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# 5: Waves

• • Johan Wevers
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## 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:

• Pressure waves in a liquid or gas: $$v=\sqrt{\kappa/\varrho}$$, where $$\kappa$$ is the modulus of compression.
• Further for pressure waves in a gas: $$v=\sqrt{\gamma p/\varrho}=\sqrt{\gamma RT/M}$$.
• Pressure waves in a thin solid bar with diameter $$<<\lambda$$: $$v=\sqrt{E/\varrho}$$
• Waves in a string: $$v=\sqrt{F_{\rm span}l/m}$$
• Surface waves on a liquid: $$\displaystyle v=\sqrt{\left(\frac{g\lambda}{2\pi}+ \frac{2\pi\gamma}{\varrho\lambda}\right)\tanh\left(\frac{2\pi h}{\lambda}\right)}$$
where $$h$$ is the depth of the liquid and $$\gamma$$ the surface tension. If $$h\ll\lambda$$ then $$v\approx\sqrt{gh}$$ holds.

## Solutions of the wave equation

### 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 imposes a phase change of $$\pi/2$$ to the reflected wave, with boundary condition $$u(l)=0$$. A loose end yields 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: $$\displaystyle\frac{f}{f_0}=\frac{v_{\rm f}-v_{\rm obs}}{v_{\rm f}}$$.

### 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:

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

### 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 that can be written as Hankel functions. For sufficient large values of $$r$$ these are approximated by:

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

### The general solution in one dimension

Starting from 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 $$b_m \in \mathbb{R}$$. 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.

## The stationary phase method

Usually the Fourier integrals of the previous section cannot be calculated exactly. If $$\omega_j(k) \in \mathbb{R}$$ 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:

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

## Green functions for the initial-value problem

This method is preferable if the solutions deviate strongly 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)&=&\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.

## Waveguides and resonating cavities

The boundary conditions for a perfect conductor can be derived from Maxwell's equations. If $$\vec{n}$$ is a unit vector $$\perp$$ the surface, pointing 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 $$\vec{B}(\vec{x},t)=\vec{\cal{B}}(x,y){e}^{i(kz-\omega t)}$$ holds. 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. $$E_z\equiv0$$: the Transverse Electric Modes (TM). Boundary condition: $$\frac{\partial \cal B}_z}{\partial n} \bigg\rvert _{surf}=$$.
For the TE and TM modes this results in 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 has 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$$_{m,n}$$ or 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 Eectro-Magnetic Modes (TEM). Then: $$k=\pm\omega\sqrt{\varepsilon\mu}$$ and $$v_{\rm f}=v_{\rm g}$$, just as if there were no waveguide. Further $$k \in \mathbb{R}$$, so no cut-off frequency exists.

In a rectangular, three 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 transverse waves have two possible polarizations $$N_{\rm T}=2N_{\rm L}$$ holds for them.

## 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 which limits 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, additional 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:

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

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 $$e^2=ad/(12b^2)$$.

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5: Waves is shared under a CC BY license and was authored, remixed, and/or curated by Johan Wevers.