6  Energy

Besides Information, waves transmit another practically important quantity: Energy. Note that waves do not transport matter. Matter may oscillate up and down or forth and back as a wave passes, but it is not swept away with the wave. But energy is. In this lecture we are going to first introduce the expression for the energy in a wave on a string as an integral over the energy density. The energy density in turn is made up out of kinetic and potential energy density. We will then calculate the energy in a few example waves, and then discuss the conservation of energy.

You can find related material in the textbooks:

6.1 Energy density

Consider an infinitesimal bit of string between \(x\) and \(x+\delta x\). Its kinetic energy is \[ \delta K = \frac12 m\,v^2 = \frac12 \rho\,\delta x\left(\partial_t y\right)^2. \tag{6.1}\] The kinetic energy of the entire string is then obtained by integrating over its infinitesimal parts: \[ K=\int \frac{\rho}{2} (\partial_t y)^2 \, dx = \int \mathcal{E}_K\,dx. \tag{6.2}\] The quantity \(\mathcal{E}_K\) is the kinetic energy density.

To derive the formula for the potential energy, we again look first at an infinitesimal segment of the string. It has been stretched from a length of \(\delta x\) to the longer length \(\delta s\). The work done to change the length from \(\delta x\) to \(\delta s\) is \(T(\delta s - \delta x)\). This gives the potential energy (we neglect the potential energy coming from gravity). We have \[ \delta s= \sqrt{1+\left(\partial_x y\right)^2} \, \delta x \approx \delta x \left(1 + \frac{\left(\partial_x y\right)^2}{2} + \cdots \right), \tag{6.3}\] where we have only kept the first two terms in the Taylor expansion because, as we did when we derived the wave equation, we assume that the slope of the string is small and thus the higher order terms in \(\partial_x y\) are negligible. Thus the potential energy in the infinitesimal segment of the string is \[ \delta V=T\left(\delta s - \delta x\right) = \frac{T}{2} \left(\partial_x y\right)^2\, \delta x . \tag{6.4}\] Summing up contributions from all small elements of the string (i.e. integrating over the whole string), we find the potential energy \[ V=\int T \, \frac{(\partial_x y)^2}{2} \, dx = \int \mathcal{E}_V\,dx. \tag{6.5}\] The quantity \(\mathcal{E}_V\) is the potential energy density.

The total energy \(E\) is the sum of the kinetic and potential energy: \[\begin{split} E&=K+V\\ &=\int \left( \frac{\rho}{2} (\partial_t y)^2+ \frac{T}{2} (\partial_x y)^2\right) dx\\ &= \int \mathcal{E}\,dx, \end{split} \tag{6.6}\] where \(\mathcal{E}=\mathcal{E}_K + \mathcal{E}_V\) is the total energy density.

6.2 Energy density of example waves

Example 6.1 Consider a localised wave \(y(x,t)=f(x-ct)\) travelling to the right with speed \(c\). Substituting this into the general expression for the energy density \[ \mathcal{E}=\frac{\rho}{2}(\partial_ty)^2+\frac{T}{2}(\partial_x y)^2 \tag{6.7}\] gives \[ \mathcal{E}(x,t)=\frac{\rho}{2}(-c)^2(f'(x-ct))^2+\frac{T}{2}(f'(x-ct))^2. \tag{6.8}\] Because \(c^2\rho = T\), we see that the kinetic and the potential energy densities are equal. This phenomenon is referred to as “equipartition” of the energy. Together we have \[ \mathcal{E}(x,t)=T\,(f'(x-ct))^2. \tag{6.9}\] Note how the energy density is travelling along with the wave profile.

Example 6.2 (Standing harmonic wave) Now we consider solutions of the form \[\begin{split} y(x,t)&=\sin(kx)(F\sin(kct)+G\cos(kct))\\ &=\alpha\cos(kct+\phi). \end{split} \tag{6.10}\] We calculate the energy densities \[\begin{split} \mathcal{E}_K&=\frac{\rho}{2}(\partial_t y(x,t))^2\\ &=\frac{\rho}{2}\alpha^2 k^2 c^2(-\sin(kct+\phi))^2\sin^2(kx) \end{split} \tag{6.11}\] and \[\begin{split} \mathcal{E}_V&=\frac{T}{2}(\partial_x y(x,t))^2\\ &=\frac{T}{2}\alpha^2 k^2 (\cos(kct+\phi))^2\cos^2(kx) \end{split} \tag{6.12}\] Again we notice that the prefactors are the same because \(c^2\rho=T\). For the energies we find \[\begin{split} K&=\frac{T}{2}\alpha^2 k^2 \sin^2(kct+\phi) \int_0^\pi\sin^2(kx)dx\\ &=\frac{T\alpha^2 k^2 \pi}{4}\sin^2(kct+\phi) \end{split} \tag{6.13}\] and \[\begin{split} T&=\frac{T}{2}\alpha^2 k^2 \cos^2(kct+\phi)\int_0^\pi\cos^2(kx)dx\\ &=\frac{T\alpha^2 k^2 \pi}{4}\cos^2(kct+\phi). \end{split} \tag{6.14}\]

