5. Waves

5.1 The wave equation

The general form of the wave equation is: \(\Box u=0\), or:

\[\nabla^2u-\frac{1}{v^2}\frac{\partial^2 u}{\partial t^2}=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}-\frac{1}{v^2}\frac{\partial^2 u}{\partial t^2}=0\]

where \(u\) is the disturbance and \(v\) the propagation velocity. In general \(v=f\lambda\) holds. By definition \(k\lambda=2\pi\) and \(\omega=2\pi f\).

In principle, there are two types of waves:

  1. Longitudinal waves: for these  \(\vec{k}\parallel\vec{v}\parallel\vec{u}\) holds.
  2. Transversal waves: for these \(\vec{k}\parallel\vec{v}\perp\vec{u}\). holds

The phase velocity is given by \(v_{\rm ph}=\omega/k\). The group velocity is given by:

\[v_{\rm g}=\frac{d\omega}{dk}=v_{\rm ph}+k\frac{dv_{\rm ph}}{dk}= v_{\rm ph}\left(1-\frac{k}{n}\frac{dn}{dk}\right)\]

where \(n\) is the refractive index of the medium. If \(v_{\rm ph}\) does not depend on \(\omega\) then: \(v_{\rm ph}=v_{\rm g}\). In a dispersive medium it is possible that \(v_{\rm g}>v_{\rm ph}\) or \(v_{\rm g}<v_{\rm ph}\), and \(v_{\rm g}\cdot v_{\rm f}=c^2\). If one wants to transfer information with a wave, e.g. by modulation of an EM wave, the information travels with the velocity at which a change in the electromagnetic field propagates. This velocity is often almost equal to the group velocity.

For some media, the propagation velocity follows from:

Solutions of the wave equation

5.2.1 Plane waves

In \(n\) dimensions a harmonic plane wave is defined by:

\[u(\vec{x},t)=2^n\hat{u}\cos(\omega t)\sum_{i=1}^n\sin(k_ix_i)\]

The equation for a harmonic traveling plane wave is: \(u(\vec{x},t)=\hat{u}\cos(\vec{k}\cdot\vec{x}\pm\omega t+\varphi)\)

If waves reflect at the end of a spring this will result in a change in phase. A fixed end gives a phase change of \(\pi/2\) to the reflected wave, with boundary condition \(u(l)=0\). A loose end gives no change in the phase of the reflected wave, with boundary condition \((\partial u/\partial x)_l=0\).

If an observer is moving w.r.t. the wave with a velocity \(v_{\rm obs}\), they will observe a change in frequency: the Doppler effect. This is given by: \(\displaystyle\frac{f}{f_0}=\frac{v_{\rm f}-v_{\rm obs}}{v_{\rm f}}\).

5.2.2 Spherical waves

When the situation is spherically symmetric, the homogeneous wave equation is given by:

\[\frac{1}{v^2}\frac{\partial^2 (ru)}{\partial t^2}-\frac{\partial^2 (ru)}{\partial r^2}=0\]

with a general solution:

\[u(r,t)=C_1\frac{f(r-vt)}{r}+C_2\frac{g(r+vt)}{r}\]

5.2.3 Cylindrical waves

When the situation has a cylindrical symmetry, the homogeneous wave equation becomes:

\[\frac{1}{v^2}\frac{\partial^2 u}{\partial t^2}-\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u}{\partial r}\right)=0\]

This is a Bessel equation, with solutions which can be written as Hankel functions. For sufficient large values of \(r\) these are approximated by:

\[u(r,t)=\frac{\hat{u}}{\sqrt{r}}\cos(k(r\pm vt))\]

5.2.4 The general solution in one dimension

Starting point is the equation:

\[\frac{\partial^2 u(x,t)}{\partial t^2}=\sum_{m=0}^{N}\left(b_m\frac{\partial ^m}{\partial x^m}\right)u(x,t)\]

where \(b_m \in  I\hspace{-1mm}R\). Substituting \(u(x,t)=A{\rm e}^{i(kx-\omega t)}\) gives two solutions \(\omega_j=\omega_j(k)\) as dispersion relations. The general solution is given by:

\[u(x,t)=\int\limits_{-\infty}^{\infty}\left(a(k){\rm e}^{i(kx-\omega_1(k)t)}+ b(k){\rm e}^{i(kx-\omega_2(k)t)}\right)dk\]

Because in general the frequencies \(\omega_j\) are non-linear in \(k\) there is dispersion and the solution cannot be written any more as a sum of functions depending only on \(x\pm vt\): the wave front transforms.

