Normal modes and waves (PS2): The wave equation and dispersive waves

For simplicity, consider $a = 1$ and $c = 1$. The key is to note that at time $t = 0$, we have a single wavelength of a sine wave from $a = -1$ to $a = 1$. However, d'Alembert's solution \[ y(x, t) = \frac{1}{2} \left[ f(x-ct) + f(x + ct)\right], \]

indicates that the initial displacement then splits into two parts, both half the initial displacement, but moving in separate directions. In steps of $t = a/2c = 1/2$, the two individual waves move a quarter of a wavelength in either direction. This is shown in the two pictures below

We use the form energy derived Question 7. By symmetry, the energy of either humps are the same. Let us focus on the one initially on the right. Then at $t = 0$, \begin{align*} \text{KE} &= \int_{\pi/k}^{2\pi/k} \frac{T}{2} \left( \pd{y_1}{x}\right)^2 \, \de{x} = \frac{Tk^2}{4} A^2 \frac{\pi}{k} \\ \text{PE} &= \int_{\pi/k}^{2\pi/k} \frac{\rho}{2} \left( \pd{y_1}{t}\right)^2 \, \de{x} = \frac{\rho\omega^2}{4} A^2 \frac{\pi}{k}, \end{align*}

so the total energy of the two humps is \[ E = 2(\text{KE} + \text{PE}) = A^2 \frac{\pi}{k} \left[ 1 + \frac{\rho}{T}\frac{\omega^2}{k^2}\right] Tk^2 = A^2 \frac{\pi}{k} Tk^2 (2), \]

since $c^2 = \rho/T = \omega^2/k^2$. Indeed, the potential and kinetic energies are zero.

At $t = 3\pi/2\omega$, both humps are centered the origin and thus sum to zero displacement. Their potential energies are zero, but they still possess kinetic energy. By conservation of energy, we expect the total energy to be the same, but simply formed completely from kinetic energy.

A traveling wave transports energy from one area of space to another. Standing waves have nodes and antinodes and its ends are fixed. Mathematically, we generally write $y(x,t) = X(x)T(t)$, so standing waves have a domain dependence that does not change with time.

Everything else is bookwork.

For simplicity assume that the solution of the system is a single wave \[ u(x,t) = A\cos(kx - \omega t + \phi), \]

where a general solution might be formed by linear combination of such waves. The dynamics of the system will will be described by a dispersion relation, \[ \omega = \omega(k), \]

which defines the frequency as a function of the wavenumber. The speed that the wave propagates is the phase velocity: \[ c_p = \frac{\omega(k)}{k}. \]

While this works for a single wave, the problem is that if the medium is dispersive, then when we add together various waves with different wavenumbers, $k$, then they will interfere with each other in a way that changes with time, so the shape of the disturbance will change. In many cases, the solution will have a dominant wavenumber, say $k = k_0 - \delta k$. The key is to consider the effect of adding a second wave which has wavenumber $k = k_0 + \delta k$. Then \[ y = A\sin[(k_0 + \delta k)x - (\omega + \delta\omega)t] + A\sin[(k_0 - \delta k)x - (\omega - \delta\omega)t] = 2A\cos(\delta k x - \delta \omega t) \sin(kx - \omega t), \]

which as we know describes a wavepacket whose envelope move with velocity $\delta \omega/\delta k$. This then motivates the definition of the group velocity \[ c_g = \dd{\omega}{k}, \]

as effectively, the speed for which the energy is carried. Note that if the medium is non-dispersive, then $c_p$ is constant, and thus $\omega$ is proportional to $k$ and $c_p = c_g$.

First recall that for a linear wave of the form \[ y = A \cos(kx - wt) = A\cos[k(x - (w/k)t], \]

the phase velocity and group velocities are defined as \[ c_p = \frac{w}{k}, \qquad c_g = \dd{w}{k}. \]

If we assume that the energy and momentums are of the form \[ E = \frac{h}{2\pi} w, \qquad p = \frac{h}{2\pi} k, \]

then we can write the velocities in the form \[ c_p = \frac{E}{p}, \qquad c_g = \dd{E}{p}. \]

As a warm-up, consider non-relativistic motion, with $p = m_0 v$ and $E = \frac{1}{2} m_0 v^2$. The key is that the energy is related to the momentum by $E = \frac{1}{2} p^2/m_0$. Thus we can simply check that the velocities are \[ c_p = \frac{E}{p} = \frac{v}{2}, \qquad c_g = \dd{E}{p} = \frac{p}{m_0} = v, \]

and the phase velocity is half the group velocity.