For standing waves, both the kinetic energy and the potential energy depend on time and are not equal. However, their averages, averaged over a period in \(t\), are equal. The total energy is constant \[ E=K+T =\frac{T\alpha^2 k^2 \pi}{4}(\sin^2(kct+\phi)+\cos^2(kct+\phi))=\frac{T\alpha^2 k^2 \pi}{4} \tag{6.15}\]

Example 6.3 (Sum of two standing harmonic waves) Consider two harmonic waves \[\begin{split} y_k&=\alpha_k\sin(kx)\cos(kct+\phi_k)~~\text{and}\\ y_l&=\alpha_l\sin(lx)\cos(lct+\phi_l) \end{split} \tag{6.16}\] with \(k\neq l\) and let us calculate the energy of \(y=y_k+y_l\). We have \[\begin{split} K&=\frac{\rho}{2}\int_0^\pi(\partial_t y)^2dx =\frac{\rho}{2}\int_0^\pi(\partial_t y_k+\partial_t y_l)^2dx \\ &=K_k+K_l+\rho\,\alpha_k\, \alpha_l\, k\,l\,c^2\cos(kct+\phi_k)\cos(lct+\phi_l) \cdot\\ &\qquad\qquad\qquad\int_0^\pi\sin(kx)\sin(lx)dx\\ &=K_k+K_l \end{split} \tag{6.17}\] where \(K_k\) and \(K_l\) are the kinetic energies of the individual harmonic waves. A similar calculation shows that also the potential energy of the sum is the sum of the potential energies and so this is also true of the total energy: \[ E[y_k+y_l]=E[y_k]+E[y_l]. \tag{6.18}\] This is one of the nice properties of harmonic waves.

Example 6.4 (Complex exponential wave) We calculate the energy density of the complex solution \[ y(x,t)=A\,e^{i(kx-\omega t)}. \tag{6.19}\] The expression for the energy density of complex solutions involves the absolute value squared: \[ \mathcal{E}[y]=\frac{\rho}{2}|\partial_t y|^2+\frac{T}{2}|\partial_x y|^2. \tag{6.20}\] This has the effect that the energy density is the sum of the energy density of the real part of the solution and the energy density of the imaginary part of the solution. We find \[\begin{split} \mathcal{E}[A\,e^{i(kx-\omega t)}]&=\frac{\rho}{2}\left|-i\omega \,A\,e^{i(kx-\omega t)}\right|^2+\frac{T}{2}\left|ik \,A\,e^{i(kx-\omega t)}\right|^2\\ &=\left(\frac{\rho}{2}\omega^2+\frac{T}{2}k^2\right)|A|^2 \end{split} \tag{6.21}\] So another miracle of these complex exponential solutions is that their energy density is constant.

6.3 Conservation equation

Let \(y(x,t)\) be a solution of the wave equation for the string and \(\mathcal{E}\) its energy density \[ \mathcal{E}=\frac{\rho}{2}(\partial_t y)^2+\frac{T}{2}(\partial_x y)^2. \tag{6.22}\] For the time derivative of the energy density we find \[\begin{split} \partial_t \mathcal{E} &= \rho \, \partial_t y \, \partial_t^2 y + T \, \partial_x y \, \partial_t\partial_x y \\ &= \partial_t y \, T\, \partial_x^2 y + T \, \partial_x y \, \partial_t\partial_x y \quad (\text{using} \ \ \rho \, \partial_t^2 y = T\, \partial_x^2 y \ ) \\ &=- \partial_x\left( -T \, \partial_t y \, \partial_x y\right) . \end{split} \tag{6.23}\] In terms of the quantity \(\mathcal{F} = - T \, \partial_t y \, \partial_x y\) this equation takes the form \[ \partial_t\mathcal{E}=-\partial_x\mathcal{F}. \tag{6.24}\] The quantity \(\mathcal{F}\) is called the energy flux.

Eq. 6.24 implies the law of conservation of energy (and therefore is called a conservation equation}). Indeed, we have \[ \frac{dE}{dt}= \int\limits_{x_1}^{x_2} \partial_t \mathcal{E} \, dx = -\int\limits_{x_1}^{x_2} \partial_x \mathcal{F} \, dx = \mathcal{F}(x_1)-\mathcal{F}(x_2). \tag{6.25}\] So if we interpret the energy flux \(\mathcal{F}(x)\) as the rate at which energy flows through a point \(x\) from left to right, then the conservation equation expresses that the rate at which energy in a region changes is equal to the difference between the rate at which energy flows in and the rate at which energy flows out of the region.

We already showed conservation of energy of a finite string with fixed boundary conditions in the previous section. We will now use the above machinery to show the conservation of the energy of a finite string with free boundary conditions. We consider a string between \(x=0\) and \(x=\pi\) satisfying the free boundary conditions \[ \partial_xy(0,t)=0=\partial_xy(\pi,t)\text{ for all }t. \tag{6.26}\] The energy flux at the left end of the string at \(x=0\) is \[ \mathcal{F}(0,t)=-T\,\partial_ty(0,t)\,\partial_xy(0,t). \tag{6.27}\] This flux is zero due to the boundary condition. Similarly the flux at the right boundary is zero: \[ \mathcal{F}(\pi,t)=-T\,\partial_ty(\pi,t)\,\partial_xy(\pi,t)=0. \tag{6.28}\] So by the conservation equation it follows that the energy is conserved.