5.3 The stationary phase method

Usually the Fourier integrals of the previous section cannot be calculated exactly. If \(\omega_j(k) \in I\hspace{-1mm}R\) the stationary phase method can be applied. Assuming that \(a(k)\) is only a slowly varying function of \(k\), one can state that the parts of the \(k\)-axis where the phase of \(kx-\omega(k)t\) changes rapidly will give no net contribution to the integral because the exponent oscillates rapidly there. The only areas contributing significantly to the integral are areas with a stationary phase, determined by \(\displaystyle\frac{d}{dk}(kx-\omega(k)t)=0\). Now the following approximation is possible:

\[\int\limits_{-\infty}^\infty a(k){\rm e}^{i(kx-\omega(k)t)}dk\approx \sum_{i=1}^{N}\sqrt{\frac{2\pi}{\frac{d^2\omega(k_i)}{dk_i^2}}} \exp\left[-i\mbox{$\frac{1}{4}$}\pi+i(k_ix-\omega(k_i)t)\right]\]

5.4 Green functions for the initial-value problem

This method is preferable if the solutions deviate much from the stationary solutions, like point-like excitations. Starting with the wave equation in one dimension, with \(\nabla^2=\partial^2/\partial x^2\)  if \(Q(x,x',t)\) is the solution with initial values \(Q(x,x',0)=\delta(x-x')\) and \(\displaystyle\frac{\partial Q(x,x',0)}{\partial t}=0\), and \(P(x,x',t)\) the solution with initial values \(P(x,x',0)=0\) and \(\displaystyle\frac{\partial P(x,x',0)}{\partial t}=\delta(x-x')\), then the solution of the wave equation with arbitrary initial conditions \(f(x)=u(x,0)\) and \(\displaystyle g(x)=\frac{\partial u(x,0)}{\partial t}\) is given by:

\[u(x,t)=\int\limits_{-\infty}^\infty f(x')Q(x,x',t)dx'+ \int\limits_{-\infty}^\infty g(x')P(x,x',t)dx'\]

\(P\) and \(Q\) are called the propagators. They are defined by:

\[\begin{aligned} Q(x,x',t)&=&\mbox{$\frac{1}{2}$}[\delta(x-x'-vt)+\delta(x-x'+vt)]\\ P(x,x',t)&=& \left\{\begin{array}{ll} \displaystyle\frac{1}{2v}&~~~\mbox{if}~~|x-x'|<vt\\ 0&~~~\mbox{if}~~|x-x'|>vt \end{array}\right.\end{aligned}\]

Further the relation: \(\displaystyle Q(x,x',t)=\frac{\partial P(x,x',t)}{\partial t}\) holds.

5.5 Waveguides and resonating cavities

The boundary conditions for a perfect conductor can be derived from the Maxwell equations. If \(\vec{n}\) is a unit vector \(\perp\) the surface, pointed from 1 to 2, and \(\vec{K}\) is a surface current density, then:

\[\begin{array}{ll} \vec{n}\cdot(\vec{D}_2-\vec{D}_1)=\sigma~~&~~\vec{n}\times(\vec{E}_2-\vec{E}_1)=0\\ \vec{n}\cdot(\vec{B}_2-\vec{B}_1)=0~~&~~\vec{n}\times(\vec{H}_2-\vec{H}_1)=\vec{K} \end{array}\]

In a waveguide because of the cylindrical symmetry: \(\vec{E}(\vec{x},t)=\vec{\cal E}(x,y){e}^{i(kz-\omega t)}\) and \(\vec{B}(\vec{x},t)=\vec{\cal{B}}(x,y){e}^{i(kz-\omega t)}\). From this one can now deduce that, if \({\cal B}_z\) and \({\cal E}_z\) are not \(\equiv0\):

\[\begin{array}{ll} \displaystyle {\cal B}_x=\frac{i}{\varepsilon\mu\omega^2-k^2}\left(k\frac{\partial {\cal B}_z}{\partial x}-\varepsilon\mu\omega\frac{\partial {\cal E}_z}{\partial y}\right)~~&~~ \displaystyle {\cal B}_y=\frac{i}{\varepsilon\mu\omega^2-k^2}\left(k\frac{\partial {\cal B}_z}{\partial y}+\varepsilon\mu\omega\frac{\partial {\cal E}_z}{\partial x}\right)\\ \displaystyle {\cal E}_x=\frac{i}{\varepsilon\mu\omega^2-k^2}\left(k\frac{\partial {\cal E}_z}{\partial x}+\varepsilon\mu\omega\frac{\partial {\cal B}_z}{\partial y}\right)~~&~~ \displaystyle {\cal E}_y=\frac{i}{\varepsilon\mu\omega^2-k^2}\left(k\frac{\partial {\cal E}_z}{\partial y}-\varepsilon\mu\omega\frac{\partial {\cal B}_z}{\partial x}\right) \end{array}\]

Now one can distinguish between three cases:

  1. \(B_z\equiv0\): the Transverse Magnetic modes (TM). Boundary condition: \({\cal E}_z|_{\rm surf}=0\).
  2. For the TE and TM modes this gives an eigenvalue problem for \({\cal E}_z\) resp. \({\cal B}_z\) with boundary conditions:
    \[\left(\frac{\partial^2 }{\partial x^2}+\frac{\partial^2 }{\partial y^2}\right)\psi=-\gamma^2\psi~~\mbox{with eigenvalues}~~ \gamma^2:=\varepsilon\mu\omega^2-k^2\]
    This gives a discrete solution \(\psi_\ell\) with eigenvalue \(\gamma_\ell^2\): \(k=\sqrt{\varepsilon\mu\omega^2-\gamma_\ell^2}\). For \(\omega<\omega_\ell\), \(k\) is imaginary and the wave is damped. Therefore, \(\omega_\ell\) is called the cut-off frequency. In rectangular conductors the following expression can be found for the cut-off frequency for modes TE\(_{m,n}\) of TM\(_{m,n}\):
    \[\lambda_\ell=\frac{2}{\sqrt{(m/a)^2+(n/b)^2}}\]
  3. \(E_z\) and \(B_z\) are zero everywhere for the Transversal Eectromagnetic (TEM). Then: \(k=\pm\omega\sqrt{\varepsilon\mu}\) and \(v_{\rm f}=v_{\rm g}\), just as if here were no waveguide. Further \(k \in \hspace{-1mm}R\), so there exists no cut-off frequency.

In a rectangular, 3 dimensional resonating cavity with edges \(a\), \(b\) and \(c\) the possible wave numbers are given by: \(\displaystyle k_x=\frac{n_1\pi}{a}~,~~k_y=\frac{n_2\pi}{b}~,~~k_z=\frac{n_3\pi}{c}\) This results in the possible frequencies \(f=vk/2\pi\) in the cavity:

\[f=\frac{v}{2}\sqrt{\frac{n_x^2}{a^2}+\frac{n_y^2}{b^2}+\frac{n_z^2}{c^2}}\]

For a cubic cavity, with \(a=b=c\), the possible number of oscillating modes \(N_{\rm L}\) for longitudinal waves is given by:

\[N_{\rm L}=\frac{4\pi a^3f^3}{3v^3}\]

Because transversal waves have two possible polarizations \(N_{\rm T}=2N_{\rm L}\) holds for them.

5.6 Non-linear wave equations

The Van der Pol equation is given by:

\[\frac{d^2x}{dt^2}-\varepsilon\omega_0(1-\beta x^2)\frac{dx}{dt}+\omega_0^2x=0\]

\(\beta x^2\) can be ignored for very small values of the amplitude. Substitution of \(x\sim{\rm e}^{i\omega t}\) gives: \(\omega=\frac{1}{2}\omega_0(i\varepsilon\pm2\sqrt{1-\frac{1}{2}\varepsilon^2})\). The lowest-order instabilities grow as \(\frac{1}{2}\varepsilon\omega_0\). While \(x\) is growing, the 2nd term becomes larger and diminishes the growth. Oscillations on a time scale \(\sim\omega_0^{-1}\) can exist. If \(x\) is expanded as \(x=x^{(0)}+\varepsilon x^{(1)}+\varepsilon^2x^{(2)}+\cdots\) and this is substituted one obtains, besides periodic, secular terms \(\sim\varepsilon t\). If it is assumed that there exist timescales \(\tau_n\), \(0\leq\tau\leq N\) with \(\partial\tau_n/\partial t=\varepsilon^n\) and if the secular terms are put to 0 one obtains:

\[\frac{d}{dt}\left\{\frac{1}{2}\left(\frac{dx}{dt}\right)^2+\mbox{$\frac{1}{2}$}\omega_0^2x^2\right\}= \varepsilon\omega_0(1-\beta x^2)\left(\frac{dx}{dt}\right)^2\]

This is an energy equation. Energy is conserved if the left-hand side is 0. If \(x^2>1/\beta\), the right-hand side changes sign and an increase in energy changes into a decrease of energy. This mechanism limits the growth of oscillations.

The Korteweg-De Vries equation is given by:

\[\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x}-\underbrace{au\frac{\partial u}{\partial x}}_{\rm non-lin}+ \underbrace{b^2\frac{\partial ^3u}{\partial x^3}}_{\rm dispersive}=0\]

This equation is for example a model for ion-acoustic waves in a plasma. For this equation, soliton solutions of the following form exist:

\[u(x-ct)=\frac{-d}{\cosh^2(e(x-ct))}\] with \(c=1+\frac{1}{3}ad\) and \(e^2=ad/(12b^2)\).