Consider now relativistic motion, with $p = m(v) v$ and $E = m(v) c^2$ where $m(v)$ is the relative mass of the particle as a function of its velocity. The phase velocity is now \[ c_p = \frac{E}{p} = \frac{c^2}{v}. \]

The group velocity is harder to compute, however, since the mass is no longer constant, and us a function of the velocity, which is linked to the momentum. Considering the energy as a function of the velocity, then by the chain rule, \[ c_g = \dd{E}{p} = \dd{v}{p}\dd{E}{v} = \frac{1}{\frac{m}{(1-v^2)^{3/2}}} \frac{mv}{(1-v^2)^{3/2}} = v. \]

Thus again in this case, the group velocity is the particle velocity.

Open question: This seems to be a very deep fact. Why is it important that when taking into account relativistic effects, that $c_p = c^2/v$ instead of $c_p = v/2$?

The first equation is found by setting $\omega = kc_p$. The second equation is found by transforming $k = \frac{2\pi}{\lambda}$. In a medium with refractive index $\mu$, we have $\mu = c/c_p$. Thus, using $c_p = c/\mu$ in the second equation yields the third. To derive the last relation, we use the refractive index to posit \[ \mu = \frac{c}{c_p} = \frac{\lambda'}{\lambda}. \]

First, \eqref{cg2} gives \[ c_g = c_p\left[ - \frac{1}{\frac{c_p}{\lambda} \dd{\lambda}{c_p}}\right], \]

where looking at the desired final form gave us the intuition of flipping the fractions. Now writing $\lambda = f(c_p, \lambda')$, we have \[ \dd{\lambda}{c_p} = \pd{f}{c_p} + \pd{\lambda'}{c_p}\pd{f}{\lambda} = \dd{}{c_p}\left(\frac{\lambda'}{c} c_p\right) = \frac{\lambda'}{c} + \frac{\lambda}{\lambda'} \pd{\lambda'}{c_p}. \]

The group velocity is then \[ c_g = \left[1 - \frac{1}{1 + \frac{c_p}{\lambda'}\pd{\lambda}{c_p}}\right]. \]

The derivation of the form of the kinetic energy is straightforward and found by summing the kinetic energy over small line segments of size $\de{x}$. The potential energy is given by \begin{align*} V = T \cdot \left[ \text{Extension of the string} \right] &= T \cdot \left[ \int_0^L \sqrt{1 + \left(\pd{y}{x}\right)^2} \, \de{x} - L \right] \\ &= T \left[\int_0^L \left(1 + \frac{1}{2} \left(\pd{y}{x}\right)^2 + \ldots \right) \, \de{x} - L \right] \\ &\sim \int_0^L \frac{T}{2} \left(\pd{y}{x}\right)^2 \, \de{x}, \end{align*}

valid assuming that the slope of string is small.

When we evaluate the energies for the given length, it is easier to put the solution into a form that is more explicit about the quantities. Recall the wave speed is $c = \omega/k$, so \[ y(x, t) = A\cos\left( \frac{2\pi}{L} x + \frac{2\pi}{L} c t + \phi\right). \]

Then we get \begin{gather*} U = \frac{\rho}{2} A^2 \left(\frac{2\pi}{L} \right)^2 \int_0^L \sin^2 \left( \frac{2\pi}{L} x + \frac{2\pi}{L} c t + \phi\right) \, \de{x} = \frac{A^2 \pi^2 c^2 \rho}{L} \\ V = \frac{T}{2} A^2 \left(\frac{2\pi}{L} \right)^2 \int_0^L \sin^2 \left( \frac{2\pi}{L} x + \frac{2\pi}{L} c t + \phi\right) \, \de{x} = \frac{A^2 \pi^2 T}{L}. \end{gather*}

Setting $T/\rho = c^2$ then gives the desired result.