# Phase Velocity Derivation

I spend few hours trying to derive phase velocity of sinusoidal wave $$\cos(kx - \omega t).$$ I know that it must be equal to $$\omega \over k$$ but after banging my head for few hours and trying to find solution on internet I gave up.

• Note that in 1-dimension \begin{equation} \text{phase}\boldsymbol{=} \phi \boldsymbol{\stackrel{def}{\equiv}} k\cdot x\boldsymbol{-}\omega\cdot t\qquad k,\omega \in \mathbb{R} \tag{01}\label{01} \end{equation} and in 3-dimensions \begin{equation} \text{phase}\boldsymbol{=} \phi \boldsymbol{\stackrel{def}{\equiv}} \mathbf{k}\boldsymbol{\cdot} \mathbf{x}\boldsymbol{-}\omega\cdot t\qquad \mathbf{k}\in \mathbb{R}^{3},\omega \in \mathbb{R} \tag{02}\label{02} \end{equation} – Frobenius Dec 22 '18 at 8:05

The wave equation is $$z(x,t)= \cos(kx - \omega t)$$ and a graph of the displacement of a particle from its equilibrium position $$z$$ against the position of the particle from an origin $$x$$ at a given time $$t$$ is shown below.
You can liken the graph to a photograph of the wave taken at one instant of time; it is called a wave profile.

The graph actually shows shows two such photographs (wave profiles) taken at a time $$t$$ and at a later time $$t+\Delta t$$. By inspection of the movement of the peaks $$A$$ and $$A'$$, or the troughs $$C$$ and $$C'$$, or the positions of zero displacement $$B$$ and $$B'$$ etc, you can surmise that the wave is travelling in the positive x-direction.

The important thing is that the displacements of the particle at different times which you considering are the same; peak and peak, trough and trough etc.

So you have $$z(x,t) = z(x+\Delta x , t + \Delta t)$$ or $$\cos(kx - \omega t) = \cos(k[x+\Delta x] - \omega [t+\Delta t])$$ and a solution of this equation is $$kx - \omega t = k[x+\Delta x] - \omega [t+\Delta t] \Rightarrow \dfrac {\Delta x}{\Delta t} = \dfrac {\omega}{k}$$ and this is called the phase velocity.

Another way of derivation is to say that you want to find a condition such that $$kx-\omega t$$ (the "phase" of the wave) is a constant ie you are tracking the passage of a crest at a given time to a crest at a later time etc.

Now differentiate $$kx-\omega t = \rm constant$$ with respect to time.

$$k \dfrac {dx}{dt} - \omega =0 \Rightarrow \dfrac{dx}{dt} = \dfrac {\omega}{k}$$ and you have the phase velocity.

• I liked that that your soulutions are quite formal because it leaves a little possibility for an error but I missing some steps in these solutions. In first solution you calculated $\Delta x \over \Delta t$. This is only approximation of speed. It is accurate only if speed is constant (so the ${dx(t) \over dt} = const$). In second solution you first taken $x$ to be variable and then you differentiated it by $t$. This should be equal to zero. If it is not then $x$ should really be function of $t$ i.e. $x(t)$ but in your solution it is not. – Trismegistos Dec 22 '18 at 12:32
• @Trismegistos : For constant $k,\omega$ \begin{equation} \dfrac{\Delta x}{\Delta t}\boldsymbol{=}\dfrac{\mathrm dx}{\mathrm dt}\boldsymbol{=}\dfrac{\omega}{k} \end{equation} – Frobenius Dec 22 '18 at 13:26
• @Trismegistos When you are looking at the peak of a travelling wave you will notice that the position of the peak $x$ changes as the time $t$ charges, so $x$ is a function of $t$. It is the displacement of the particle from its equilibrium position $z$ ie the peak being observed which is kept constant. – Farcher Dec 22 '18 at 14:02

The phase velocity denotes the velocity at which a peak of the sinusoidal pattern is moving. Consider the wave as $$\cos[\psi(x,t)]$$, where $$\psi=kx-\omega t$$. There will always be a peak at $$\psi=0$$, since that corresponds to $$\cos0=1$$. We want to know how fast that peak at $$\psi=0$$ is moving.* But the condition $$\psi=0$$ is just $$kx-\omega t=0\\ x-\frac{\omega}{k}t=0.$$ And the last is the just the equation of something moving with velocity $$\omega/k$$.

*It can be any other peak as well; that just shifts the relation between $$x$$ and $$t$$ by an overall constant, leaving the velocity the same.