I was struck by an answer to this question that made me think about various ways to write a solution to a wave equation. One can write $f(x-vt)$ or $f(t-x/v)$ (as noted in an answer to the above) or even $f(kx-\omega t)$.

I'm curious about how to connect interpretations of these three forms. I see the last one has an argument which is explicitly dimensionless while the other two arguments seem to have dimensions of distance or time. My instinct tells me that $kx-\omega t$ ought to be "better" in some fashion, but I feel fuzzy about that.

What is the right way to think of these three different ways to write a wave function?


2 Answers 2


This is a surprisingly subtle issue.

It's straightforward enough in the simple case: $$-\frac{\partial^2 f}{\partial t^2} + v^2\frac{\partial^2 f}{\partial x^2} = 0$$ You can show that solutions to this take the form $f(x \pm vt)$, i.e. that they depend only on the combination $x\pm vt$ but not on $x$ and $t$ separately. Note that $v$ is a constant. If you would rather write your solution as a function of $t \pm x/v$, just factor out $-v$. Or if you would rather write it in terms of $kx - \omega t$, just factor out $k$, taking note of the fact that $k = \omega/v$. For example: $$\sin\biggl(\frac{1}{x_0}(x - vt)\biggr) = \sin\biggl[-\frac{v}{x_0}\biggl(t - \frac{x}{v}\biggr)\biggr] = \sin\biggl(\frac{1}{k x_0}(k x - \omega t)\biggr)$$ In each case, the solution takes the form $\sin(\text{constant}\times\text{argument})$, so there's no real difference between these. However, many people do prefer $kx - \omega t$ simply because it has no units.

Things get tricky once you introduce dispersion, which basically entails making the velocity dependent on frequency. $$-\frac{\partial^2 f}{\partial t^2} + v(\omega)^2\frac{\partial^2 f}{\partial x^2} = 0$$ More precisely: when you solve the wave equation, as you may know, you get functions that describe plane waves, like $\sin(kx - \omega t)$, with many different possible frequencies $\omega$. With the normal wave equation, all these waves move at the same speed, $v$, so a batch of plane waves of different frequencies will nevertheless maintain their composite shape as they pass by. That no longer holds true for dispersion. Waves of different frequencies travel with different velocities, so the shape of the overall waveform changes with time as the wave travels.

It turns out that these changes in the shape of the waveform have the effect of making overall features of the shape move with a different speed than any of the individual plane waves. So there are two different kinds of velocity involved: the group velocity, which is the speed at which overall features of the waveform travel, and the phase velocity (a function of frequency), which is the speed at which individual plane wave components travel. These can be expressed by the equations $$\begin{align}v_g &= \frac{\partial \omega}{\partial k} & v_p &= \frac{\omega}{k}\end{align}$$ When dealing with dispersion, it's especially useful to write your plane wave solutions as functions of $kx - \omega t$ for a couple of reasons:

  • It avoids confusion between group velocity and phase velocity by leaving velocity out of the picture entirely.
  • It emphasizes that $k$ and $\omega$ are really separate quantities which can have a rather complicated relationship; they're not necessarily just proportional.

If you want to be very rigorous, then yes, your function should have a dimensionless argument. There are several reasons "why" but I like to imagine if I had a function, $f(x)= e^x$, then the Taylor series wouldn't make sense, $=\Sigma \frac{x^i}{i!}$. This results because if x had units, of say distance, the Taylor series would be adding $1+m+m^2+....$.

That being said if you don't remove dimensions from your ODE/PDE as well as solution all is not lost. You can of course discuss the functions $f(x-vt)$ and $f(t-\frac{x}{v})$. The arguments are related to each other of course. $$f(x-vt)=f(-v(t-\frac{x}{v}))$$ $$f(\tau)=f(c\tau)$$ The details about comparing one to the other depends on the specific function, but we can see that the argument scales and can be discussed in the same way that functions are discussed in pre-calc.

  • $\begingroup$ It's unclear to me what your $f(τ)=f(cτ)$ equation means. What are you representing there? Or was it supposed to be $f(t)=f(cτ)$? $\endgroup$
    – user55515
    Commented Apr 2, 2016 at 12:50

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