+\omega t) = 0$$ for the second term;

So we get:

$$-kdx + \omega dt = 0$$ for the first term;

$$kdx + \omega dt = 0$$ for the second term;

and finally:

$$v_{1} = dx/dt = \omega/k$$ for the first term;

$$v_{2} = dx/dt = - \omega/k$$ for the second term;

So the phase velocity of the first term is positive (so, towards x axis), while that of the second term is negative (so, opposite to x axis).

What does not convince me of this explanation is the fact that as far as I know the phase velocity does not represent the propagation velocity of the wavefront, but is simply the velocity corresponding to a zero phase difference. And thanks to the fact that it is not a propagation speed, it follows that it can be even greater than the speed of light (it is a known result of the analysis of waveguides, for instance).


3 Answers 3


I believe you should plot these two waves starting with $t=0$ and then slowly increasing the time. enter image description here

The red is the incident wave, which travels into the positive $x$-direction. It is reflected at $x=1.2 \lambda$. The green wave is the reflected wave. Thus it travels into the negative $x$-direction. The blue wave is the superposition of these two waves.

  • $\begingroup$ Can you please tell me what program did you use yo make this animation? $\endgroup$
    – RedGiant
    Commented Apr 9, 2020 at 13:09
  • $\begingroup$ I used R in combination with the packages ggplot2 and gganimate. However, this is a standard feature in almost all math programs, e.g. matlab, mathematica, python etc. Alternatively, you could probably generate "animated GIFs" using an external program as well. $\endgroup$
    – Semoi
    Commented Apr 9, 2020 at 13:57

I see nothing wrong with your demonstration. For a monochromatic wave in an isotropic medium I think a wavefront may always indifferently by defined as a locus of constant phase or constant amplitude. Things only become more complicated with a propagating pulse (wave paket) in a dispersive medium and/or in a anisotropic medium. Anyway, the only other alternative I know is a classical (and very related) reasoning that you may already know:

Suppose you freeze the wave at an instant $t_{0}$ and measure the phase of the vibration $\psi$ at a particular location $x_{0}$. If you now let the wave run again for a short $\Delta t < 2\pi/ \omega$ and freeze the wave again at $t_{0}+\Delta t$. Where in the nearest neighborhood of $x_{0}$ will you find the same phase $\psi$?

Case 1 : $\psi=\omega t_{0} -kx_{0}$
Solving for $x$ such that $\omega (t_{0}+\Delta t)-kx = \psi $, yields $x=x_{0}+v_{\psi}\Delta t$, with $v_{\psi}=\frac{\omega}{k}$

Case 2 : $\psi=\omega t_{0}+kx_{0}$
Solving for $x$ such that $\omega (t_{0}+\Delta t)+kx = \psi $, yields $x=x_{0}-v_{\psi}\Delta t$.

Conclusion: In case 1 the wave propagates in the +$x$ direction, in case 2, the wave propagates in $-x$ direction.


I think your concern is "because the negative sign applies to the spatial bit within the cosine and not to the time part, then how can it be that such becomes a time reversal (i.e. backward propagation)?"

This is simple. Note that $\cos$ is an even function.

$$\begin{align} u_\mathrm{dir}(x, t) &:= A \cos(-kx + \omega t)\\ &= A \cos(-[-kx + \omega t])\\ &= A \cos(kx - \omega t)\end{align}$$

Thus, it is quite evident now that the "reflected wave"

$$u_\mathrm{ref}(x, t) := A \cos(kx + \omega t)$$

is the time reversal of the "direct wave", i.e.

$$u_\mathrm{ref}(x, t) = u_\mathrm{dir}(x, -t)$$

(ignoring the amplitude difference).

viz. it is travelling the other way. As for why we call the negative one reverse instead of calling the positive one - it is because there is a tendency to imagine the wave travelling toward the right. The term with negative $kx$ (or negative $\omega t$) term travels to the right.


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